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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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