**CAUTION**: this article is my attempt to clarify, as much for my benefit as anyone else, the assumptions involved in certain probability calculations. This is really only of theoretical interest. We are trying to find the probability of a coincidence, which (almost) everyone agrees is very small. In an attempt to keep things simple, I am working with an idealized problem. Several of my previous posts have been either misguided or outright wrong. I hope I've got it right now.

A blameless Mr B has occasional physical problems which cause him to sometimes stretch his hand out flat on the table. He is going to play 4 deals. Mr B has been suspected of having a way of signalling to his partner the number of aces he has. We want to work out the probability that the unfortunate Mr B will have exactly one ace every time he stretches his hand. I will use the approximation.44 for the chance that a hand has exactly one ace.

**Example 1**: Assume there is a .5 chance that Mr B will stretch his hand on each board, with the probabilities independent. Then the chance of his being wrongly accused is approximately .206:

Pr(1 stretch) (.44) + Pr(2 stretches) (.44)^2+ Pr(3 stretches)(.44)^3 + Pr(4 stretches)(.44)^4

**Example 2**: Mr B's physiology will cause him to stretch his hand exactly twice during the 4 boards. We make no assumption about what determines when this happens--- it could be the first two boards, or the last two, or a coin-flip between the first two and the last two, etc. Under this assumption, the lynching probability is .1936:

(.44)^2

[I believe this is the model underlying Kit Woolsey's calculation]

**Example 3**: We again assume that Mr B will stretch his hand twice, but we now make the further assumption that each pair of boards will be equally likely to be the ones on which the stretches occur. For some reason, we have become suspicious of the procedure by which the boards have been dealt. Instead of our previous .44 assumption, all we are prepared to assume is that exactly 2 of the 4 boards give Mr B one ace. It may even be that some demon has stacked the decks. Since there are 6 pairs of boards (assumed equally likely) on which Mr B may stretch, the chance that Mr B will be lynched is 1/6 = .1666...

[This admittedly contrived model is the basis for the calculation I (and a few others) have posted previously,although I believe the underlying assumptions can be stated in a more convincing way.]

I conclude by thanking my patient readers (if any), and apologizing for all past, present, and future mistakes.

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