A birthday puzzle

[jason_t]

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You should stand 23rd in line. The reason being is that the more people in front of you, the more likely it is to have already occurred. Since 23 people is the number required for there to be at least 50% chance of two people having the same birthday, you expect that any spot after that in line it has already occurred.

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No.

[bobman0330]

your chance of winning with N other people = N/365 [you match someone] * (365!/((365-N)! * 365^N)) [no other matches]
Using this table of results for the latter expression (link):
N 365*P (for calculation ease)
18 11.754
19 11.799
20 11.780
21 11.676

This implies that you should want 19 people in front of you, which will allows you to win 3.23% of the time. This seems low, but I think it is right nonetheless.

You people should be ashamed of yourselves for Monte-Carloing this.

Note: the correct answer has nothing to do with when there's a 50% likelihood of someone having won before you. P(someone else winning) increases in an S-curve shape, while P(you winning, assuming no one else has won) increases linearly. The idea is to take advantage of as much of the linear increase as you can before the steep portion of the S starts killing your equity.

[FouTight]

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Why would the number of people in line affect the likelihood of the 2nd person's birthday matching the first person's?

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It wouldn't. But it would effect the chance of any person in line getting more than one shot at it. The game will be over as soon as the first ticket is bought. If you don't match birthdays with the first in line you are just out of luck because you won't get any more chances. But if you do match birthdays your most advantageous place to be is second in line so that someone in front of you who also matches won't take the prize away from you.

PairTheBoard

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I can't pretent to want to figure out the right answer to this question, but this, without a doubt, is not it.

[LetYouDown]

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You people should be ashamed of yourselves for Monte-Carloing this.

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LOL, normally I would be...but this computer doesn't have excel and I didn't feel like doing any computation. Also, I wanted to write a "monte-carloing" program because I was bored. Apparently I suck at doing it quickly.

[SheetWise]

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If it matched, saved it and continued to the next trial. Then I divided by the number of trials.

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I just looked at frequency. 20 has the highest class frequency in every trial.

[SossMan]

it's not the chances of the guy in front of you having your birthday, it's anyone in front of you having your birthday.

[PairTheBoard]

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Why would the number of people in line affect the likelihood of the 2nd person's birthday matching the first person's?

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It wouldn't. But it would effect the chance of any person in line getting more than one shot at it. The game will be over as soon as the first ticket is bought. If you don't match birthdays with the first in line you are just out of luck because you won't get any more chances. But if you do match birthdays your most advantageous place to be is second in line so that someone in front of you who also matches won't take the prize away from you.

PairTheBoard

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I can't pretent to want to figure out the right answer to this question, but this, without a doubt, is not it.

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Under the assumption that there are a billion people in line, what's wrong with it?

PairTheBoard

[bobman0330]

The explanation of the game stated that you win if someone in front of you has the same birthday...

[BruceZ]

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Why would the number of people in line affect the likelihood of the 2nd person's birthday matching the first person's?

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It wouldn't. But it would effect the chance of any person in line getting more than one shot at it. The game will be over as soon as the first ticket is bought. If you don't match birthdays with the first in line you are just out of luck because you won't get any more chances. But if you do match birthdays your most advantageous place to be is second in line so that someone in front of you who also matches won't take the prize away from you.

PairTheBoard

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I can't pretent to want to figure out the right answer to this question, but this, without a doubt, is not it.

[/ QUOTE ]

Under the assumption that there are a billion people in line, what's wrong with it?

PairTheBoard

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It sounds like you are only taking one factor into account. The further back you are in line, the more chance there is that someone in front of you will match and take the prize away from you, that is true. The other factor is that the earlier you are in line, the less chance you have of matching anyone ahead of you. These two competing factors produce a maximum probability of first matching which happens to occur at the 20th position.

To win, two things must happen. First, everyone ahead of you must have different b-days, or else someone else will win. The number of ways that n people can have different b-days is 365*364*363*... (n terms) = 365!/(365-n)! = P(365,n). In Excel, P is the =PERMUT function. The total number of ways the n people can have b-days is 365^n, so the probability that no one ahead of you has the same b-day is P(365,n)/365^n. Now, when the n people in front of you have different b-days, the probability that one of them has your b-day is n/365. So all together, the probability that you win is:

P(365,n)/365^n * n/365.

This has a maximum of 3.232% at n=19, so we want 19 people ahead of us, meaning that we want to be 20th.

[jason_t]

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Why would the number of people in line affect the likelihood of the 2nd person's birthday matching the first person's?

[/ QUOTE ]

It wouldn't. But it would effect the chance of any person in line getting more than one shot at it. The game will be over as soon as the first ticket is bought. If you don't match birthdays with the first in line you are just out of luck because you won't get any more chances. But if you do match birthdays your most advantageous place to be is second in line so that someone in front of you who also matches won't take the prize away from you.

PairTheBoard

[/ QUOTE ]

I can't pretent to want to figure out the right answer to this question, but this, without a doubt, is not it.

[/ QUOTE ]

Under the assumption that there are a billion people in line, what's wrong with it?

PairTheBoard

[/ QUOTE ]

It sounds like you are only taking one factor into account. The further back you are in line, the more chance there is that someone in front of you will match and take the prize away from you, that is true. The other factor is that the earlier you are in line, the less chance you have of matching anyone ahead of you. These two competing factors produce a maximum probability of first matching which happens to occur at the 20th position.

To win, two things must happen. First, everyone ahead of you must have different b-days, or else someone else will win. The number of ways that n people can have different b-days is 365*364*363*... (n terms) = P(365,n). In Excel, P is the =PERMUT function. The total number of ways the n people can have b-days is 365^n, so the probability that no one ahead of you has the same b-day is P(365,n)/365^n. Now, when the n people in front of you have different b-days, the probability that one of them has your b-day is n/365. So all together, the probability that you win is:

P(365,n)/365^n * n/365.

This has a maximum of 3.232% at n=19, so we want 19 people ahead of us, meaning that we want to be 20th.

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This solution is correct.