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In the gambling game "craps," a pair of dice is rolled and the outcome of the experiment is the sum of the points on the up sides of the six-sided dice. The bettor wins on the first roll if the sum is 7 or 11. The bettor loses on the first roll if the sum is 2, 3 or 12. If the sum is 4, 5, 6, 8, 9, or 10, that number is called the bettor's "point." Once the point is established, the rule is as follows: If the bettor rolls a 7 before the point, the bettor loses; but if the point is rolled before a 7, the bettor wins.

Given that 8 is the outcome on the first roll, find the probability that the bettor now rolls the point 8 before rolling a 7 and thus wins. Note that at this stage in the game the only outcomes of interest are 7 and 8. Thus, find P(8|7 or 8)

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  • $\begingroup$ Is this supposed to be related to Craps? $\endgroup$ – Henry Feb 4 '16 at 0:31
  • $\begingroup$ @Henry yes, I just figured that wasn't really necessary for the question (since I feel the basic idea is the same) $\endgroup$ – secondubly Feb 4 '16 at 0:33
  • $\begingroup$ Because if your first roll in Craps is $8$ then it becomes the "point" and the next stage is to keep rolling until you get a $7$ or a second $8$. Your question then starts to make sense if understood as being "Given you initially rolled an $8$ with two six-sided dice and then keep rolling them, what is the probability you roll a second $8$ before you roll a $7$?" $\endgroup$ – Henry Feb 4 '16 at 0:39
  • $\begingroup$ Well, I don't understand the question and I imagine there was something in the wording that gave more information that is lacking in the post. Having rolled an eight or a seven already has as much bearing on whether you roll a seven as whether you ate breakfast that morning. $\endgroup$ – fleablood Feb 4 '16 at 0:41
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    $\begingroup$ if the question is: Given that you rolled either an 7 or an 8, what is the probability that you rolled (the very same roll) a 8. THEN the answer is 5/11. So the real question is what is the actual question. $\endgroup$ – fleablood Feb 4 '16 at 0:51
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I suppose we are rolling two dices with $6$ faces, and we know we have rolled a $7$ or an $8$ (summing the two scores of the dices), I'm not sure this is the case. Given that:

Every combo has the same probability to be rolled, ie $\frac{1}{36}$.

You can roll $7$ in 6 different ways, and $8$ in 5 different ways. They sum up to 11. Only $5$ of them are good so the probability is $\frac{5}{11}$.

If you don't understand that you can blindly apply the Bayes Formula:

$A$ is the event: you get a $8$

$B$ is the event: you get a $7$ or a $8$.

Of course $P(A)=\frac{5}{36}$, $P(B)=\frac{11}{36}$ and $P(B|A)=1$.

The Bayes Formula says:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

Thus $P(A|B)=\frac{5}{11}$.

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  • $\begingroup$ My question here is "only five of them are good" - what does that mean? I mean, are we not rolling the dice again? Why is the result out of 11 instead of out of 36 or something else? I understand that on the first roll, you have 11/36 chance to get a seven or an eight - but how does that reduce the possibilities of the second roll? $\endgroup$ – secondubly Feb 4 '16 at 0:46
  • $\begingroup$ I edited adding Bayes Formula! $\endgroup$ – Maffred Feb 4 '16 at 0:48
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    $\begingroup$ There are no two rolls! Someone rolled two dices in a secret room, and someone told you he rolled a $7$ or an $8$ but he is not sure which one of them. $\endgroup$ – Maffred Feb 4 '16 at 0:49
  • $\begingroup$ I edited the question so maybe now you'll see why I'm a bit confused. $\endgroup$ – secondubly Feb 4 '16 at 1:18
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In view of this answer, I think the question should be worded differently. It should be "Assuming that you have rolled 7 or 8, what is the probably that you have rolled 8?) "

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    $\begingroup$ This actually makes more sense - reading the question it was unclear top me what exactly was happening. Because they made it sound like there were two rolls, not just the one $\endgroup$ – secondubly Feb 4 '16 at 1:02
  • $\begingroup$ This is not an answer. It should be a comment. It is a good comment. $\endgroup$ – Ross Millikan Oct 5 at 3:45

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