Four friends passing a ball

Question

Four friends stand in a group. Whoever holds the ball passes it to one of the other three friends, chosen uniformly at random (probability $\tfrac13$ each). This repeats many times.

In the long run, what fraction of passes does any one chosen friend hold the ball?

Accepted Answer

We could set up four equations, but the arrivals are symmetric. A given friend receives the ball only from the other three, and each of those three passes to her with probability 	frac13. So the total chance of the ball arriving at any one friend is 	frac13 + 	frac13 + 	frac13 = 1, the same for everyone. Each row sums to 1 (the holder always passes) and now each column sums to 1, making the chain doubly stochastic. That forces an even long-run split. With four friends, \pi = 	frac14 	ext{ per friend}. So any one friend holds the ball one quarter of the time. Takeaway. When arrivals into…

Four friends passing a ball

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