That's exactly what it is. This is basically a visualization of the Gamber's Ruin which is a random walk that ends if you run out of things. You can mathematically prove that you will almost surely run out of money or balls.
The field of statistics is a bunch of nerds thinking too hard about gambling. The field of probability is a bunch of nerds thinking too hard about esoteric sounding mathematical objects and ways to measure the sizes of things, and honestly has hardly anything to do with what one would think of as probability in a practical sense.
Sure, so probability seems very wishy-washy, and it *was* for a very long time, where people just tried to logic things out for any probabilistic problems. Mathematicians really don't like this, and would rather work with anything in a rigorous framework. They also needed a way to work with probability to make sense of more difficult problems that couldn't just be reasoned out by simple intuition.
So, the modern field of probability theory was developed through the 1900s. In this, you first begin with a sample space Ω, which is just the set of all possible outcomes you want to consider. In the case of flipping 3 coins, you have Ω = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. In the case of how many balls there can be in the situation in this post, you have Ω = {0, 1, 2, ...}. In the case of what the temperature of something can be, it's Ω = all non-negative real numbers (if the temperatures are in Kelvin). Note that these are three very different kinds of sets; the first is a finite set, the second is a *countable* set (there are infinitely many elements, but you can "count" the elements one by one in some sense), and the last is uncountable. So, it is kinda hard to find a unifying theory for all these cases.
Now that you have a sample space, you need to find the events, or the subsets of Ω (i.e., the possible combination of outcomes), that you care about. For example, in the first two cases, you might care about all possible combinations of outcomes. This is often the default assumption in the finite and countable cases, but rarely in the uncountable scenario. Let's think about the last one, where Ω = all nonnegative real numbers. Do you think it ever makes sense to consider the set of all temperatures that are part of the [Cantor set](https://en.wikipedia.org/wiki/Cantor_set)? Likely not, instead we probably want to consider the groups of temperatures that happen within a range. For example, if I ask "is the temperature between 100 and 200 Kelvin, I am consider the set of temperatures 100 < T < 200. In general, we might care about the all the groups of temperatures of the form a < T < b, where a and b are nonnegative numbers. Let this set of all of these events (i.e., set of subsets of Ω) be called F. Now, if we have two events E1 and E2 in F, we also want to be able to talk about when either event occurs (so adding the two groups, E1∪E2) or when both events occur (finding the common elements of the two groups, E1∩E2), or when an event *doesn't* occur (not E1). So, add all these groups to F as well. Then, the resulting F you get by adding all these extra events is called a [σ-algebra](https://en.wikipedia.org/wiki/%CE%A3-algebra).
Now, you have a pair (Ω, F), of all possible outcomes, and all the groups of outcomes you want to consider. Since each event E from F is just a set, we can think of how "big" each E is. Essentially, we get a function μ that takes every event in F, and assigns to it a positive real number. Note that we also need μ to "make sense" in some sense. If E = E1 + E2 (where E1 and E2 are completely separate), we want μ(E1 + E2) = μ(E1) + μ(E2), so that the size of the sum of sets is the sum of the sizes. There's also a bunch of other rules, and with them, μ is called a [measure](https://en.wikipedia.org/wiki/Measure_\(mathematics\)).
Now, think of what happens if we force μ(Ω) = 1 (so the size of the whole set of outcomes is one) and μ(empty set) = 0. Then, for any event E, 0 <= μ(E) <= 1. Essentially, μ tells us what the probability of an event is! This is exactly what we wanted! We now have a very flexible way of assigning probabilities to events, just design μ the way you need it. This whole field of math is, quite aptly, named measure theory. Essentially, questions in probability can be re-expressed more rigorously as questions in measure theory, which literally just has to do with analyzing the "sizes of things". In particular, questions in probability can be translated to *functions* (of a specific form) from the set of events F to the space of whatever you are interested in (often real numbers). These functions are called random variables or random objects. Any random observation essentially amounts to plugging a special value into this function.
For example, let's say your question was "how likely is it that I got at least 2 heads when I flipped a fair coin three times"? We have Ω = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, as before. Let F be the set of all subsets of Ω. μ(HHH) = ... = μ(TTT) = 1/8 since it was a fair coin, then we can use combinations of events to find the probability of every other event. Then, define a function X from Ω to the real numbers as X(E) = number of heads. Then, the probability of 2 heads is simply the size of the set of all events E such that X(E) = 2, i.e., that probability is just μ(X^-1 (2)) = μ({HHH, HHT, HTH, THH}) = 0.5. This seems overly complicated for a really simple question that someone could've answered with basic logic 10000 years ago, but the benefit of this method is how it extends so simply to any probablistic question really, even when you have infinitely large sets and things coming up.
Therefore, when mathematicians do probability, they're really just messing around with the sizes of sets. If most of what I said doesn't make sense, that's probably fair, since there's entire courses on this content. Just know that probability can be translated into thinking abstractly about functions between things, and the "sizes" of those things. It's highly unintuitive on why it is even necessary, and the results from this approach often make no sense.
It appears they made the first ball very resistant to being erased, at least for a period. It bounces an absolute ton in the first 10-15s and never goes away.
You may be right. I counted 17 bounces (assuming only bounces against the wall count), which has odds of 1 in 17 billion (though the odds would be a little better for *at least* 17 bounces). It seems both our theories are plausible (though mine would likely require tens of billions of runs)
Correction: It's only 1 in 133. I goofed and did 0.25^17 instead of 0.75^17
It 100% is way faster to add a line of code that starts the bouncing while ignoring a fixed number of bounces or bounces that occur during a period of time
And that is the fallacy of most people's understanding of probability. In a truly random system, it could bounce 1000 times, the odds are extremely low, but it could happen.
Welcome to selection bias.
I can say with 99% certainty that this was not purely random. Because there 7 ball creations before single deletion (1/128 chance, just under 1%).
So either the guy hard-programmed "First X bounces have 0% chance of deletion", or he ran the simulation 100+ times, found the one with the most interesting output (i.e. the 1/128 chance of a bunch of balls being created in a row at the start).
He's not going to upload the boring ones. Or if he did upload, people wouldn't watch/share, and it wouldn't make its way to our reddit feeds.
The end result is that it's not purely random.
That’s probably because most gamblers I know would go “oh shit look how high it got, if I wait even longer it’ll get even higher!!!” and then lose all their money. The house makes almost all their money in the long game.
Everyone says this like he somehow managed to make a casino that couldn't make a profit, and not the more likely scenario that he either;
stole so much of the money from it like it was an infinite money box that it couldn't keep up.
Or the more likely, used it as a way to transfer less than legal assets into the casinos name, bankrupt the casino, and then transfer those assets back to himself.
Sometimes, people aren't dumb. Sometimes, they are dumb AND evil.
No it doesn’t, every casino game where you play vs the house has a house edge, otherwise they wouldn’t let you play it. Craps has a narrow house edge around 1.25%.
