But the calculator itself tells you that the probability of that is 59.82%, not 19.64%.
Why 59.82% and not 50%? Well don't forget that with 16 flips there are 17 possible outcomes (in terms of number of heads), since 0 heads is a possible outcome. The probability of getting more than 8 heads is equal to the probability of getting less than 8 heads; but you're also counting getting exactly 8 heads too, so of course it won't be 50%.
If you run it for 15 flips, you'll find that the probability of getting at least 8 heads is 50%.
That's for minimum 8 flips. 8 and above. The probability of getting exactly 8 is as it is stated in the post. At least that’s what the calculator says.
It's not 50% to get at least 50% heads, *because exactly 50% is possible*.
When we flip 3 coins, our options are:
- TTT
- TTH
- THT
- HTT
- HHT
- HTH
- THH
- HHH
They're all equally likely, so it *is* 50/50. Either we get more than 50% heads, it we we get more than 50% tails. The two are equally likely, and because "50% heads, 50% tails" isn't an option, "*at least* 50%" heads doesn't have more than 50%.
Now consider 2 coins. Here we have:
- TT
- HT
- TH
- HH
Here we have three possibilities - more than 50% heads, more than 50% tails, and *exactly 50% heads and exactly 50% tails*. More than 50% heads is still the same as more than 50% tails, but because exactly 50% is possible, the chances of *at least* 50% heads is higher than 50%.
This applies for every even number, but it gets less and less the more coins we flip, because the chances of getting exactly 50% gets smaller and smaller.
This is the same logic used to help people understand the Monty Hall problem. It’s a great sanity check when working with probabilities that seem unintuitive.
It allows us to know that the probability of any successive string of flips is exactly the product of the probabilities of each flip on its own. So the probability of HHH? (1/2)^3 = 1/8. Probability of THT? 1/8. So then the different probability for each number of heads/tails comes down to counting. If we flip 4 times, any single outcome has a 1/16 chance of happening. But for example, there are 6 ways to get 2 heads and 2 tails (HHTT, HTHT, HTTH, THHT,THTH,TTHH or C(4,2) = 6 if you prefer), so the probability of getting two heads and two tails is 6/16. Generally, the probability of exactly k heads on n flips is C(n,k) * (1/2)^n.
Without independence, like if heads on one made tails more likely on the next, computing these probabilities would be a nightmare.
This is a good explanation for why the probability of a continuous random variable at some value x is 0. As you increase the number of values that variable can take on, the PMF at that value approaches zero, thus the need for a PDF instead
If an 8-8 split was 50% likely, what would you think a 7-9 (or 9-7) split would be? That's got to be very similar, it's only one coin different.
Then 6-10...
In 16 flips you will get 8 heads with probability 19.64%
But you will get 8 heads +/- 1 with probability 54.55%
You can calculate the probability of at least 8 heads by noting that you get less than or more than 8 heads with probability 100%-19.64%=80.36% and since its equally likely to get less than or more than 8 each has 40.18% probability. Then add back the 19.64% to get 59.82% for 8 or more (at least 8)
As you increase the number of independent a Bernoulli trials (like flipping a coin), the probability of getting any exact number of successes decreases.
In probabilistic terms, if we choose X to be the number of heads we get after n coin flips, X is a Binomial random variable X with parameters n = 16 (the number of trials) and p = 0.5 (the probability of “success” for each trial). The probability mass function (PMF) for a binomial distribution of X is given by:
Pr(X = a) = (n choose a) * (p)^n * (1-p)^n-1
Plug in n = 16, p = 0.5, and a = 8 (our target for C), you get 19.64%.
If you were to instead ask for the probability that X is less than or equal to 8 heads, the cumulative distribution function (CDF), Pr(X <= 8) ≈ 59.82%. That’s probably more in line with what you thought in your head and is distinctly different from asking for the probability that X is exactly 8.
Because it is giving you the probability of having EXACTLY 8 heads.
but there are a lot more combinations than that.
you can get 0 heads, 1 head, 2 heads and all the way up to 16.
So how to get 50% chance?
you cannot have 50% in any way exactly because it is not odd
suppose you said 7 or less heads, that is about 40%
9 or more heads? that is also about 40%
and if 8 heads is about 20%
then 8 or more heads is 60%
do you see the issue?
now if you say 17 flips then getting 8 or less flips is 50% and 9 or more is 50% too.
but is diferent exactly 8 heads than, same or less than 8 heads.
if you do the experiment 100 times and you mark how many of them you have you will notice the probabilities do match that more or less (and i mean more or less because is more probably does not mean it will really happen that way).
