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Re: [OT] Math

From: Sean Bayan Schoonmaker <s_schoon@p...>
Date: Sun, 21 Jan 2001 09:41:52 -0800
Subject: Re: [OT] Math

Here we go.

Granted that my initial answer of p=(1/6)^4=7.716*10^(-4) only takes 
into account four hits and totally disregards the other two dice.

The chance of the other two dice missing is p=(5/6)^2.

Thus EXACTLY 4 dice hitting is p=(1/6)^4*(5/6)^2=5.358*10^(-4)

...or .05358%

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