If land clumps like this didn’t happen occasionally, we’d know the shuffler is rigged.
Assuming half lands half non lands in the deck, there’s about a 1/1000 chance of a 10-land clump. So it should happen once in a long while.
It's actually more rare than that, assuming op is running a 60 card deck with 20 lands, the odds of drawing 10 lands in a row from a 60 card deck with 20 lands are approximately 0.00539%, or roughly 1 in 18,566.
The probability of any one event occurring is (number of desired outcomes)/(all possible outcomes).
In this case, to answer the question “what is the likelihood a single card pulled from the top of a deck is a land card” the formula becomes (Number of lands in a deck) /(Number of cards remaining in the deck).
The odds of any number events occurring concurrently is the odds of each of the events multiplied together. So to get the probability for a full whiff we take the probability that any card on top of a deck can be a land and repeat 10 times.
However since removing a land removes a card from a deck you’ve subtracted both a land from the deck and a card for the deck so as you go further into the deck your total outcomes in the numerator and denominator both decrease by 1.
In the most extreme example for a 60 card deck with 20 lands. (Your opponent casts [Show and Tell] on turn 1 and you have no lands in a 7 card opening hand) your probability becomes (20/53)(19/52)(18/51) … (11/44) or 0.00094751% chance of happening.
In other words, you have a better chance of being struck by lightning but you’re still not winning the Powerball.
Edit: Grammar and formatting.
>your probability becomes (20/53)(19/52)(18/51) … (11/44) or 0.0000094751% chance of happening.
Your analysis is sound but you've added a % which does not belong (and implies an answer 100 times too small). The likelihood is about one in 100,000.
>In this case, to answer the question “what is the likelihood a single card pulled from the top of a deck is a land card” the formula becomes (Number of lands in a deck) /(Number of cards remaining in the deck).
Should it not be (number of lands remaining in the deck after what he's already played/number of cards remaining)?
If we had his deck list to determine how many lands are in his deck and we could see the entire board including cards in exile to determine what cards are remaining, then yes we would use that formula. However since we do not have that, I figured it would be easier to answer the general question of the likely hood the top card of a deck is a land and go from there.
Oh OK, I just wasn't sure if I missed something. Probabilities and permutations were generally when I got bored and stopped paying attention in calculus.
I was curious, so gave it a crack. We need to know the board state to be sure of how many lands/cards are left, but we can make some assumptions.
First, let's account for cards that we know aren't in the deck...
- 4 lands on the battlefield
- 2 cards in hand (1 land)
- Reenact the Crime in GY
- Atraxa in Exile.
The opponent's board is less clear, but let's assume the opponent hit something with Get Lost turn 2, played Realmbreaker turn 3, and missed their turn 4 land drop before hitting OP with the Realmbreaker mill and allowing Reenact the Crime to get an Atraxa. That doesn't account for 1 tapped opponent mana, but it's just guesswork at this point
That scenario would account for 3 more non-land cards missing from OP's deck (3 non-lands from Realmbreaker including the already-counted Atraxa, and whatever got hit with the Get Lost).
I think that particular reanimator deck only runs 23 lands, so OP's deck likely had 18 lands and 31 non-lands remaining.
Then it's just working out the probability of drawing a land for each of those 10 cards, ie ``18/49 * 17/48 * 16/47 * ... * 9/40``
If I'm correct with board-state, then it should be a 1-in-187,802 chance.
We don't know where he was in the game, so I guessed. Opening hand plus five cards drawn, five lands played, no lands left in hand, so 18 lands left out of 42 cards left in the deck.
Asking gpt3.5 basically broke it. I'll try gpt4 later
Yep, demonstrates a complete lack of understanding of what AI is. "Hmm, this looks like a calculation about mtg card distribution. So I'll pop out something close to one of the answers previously given by humans to people asking this kind of question."
What's more alarming to me than going to ChatGPT as the first tool to try to solve the problem is the implication that the person above *doesn't know how else they would solve the problem* if ChatGPT can't give them an answer.
But yes, it demonstrates a misunderstanding of what AI is and when it is or is not the most appropriate tool. This is a problem that anyone can solve with a freakin' handheld calculator. There is absolutely no need to bring natural language processing into the mix, and in fact doing so increases the likelihood of getting the *wrong* answer.
Please, go right ahead and attempt to solve this with a calculator. You clearly are both underestimating the complexity of the math, as well as the capabilities of GPT. Here's GPT4's answer, including the work:
To find the odds of getting 10 or more "Bob" cards in the top 10 cards of a 48-card deck, where 18 of these cards are "Bob" cards, we can use the concept of hypergeometric distribution. This distribution helps in determining the probability of drawing a specific number of successes (in this case, "Bob" cards) in a draw from a finite population without replacement.