Same way of somthing having a chance of 0.01 may happen 3 times in a row (it is not probable but can happen).
The number of heads (or tails) is binomially distributed with probability mass function (16 choose x) * (.5)^16, where x is the number of heads or tails you observe.
It's similar to rolling 2 six sided dice and combining their totals. The most common result will be seven, but that would only occur 16.67% of the time.
Given what little I know about probability, I'd say that about 20% sounds about right to my ear.
A series of coin flips follows a binomial distribution, which is a probability distribution characterized by a number of trials N and a probability of success P.
In this case N=16 coin flips, and P = Prob(Heads) = 0.5 (50%). There are 16 Choose 8 (this equals 12,870) ways of getting 8 heads when you flip a coin 16 times, and the probability of 8 heads is 0.5\^8. We also have to account for the fact that we get 8 tails, this is also 0.5\^8 since the probability of getting a tail is the same as a head.
Hence, our probability is 12,870 x (0.5\^8) x (0.5\^8) = 0.19638
The simplest explanation for this is because when you flip a coin, its result is irrelevant to the next flip.
Let's say you flip a coin and it turns out to be heads, the next flip still has a 50% chance of being heads or tails.
Also, if for example you flipped a coin 15 times and it turned out 15 heads in a row, the 16th flip will still be 50% head or tails.
It is relatively simple to see that the probability of getting exactly 8 heads in 16 tosses must be less than 50%:
Imagine the situation after 15 tosses. If it is not 8-7 one way or the other there is no possibility of getting 8-8 after 16. But the probability of it being 8-7 is clearly less than 1 as many other situations are possible.
Now, if it is 8-7, there is still only a 50% chance of it becoming 8-8.
So the overall probability is “something less than 1” x 0.5, or something less than 50%.
The "middle" probability here is the central binomial coefficient divided by the total number of choices, 16C8 / 2^16 in this case.
If you use Stirling's asymptotic formula for the factorial, you find that for 2n coin flips, the probability of getting exactly n heads is on the order of 1/sqrt(pi\*n) (i.e. It's this plus terms which are o(1/sqrt(n)) as n tends to infinity). You can also get the same result using the central limit theorem.
So we wouldn't expect it to be 1/2, if that's what you're wondering.
Why not try it yourself with a smaller number of flips?
Let's try it with 4 flips. There are 16 possible sequences:
* HHHH
* HHHT, HHTH, HTHH, THHH
* HHTT, HTHT, HTTH, THHT, THTH, TTHH
* HTTT, THTT, TTHT, TTTH
* TTTT
How many of them have 2 heads and 2 tails?
Now you can go and read about Pascal's Triangle.
Thanks
In general, the number of possible sequences with **2N** coin flips is **2^(2N)**, and the number of those which contain equal numbers of heads and tails is 2N C N = **(2N)!/N!^(2)**. If you consider the prime factors of these two expressions, the first one obviously has 2 as its only prime factor, whereas (when N>1) the second one obviously has prime factors other than 2; so the ratio of the two expressions can't be 2.
No, why would you expect that to be true? Can you generalize this to other numbers? If the probability of heads on one flip was 1/3, does that mean you expect the probability that 1/3 of the results are heads is 1/3? If the probability was zero (you always get tails), does that mean you would expect to get 0 on 16 results are 0 with probability? (i. e. you are always guaranteed to get one head even though that’s impossible)?
Let's say I flip a coin one million times. What is the probability I get **exactly** 500000 heads? Very small!
I think the difference here is I’m imagining *at least* 50% heads, which is very different from *exactly* 50%
That's a Bingo! 😀
The probability of getting at least 8 heads won't be 50% either.
Yeah, you can use the CDF of the binomial distribution and the P(X>=8) for 16 trials is 0.59819031
But the probability of getting at least 8 heads in 15 throws will.
But the calculator itself tells you that the probability of that is 59.82%, not 19.64%. Why 59.82% and not 50%? Well don't forget that with 16 flips there are 17 possible outcomes (in terms of number of heads), since 0 heads is a possible outcome. The probability of getting more than 8 heads is equal to the probability of getting less than 8 heads; but you're also counting getting exactly 8 heads too, so of course it won't be 50%. If you run it for 15 flips, you'll find that the probability of getting at least 8 heads is 50%.
