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Determine the probability of winning a game of craps by rolling a sum of 6

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
I'm aware that the

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.
 

awkward

Member
Feb 18, 2012
36
Hi anemone,

I'm not sure I understand exactly which part you're having trouble with, but I'm going to assume that it's how you can avoid summing the infinite series-- i.e., how we can find that the probability of rolling a 6 before rolling a 7 is 5/11. (I'm skipping the part about rolling a 6 on your first roll.)

So let's say \( p \) is the probability of rolling 6 before you roll a 7. On the first roll, there are three possibilities:
(1) With probability 5/36, you roll a 6. Hurray! If this happens, your probability of success is 1.
(2) With probability 6/36, you roll a 7. Sob! If this happens, your probability of success is 0.
(3) With probability 25/36, you roll something else. Now you are right back where you started, and your probability of (eventual) success is \( p \).

So $$ p = (5/36) \cdot 1 + (6/36) \cdot 0 + (25/36) \cdot p $$
This is an equation which you can solve easily, with the result \( p = 5/11 \).
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
:)
Yes, I mean to ask how one can avoid summing the infinite series.
It makes a lot of sense now!
Thanks for the kind and great explanation!