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

[anemone]

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]

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 \( p = 5/11 \).

 

[anemone]

:)
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!