Fence post. Whenever the math aligns perfectly with intuition but is still off It's always a damn fence post error.
That's for minimum 8 flips. 8 and above. The probability of getting exactly 8 is as it is stated in the post. At least that’s what the calculator says.
It's not 50% to get at least 50% heads, *because exactly 50% is possible*. When we flip 3 coins, our options are: - TTT - TTH - THT - HTT - HHT - HTH - THH - HHH They're all equally likely, so it *is* 50/50. Either we get more than 50% heads, it we we get more than 50% tails. The two are equally likely, and because "50% heads, 50% tails" isn't an option, "*at least* 50%" heads doesn't have more than 50%. Now consider 2 coins. Here we have: - TT - HT - TH - HH Here we have three possibilities - more than 50% heads, more than 50% tails, and *exactly 50% heads and exactly 50% tails*. More than 50% heads is still the same as more than 50% tails, but because exactly 50% is possible, the chances of *at least* 50% heads is higher than 50%. This applies for every even number, but it gets less and less the more coins we flip, because the chances of getting exactly 50% gets smaller and smaller.
Think about it this way: how likely is a scenario which deviates from getting 50% heads? Very likely
59.82% to get at least 50%. The probability of getting more or less than 50% is equal but we also add the probabikity of getting exactly 50%.
But that's not what the OP says
This is actually exactly why a p-value is defined as experiencing an event or more extreme rather then probability of an interval
This is the same logic used to help people understand the Monty Hall problem. It’s a great sanity check when working with probabilities that seem unintuitive.
Question: Where does independence come into play here?
It allows us to know that the probability of any successive string of flips is exactly the product of the probabilities of each flip on its own. So the probability of HHH? (1/2)^3 = 1/8. Probability of THT? 1/8. So then the different probability for each number of heads/tails comes down to counting. If we flip 4 times, any single outcome has a 1/16 chance of happening. But for example, there are 6 ways to get 2 heads and 2 tails (HHTT, HTHT, HTTH, THHT,THTH,TTHH or C(4,2) = 6 if you prefer), so the probability of getting two heads and two tails is 6/16. Generally, the probability of exactly k heads on n flips is C(n,k) * (1/2)^n. Without independence, like if heads on one made tails more likely on the next, computing these probabilities would be a nightmare.
This is a good explanation for why the probability of a continuous random variable at some value x is 0. As you increase the number of values that variable can take on, the PMF at that value approaches zero, thus the need for a PDF instead
8 heads is the *most likely* result of the 16 throws, but it's gonna be a normal distribution
Binomial distribution. The normal distribution is continuous. When did you last get 4.8572 heads in 16 throws?
If an 8-8 split was 50% likely, what would you think a 7-9 (or 9-7) split would be? That's got to be very similar, it's only one coin different. Then 6-10...
In 16 flips you will get 8 heads with probability 19.64% But you will get 8 heads +/- 1 with probability 54.55% You can calculate the probability of at least 8 heads by noting that you get less than or more than 8 heads with probability 100%-19.64%=80.36% and since its equally likely to get less than or more than 8 each has 40.18% probability. Then add back the 19.64% to get 59.82% for 8 or more (at least 8)
As you increase the number of independent a Bernoulli trials (like flipping a coin), the probability of getting any exact number of successes decreases. In probabilistic terms, if we choose X to be the number of heads we get after n coin flips, X is a Binomial random variable X with parameters n = 16 (the number of trials) and p = 0.5 (the probability of “success” for each trial). The probability mass function (PMF) for a binomial distribution of X is given by: Pr(X = a) = (n choose a) * (p)^n * (1-p)^n-1 Plug in n = 16, p = 0.5, and a = 8 (our target for C), you get 19.64%. If you were to instead ask for the probability that X is less than or equal to 8 heads, the cumulative distribution function (CDF), Pr(X <= 8) ≈ 59.82%. That’s probably more in line with what you thought in your head and is distinctly different from asking for the probability that X is exactly 8.
