Coincident Birthdays

**milleran** · 01-16-2010, 03:22 PM

Hi, I wanted to know how to both do this probability problem and how to do it in excel.

A department store is celebrating their 100th anniversary. They are offering a $100,000 prize to one lucky customer on the date of their anniversary. Customers are asked to line up outside the store, and the first person to have a birthday that matches someone in front of them will win the prize (this doesn't have to be directly in front of them; it can be anywhere in front). Which position in line has the highest probability of winning?

Thank you in advance.

**daddylonglegs** · 01-16-2010, 03:45 PM

Hello milleran, that's an interesting question.

Is this some sort of assignment question or just for general interest....or something else?

**zeke varg** · 01-16-2010, 04:09 PM

Not a problem to do in Excel. A fairly basic statistics assignment. Which class are you taking?

**shg** · 01-16-2010, 04:12 PM

Pssst: Be 12th in line for a 9.1% chance.

**milleran** · 01-18-2010, 04:53 PM

Yeah, it is for my honors math analysis class and it is due tomorrow. I kind of know what to do, but I really don't know where to start, nor can I figure out how to do it in excel.

**milleran** · 01-18-2010, 04:57 PM

shg, how did you get that answer, because I need to write a step-by-step mathematical justification.

thanks.

**milleran** · 01-18-2010, 05:02 PM

And how exactly do you do in excel, zeke varg?

**shg** · 01-18-2010, 06:07 PM

So why don't you post the math, and we'll help with the Excel part?

First, calculate the probability of all birthdays being unique for each position 1, 2, 3, ...

Then, for a person in position N+1, the probability that they have the first common birthday is the probability that all birthdays are unique for the person immediately ahead of them, times the probability that they share a birthday with any of the people ahead.

And my prior answer was wrong

**zeke varg** · 01-18-2010, 06:52 PM

shg, how'd you get to that? I'd stay 20th for a measly 3.23%! Does that mean I won´t win?

Please, let me know how you got there...

milleran, to give you a hint were to start, remember that to be the first with a same birthday, noone in front of you may have already won (1-p)! Good Luck!

**shg** · 01-18-2010, 07:13 PM

Same as you did. Here are my results with 50,000 Monte Carlo trials.
Clip1.jpg

**Richard Buttrey** · 01-18-2010, 07:38 PM

Hi,

This seems remarkably similar to the fairly well known puzzle which asks how many people are there needed in a group to ensure that the chance that two people will have the same birthday is better than 50%.

The answer to that problem is to turn the question on its head and ask what are the chances that the 2nd, 3rd, 4th etc. people in the group don't have the same birthday as any of the others.

The chance of the 2nd person not having the same birthday as the first is 364/365.
The chance that the 3rd doesn't have the same birthday as 1 or 2, is 364/365 x 363/365.
The chance that the 4th doesn't have the same birthday as 1, 2 or 3, is 364/365 x 363/365 x 362/365.

etc.

If you continue those expressions you find that you need 23 people to ensure that the probability of NOT having the same birthday drops below 50%.

However your question seems oddly phrased. Each succeeding person in the line has a better chance of having the same birthday as someone in front of them. Hence the 366th person is certain to have the same birthday as someone in front of them whereas all the others before him have a less than certain chance. Thus the answer to the question you asked is (trivially) the 366th person.

Are you sure the question has been posed correctly?

Regards

**shg** · 01-18-2010, 07:42 PM

The question is, which position in line has the best chance of being the FIRST with a birthday coincident to someone in front of them.

**milleran** · 01-18-2010, 07:51 PM

What I ended up doing is finding the Probability of Matching P(M) times the Probability of Winning Given that You Match P(W|M) = the Probability of Matching and Winning.

So, P(M) X P(W|M) = P(M and W)

To get P(M), I did n people divided by 365.25 to account for leap years to get P(M).
To get P(W|M), I did 1- P(M).

Then, I multiplied the two together to get P(M and W).

Because I got that the 17th person in line has a .0287 chance of winning.

Is that right?

Thanks.

**shg** · 01-18-2010, 08:41 PM

That's not what I get. How did you compute P(M) and P(W|M)?

**milleran** · 01-18-2010, 09:18 PM

P(M) = n people divided by 365.25

P(W|M) = 1- P(M), where P(M) is = to the sum of all the P(M)s before that person,

so for the 15th person, it is 15/365.25 X (1-P(M) for persons 1-14) = .02879.

What did you get?

**shg** · 01-18-2010, 09:31 PM

The chance of the first person being unique is 365/365

The second is 365/365 * 364/365

The third is 365/365 * 364/365 * 363/265

The Nth is 365 * 364 * 363 * ... * (365 - N + 1) / 365^N

**milleran** · 01-18-2010, 09:43 PM

That is what I did at first; however, that doesn't take into account the probability that someone else might have already won.

**shg** · 01-19-2010, 01:09 AM

That's the probablility that the Nth person has the Nth unique birthday.

