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# RE: [GZG] lessons in dice

From: le morpion <morpion_1@y...>
Date: Tue, 25 Oct 2005 18:54:28 +0200 (CEST)
Subject: RE: [GZG] lessons in dice

for thoses interessted in probabilities in FT I'd
calculate long ago the math expectancy and standard
gap of some weapons :

the reroll possibilies are included and were dealt as
infinite limit of serie's sum (don't know how to say
that in proper english lool)

/L

> One quick point:
>
> Rolling 1 dice for 2 sixes or 3 sixes versus rolling
> 2 or 3 dice for the same result -
> equal probability!
>
> Proof:
>
> 1 dice for two sixes in 2 rolls:
> Chance of getting a 6: 1/6.
> Change of getting 2 in a row: 1/6 * 1/6 = 1/36
>
> 2 dice for two sixes in one roll:
> Chance of getting a single 6 on one die: 1/6
> Chance of getting both dice as sixes: 1/6 * 1/6 =
> 1/36
>
> Expand this to three dice, and you have:
> 1 dice, rolled 3 times, for 6, 6, 6: 1/6 * 1/6 * 1/6
> = 1/216
> 3 dice, rolled once, for 6, 6, 6: 1/6 * 1/6 * 1/6 =
> 1/216
>
> I figured most of us know that rolling two dice
> yields 36 possible combos, only one of
> which is double six. Similarly, roling 3 dice yields
> only one way to generate an 18 for
> possibilities. That works out to be
> just the same as the odds of rolling 1 die for two
> sixes in a row or for three sixes in a
> row (respectively). So it doesn't matter if you
> throw Nd6 or d6 N times to generate X
> number of sixes. This logic applies beyond the
> number of dice = 3.
>
> Tom B
> (just to settle that side point)
> (Can we stop now, BTW, or do we need to open a new
> gzg-math@csua.berkeley list...?)
>
>
>
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>
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