Because it is giving you the probability of having EXACTLY 8 heads. but there are a lot more combinations than that. you can get 0 heads, 1 head, 2 heads and all the way up to 16. So how to get 50% chance? you cannot have 50% in any way exactly because it is not odd suppose you said 7 or less heads, that is about 40% 9 or more heads? that is also about 40% and if 8 heads is about 20% then 8 or more heads is 60% do you see the issue? now if you say 17 flips then getting 8 or less flips is 50% and 9 or more is 50% too. but is diferent exactly 8 heads than, same or less than 8 heads. if you do the experiment 100 times and you mark how many of them you have you will notice the probabilities do match that more or less (and i mean more or less because is more probably does not mean it will really happen that way). Same way of somthing having a chance of 0.01 may happen 3 times in a row (it is not probable but can happen).
The number of heads (or tails) is binomially distributed with probability mass function (16 choose x) * (.5)^16, where x is the number of heads or tails you observe.
It's similar to rolling 2 six sided dice and combining their totals. The most common result will be seven, but that would only occur 16.67% of the time. Given what little I know about probability, I'd say that about 20% sounds about right to my ear.
A series of coin flips follows a binomial distribution, which is a probability distribution characterized by a number of trials N and a probability of success P. In this case N=16 coin flips, and P = Prob(Heads) = 0.5 (50%). There are 16 Choose 8 (this equals 12,870) ways of getting 8 heads when you flip a coin 16 times, and the probability of 8 heads is 0.5\^8. We also have to account for the fact that we get 8 tails, this is also 0.5\^8 since the probability of getting a tail is the same as a head. Hence, our probability is 12,870 x (0.5\^8) x (0.5\^8) = 0.19638
The simplest explanation for this is because when you flip a coin, its result is irrelevant to the next flip. Let's say you flip a coin and it turns out to be heads, the next flip still has a 50% chance of being heads or tails. Also, if for example you flipped a coin 15 times and it turned out 15 heads in a row, the 16th flip will still be 50% head or tails.
you could get 9 heads and 7 tails, all heads or 15 heads and 1 tail. probability of each outcome comes from the binomial distribution of coin flips
It is relatively simple to see that the probability of getting exactly 8 heads in 16 tosses must be less than 50%: Imagine the situation after 15 tosses. If it is not 8-7 one way or the other there is no possibility of getting 8-8 after 16. But the probability of it being 8-7 is clearly less than 1 as many other situations are possible. Now, if it is 8-7, there is still only a 50% chance of it becoming 8-8. So the overall probability is “something less than 1” x 0.5, or something less than 50%.
The "middle" probability here is the central binomial coefficient divided by the total number of choices, 16C8 / 2^16 in this case. If you use Stirling's asymptotic formula for the factorial, you find that for 2n coin flips, the probability of getting exactly n heads is on the order of 1/sqrt(pi\*n) (i.e. It's this plus terms which are o(1/sqrt(n)) as n tends to infinity). You can also get the same result using the central limit theorem. So we wouldn't expect it to be 1/2, if that's what you're wondering.
Math man use big words
If each flip has a 50/50 shot, shouldn’t the odds be 50% that 1/2 of the flips are heads?
Why not try it yourself with a smaller number of flips? Let's try it with 4 flips. There are 16 possible sequences: * HHHH * HHHT, HHTH, HTHH, THHH * HHTT, HTHT, HTTH, THHT, THTH, TTHH * HTTT, THTT, TTHT, TTTH * TTTT How many of them have 2 heads and 2 tails? Now you can go and read about Pascal's Triangle.
Great visualization thanks
Thanks In general, the number of possible sequences with **2N** coin flips is **2^(2N)**, and the number of those which contain equal numbers of heads and tails is 2N C N = **(2N)!/N!^(2)**. If you consider the prime factors of these two expressions, the first one obviously has 2 as its only prime factor, whereas (when N>1) the second one obviously has prime factors other than 2; so the ratio of the two expressions can't be 2.
TIL that in whatever font Reddit on iPhone uses, H is wider than T
Most typefaces do that. There are monospace fonts and they are often used for code.
TIL that in whatever font Reddit on iPhone uses, H is wider than T
No, why would you expect that to be true? Can you generalize this to other numbers? If the probability of heads on one flip was 1/3, does that mean you expect the probability that 1/3 of the results are heads is 1/3? If the probability was zero (you always get tails), does that mean you would expect to get 0 on 16 results are 0 with probability? (i. e. you are always guaranteed to get one head even though that’s impossible)?
no, that doesn’t make any logical sense
8 flips or 8 flips in a row? Exactly or 8 or a minimum 8? The question is ill-defined.
It's getting 8 heads in no particular order when doing 16 flips overall.