# The birthday problem revisited?

This concerns the classic birthday problem. Supposing that leap-day birthdays are not possible, 23 people are asked their birthdays. The probability that at least two of them match is greater than 1/2.

Now I cannot recall the source, but long ago I read an account (supposedly historical) in which a mathematician was demonstrating it as a parlor trick. After 22 party guests had given their birthdays, there were no matches. According to the source, the mathematician was still confident that there would be a match, because the probability favored it. Sure enough, the last guest shared a birthday with one of the others.

Explain why the mathematician had no reason to be confident in success after there were no matches among the first 22. After how many birthdays with no match would the mathematician lose the advantage? That is, at what point would he first be in a position in which his probability of winning would be less than 1/2?

Sorry, friends. I thought it was clear. There are 23 people in the sample. That number is not variable. For the mathematician to succeed there must be at least one match among the 23. After 22 birthdays without a single match, he inexplicably believes that he still holds the advantage. Of course he does not. At some earlier point in the game, he should have been able to see that he probably would not succeed.

### 3 Answers

- ?Lv 710 years agoFavourite answer
Hm. Interesting.

It was lucky for the mathematician that the 23rd guest did share a birthday with one of the others; in fact, after it was discovered that the first 22 guests all had different birthdays, the probability that the 23rd shared a birthday, and that the mathematician would win, was only 22/365. So the mathematician would lose with probability 1 - 22/365 = 343/365

Going backwards in time, after the first 21 guests declared no matches, the 22nd guest could have matched one of them with a probability of 21/365. He will fail to do so with a probability of 1 - 21/365 = 344/365, in which case the 23rd guest may still match any of the previous 22, as above, with probability 22/365. So, the mathematician loses with probability (344/365)(343/365).

Continuing in this way, the mathematician loses with probability > 0.5 after the 4th guest exhibits no match. That is, (361! / (342! 365^(361 - 342)) > 0.5.

Hmm ... this doesn't seem exactly right to me. I'll be interested to see what other solutions people come up with. Sorry :(

- BigBrainOnBradLv 510 years ago
The question is a bit confusing... Maybe it's just me...

The solution to the birthday problem is:

1 - [(365-1)(365-2)(365-3)...(365-n)]/[365^n] where n represents the number of people at the gathering.

Actually, the more people come the better the chances of a "hit". As they say: the more the merrier. The math guy had every reason to stay confident in his chances. He may lose the attention of the crowd, but not the mathematical chance.

For n=23 the chances go just above 0.5 .

For any n<23 the chances stay below 0.5 .

Maybe you wanted to ask something else...

Best regards!

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