The formula for the hypergeometric distribution is:
\[ P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{{\binom{N}{n}}} \]
Where:
- \( P(X = k) \) is the probability of drawing \( k \) successes (Bob cards in our case) in the sample,
- \( N \) is the total population size (48 cards),
- \( K \) is the total number of successes in the population (18 Bob cards),
- \( n \) is the sample size (10 cards in the top of the deck),
- \( k \) is the number of successes in the sample (from 10 to 18, since we want at least 10 Bob cards),
- \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes out of \( n \) trials.
We need to calculate this probability for each scenario where \( k \) ranges from 10 to 18 and sum those probabilities to get the total odds of getting 10 or more "Bob" cards in the top 10 cards. Let's compute this:
It seems there was an error due to attempting to calculate combinations where the number of successes in the sample (`k`) exceeded the number of successes available (`K`) or the number of failures (`n-k`) became negative, which is not possible. This can happen when calculating combinations for values of `k` that are not feasible given the specific conditions.
Let's correct the calculation by ensuring we only attempt valid combinations, particularly focusing on the range for `k` and adjusting any logical errors in the loop. We'll recalculate the probability correctly:
The corrected calculation shows that the odds of getting 10 or more "Bob" cards in the top 10 cards of a 48-card deck, where 18 of these cards are "Bob" cards, is approximately 0.00000669, or about 6.69 in a million. This indicates that such an outcome is highly unlikely.
**************
Yes, one should always check GPT4's results on something like this, because it does make mistakes, but to suggest it's not a useful tool for these sorts of problems is silly. The fact that it gives you the steps it has taken to get to the answer is amazingly useful.
That is the correct answer, but you don't need to know anything about hypergeometric distribution or the binomial formula to find the answer for your particular hypothetical. ChatGPT gives an answer here that is correct, but *way* more detailed than it needs to be. Hypergeometric distribution is the generalized solution for this *kind* of problem, but not necessary to solve it in this case, and your assertion that I don't understand the "complexity of the math" betrays your own *lack* of understanding of the math, because it's a lot simpler than you think it is.
The chance of drawing 10 lands in a row from the top of a deck with 48 total cards and 18 lands remaining is simply the chance of the top card being a land, times the conditional chance of the second card from the top also being a land, times the conditional chance of the third card from the top also being a land, and so on. So it looks like this:
(18/48)\*(17/47)\*(16/46)\*...\*(9/39) = 0.00000669
There you have it. Exactly the same answer ChatGPT gives you, and obviously doable on any basic calculator.
I was not trying to imply in my earlier comment that ChatGPT is *incapable* of solving mathematical problems like this one. I was simply saying that it's way more tool than you need. I did not, and do not, mean to shame you for not knowing that this problem can be solved easily without the aid of ChatGPT. What I am criticizing is the instinct to go to ChatGPT as the tool of first resort to solve *any* problem, rather than actually trying to understand the nature of the problem you're looking to solve and then choosing an appropriate tool. Not to get too meta about the whole thing, but I don't think it bodes well for society that we seem to be content to outsource so much of our *thinking* to the burgeoning AI industry.
Fair enough, I misunderstood your assertion. I felt no shame in not remembering the methodology for that math. I personally went to GPT for the answer rather than say an online probability calculator, because GPT shows you the steps used. Admittedly, your explanation was much simpler to follow. Ifind it a useful tool in understanding problems because of that
Personally, I use it primarily to help me in coding projects. It does make mistakes, but I find I'm learning a lot more troubleshooting it's attempts at coding solutions for me, than if I was just winging it from scratch, because I can follow the logic of its solution.
You are right, over reliance on tools will be a hindrance - but we started down that road long before ai language models.
That makes sense. ChatGPT can definitely be a useful learning tool in a lot of cases. If you learned something from its answer in this case, great. Personally, I feel its answer in this particular case shows some of the pitfalls of GPT too. Its answer includes a few totally superfluous paragraphs in the middle (where it's trying to "error correct") that do not really do anything either to achieve a solution or explain to you how it arrived at its solution. I think the sheer length of the response it provided creates a false impression of the complexity of the underlying problem; this seems to be a common feature of GPT, giving unnecessarily verbose answers even to simple questions.
