pr.probability - probability puzzle - selecting a person

I have written an article about this game, and solved it numerically for the case of 10 person. I computed that the probability is the same for each of the 9 person.

In short, I have defined the following:

Let P(n,i,j,k) be the probability of at the n-th round, the k person is having the coin, while the people from i counting clockwise to k have all received the coin before.

And 3 recurrence equations are formulated:

$P (n+1,i,j,k)= frac{1}{2} P (n,i,j,k-1)+ frac{1}{2} P (n,i,j,k+1)$ ........(1)

$P (n+1,i,j,j)=frac{1}{2} P (n,i,j-1,j-1)+ frac{1}{2} P(n,i,j,j-1)$ ........(2)

....(3)

For details, please refer to:

Solving a probability game using recurrence equations and python

In this article, a(n,i) which denotes the probability of the i-th person being the head in the n-th round is found and plotted out as well.

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