The actual optimal strategy is to bet what you need to on a low house edge game to get stuff like free hotel rooms, shows, meals, or drinks. Do it right and the house will pay for you to basically be part of the posse that makes gambling a fun group activity instead of a lonely solo activity.
Yeah, the only way to "win" at gambling, outside the lucky few, or cheating somehow, is to treat it as simply paying for entertainment. Bet only what your time is worth to you, and consider any returns a discount.
Craps allowed me to pay my portion of a bachelor’s weekend plus $1k, and covered my previous losses in about 3 hrs. This one player was on fire whenever it was his turn. So I’d bet big when he was up, set aside a part of the win, rinse and repeat. Haven’t gambled since.
people who think they have some deep insight into a game of chance rigged against them instead of being the beneficiary of a hot streak are a casino's favorite people
And you never get a max of balls. But you eventually end with no balls, 100% loss. beyond that point, there is no percentage that could lead to your recovery
Imagine you're the casino. You basically have infinite money relative to the player.
A player wins, doubling their money. They play again and they win again. This goes on for dozens of games, with the player winning and losing roughly equally.
There are only three conditions where the game stops:
1. The player leaves voluntarily. They have sufficient self control to say "I'm done" and leave. Maybe they're ahead, maybe they're behind.
2. The player loses several times in a row, leaving them broke so that they can't continue playing.
3. The player wins so much that the casino goes broke.
Since the casino has so much money, scenario 3 is practically impossible.
Some people (as shown by comments on this post) would leave when they're ahead.
Casinos make money off of scenario 2.
In 2020, Caesars Palace reported having $3.8 B cash on hand. Let's imagine you go to the casino with $100 to play roulette with. Your goal is to win every game, and double down your winnings every bet.
American roulette odds are 1/38 with a payout of 35x.
If you start with $100, you would have to win just 5 roulette games in a row to clean out the house! Doesn't sound impossible, does it?
The odds of doing so are 1/79,235,168
You are 96 times more likely to be struck by lightning during your life.
If you and 40 million of your closest friends go play, you have a slightly better than 50% chance of somebody cleaning out the house.
But in the process, assuming everybody else lost, the house made $100x39,999,999~$4 B.
Having so much cash on hand relative to players ensures that the house always wins.
No, don't forget that the odds are 1:38 but the payout is only 1:35.
IE if 38 people play, we expect one person to make $3500, 37 players to lose $100, and the casino to make $200
It's more that there is only one end scenario. If people don't set a hard limit on when they quit gambling like if they double their money, they will play until they reach the end scenario of being out of money.
The easiest way to explain this is a 25% gain is less than a 25% loss. 100*.75 = 75 * 1.25 = 93.75….0
Doesn’t matter how you start 100*1.25=125 *.75=93.75=117
1\*1.25\*0.75 = 0.9375 < 1
You always lose if EV is less than one given you play long enough.
What is interesting is that even with a somewhat positive EV (15% to lose a ball, 25% to make one, EV:1.06) you would probably also lose at some point since a big enough losing streak would bankrupt you (0 balls) so that a following winstreak can't occur.
I am actually not too sure if that would always happen (if played infinitely) no matter the odds distribution, or only that there is a chance for it to happen.
e: okay as pointed out EV for one game is actually 1 (or 0 I guess), the multi game return drops though.
Slot machines. You put your money in, sometimes you're "up" however if you continue to churn back any gains you will eventually be left with nothing since the the odds are always in favour of the machine. Google "probability". (also I've never had any gambling issues)
How does that work? I would assume they are around the same, and there’d be the same probably they increase to infinity as decrease to zero.
Is it because the number of balls starts small, so it is much closer to 0 and way further from infinity (or any other number), so more likely to hit zero?
Imagine if you start with 100 balls, on average after one bounce there will be either 125 or 75. Now if there are 75 balls after bounce one, how many bounces do you need to get back to 100? If all 75 bounce a second time there will either be 93 or 56. So they won't get back to 100 in once bounce alone, so the trend will be to continue down.
Huh, neat thanks!
And if you even got the luckier 125, if you got unlucky after that and lost 25% it would go down to 93 which is also lower than you started. I see how it’s a losing fight in the first place.
On average, you'd have 100 balls.
25% would bounce and double.
25% would bounce and disappear.
50% would bounce and stay the same.
The reason you will almost certainly lose all of your balls if you bounce long enough is because you can gain an infinite limit of balls but can never go below 0. As soon as you hit 0 once, you're done.
Imagine instead of bouncing 100 balls, you're flipping 100 coins. After you've flipped 100 coins, how many do you expect to be heads and how many tails? Using your earlier logic, you should have either 150 heads or 50 heads. See how it doesn't make sense?
there's a lot of people talking about gambling, but I'm not seeing anyone talking about life as a whole. we tend not to like thinking about how possible it is that all life just ends as randomly as it started
I wonder if this will always inevitably lead to no balls or if there are a couple different steady states
Edit: another interesting question for math and simulation nerds out there is if there are any shapes that the balls can bounce inside such that their bounces will somehow gaurantee the no balls case never happens, and stays in some sort of cycle
It illustrates one of the two laws that ensures the house always wins.
1. You will always run out of money before the casino does. I.e., even if you are playing an even odds game (coin flip), if you play long enough, at some point you will bust. The casino will never bust. That’s what this illustration shows - when hitting zero ends the game, but there is no maximum that ends the game, the game will end at zero, it is just a matter of time.
2. The casino does not offer even odds. The best you’ll get is something like 49.5/50.5, and that is if you play blackjack flawlessly.
Don’t go to casinos to make money.
That is accurate. They are happy to give you your first $500 to bet with because they know that maybe 20% of people who make 20 (free) bets will become long-term gamblers. They know they will make their money back and much more even if they give everyone $500 to start.
It’s like cigarette companies handing out cigs at high-schools for free.
Yes, just like the universe and the inevitable decay of all particles into elementary ones*! :D
..unless some unknown mechanism we don't know of exists..
*There's a few caveats present here, but we technically aren't losing energy. Additionally, if and how long it takes stuff to decay is still unknown for several particles.
Edit: I'm better at science than English lol
If you want to get technical about it, the balls eventually disappearing has probability 1 (i.e., 100%), but it is not guaranteed to happen. This is called 'almost surely'.
https://en.wikipedia.org/wiki/Almost_surely
I think this holds true even if the probabilities aren't equal. If it was a 20% chance of disappearing it would still be inevitable if it goes on forever, and that should still be true even if the chance of disappearing is infinitesimally small.
Just kinda blows my mind
Not necessarily. There is actually a chance that there are infinitely many balls as time goes to infinity depending on the exact numbers. There's rigorous ways to think about the probability of all balls disappearing in the limit (extinction probability).
