More seconds?

In the last blog post, in our attempt to find the probability that a player going second finishes second in a game with four players, we found that the probability that the player going first finishes first and the player going second finishes second (in a game with four players) is equal to the probability that a player going first finishes first in a game with four players times the probability that a player going first finishes first in a game with three players. Of course, it's also possible for the players going third or fourth to finish in first and the player going second to finish second. Since the player going in first is still in the game if this happens, the second condition then changes to F(2,3,1), since the player then has the second position.

The total probability that the player going second finishes the game in second place is therefore:

F(1,4,1)*F(1,3,1) + F(3,4,1)*F(2,3,1) + F(4,4,1)*F(2,3,1)

Now we can take a step closer towards a generalization. The probability that a player going in position n gets second place in a game with p players is:

F(1,p,1)*F(n-1,p-1,1) + F(2,p,1)*F(n-1,p-1,1) + ... + F(n-1,p,1)*F(n-1,p-1,1) +   F(n+1,p,1)*F(n,p-1,1) + ... + F(p,p,1)*F(n,p-1,1)

Let's actually try this out. I wrote a computer program that performs this calculation for a given n and p. Letting p be equal to four and varying n between the integers between 1 and 4 (inclusive), we get:

When n = 1: 0.2537844586561435
When n = 2: 0.2513862054039849
When n = 3: 0.24879532739096866
When n = 4: 0.24603400854886412

These values, when added, sum to 1, which is at least a somewhat decent verification that they are correct (more precisely, it is a necessary but not sufficient condition to it being correct). Interestingly the first player still has an edge over the other players.

More to come on this later, but this seems like a good place to stop for now.