Problem of Points - The Solution

The solution to the problem of points tells you how you would divide up the stakes in a fair game (fair in the sense of each step outcome being equally likely to favour any player) between two players A and B if A needed $n_A$ more wins and B needs $n_B$.  Pascal and Fermat both end up counting the set of all possibilities and comparing the respective counts to each other and come up with the ratio 

$\sum_{k=0}^{n_B-1}\frac{(n_A+n_B-1)!}{k!(n_A+n_B-1-k)!}$ to $\sum_{k=n_B}^{n_A+n_B-1}\frac{(n_A+n_B-1)!}{k!(n_A+n_B-1-k)!}$.  

This rather ugly looking formulation is something I'll be looking at over the next couple of posts, in a way mathematicians usually don't.  Enamoured of Euclid, they think interesting maths involves proof and concise statement.  That does not work for me.  I want to unpack it and see some examples with real numbers.  Get a feel for it in use.  And after those posts, I'll be doing the same for gambler's ruin, which I personally think has caused me to think a lot more generally than this solution to the problem of points.

Before I finish on this short post, I'd like to say that this solution to the problem of the division of stakes, if you think about it, is the price of the seat if somebody wanted to buy you out of the game.  This is the fair price of your seat at that moment, or the fair price of your position in the same.  And given that the moment in question can analysed at any point in the game, including the moment before the game starts, it also represents an algorithm for working out the fair price of the game for both players, at all points.  That is, it tells you fully at all moments in the game the expected value of each hand.

If the stakes are a value of S then the expected value of one player is 

$S\frac{\sum_{k=0}^{n_B-1}\frac{(n_A+n_B-1)!}{k!(n_A+n_B-1-k)!}}{\sum_{k=0}^{n_B-1}\frac{(n_A+n_B-1)!}{k!(n_A+n_B-1-k)!} + \sum_{k=n_B}^{n_A+n_B-1}\frac{(n_A+n_B-1)!}{k!(n_A+n_B-1-k)!}}$

and for the other just has the other sum on the numerator.