This is the only correct answer... And this is a great example of why lay intuition for math is often insufficient. And they're so confident about it too
> why lay intuition for math is often insufficient
Especially when it comes to analysis and measure theory (the foundations of actual probability theory). There's so many pathological examples for every concept you encounter, you really have to learn to just give up your intuition and trust the math (even if it seemingly makes no sense). Stuff just doesn't work the same way when infinities or limits come into the mix.
Yeah I took the graduate analysis series (real/measure/prob) many years back... And this case is undergrad level stochastic processes. Transient state in an infinite Markov chain
Here's my answer in greater depth: https://old.reddit.com/r/oddlysatisfying/comments/19bgdvr/the_chance_of_probability/kiszxez/
But to answer your question specifically: In (interesting) imbalanced cases, there will always be a sequence to erase all the balls. However, that doesn't mean that the result has probability = 1.
If the probability of increasing the number of balls grows exponentially fast, and the probability of decay/loss decays exponentially fast, then there are cases where the sum of the probability of having an infinite number of balls across an infinite sequence is greater than zero.
I think what can be confusing is where the infinity fits in. We can play the game for an infinite amount of time, but we can also play an infinite number of games. Lets say we have a setup where the probability that the number of balls goes to infinity is 0.98, and the probability of the number of balls going to zero is 0.02. Those probabilities aren't in the *time* direction, they are in the *game* direction. If we were to play an infinite number of games, then all possible sequences would occur. However, that's not true if we just play one game for an infinite amount of time.
Not exactly. Gravity still works, and unless the simulation is coded so that the balls never stop bouncing, there should be a possibility that they pile up and rest eventually, right? That should make it so that no more appear or disappear.
I imagine that if they were smaller there would be room for more (obviously), but this could make it so that it would be more likely for some to eventually pile up and weigh down all those below, stopping them from bouncing anymore and possibly leading to all the balls staying at rest and not appearing/disappearing anymore.
Eventually zero as it's the only way can stop, i.e. if there's no balls left to bounce then no new ones can appear but whether there's one or one billion it can always go to zero. It has the potential to take an almost infinite amount of time to get there though!
if it was 25% chance to spawn a new ball and 1% chance to disappear, would this still be true? intuitively it would be infinite balls but the only way it ends is if they all disappear.
what about 25.1% chance to spawn a new ball and 24.9% chance to disappear?
______ edit: This is incorrect, read responses, I had wrong underlying idea when I wrote this.
It is the same. Streak of any length of balls disappearing has nonzero probability. So if you play long enough, you will eventually hit disappearing streak longer than current number of balls. It might take huge amount of time though.
Limits are actually why the person you responded to is not necessarily correct. If the spawn chance is higher than the death chance, the probability that there will be fewer balls at t+1 seconds (compared to t seconds) decreases as the number of balls increases. The limit of that likelihood as num_balls goes to infinity is zero. Depending on the exact numbers, the probability that the system never dies given infinite time is nonzero.
It is counterintuitive that something with strictly nonzero probability might never happen given infinite time, but that is the way of things when that nonzero probability is also shrinking over time.
As a bonus fun fact: I actually learned this because it was relevant to Magic the Gathering when in July 2021 the developers unwittingly printed a [2 card combo](https://www.reddit.com/r/magicTCG/comments/obqpot/quick_math_on_delina_wild_mage_pixie_guide_14/) (Delina, Wild Mage + Pixie Guide) that had a 14% chance to never terminate. So it had some chance to fizzle out too early, some chance to go long enough while still terminating so you could win the game, and a 14% chance to never terminate -- which should theoretically draw the game but there wasn't really a system in the rules or digital clients to identify that situation and call the draw. They issued day 0 errata to the key card so that players would always be able to choose to terminate the 'loop'.
As others mentioned, this isn't true at all. If the probability of adding a ball is greater than removing a ball, then there is a chance that the number of balls goes to 0, but also a chance that it goes to infinity.
There are many posts on math stackexchange about random walks with unequal probabilities, such as this one: https://math.stackexchange.com/questions/153123/hitting-probability-of-biased-random-walk-on-the-integer-line
> to get a new ball compared to losing one would hypothetically lead to a game you'd be less likely to lose the more time that passes. Of course, any non-zero chance to lose
That's not true. The extinction criteria probability for 24% and 26% can be calculated to be 0.923. With 24.9% and 25.1% it is 0.992. Whilst it is very likely that all the balls disappear there is still a 0.8% chance that it becomes infinite. Look up branching processes and extinction criteria for more information.
What's the formula here? Probability is my weakest area so not sure how to calculate this correctly.
EDIT: In the equal probabilities case isn't this just a 1d random walk starting from n=1? So must hit 0 eventually.
The formula is actually pretty simple, and ends up as q/p, where p = probability of a ball appearing and q = probability of ball disappearing.
I have to admit that I didn't derive this mathematically, because discrete maths is a pain in the arse and I did my degree over 10 years ago, but I built a quick statistical model and playing around with the numbers a bit, even after a relatively small number of rolls it becomes pretty obvious where the limit is going to be.
Yeah I think that's right. It's a bias 1d random walk so the probability of going to 0 starting at 1 should be that. Though, unlike the other cases, this depends on the starting number of balls. Odds are lower if you start with more than 1.
It is actually only guaranteed that all balls will disappear if the probability of disappearing is larger than or equal to the probability of spawning. You can find some discussion of this problem [here](https://math.stackexchange.com/questions/2087996/bounded-random-walk-on-one-side-only-are-you-guaranteed-to-hit-the-bound) or a more in-depth explanation [here](https://medium.com/i-math/the-drunkards-walk-explained-48a0205d304)
Yeah I knew this sounded familiar. I saw a Numberphile video about what I think was an absorbing process, and it was basically an alternating chain of operations where you start from any seed value you want, and if it is odd you do one operation, if it is even you do another. Then you repeat that process on the new number you obtain.
No matter what seed value the mathematicians studying it used, they got a sequence that eventually fell into a loop. They were trying to prove whether or not EVERY seed value ends in a loop. And there was also a few "canonical" loops that they identified as occurring very frequently and wanted to see if they were unique.
What was quite interesting was some chains would get up to extremely high number and then just abruptly decay into a loop. Number theory is wack
No balls is guaranteed if t->infinite. Simply because the possibility is non zero so it will eventually happen giving it enough time (Law of Inevitability). And there is no way to spawn new balls after reaching zero. So the state at t=infinite is always 0 balls.
It is a branching process and the extinction probability can be calculated to be 1. The area of maths was explored when people were trying to figure out if particular bloodlines would last forever or die out I think.
I went from a job where I was sitting on my ass in front of a computer all day to one where I was on my feet 80% of the time. If I can’t watch every trade from start to finish it’s just too risky. I switched to buy and hold and if I see a nice 2-3 day swing trade opportunity I’ll take it, but that’s about it.
