Timeline for Probability of 3 people in a room of 30 having the same birthday
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
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| Jul 11, 2022 at 16:52 | comment | added | Frei Zhang | Numerical solution for the triple case compared with approximation here ``` With[{M = 2^32}, {NSolve[1 - Exp[-Binomial[k, 3]/M^2] == 0.5, k, Reals], NSolve[ n E^(-n/( d k )) - (d^(k - 1) k! Log[1/(1 - p)] (1 - n/(d (k + 1))))^(1/ k) == 0 /. { p -> 0.5 , k -> 3, d -> M}, n, Reals]}] ``` result: ``` {{{k -> 4.24912*10^6}}, {{n -> 4.25017*10^6}}} ``` | |
| Jul 13, 2021 at 3:26 | comment | added | vish | (late comment) probabilities are not independent when selecting triplets (N Choose K for K = 3) if you've randomly gone through half the triplets and had no matches, any triplet composed of a pair that didn't match and a third element will also not triple match @A.S. | |
| Jan 14, 2020 at 19:46 | comment | added | user35834 | Let me try to implement it in R # probability of at least b people in a room with r people having birthday on the same day. p=function(r,b) 1-exp(-choose(r,b)/365^(b-1)) So in a room with 30 people at least 3 will have a birthday on the same day with probability 3% > p(25,2) [1] 0.5604122 > p(30,3) [1] 0.03001509 > p(171,4) [1] 0.5069844 Is it right? | |
| Jan 25, 2016 at 21:58 | comment | added | Olivier Oloa | @ByronSchmuland Nice answer. | |
| Dec 13, 2015 at 17:16 | comment | added | A.S. | I have a question. Why does the simple Poisson approximation in your post overcounts probability of triple-shared birthday (in the first two numerical examples) but the same approach undercounts probability of the classical birthday problem with $23$ people and pair of birthdays? | |
| Dec 13, 2015 at 8:37 | comment | added | A.S. | You are over-counting because triples of birthdays are positively correlated and attract each other. How much do they attract? Given a birthday "B" triple, each of the other $n-3$ people would form $3$ more triplets if their birthday was $B$. Divide by $2$ to account for each pair of influences being counted twice to get $(100-3)p/2$ expected affected pairs. Using $\frac {p^2} {1+\frac{(100-3)p}2}$ instead of $p^2$ in Poisson formula, we get $P\approx 0.5801$. But this approximation undercounts since clumps of people sharing a birthday repel. Average the two to get 0.6415 for a $<0.7\%$ error. | |
| Jul 26, 2015 at 14:13 | comment | added | user940 | @ShitalShah I am using $P(X\geq 1)=1-P(X=0)$, so I only need to approximate $P(X=0).$ | |
| Jul 26, 2015 at 8:57 | comment | added | Shital Shah | One final note, shouldn't we be using Poisson CDF instead of PDF because we want to find at least one pair, not just exactly 1 pair? The CDF actually gives probability of 0.6577 which is closer to @Henry's observation. Calculation is here: bit.ly/1CYwNfI | |
| Jul 26, 2015 at 8:47 | comment | added | Shital Shah | Also it would be helpful to justify why Poisson and not Normal distribution. Here justification is that p = 1/365^2 is very small and so mean ~ variance. If that wasn't the case then normal distribution would actually have been better choice. | |
| Jul 26, 2015 at 6:43 | comment | added | Shital Shah | Poisson approximation is good but your answer has a bug. You need to do 1 - u * exp(-u) instead of 1 - exp(-u). Remember Poisson distribution is P(k) = u * exp(-u) / k! Here we take k=1 as we are interested in finding just one pair. Making this change gives answer 0.6394, pretty close to true value. | |
| Jan 28, 2014 at 23:09 | comment | added | user940 | @Henry You are right, the exact probability is .6458645065. This Poisson approximation is not very good. | |
| Sep 6, 2013 at 7:43 | comment | added | Henry | Empirically it seems the probability of at least one triple birthday with $100$ people is closer to $0.64$ or $0.65% than $0.70$ | |
| Mar 9, 2011 at 16:26 | vote | accept | irl_irl | ||
| Mar 9, 2011 at 16:26 | comment | added | irl_irl | Ah, that makes sense! Thanks for the great answer. | |
| Mar 9, 2011 at 16:25 | comment | added | user940 | Take the three random people one at a time. The first guy has some birthday, say March 9. The chance that the second guy has the same birthday is 1/365, and the chance that the third guy has the same birthday is also 1/365. Multiplying these gives 1/365^2. | |
| Mar 9, 2011 at 16:22 | comment | added | irl_irl | " and 1/365^2 is the chance that any particular triple is a success." - Hmm, I don't understand that part. Why squared? | |
| Mar 9, 2011 at 1:52 | history | answered | user940 | CC BY-SA 2.5 |