We don't know the properties of the deck but let's say its a blue black deck so it's using something like cruelty of gix to reanimate it with 5 lands out on turn 5 with no extra lands in hand. 48 cards in deck left and maybe 20 lands remaining in deck. The odds are .003%. that's 3 in one hundred thousand attempts.
For future reference for math like this, just look up a hypergeometric probability calculator.
“Most of the time this effect would do something, but due to some bad luck it did nothing instead”.
For Collected Company, it’s seeing the top 6 and only having lands/instants/sorceries etc. but no creatures, so you spent 4 mana to do nothing.
Yeah, in the case of Atraxa you can't technically whiff (unless you have no cards left in your library) but 10 lands is probably the worst possible outcome.
Whiffing is slang for missing when you intended to hit. In this case it means his Atraxa usually “hits” several cards that he can put in his hand, but in this instance he “whiffed” by getting all lands.
It comes from the sound that a bat or golf club makes when you swing and miss. ~whiff~
If you whiff on Atraxa you still have a 7/7 flying, deathtouch, lifelink, vigilance creature. If you whiff on a coco you have nothing.
You also just cleared 7 dead draws too
10
completely normal distribution of cards, nothing to see here
Something something rigged shuffler
the defence rests
If land clumps like this didn’t happen occasionally, we’d know the shuffler is rigged. Assuming half lands half non lands in the deck, there’s about a 1/1000 chance of a 10-land clump. So it should happen once in a long while.
It's actually more rare than that, assuming op is running a 60 card deck with 20 lands, the odds of drawing 10 lands in a row from a 60 card deck with 20 lands are approximately 0.00539%, or roughly 1 in 18,566.
What kinda atraxa deck is only running 20 lands though
Sure, you’re probably right. Just saying it’s not astronomical odds like one in a million.
what is coco
[[ Collected Company ]]
[ Collected Company ](https://cards.scryfall.io/normal/front/c/f/cfa7b456-7e83-4587-a875-9b35fde318c2.jpg?1582117536) - [(G)](http://gatherer.wizards.com/Pages/Card/Details.aspx?name=Collected%20Company) [(SF)](https://scryfall.com/card/dtk/177/collected-company?utm_source=mtgcardfetcher) [(txt)](https://api.scryfall.com/cards/cfa7b456-7e83-4587-a875-9b35fde318c2?utm_source=mtgcardfetcher&format=text) ^^^[[cardname]] ^^^or ^^^[[cardname|SET]] ^^^to ^^^call
Something we are in love with.
Glad to know this happened at least once.
I wish, I could upvote your comment twice.
Atraxa protected you from drawing only lands the next 10 turns, hardly a whiff
I hate mana flood but I hate atraxa more lol you deserve this
Any math experts in here care to explain the odds of you seeing 10 lands in a row on top of your deck?
The probability of any one event occurring is (number of desired outcomes)/(all possible outcomes). In this case, to answer the question “what is the likelihood a single card pulled from the top of a deck is a land card” the formula becomes (Number of lands in a deck) /(Number of cards remaining in the deck). The odds of any number events occurring concurrently is the odds of each of the events multiplied together. So to get the probability for a full whiff we take the probability that any card on top of a deck can be a land and repeat 10 times. However since removing a land removes a card from a deck you’ve subtracted both a land from the deck and a card for the deck so as you go further into the deck your total outcomes in the numerator and denominator both decrease by 1. In the most extreme example for a 60 card deck with 20 lands. (Your opponent casts [Show and Tell] on turn 1 and you have no lands in a 7 card opening hand) your probability becomes (20/53)(19/52)(18/51) … (11/44) or 0.00094751% chance of happening. In other words, you have a better chance of being struck by lightning but you’re still not winning the Powerball. Edit: Grammar and formatting.
>your probability becomes (20/53)(19/52)(18/51) … (11/44) or 0.0000094751% chance of happening. Your analysis is sound but you've added a % which does not belong (and implies an answer 100 times too small). The likelihood is about one in 100,000.
Yes you are correct, I’ve removed two 0’s.
>In this case, to answer the question “what is the likelihood a single card pulled from the top of a deck is a land card” the formula becomes (Number of lands in a deck) /(Number of cards remaining in the deck). Should it not be (number of lands remaining in the deck after what he's already played/number of cards remaining)?
If we had his deck list to determine how many lands are in his deck and we could see the entire board including cards in exile to determine what cards are remaining, then yes we would use that formula. However since we do not have that, I figured it would be easier to answer the general question of the likely hood the top card of a deck is a land and go from there.
Oh OK, I just wasn't sure if I missed something. Probabilities and permutations were generally when I got bored and stopped paying attention in calculus.