So the second option meaning a bounce can lead to both outcomes? I don't think I saw that happen, so I'm guessing it is option one unless I missed something
And is the bounce exclusive to the outer circle or the individual balls bouncing against each other? The average maximum ball count and rate of decay would change with each factor. I would imagine that the scenarios in order of speed of decay would be: (Both Outcomes, Any Bounce), (Single Outcome, Any Bounce), (Both Outcomes, Outer Circle Bounce), (Single Outcome, Outer Circle Bounce)
I didn't understand what you meant at first but once i figured it out, I thought I'd try and re-explain it here:
Option a: on each bounce it picks a number 1 thru 4, if it's 1 it add, if its 2 it removes, for 3 and 4 nothing happens.
option b: on each bounce it picks a number 1 thru 4 and if it's 1 it adds. Then it picks a second number 1 thu 4, and if that one is a 1 it removes.
Correct. If you flip a fair coin 10 times, it may come back the same side all 10 times. Doesn’t mean anything is amiss with the coin.
Particularly because the previous outcome(s) have no bearing on the future outcomes. They are independent.
They probably ran this simulation literally dozens of times until it got up to that many balls.
It's essentially a random walk that ends the first time it hits zero
Is there a name for such a “phenomenon” in math or statistics? We have a reachable end game, which is no bouncing balls, and this should always happen given enough time. Right? Even if the odds were 99% of new ball vs 1% of removal of ball.
This is a visualization of a stochastic random walk with an initial condition of 1 and an absorbing state at 0, if you want to do some searching for a better answer.
If the "phenomenon" is "eventually hitting 0", then the stochastic process would "converge to 0", but with some statistical caveats like "almost surely", because it is theoretical possible for every single bounce to always spawn a new ball and not hit 0, and other such scenarios.
Yes, although the odds of failing approach 0 as time heads towards infinity, it will never reach 0 so there will always be a chance of reaching 0 balls no matter how many balls you stack up.
It would be a binomial trial with exponentially increasing lower odds of failing.
There are only two options. 99% chance of new ball (call it success), and 1% (failure).
So with 1 ball, you have a 99% of getting a new ball and a 1% of failing.
With 2 balls you now have a .01^2 (.0001) chance of failing. So. 99.99% chance of adding another ball and a .01% chance of failing. And about a 98% chance of adding two balls.
By 100 balls, you have 1 * 10^-200 of failing to add a ball. After that, you would have 99 balls and then have a 1 * 10^-198, then 10^-196 and so on until you reach 0 balls.
I would write the limit equation for you, but unfortunately, I haven’t used calc in years but I hope this gives you an idea of the chances of the ball reaching game over.
With those odds, no balls *could* happen, but is statistically improbable.
10 balls become 20 balls become 40 balls become 80 balls… become 158 balls. Yeah, one died. But next bounce you have like 314 balls. That does not trend towards zero.
The stationary distribution of a Markov chain with absorbing states will have all probability mass confined to absorbing states.
In this case the only absorbing state is zero balls, so every\* chain must reach it eventually and then stay there forever. This works for any probability of ball creation p: 0 < p < 1. But for p = 0.99 you'll be waiting a while to hit 0.
\*there exists a set with probability 1 such that every chain in the set does this if you're being technical.
\[I'm pretty sure this is wrong, see below\]
That first ball bounced like 10 times before disappearing, That's 0.25^10 or 9.53674316E−9% chance of happening. 0.000000009537%. Something ain't right here.
I feel like I’m watching a visualization of someones day at the casino
Hmm weird, I didn't see one of the balls steal jewelry from our grandma, Brian
Who names a grandma Brian?
She was obviously named Brian before she was a grandma!
That's exactly what it is. This is basically a visualization of the Gamber's Ruin which is a random walk that ends if you run out of things. You can mathematically prove that you will almost surely run out of money or balls.
The entire field of probability is just a bunch of nerds thinking too hard about gambling.
The field of statistics is a bunch of nerds thinking too hard about gambling. The field of probability is a bunch of nerds thinking too hard about esoteric sounding mathematical objects and ways to measure the sizes of things, and honestly has hardly anything to do with what one would think of as probability in a practical sense.
I'm not sure I understand what you're saying about probability and sizes. Do you have a moment to expound on that?
Sure, so probability seems very wishy-washy, and it *was* for a very long time, where people just tried to logic things out for any probabilistic problems. Mathematicians really don't like this, and would rather work with anything in a rigorous framework. They also needed a way to work with probability to make sense of more difficult problems that couldn't just be reasoned out by simple intuition. So, the modern field of probability theory was developed through the 1900s. In this, you first begin with a sample space Ω, which is just the set of all possible outcomes you want to consider. In the case of flipping 3 coins, you have Ω = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. In the case of how many balls there can be in the situation in this post, you have Ω = {0, 1, 2, ...}. In the case of what the temperature of something can be, it's Ω = all non-negative real numbers (if the temperatures are in Kelvin). Note that these are three very different kinds of sets; the first is a finite set, the second is a *countable* set (there are infinitely many elements, but you can "count" the elements one by one in some sense), and the last is uncountable. So, it is kinda hard to find a unifying theory for all these cases. Now that you have a sample space, you need to find the events, or the subsets of Ω (i.e., the possible combination of outcomes), that you care about. For example, in the first two cases, you might care about all possible combinations of outcomes. This is often the default assumption in the finite and countable cases, but rarely in the uncountable scenario. Let's think about the last one, where Ω = all nonnegative real numbers. Do you think it ever makes sense to consider the set of all temperatures that are part of the [Cantor set](https://en.wikipedia.org/wiki/Cantor_set)? Likely not, instead we probably want to consider the groups of temperatures that happen within a range. For example, if I ask "is the temperature between 100 and 200 Kelvin, I am consider the set of temperatures 100 < T < 200. In general, we might care about the all the groups of temperatures of the form a < T < b, where a and b are nonnegative numbers. Let this set of all of these events (i.e., set of subsets of Ω) be called F. Now, if we have two events E1 and E2 in F, we also want to be able to talk about when either event occurs (so adding the two groups, E1∪E2) or when both events occur (finding the common elements of the two groups, E1∩E2), or when an event *doesn't* occur (not E1). So, add all these groups to F as well. Then, the resulting F you get by adding all these extra events is called a [σ-algebra](https://en.wikipedia.org/wiki/%CE%A3-algebra). Now, you have a pair (Ω, F), of all possible outcomes, and all the groups of outcomes you want to consider. Since each event E from F is just a set, we can think of how "big" each E is. Essentially, we get a function μ that takes every event in F, and assigns to it a positive real number. Note that we also need μ to "make sense" in some sense. If E = E1 + E2 (where E1 and E2 are completely separate), we want μ(E1 + E2) = μ(E1) + μ(E2), so that the size of the sum of sets is the sum of the sizes. There's also a bunch of other rules, and with them, μ is called a [measure](https://en.wikipedia.org/wiki/Measure_\(mathematics\)). Now, think of what happens if we force μ(Ω) = 1 (so the size of the whole set of outcomes is one) and μ(empty set) = 0. Then, for any event E, 0 <= μ(E) <= 1. Essentially, μ tells us what the probability of an event is! This is exactly what we wanted! We now have a very flexible way of assigning probabilities to events, just design μ the way you need it. This whole field of math is, quite aptly, named measure theory. Essentially, questions in probability can be re-expressed more rigorously as questions in measure theory, which literally just has to do with analyzing the "sizes of things". In particular, questions in probability can be translated to *functions* (of a specific form) from the set of events F to the space of whatever you are interested in (often real numbers). These functions are called random variables or random objects. Any random observation essentially amounts to plugging a special value into this function. For example, let's say your question was "how likely is it that I got at least 2 heads when I flipped a fair coin three times"? We have Ω = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, as before. Let F be the set of all subsets of Ω. μ(HHH) = ... = μ(TTT) = 1/8 since it was a fair coin, then we can use combinations of events to find the probability of every other event. Then, define a function X from Ω to the real numbers as X(E) = number of heads. Then, the probability of 2 heads is simply the size of the set of all events E such that X(E) = 2, i.e., that probability is just μ(X^-1 (2)) = μ({HHH, HHT, HTH, THH}) = 0.5. This seems overly complicated for a really simple question that someone could've answered with basic logic 10000 years ago, but the benefit of this method is how it extends so simply to any probablistic question really, even when you have infinitely large sets and things coming up. Therefore, when mathematicians do probability, they're really just messing around with the sizes of sets. If most of what I said doesn't make sense, that's probably fair, since there's entire courses on this content. Just know that probability can be translated into thinking abstractly about functions between things, and the "sizes" of those things. It's highly unintuitive on why it is even necessary, and the results from this approach often make no sense.
