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Alternative formulation of the recurrence

Posted By: Maik Stiebler
Date: Friday, 12 April 2019, at 6:51 a.m.

In Response To: Two ways to get the limiting probability (Bob Koca)

If there are 2^N players, the probability that they meet in in the first round is 1 / (2^N - 1). If they don't - probability (2^N - 2) / (2^N - 1) - they split the pot if both lose in the first round - probability 1/4 - or if they both win - also probability 1/4 - and are eliminated in a later round - which is just the (N-1)-case.

Thus P(N) = [(2^N - 2) / (2^N - 1)] * (1/4) * [1 + P(N-1)].

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