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- Feb 14, 2012

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P( winning a game of craps by rolling a sum of 6)

=P(getting 6 the first time and 6 the second time) or

P(getting 6 the first time, get some number other than 6 and 7 the second time and get 6 again the third time) or

P((getting 6 the first time, get some number other than 6 and 7 the second and third time and get 6 again the fourth time)...

the sequence continues in such a manner infinitely (this is to say that we're having an infinite sum of a geometric sequence)

= $((\frac{2!}{36}\times2)+\frac{1}{36})+ S_\infty$

=$\frac{5}{36}+ \frac{(\frac{5}{36})^2}{1-\frac{25}{36}}$

=$\frac{5}{36}+ \frac{25}{396}$

=$\frac{20}{99}$

But a quick goggle search shows that I don't have to consider the infinite sequence and the probability of winning a game of craps by rolling a sum of 6 for the third time and so on is simply

$\frac{P(winning \;a \;game \;f \;craps \;by \;rolling \;a \;sum \;of \;6 }{P(winning \;or \;losing)}=\frac{\frac{5}{36}}{\frac{5}{36}+\frac{1}{36}}=\frac{5}{11}$

Therefore,

P( winning a game of craps by rolling a sum of 6)

=P( winning a game of craps by rolling a sum of 6 the second time) and P(winning a game of craps by rolling a sum of 6 on the third time and so on)

=$\frac{5}{36}\times\frac{5}{11}=\frac{25}{369}$

Finally,

P( winning a game of craps by rolling a sum of 6)=$\frac{5}{36}+ \frac{25}{396}=\frac{20}{99}$

I can't quite get my head around that fact. It has confused rather than enlighten me.

Could someone please explain to me why this works?

Thanks.