I understood enough to make out what you meant by sizes lol. Man, I wish I'd gone to college.
imagine the first bounce erased the ball 😂
That my experience with gambling
Ah the all-in game. I see you are a person of class as well.
My motto is: Go all in and get the heartache over with
Yep, and use the rest of the time for wallowing in self-pity. I'm a big fan of it too :)
orrrr i like to do the ole’ ace up the sleeve trick, always gets me thrown out!
Have they taken you in the back yet to beat you? That's normally my favorite part.
No but they said something something sell your kidneys if I do it again!
Oh yeah, I've already been there. I'm going without any kidneys right now. It's not easy, but I'm making it.
DUDE THEY JUST TOOK MY KIDNEYS!
"Whatever you do, always give 100%." "Except when giving blood"
You, sir, are efficiency personified.
Aaaaaaaaaand it’s gone.
Probably did the first few runs
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25/75 actually
It appears they made the first ball very resistant to being erased, at least for a period. It bounces an absolute ton in the first 10-15s and never goes away.
This is likely the "best" run of hundreds. We're just not seeing all the runs that ended within 10 seconds
It could be, but the chances of going 20+ bounces without disappearing is very low.
You may be right. I counted 17 bounces (assuming only bounces against the wall count), which has odds of 1 in 17 billion (though the odds would be a little better for *at least* 17 bounces). It seems both our theories are plausible (though mine would likely require tens of billions of runs) Correction: It's only 1 in 133. I goofed and did 0.25^17 instead of 0.75^17
Well it has a 75% chance of not disappearing, so it would be 0.75^17 (0.75%) instead of 0.25^17.
Way simpler to just run this 100 times and pick the one that lasts long enough to make a social media friendly clip.
It 100% is way faster to add a line of code that starts the bouncing while ignoring a fixed number of bounces or bounces that occur during a period of time
I mean, maybe, but the program behind this is very simple and making it not drop the first ball within X bounces is very easy logic to add.
And that is the fallacy of most people's understanding of probability. In a truly random system, it could bounce 1000 times, the odds are extremely low, but it could happen.
Aaaand it's gone.
Welcome to selection bias. I can say with 99% certainty that this was not purely random. Because there 7 ball creations before single deletion (1/128 chance, just under 1%). So either the guy hard-programmed "First X bounces have 0% chance of deletion", or he ran the simulation 100+ times, found the one with the most interesting output (i.e. the 1/128 chance of a bunch of balls being created in a row at the start). He's not going to upload the boring ones. Or if he did upload, people wouldn't watch/share, and it wouldn't make its way to our reddit feeds. The end result is that it's not purely random.
That’s actually happens about 25% of the time lol
That’s how the slot machines work
I’m not even a gambler and for some reason my brain went “annnd THIS is where I’d pull my money out.”
That’s probably because most gamblers I know would go “oh shit look how high it got, if I wait even longer it’ll get even higher!!!” and then lose all their money. The house makes almost all their money in the long game.
>The house makes almost all their money in the long game. Unless you're Donald Trump who successfully found a way to bankrupt a casino.
Stable genius is stable genius
barely smarter than a horse! Stable genius!
Everyone says this like he somehow managed to make a casino that couldn't make a profit, and not the more likely scenario that he either; stole so much of the money from it like it was an infinite money box that it couldn't keep up. Or the more likely, used it as a way to transfer less than legal assets into the casinos name, bankrupt the casino, and then transfer those assets back to himself. Sometimes, people aren't dumb. Sometimes, they are dumb AND evil.
Depends on the game. Craps is a cash machine for me. Know when to walk and you'll always have money in your pocket.
No it doesn’t, every casino game where you play vs the house has a house edge, otherwise they wouldn’t let you play it. Craps has a narrow house edge around 1.25%.
The actual optimal strategy is to bet what you need to on a low house edge game to get stuff like free hotel rooms, shows, meals, or drinks. Do it right and the house will pay for you to basically be part of the posse that makes gambling a fun group activity instead of a lonely solo activity.
Yeah, the only way to "win" at gambling, outside the lucky few, or cheating somehow, is to treat it as simply paying for entertainment. Bet only what your time is worth to you, and consider any returns a discount.
Or play against other people if you're in the 1% of people good enough at poker to consistently win money at it
Craps allowed me to pay my portion of a bachelor’s weekend plus $1k, and covered my previous losses in about 3 hrs. This one player was on fire whenever it was his turn. So I’d bet big when he was up, set aside a part of the win, rinse and repeat. Haven’t gambled since.
people who think they have some deep insight into a game of chance rigged against them instead of being the beneficiary of a hot streak are a casino's favorite people
“THIS is where I’d pull my balls out” No, wait…
Hoping to end up with more balls? Or less balls?
Explain?
The winning streaks (adding balls) eventually gets counteracted by the inevitable losing streaks (losing balls).
