A game is played with one standard die. On a turn, if a player rolls a 1, he gets two more rolls. If he rolls a 2, he gets one more roll. If he rolls anything else, his turn ends. His score is the sum of all the rolls on his turn (up to 3 rolls). If a player scores 5 points on a turn, what is the probability that it took 2 rolls?

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Bucharly9 · Accepted Answer

Step 1  One form of Bayes

Iacopelli5co · Accepted Answer

Step 1  Well, we could roll:  a 5 right away;   a 2, then a 3;   a 1, then a 1, then a 3;   a 1, then a 2, then a 2; or   a 1, then a 3, then a 1.  One roll:  P(5)=16  Two rolls:  P(2,3)=16×16=136  Three rolls:  P(1,1,3)=16×16×16=1216  P(1,2,2)=16×16×16=1216  P(1,3,1)=16×16×16=1216  So the probability of scoring 5 points is simply the sum of the probabilities above. The final answer we seek is the fraction of this sum belonging to the two-roll method - in other words, the relative probability of &quot;scoring 5 in two rolls&quot; to &quot;scoring 5 in any way&quot;.  Using conditional probability, we get:  P(2 rolls∣scored 5)=P(2 rolls∩scored 5)P(scored 5)  =P(2,3)P(5)+P(2,3)+P(1,1,3)+P(1,2,2)+P(1,3,1)  =13616+136+1216+1216+1216×216216  =636+6+1+1+1  =645=215  which is the same answer as before.  This method is easier in this case; however, the Bayesian method is more general and has much broader applicability.

A game is played with one standard die. On a turn, if a player rolls a 1, he gets two more rolls. If

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