3 ms·
The chances of you sharing a birthday with one of 22 other people is pretty low. the chances of at least two people in a group of 23 is 50%. In the first case,
by jfoutz 15y ago
The chances of you sharing a birthday with one of 22 other people is pretty low. the chances of at least two people in a group of 23 is 50%. In the first case, you're fixing the day that some other person must have.
- sliverstorm 15y agoIn other words, the odds that a deck you are holding is unique are high; the odds that all decks ever shuffled are unique is low.
- jgershen 15y agoActually, the probability that all decks ever shuffled are unique is also very high. We can approximate the probability that any two of the n decks shuffled in human history were identical as p=1-n^2/52! Using the same estimate as the OP for n (1.56x10^23) gives p=3.02x10^-22. Still fantastically low.
- sliverstorm 15y agoAlright, fine, it is comparatively lower. Better? :)
- gregable 15y agoNice, I came to the comments to see if someone knew how to calculate this. Related question. If there are exactly 2N people who vote in a binary election (ie: for presidential candidates) and they have an even 50/50% chance of voting either way, how do I compute the odds that they will have a even split? This is a generous estimate for the probability my vote will matter.
- flourpower 15y agoThat's the same as the probability that 2N flips of a fair coin result in exactly heads: (2N choose N)(.5^N)(.5^N). Check out: http://en.wikipedia.org/wiki/Binomial_distribution http://en.wikipedia.org/wiki/Binomial_distribution
- wnoise 15y agoThe exact answer is (2n choose n) * (1/2)^(2n). This is approximately sqrt(1/Pi n) as n grows large, with error O(n^(-3/2))
- JadeNB 15y ago> We can approximate the probability that any two of the n decks shuffled in human history were identical as p=1-n^2/52! I think that this is way too low. Shouldn't it be the quite large number 1 - \prod_{i = 1}^n (1 - (i - 1)/52!) (a la the birthday paradox)?
- JadeNB 15y agoSorry, I was thinking of the complementary probability (that there has been a coincidence). Mathematica overflowed when I tried to compute this by brute force. The next best thing I can think of is to use the exponential approximation 1 - x ≈ e^{-x}, good for very small `x`, such as ours. Ignoring the cascading errors gives \prod_{i = 1}^n (1 - (i - 1)/52!) ≈ \prod_{i = 1}^n e^{-(i - 1)/52!} = e^{-n(n - 1)/52!} ≈ 1 - n(n - 1)/52!. The error should be roughly of the size \frac1 2\sum_{i = 1}^n [(i - 1)/52!]^2 ≈ n^3/(2(52!)^2), which is relatively small. (That's stronger, here, than just saying that it is small.) That is to say: I guess I agree with jgershen after all!