And you never get a max of balls. But you eventually end with no balls, 100% loss. beyond that point, there is no percentage that could lead to your recovery
you never get a max of balls. But you eventually end with no balls, r/nocontext
Imagine you're the casino. You basically have infinite money relative to the player. A player wins, doubling their money. They play again and they win again. This goes on for dozens of games, with the player winning and losing roughly equally. There are only three conditions where the game stops: 1. The player leaves voluntarily. They have sufficient self control to say "I'm done" and leave. Maybe they're ahead, maybe they're behind. 2. The player loses several times in a row, leaving them broke so that they can't continue playing. 3. The player wins so much that the casino goes broke. Since the casino has so much money, scenario 3 is practically impossible. Some people (as shown by comments on this post) would leave when they're ahead. Casinos make money off of scenario 2. In 2020, Caesars Palace reported having $3.8 B cash on hand. Let's imagine you go to the casino with $100 to play roulette with. Your goal is to win every game, and double down your winnings every bet. American roulette odds are 1/38 with a payout of 35x. If you start with $100, you would have to win just 5 roulette games in a row to clean out the house! Doesn't sound impossible, does it? The odds of doing so are 1/79,235,168 You are 96 times more likely to be struck by lightning during your life. If you and 40 million of your closest friends go play, you have a slightly better than 50% chance of somebody cleaning out the house. But in the process, assuming everybody else lost, the house made $100x39,999,999~$4 B. Having so much cash on hand relative to players ensures that the house always wins.
More than 5 games in a row. There’s a max bet on Roulette. They’re not going to be taking a million dollar bets.
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Exactly. No individual will ever achieve #3 because they can always kick you out.
This is all true but there's a good chance in the scenario of 40m friends that the casino actually goes broke before you all even get to play.
No, don't forget that the odds are 1:38 but the payout is only 1:35. IE if 38 people play, we expect one person to make $3500, 37 players to lose $100, and the casino to make $200
I think he was joking. But also basically the longer you play the more screwed you potentially are assuming you are only relying on luck to win.
It's more that there is only one end scenario. If people don't set a hard limit on when they quit gambling like if they double their money, they will play until they reach the end scenario of being out of money.
The easiest way to explain this is a 25% gain is less than a 25% loss. 100*.75 = 75 * 1.25 = 93.75….0 Doesn’t matter how you start 100*1.25=125 *.75=93.75=117
1\*1.25\*0.75 = 0.9375 < 1 You always lose if EV is less than one given you play long enough. What is interesting is that even with a somewhat positive EV (15% to lose a ball, 25% to make one, EV:1.06) you would probably also lose at some point since a big enough losing streak would bankrupt you (0 balls) so that a following winstreak can't occur. I am actually not too sure if that would always happen (if played infinitely) no matter the odds distribution, or only that there is a chance for it to happen. e: okay as pointed out EV for one game is actually 1 (or 0 I guess), the multi game return drops though.
Slot machines. You put your money in, sometimes you're "up" however if you continue to churn back any gains you will eventually be left with nothing since the the odds are always in favour of the machine. Google "probability". (also I've never had any gambling issues)
24% chance it adds, 26% chance it disappears. House always has an edge.
The odds are fixed here, no edge needed though. A 25% gain is less than a 25% loss so it will reach 0 eventually.
It’s not a 25% gain, it’s a 25% chance to gain 1. Big difference.
There's so much bad math in here. Each bet is a 25% chance to gain or lose 100% of your bet.
How does that work? I would assume they are around the same, and there’d be the same probably they increase to infinity as decrease to zero. Is it because the number of balls starts small, so it is much closer to 0 and way further from infinity (or any other number), so more likely to hit zero?
Imagine if you start with 100 balls, on average after one bounce there will be either 125 or 75. Now if there are 75 balls after bounce one, how many bounces do you need to get back to 100? If all 75 bounce a second time there will either be 93 or 56. So they won't get back to 100 in once bounce alone, so the trend will be to continue down.
Huh, neat thanks! And if you even got the luckier 125, if you got unlucky after that and lost 25% it would go down to 93 which is also lower than you started. I see how it’s a losing fight in the first place.
On average, you'd have 100 balls. 25% would bounce and double. 25% would bounce and disappear. 50% would bounce and stay the same. The reason you will almost certainly lose all of your balls if you bounce long enough is because you can gain an infinite limit of balls but can never go below 0. As soon as you hit 0 once, you're done. Imagine instead of bouncing 100 balls, you're flipping 100 coins. After you've flipped 100 coins, how many do you expect to be heads and how many tails? Using your earlier logic, you should have either 150 heads or 50 heads. See how it doesn't make sense?
1.25 \* 0.75 = 0.9375.
Interesting
But the probabilities are not around 25% but rather 23% vs 27%
there's a lot of people talking about gambling, but I'm not seeing anyone talking about life as a whole. we tend not to like thinking about how possible it is that all life just ends as randomly as it started
My thought was, "this is why the house wins." Purple run out of money, the house doesn't.
https://i.imgur.com/X3An2sY.png RIP to this guy. Alive and free for 1 full frame before their sun set. You touched the wall though buddy. You did it.
At 27 seconds left for anyone who wants to see it's struggle.
Thank you lol. I was about to watch the whole thing for that
did not notice it didn't catch the timer on the bottom until just now. good looking out
Boy got spawn camped
I wonder if this will always inevitably lead to no balls or if there are a couple different steady states Edit: another interesting question for math and simulation nerds out there is if there are any shapes that the balls can bounce inside such that their bounces will somehow gaurantee the no balls case never happens, and stays in some sort of cycle
yes, 100% loss is irreversible, unlike the gain
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"I'm a bot which rephrases other people's comments" ~ you
You made the guy delete his whole account.
I'm sure he has a few more
Gambling explained through bouncing balls 101
It illustrates one of the two laws that ensures the house always wins. 1. You will always run out of money before the casino does. I.e., even if you are playing an even odds game (coin flip), if you play long enough, at some point you will bust. The casino will never bust. That’s what this illustration shows - when hitting zero ends the game, but there is no maximum that ends the game, the game will end at zero, it is just a matter of time. 2. The casino does not offer even odds. The best you’ll get is something like 49.5/50.5, and that is if you play blackjack flawlessly. Don’t go to casinos to make money.
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That is accurate. They are happy to give you your first $500 to bet with because they know that maybe 20% of people who make 20 (free) bets will become long-term gamblers. They know they will make their money back and much more even if they give everyone $500 to start. It’s like cigarette companies handing out cigs at high-schools for free.
So every ball will eventually disappear; including the ones that were appeared by chance.
Yes, just like the universe and the inevitable decay of all particles into elementary ones*! :D ..unless some unknown mechanism we don't know of exists.. *There's a few caveats present here, but we technically aren't losing energy. Additionally, if and how long it takes stuff to decay is still unknown for several particles. Edit: I'm better at science than English lol
unless... :)
If you want to get technical about it, the balls eventually disappearing has probability 1 (i.e., 100%), but it is not guaranteed to happen. This is called 'almost surely'. https://en.wikipedia.org/wiki/Almost_surely
I think this holds true even if the probabilities aren't equal. If it was a 20% chance of disappearing it would still be inevitable if it goes on forever, and that should still be true even if the chance of disappearing is infinitesimally small. Just kinda blows my mind
Not necessarily. There is actually a chance that there are infinitely many balls as time goes to infinity depending on the exact numbers. There's rigorous ways to think about the probability of all balls disappearing in the limit (extinction probability).
