show/hide this revision's text 3 added 219 characters in body

First: we are ignoring leap years.

This means that if we have a line of 366 people, two of them must share a birthday.

Also, the first person in line cannot win.

So the sum of probabilities for the remaining 365 positions must total to 1.

Meanwhile, since only one person can win, we know that all positions ahead of the winner have distinct birthdays (and we do not know or care about the positions following the winner).

The base chance to win (ignoring other possibilities) thus increases linearly.

The base chance to be continuedbeaten increases quadratically (it's the sum of the base chance to be beaten, by each of your predecessors).

(to be continued...)
show/hide this revision's text 2 left out a sum, scratching most of post for now

First: we are ignoring leap years.

This means that if we have a line of 366 people, two of them must share a birthday.

Also, the first person in line cannot win.

So the sum of probabilities for the remaining 365 positions must total to 1.

Meanwhile, since only one person can win, we know that all positions ahead of the winner have distinct birthdays (and we do not know or care about the positions following the winner).

This gives us a sequence like this (and I am going to use y for what the OP expressed as N because it's a year that we are talking about here):

y=365

P(1): 0

P(2): 1/y

P(3): (2/y)*(1-1/y) = (2y-2) / (y^2)

P(4): (3/y)(1-(2/y)(1-1/y)) = (3y^2-6y+6)/ (365^3)

P(5): (4/y)*(1-...) = (4y^3 - 12y^2 + 24y - 24) /

(365^4) ...

In other words, our probabilities can be expressed as the absolute values of a sequence of polynomials:

$P(k) = \frac{k!}{y^k}\sum_{s=0}^{k-1} \frac{(-y)^s}{s!}$

(to be continued...continued)
show/hide this revision's text 1

First: we are ignoring leap years.

This means that if we have a line of 366 people, two of them must share a birthday.

Also, the first person in line cannot win.

So the sum of probabilities for the remaining 365 positions must total to 1.

Meanwhile, since only one person can win, we know that all positions ahead of the winner have distinct birthdays (and we do not know or care about the positions following the winner).

This gives us a sequence like this (and I am going to use y for what the OP expressed as N because it's a year that we are talking about here):

y=365

P(1): 0

P(2): 1/y

P(3): (2/y)*(1-1/y) = (2y-2) / (y^2)

P(4): (3/y)(1-(2/y)(1-1/y)) = (3y^2-6y+6)/ (365^3)

P(5): (4/y)*(1-...) = (4y^3 - 12y^2 + 24y - 24) / (365^4) ...

In other words, our probabilities can be expressed as the absolute values of a sequence of polynomials:

$P(k) = \frac{k!}{y^k}\sum_{s=0}^{k-1} \frac{(-y)^s}{s!}$

(to be continued...)