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eXtreme Gammon question

Posted By: Tom Keith
Date: Thursday, 27 August 2009, at 5:13 p.m.

In Response To: eXtreme Gammon question (MaX)

Here is a counterexample.

Suppose we somehow know the correct answer for Play 1 -- maybe we rolled it out a gazillion times. In other words, the uncertainties are all in Plays 2 and 3.

There is an 85% chance that Play 2 < Play 1 and a 90% chance that Play 3 < Play 1. If the uncertainties of Plays 2 and 3 are independent (what we're assuming, I think), then the probability that both plays are < Play 1 is .85 x .90 = .765.

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