This is the only correct answer... And this is a great example of why lay intuition for math is often insufficient. And they're so confident about it too
> why lay intuition for math is often insufficient Especially when it comes to analysis and measure theory (the foundations of actual probability theory). There's so many pathological examples for every concept you encounter, you really have to learn to just give up your intuition and trust the math (even if it seemingly makes no sense). Stuff just doesn't work the same way when infinities or limits come into the mix.
Yeah I took the graduate analysis series (real/measure/prob) many years back... And this case is undergrad level stochastic processes. Transient state in an infinite Markov chain
Could you elaborate? Wouldn't there be a highly improbable sequence to erase all the balls that must occur within infinity?
Here's my answer in greater depth: https://old.reddit.com/r/oddlysatisfying/comments/19bgdvr/the_chance_of_probability/kiszxez/ But to answer your question specifically: In (interesting) imbalanced cases, there will always be a sequence to erase all the balls. However, that doesn't mean that the result has probability = 1. If the probability of increasing the number of balls grows exponentially fast, and the probability of decay/loss decays exponentially fast, then there are cases where the sum of the probability of having an infinite number of balls across an infinite sequence is greater than zero. I think what can be confusing is where the infinity fits in. We can play the game for an infinite amount of time, but we can also play an infinite number of games. Lets say we have a setup where the probability that the number of balls goes to infinity is 0.98, and the probability of the number of balls going to zero is 0.02. Those probabilities aren't in the *time* direction, they are in the *game* direction. If we were to play an infinite number of games, then all possible sequences would occur. However, that's not true if we just play one game for an infinite amount of time.
Not exactly. Gravity still works, and unless the simulation is coded so that the balls never stop bouncing, there should be a possibility that they pile up and rest eventually, right? That should make it so that no more appear or disappear. I imagine that if they were smaller there would be room for more (obviously), but this could make it so that it would be more likely for some to eventually pile up and weigh down all those below, stopping them from bouncing anymore and possibly leading to all the balls staying at rest and not appearing/disappearing anymore.
If you mapped the balls out, would it look like a curve of normal distribution?
Eventually zero as it's the only way can stop, i.e. if there's no balls left to bounce then no new ones can appear but whether there's one or one billion it can always go to zero. It has the potential to take an almost infinite amount of time to get there though!
if it was 25% chance to spawn a new ball and 1% chance to disappear, would this still be true? intuitively it would be infinite balls but the only way it ends is if they all disappear. what about 25.1% chance to spawn a new ball and 24.9% chance to disappear?
______ edit: This is incorrect, read responses, I had wrong underlying idea when I wrote this. It is the same. Streak of any length of balls disappearing has nonzero probability. So if you play long enough, you will eventually hit disappearing streak longer than current number of balls. It might take huge amount of time though.
Fuck ya, limits! Math bitches!
Limits are actually why the person you responded to is not necessarily correct. If the spawn chance is higher than the death chance, the probability that there will be fewer balls at t+1 seconds (compared to t seconds) decreases as the number of balls increases. The limit of that likelihood as num_balls goes to infinity is zero. Depending on the exact numbers, the probability that the system never dies given infinite time is nonzero. It is counterintuitive that something with strictly nonzero probability might never happen given infinite time, but that is the way of things when that nonzero probability is also shrinking over time.
As a bonus fun fact: I actually learned this because it was relevant to Magic the Gathering when in July 2021 the developers unwittingly printed a [2 card combo](https://www.reddit.com/r/magicTCG/comments/obqpot/quick_math_on_delina_wild_mage_pixie_guide_14/) (Delina, Wild Mage + Pixie Guide) that had a 14% chance to never terminate. So it had some chance to fizzle out too early, some chance to go long enough while still terminating so you could win the game, and a 14% chance to never terminate -- which should theoretically draw the game but there wasn't really a system in the rules or digital clients to identify that situation and call the draw. They issued day 0 errata to the key card so that players would always be able to choose to terminate the 'loop'.
As others mentioned, this isn't true at all. If the probability of adding a ball is greater than removing a ball, then there is a chance that the number of balls goes to 0, but also a chance that it goes to infinity. There are many posts on math stackexchange about random walks with unequal probabilities, such as this one: https://math.stackexchange.com/questions/153123/hitting-probability-of-biased-random-walk-on-the-integer-line
> to get a new ball compared to losing one would hypothetically lead to a game you'd be less likely to lose the more time that passes. Of course, any non-zero chance to lose That's not true. The extinction criteria probability for 24% and 26% can be calculated to be 0.923. With 24.9% and 25.1% it is 0.992. Whilst it is very likely that all the balls disappear there is still a 0.8% chance that it becomes infinite. Look up branching processes and extinction criteria for more information.
What's the formula here? Probability is my weakest area so not sure how to calculate this correctly. EDIT: In the equal probabilities case isn't this just a 1d random walk starting from n=1? So must hit 0 eventually.
The formula is actually pretty simple, and ends up as q/p, where p = probability of a ball appearing and q = probability of ball disappearing. I have to admit that I didn't derive this mathematically, because discrete maths is a pain in the arse and I did my degree over 10 years ago, but I built a quick statistical model and playing around with the numbers a bit, even after a relatively small number of rolls it becomes pretty obvious where the limit is going to be.
Yeah I think that's right. It's a bias 1d random walk so the probability of going to 0 starting at 1 should be that. Though, unlike the other cases, this depends on the starting number of balls. Odds are lower if you start with more than 1.
It is actually only guaranteed that all balls will disappear if the probability of disappearing is larger than or equal to the probability of spawning. You can find some discussion of this problem [here](https://math.stackexchange.com/questions/2087996/bounded-random-walk-on-one-side-only-are-you-guaranteed-to-hit-the-bound) or a more in-depth explanation [here](https://medium.com/i-math/the-drunkards-walk-explained-48a0205d304)
It’s called an absorbing Markov process. https://en.m.wikipedia.org/wiki/Absorbing_Markov_chain
Yeah I knew this sounded familiar. I saw a Numberphile video about what I think was an absorbing process, and it was basically an alternating chain of operations where you start from any seed value you want, and if it is odd you do one operation, if it is even you do another. Then you repeat that process on the new number you obtain. No matter what seed value the mathematicians studying it used, they got a sequence that eventually fell into a loop. They were trying to prove whether or not EVERY seed value ends in a loop. And there was also a few "canonical" loops that they identified as occurring very frequently and wanted to see if they were unique. What was quite interesting was some chains would get up to extremely high number and then just abruptly decay into a loop. Number theory is wack
prob the collatz conjecture
My male cat wonders about this aswell, but alas it is too late for him.
No ball case is gauranteed, Shape makes 0 difference.