**Richard Buttrey** · 01-19-2010, 06:08 AM

Originally Posted by shg

The question is, which position in line has the best chance of being the FIRST with a birthday coincident to someone in front of them.

Hi shg,

That's not my reading of the OP which specifically simply said,

Which position in line has the highest probability of winning?

All the customers have a probability of winning, and before all the birthdays are noted in order to find who actually has a birthday co-incident with someone before them, the highest probability is surely the 366th person, since this is the only person certain to have a birthday the same as someone before them.

Or am I missing something?

Rgds

**daddylonglegs** · 01-19-2010, 08:19 AM

I believe shg is correct

The question is "which position in line has the highest probablity of winning".

How do you win? By being the first person to have the same birthday as somebody in front.

The trick is that, of course, as you go down the line each successive person has a better chance of matching birthdays with somebody in front......but at the same time there's a greater chance that you won't win because somebody in front will have won before you.

This means that in terms of best position the probablilty of winning increases from zero at position 1 until some point when it starts to decrease again (when the probablity of somebody in front having won outweighs your advantage in being further back).

I started doing the actual maths and my initial thoughts were that position was around 10th but I think there was a flaw in my thinking......more work required

**Richard Buttrey** · 01-19-2010, 08:39 AM

H ddl,,

I don't disagree with your logic, but the question asked wasn't how do you win? (We know how you win - by having the same birthday). The question was 'which position in line has the highest probability of winning? Since it was asking about probabilities that suggests to me that we're looking to evaluate the starting position and give an answer before we even start noting the individual birthdays to find out who has actually won.

I think it just shows how careful one has to be with the words used in these sort of things. Not directly related to this, but amongst the classic logic problems like the prisoner's dilemma etc. one of my favourites is the Monty Hall problem. But maybe this is now a water cooler topic.

Rgds

**zeke varg** · 01-19-2010, 03:48 PM

I have to agree with shg on this one. Remember that winning is all about being the first to have a coincident birthday! For that to be fulfilled, it must be conditional that no two persons in front of you already have same birthdays, as hence you win no longer be first with a coincident birthday and thus not the winner. It is a compounded probability - In order to be first with a second birthday on the same date, no one else may have had that before. You may not ignore that part of the probability.

The position with the highest probability to win is 20.

**Richard Buttrey** · 01-19-2010, 04:23 PM

Hi,

We'll have to agree to disagree. I may be being Percy Pedant and maybe you think this is symantics, but the question was which person in line has the highest probability of winning, NOT who wins. Are the two the same?

The highest probability which satisfies the question is self evidently 1, i.e. certainty. the only person in the line who has that probability is number 366, everyone else before them has a lower probability.

But out of interest how do you get the answer 20? I would agree that there's a better than 50% chance that someone in the first 23 people has the same birthday as someone else in that population of 23, (the oft quoted birthday paradox) but I don't think that is relevant here.

Regards

**shg** · 01-19-2010, 05:45 PM

... but the question was which person in line has the highest probability of winning,

Winning is defined as being the FIRST person in line with a coincident birthday, Richard. Given the choice, in which position would you stand?

... how do you get the answer 20?

See attached. The macro is just for running the Monte Carlo simulation, need not be enabled.

**zeke varg** · 01-20-2010, 03:38 PM

milleran, did you see how it is done now?

I did basically what shg did, with a nuance. It may be easier to follow (?).

At first and consider that any person entering the store before you will NOT have coincident birthdays with anyone else. What is your probability of winning for each person that enters the store? Well, after 1 person has entered, your probability p1 will be 1/365 (ignoring leap years). After 2 persons have entered, p2 = 2/365 and so on. In general, pn = n/N, where n = number of persons already entered the store and N = total possible different birthdays (and pn is p1, p2 etc.).

Ok, so that's your chance of winning, conditionally that noone has already won! So, what is this probability? Well, look at yourself entering again. For each person entering your probability to win was pn = n/N. So, your probability of NOT winning is (1-pn), right? Remember p is changing, depending on how many have already entered the store!. After, say, 5 have entered the chance of having an equal birthday with anyone else is p5. But, in order to win, noone else may have already had won, i.e. had an equal birthday. The probability the second guy who entered didn't win is (1-p1), the probability the third guy entering didn't win is (1-p2) The probability neither of them won is (1-p1)(1-p2)! Thus, the probability neither of the second, third, fourth and fifth person entering didn't win is (1-p1)(1-p2)(1-p3)(1-p4). After this, you just multiply the probability that you will win, standing in the sixth position, i.e. p5, thus...

The probability to win in the sixth position in the line is: p5(1-p1)(1-p2)(1-p3)(1-p4), i.e.
(you have the same birthday with someone else)*(noone in front of you has the same birthday with someone else) = (you win)*(noone else has already won)

Does this make it clear how I did it?

And Richard, do you now see why your reasoning is incorrect (not intending to be rude here)? Your probability of winning will NOT increase by being farther back, as your probability of winning is DEPENDENT on the probability noone else has already won.

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