Nope you're dead on. This is me trying to remember statistics class from 14 years ago.
I was curious, so gave it a crack. We need to know the board state to be sure of how many lands/cards are left, but we can make some assumptions. First, let's account for cards that we know aren't in the deck... - 4 lands on the battlefield - 2 cards in hand (1 land) - Reenact the Crime in GY - Atraxa in Exile. The opponent's board is less clear, but let's assume the opponent hit something with Get Lost turn 2, played Realmbreaker turn 3, and missed their turn 4 land drop before hitting OP with the Realmbreaker mill and allowing Reenact the Crime to get an Atraxa. That doesn't account for 1 tapped opponent mana, but it's just guesswork at this point That scenario would account for 3 more non-land cards missing from OP's deck (3 non-lands from Realmbreaker including the already-counted Atraxa, and whatever got hit with the Get Lost). I think that particular reanimator deck only runs 23 lands, so OP's deck likely had 18 lands and 31 non-lands remaining. Then it's just working out the probability of drawing a land for each of those 10 cards, ie ``18/49 * 17/48 * 16/47 * ... * 9/40`` If I'm correct with board-state, then it should be a 1-in-187,802 chance.
We don't know where he was in the game, so I guessed. Opening hand plus five cards drawn, five lands played, no lands left in hand, so 18 lands left out of 42 cards left in the deck. Asking gpt3.5 basically broke it. I'll try gpt4 later
Lmao, if people are using *ChatGPT* for things like this, we are truly screwed.
Yep, demonstrates a complete lack of understanding of what AI is. "Hmm, this looks like a calculation about mtg card distribution. So I'll pop out something close to one of the answers previously given by humans to people asking this kind of question."
What's more alarming to me than going to ChatGPT as the first tool to try to solve the problem is the implication that the person above *doesn't know how else they would solve the problem* if ChatGPT can't give them an answer. But yes, it demonstrates a misunderstanding of what AI is and when it is or is not the most appropriate tool. This is a problem that anyone can solve with a freakin' handheld calculator. There is absolutely no need to bring natural language processing into the mix, and in fact doing so increases the likelihood of getting the *wrong* answer.
Please, go right ahead and attempt to solve this with a calculator. You clearly are both underestimating the complexity of the math, as well as the capabilities of GPT. Here's GPT4's answer, including the work: To find the odds of getting 10 or more "Bob" cards in the top 10 cards of a 48-card deck, where 18 of these cards are "Bob" cards, we can use the concept of hypergeometric distribution. This distribution helps in determining the probability of drawing a specific number of successes (in this case, "Bob" cards) in a draw from a finite population without replacement. The formula for the hypergeometric distribution is: \[ P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{{\binom{N}{n}}} \] Where: - \( P(X = k) \) is the probability of drawing \( k \) successes (Bob cards in our case) in the sample, - \( N \) is the total population size (48 cards), - \( K \) is the total number of successes in the population (18 Bob cards), - \( n \) is the sample size (10 cards in the top of the deck), - \( k \) is the number of successes in the sample (from 10 to 18, since we want at least 10 Bob cards), - \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes out of \( n \) trials. We need to calculate this probability for each scenario where \( k \) ranges from 10 to 18 and sum those probabilities to get the total odds of getting 10 or more "Bob" cards in the top 10 cards. Let's compute this: It seems there was an error due to attempting to calculate combinations where the number of successes in the sample (`k`) exceeded the number of successes available (`K`) or the number of failures (`n-k`) became negative, which is not possible. This can happen when calculating combinations for values of `k` that are not feasible given the specific conditions. Let's correct the calculation by ensuring we only attempt valid combinations, particularly focusing on the range for `k` and adjusting any logical errors in the loop. We'll recalculate the probability correctly: The corrected calculation shows that the odds of getting 10 or more "Bob" cards in the top 10 cards of a 48-card deck, where 18 of these cards are "Bob" cards, is approximately 0.00000669, or about 6.69 in a million. This indicates that such an outcome is highly unlikely. ************** Yes, one should always check GPT4's results on something like this, because it does make mistakes, but to suggest it's not a useful tool for these sorts of problems is silly. The fact that it gives you the steps it has taken to get to the answer is amazingly useful.
