Fifty people are in a line. The first person in the line to have a birthday matching someone in front of them will win a prize. Of course, this means the first person in the line has no chance of winning. Which person has the highest likelihood of winning? Defend your answer with a statistically sound argument.

Answer:

20th person

Step-by-step explanation:

Before proceeding into problem,assume that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout a 365 day year.

The probability, p(n), of winning a prize when you are the nth person is line is:

(probability that none of the first n−1 people share a birthday)*( probability that you share a birthday with one of the first n−1 people)

So,The first person can't have the same birthday as anyone less by themselves, so lets look at the second person. At the point where we add the second person there is a 364/365 chance of them not sharing a birthday with the previous person. When we add the 3rd person there is there is a 363/365 chance of them not sharing a birthday and so on such that the (n-1)th person when added has a probability of not forming a share of (365 -n -2)/365. We can multiply these together to get the probability that none of first (n-1) people share their birthday.

Therefore the position in line that has highest chance of winning is 20th.

We want to see which position has the largest probability of winning the prize. We will find that the 18th person has the largest probability of winning.

We know that a person only will win the prize if that person has the same birthday that some in front of him/her.

So, the probability that the n-th person wins the prize is given by:

(probability that none of the persons in front share a birthday)*(probability of sharing a birthday with some of the n - 1 people in front).

Let's find the first one.

The first person can have a birthday in any of the 365 days.

The second person must not have the same birthday, so there are 364 options remaining.

The probability of this is just

p₂ = 364/365

For the third person, we have the same thing, this time, 363 remaining options:

p₃ = 363/365

For the (n - 1) person the probability is:

pₙ₋₁ =  (365 - n + 2)/365

And the n-th person must share a birthday with some of the n persons in front of him, so there are (n - 1) options to choose, so the probability is:

pₙ = (n - 1)/365

Then, the joint probability is:

P = p₂*p₃*...*pₙ

[tex]P = \frac{364*363*...*(365 - n + 2)*(n - 1)}{365^{n-1}}[/tex]

Now we have the equation, we just need to solve it for n such that we maximize the value of P in the range n ∈ [1, 50]

This is actually a really hard thing to find, but what you can notice, is that each time that n increases by 1, the denominator increases 365 times.

While the numerator increases by a factor:

(365 - n - 1)*n/(n - 1)

So we need to find the value of n such that the thing above is equal or smaller than 365.

Let's solve:

(365 - n - 1)*n/(n - 1) = 365

(365 - n - 1)*n = 365*n - 365

364*n - n^2 - 365*n + 365 = 0

-n^2 - n + 365 = 0

This is just a quadratic equation, the solutions are:

[tex]n = \frac{1 \pm \sqrt{1 + 4*365} }{-2}[/tex]

We only care for the positive solution, so we take:

[tex]n = \frac{1 - \sqrt{1 + 4*365} }{-2} = 18.6[/tex]

So, for the 19th person, the probability of winning will be smaller than for the 18th person, this happens because the numerator increased by a smaller factor than the denominator.

This means that the 18 person has the largest probability of winning.

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