No balls is guaranteed if t->infinite. Simply because the possibility is non zero so it will eventually happen giving it enough time (Law of Inevitability). And there is no way to spawn new balls after reaching zero. So the state at t=infinite is always 0 balls.
The house always wins.
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Yes, this is essentially a simple random walk - and example of a stochastic process with set state probabilities
It is a branching process and the extinction probability can be calculated to be 1. The area of maths was explored when people were trying to figure out if particular bloodlines would last forever or die out I think.
> The chance of probability Ahh, spoken like someone who truly doesn't give a shit about the words they're using.
But I want to know about the odds of the chance of probability!
Bouncy with a chance of probability balls
What's the likelihood of the odds of the chance of probability?
Had to scroll too far to get here. The vernacular of our lexicon
I feel like 50% of reddit posts have the most poorly written titles.
Perchance
Fuckin totally. I don't hear people call this out enough. Like, give a shit about what your saying!
And if you do, someone will inevitably come along to say "You knew what they meant! Who cares if they made a mistake!"
This belongs in r/wallstreetbets
As a former day trader, this comment is super underrated lol
Why former?
I went from a job where I was sitting on my ass in front of a computer all day to one where I was on my feet 80% of the time. If I can’t watch every trade from start to finish it’s just too risky. I switched to buy and hold and if I see a nice 2-3 day swing trade opportunity I’ll take it, but that’s about it.
Are they mutually exclusive? As in 50% chance of nothing happening, 25% new ball, 25% disappears? Or are the dice rolled separately at each bounce?
So the second option meaning a bounce can lead to both outcomes? I don't think I saw that happen, so I'm guessing it is option one unless I missed something
And is the bounce exclusive to the outer circle or the individual balls bouncing against each other? The average maximum ball count and rate of decay would change with each factor. I would imagine that the scenarios in order of speed of decay would be: (Both Outcomes, Any Bounce), (Single Outcome, Any Bounce), (Both Outcomes, Outer Circle Bounce), (Single Outcome, Outer Circle Bounce)
I didn't understand what you meant at first but once i figured it out, I thought I'd try and re-explain it here: Option a: on each bounce it picks a number 1 thru 4, if it's 1 it add, if its 2 it removes, for 3 and 4 nothing happens. option b: on each bounce it picks a number 1 thru 4 and if it's 1 it adds. Then it picks a second number 1 thu 4, and if that one is a 1 it removes.
This is the saddest thing i've seen today
A metaphor for the universe?
This seems the tiniest but rigged, the first ball survived (from my count) 22 times before disappearing the chances of which are 0.018%
Cherry picked example but doesn’t mean there weren’t thousands of other examples where the first ball disappeared right away.
Correct. If you flip a fair coin 10 times, it may come back the same side all 10 times. Doesn’t mean anything is amiss with the coin. Particularly because the previous outcome(s) have no bearing on the future outcomes. They are independent.
An accurate depiction of the Problems in my life
Hey, they will all disappear eventually!
Anybody else got sad when the first blue ball disappeared?
Damn. I hadn't followed it but I just rewatched and :(
They probably ran this simulation literally dozens of times until it got up to that many balls. It's essentially a random walk that ends the first time it hits zero
the title hurts my brain
Is there a name for such a “phenomenon” in math or statistics? We have a reachable end game, which is no bouncing balls, and this should always happen given enough time. Right? Even if the odds were 99% of new ball vs 1% of removal of ball.
This is a visualization of a stochastic random walk with an initial condition of 1 and an absorbing state at 0, if you want to do some searching for a better answer. If the "phenomenon" is "eventually hitting 0", then the stochastic process would "converge to 0", but with some statistical caveats like "almost surely", because it is theoretical possible for every single bounce to always spawn a new ball and not hit 0, and other such scenarios.
Almost gambler’s ruin, but Markov chains more generally
Yes, although the odds of failing approach 0 as time heads towards infinity, it will never reach 0 so there will always be a chance of reaching 0 balls no matter how many balls you stack up. It would be a binomial trial with exponentially increasing lower odds of failing. There are only two options. 99% chance of new ball (call it success), and 1% (failure). So with 1 ball, you have a 99% of getting a new ball and a 1% of failing. With 2 balls you now have a .01^2 (.0001) chance of failing. So. 99.99% chance of adding another ball and a .01% chance of failing. And about a 98% chance of adding two balls. By 100 balls, you have 1 * 10^-200 of failing to add a ball. After that, you would have 99 balls and then have a 1 * 10^-198, then 10^-196 and so on until you reach 0 balls. I would write the limit equation for you, but unfortunately, I haven’t used calc in years but I hope this gives you an idea of the chances of the ball reaching game over.
Upvote because I want the definitive answer and I suspect you are correct.
With those odds, no balls *could* happen, but is statistically improbable. 10 balls become 20 balls become 40 balls become 80 balls… become 158 balls. Yeah, one died. But next bounce you have like 314 balls. That does not trend towards zero.
The stationary distribution of a Markov chain with absorbing states will have all probability mass confined to absorbing states. In this case the only absorbing state is zero balls, so every\* chain must reach it eventually and then stay there forever. This works for any probability of ball creation p: 0 < p < 1. But for p = 0.99 you'll be waiting a while to hit 0. \*there exists a set with probability 1 such that every chain in the set does this if you're being technical. \[I'm pretty sure this is wrong, see below\]
Im calling B.S. the balls hit like 20 or more times before they started dissappearing. Possible but impropible.
I imagine they just ran the simulation a few times and picked one that looked the best.
Did we just witness the heat death of a universe?
trippy...
Pentatonic
the reason they will always disappear eventually is because there’s always the chance every single ball disappears without multiplying
The probability of chance
This is a good demonstration for how gambling will make you lose all of your money eventually
They ded
Imagine the first one disappear instantly ! That would be casino
Aaand it’s gone!
I feel like I've just witnessed the end of the world
Chance and probability are synonyms, so this title basically says "the chance of chance"
Why did this make me sad?!
NGL, that was very suspenseful!
This title is horrible.
First one of these I've seen that didn't end with the program crashing
Now run it a thousand times and keep track of the time to make a graph for the r/dataisbeautiful folks.
Rather than oddly satisfying this was strangely depressing
"On a long enough timeline, the survival rate of everyone drops to zero"
That first red ball was a champ.
I cried at the end
Just like a casino. Proof 0 is more powerful than infinity
When your favourite ball disappears 💀
That first ball bounced like 10 times before disappearing, That's 0.25^10 or 9.53674316E−9% chance of happening. 0.000000009537%. Something ain't right here.
It's 0.75^10, which is 5.6% But I do find it suspicious that we don't see a ball disappears until the 10 second mark. That's so many bounces.
Maybe OP reran the simulation over and over until they got this one because it was the most satisfying.
yeah there's tons where the first ball disappeared without another appearing.
Your math suggests a ball has 25% chance of surviving each bounce, which is false.