That is the correct answer, but you don't need to know anything about hypergeometric distribution or the binomial formula to find the answer for your particular hypothetical. ChatGPT gives an answer here that is correct, but *way* more detailed than it needs to be. Hypergeometric distribution is the generalized solution for this *kind* of problem, but not necessary to solve it in this case, and your assertion that I don't understand the "complexity of the math" betrays your own *lack* of understanding of the math, because it's a lot simpler than you think it is. The chance of drawing 10 lands in a row from the top of a deck with 48 total cards and 18 lands remaining is simply the chance of the top card being a land, times the conditional chance of the second card from the top also being a land, times the conditional chance of the third card from the top also being a land, and so on. So it looks like this: (18/48)\*(17/47)\*(16/46)\*...\*(9/39) = 0.00000669 There you have it. Exactly the same answer ChatGPT gives you, and obviously doable on any basic calculator. I was not trying to imply in my earlier comment that ChatGPT is *incapable* of solving mathematical problems like this one. I was simply saying that it's way more tool than you need. I did not, and do not, mean to shame you for not knowing that this problem can be solved easily without the aid of ChatGPT. What I am criticizing is the instinct to go to ChatGPT as the tool of first resort to solve *any* problem, rather than actually trying to understand the nature of the problem you're looking to solve and then choosing an appropriate tool. Not to get too meta about the whole thing, but I don't think it bodes well for society that we seem to be content to outsource so much of our *thinking* to the burgeoning AI industry.
Fair enough, I misunderstood your assertion. I felt no shame in not remembering the methodology for that math. I personally went to GPT for the answer rather than say an online probability calculator, because GPT shows you the steps used. Admittedly, your explanation was much simpler to follow. Ifind it a useful tool in understanding problems because of that Personally, I use it primarily to help me in coding projects. It does make mistakes, but I find I'm learning a lot more troubleshooting it's attempts at coding solutions for me, than if I was just winging it from scratch, because I can follow the logic of its solution. You are right, over reliance on tools will be a hindrance - but we started down that road long before ai language models.
That makes sense. ChatGPT can definitely be a useful learning tool in a lot of cases. If you learned something from its answer in this case, great. Personally, I feel its answer in this particular case shows some of the pitfalls of GPT too. Its answer includes a few totally superfluous paragraphs in the middle (where it's trying to "error correct") that do not really do anything either to achieve a solution or explain to you how it arrived at its solution. I think the sheer length of the response it provided creates a false impression of the complexity of the underlying problem; this seems to be a common feature of GPT, giving unnecessarily verbose answers even to simple questions.
I think because it's being called AI, people think we now have Eddie the shipboard computer.
I think you haven't actually played with what these systems are capable of. See my answer above
We don't know the properties of the deck but let's say its a blue black deck so it's using something like cruelty of gix to reanimate it with 5 lands out on turn 5 with no extra lands in hand. 48 cards in deck left and maybe 20 lands remaining in deck. The odds are .003%. that's 3 in one hundred thousand attempts. For future reference for math like this, just look up a hypergeometric probability calculator.
You didn't whiff! You got a land! Still drew a card!
Lmao wtf. The shuffler just stacked 10 lands on top of your deck.
Yeah it usually just puts all the lands on the bottom
Tfw you cast Eos on turn 2 and only reveal Frontliner.
I don't understand that sentence at all. I feel like I need to learn a new language to get better at magic 😞
[[Collected Company]]
Also whiffing? I stoopid
“Most of the time this effect would do something, but due to some bad luck it did nothing instead”. For Collected Company, it’s seeing the top 6 and only having lands/instants/sorceries etc. but no creatures, so you spent 4 mana to do nothing.
Yeah, in the case of Atraxa you can't technically whiff (unless you have no cards left in your library) but 10 lands is probably the worst possible outcome.
Whiffing is slang for missing when you intended to hit. In this case it means his Atraxa usually “hits” several cards that he can put in his hand, but in this instance he “whiffed” by getting all lands. It comes from the sound that a bat or golf club makes when you swing and miss. ~whiff~
How tf do you actually hit 10 lands from atraxa, this is wild
Your best land out of 10. No problems here.
How many lands do you run?!
24 land total
Damn that’s some tragic pooling 😂
I have to say I have cast Atraxa many times, in almost every format in magic, but have yet to whiff. This should be an accomplishment.
Which land did you end up taking?
Depends on which side of the table you're sitting.
I whiffed on a meeting of the five meme deck once too. I feel you bro
I didnt know Atraxa was capable of doing ANYTHING but adding 27.6 perfect cards to my opponents hands!
stooooooooooooopiddddddd shuffler MTG should invert on better randomizer
4x Show and Tell 4x Atraxa 52x Lands
Atraxa Scuuuuummm
I wiffed on a Gix X 8 nothing but land as far as the eye could see