Conditional probability question?

Warning: Long answer ahead!

One form of Bayes' rule states the following:

#Pr(B|A)=(Pr(AB))/(Pr(A))=(Pr(A|B)*Pr(B))/(Pr(A))#

This allows us to write a conditional probability of "B given A" in terms of "A given B", which may be easier to calculate. In this question,

B = "die rolled twice", and
A = "score is 5".

Let's write this into the equation:

#Pr("2 rolls"|"score is 5")=(Pr("score is 5"|"2 rolls")*Pr("2 rolls"))/(Pr("score is 5"))#

The numerator on the RHS is easy to deduce; however, the denominator, in its current state, is not. But we can make it easier. We recall that, if #A# can be partitioned into #k# disjoint events #AnnC_1, AnnC_2, ..., AnnC_k# (where none of these intersected regions overlap, and together they all form #A#), then

#Pr(A)=sum_(i=1)^kPr(A|C_i)*Pr(C_i)#

We actually have that here; the #C_i#'s will be the number of possible rolls on a turn. What we're saying is that

#Pr("score is 5")=Pr("score is 5"|"1 roll")*Pr("1 roll")#
#color(white)"XXXXXXXXX"+Pr("score is 5"|"2 rolls")*Pr("2 rolls")#
#color(white)"XXXXXXXXX"+Pr("score is 5"|"3 rolls")*Pr("3 rolls")#

Kind of a "the whole is equal to the sum of its parts" thing.

From hereon, let's use the shorthand #S_5# for "the score was 5" and #R_n# for "#n# number of rolls". Putting this all together, we have

#Pr(R_2|S_5)=(Pr(S_5|R_2)Pr(R_2))/(Pr(S_5|R_1)Pr(R_1)+Pr(S_5|R_2)Pr(R_2)+Pr(S_5|R_3)Pr(R_3))#
#=[Pr(S_5nnR_2)]/[Pr(S_5nnR_1)+Pr(S_5nnR_2)+Pr(S_5nnR_3)]#

Now, we do the calculations!

#Pr(S_5nnR_1)=Pr("roll a 5")#
#=1/6#
#Pr(S_5nnR_2)=Pr("roll a 2, then a 3")=1/6*1/6#
#=1/36#
#Pr(S_5nnR_3)=Pr["roll a 1; then (1,3), (2,2), or (3,1)"]=1/6*3/36#
#=1/72#

Finally, we place these values back into the equation for #Pr(R_2|S_5):#

#Pr(R_2|S_5)=(1/36)/(1/6+1/36+1/72)color(blue)(*72/72)#

#color(white)(Pr(R_2|S_5))=(2)/(12+2+1)#

#color(white)(Pr(R_2|S_5))=2/15#

For completeness, #Pr(R_1|S_5)=12/15=4/5,# and #Pr(R_3|S_5)=1/15.# These three probabilities sum to 1, which is what we'd expect, since if the player's score was 5, it had to take either 1, 2, or 3 rolls.

Bonus:

There's a great Numberphile video on YouTube discussing Bayes' rule here:

What are the different ways to score 5 points, and what are their probabilities? 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.

#ul("One roll:")#
#P(5)=1/6#

#ul("Two rolls:")#
#P(2, 3)=1/6*1/6=1/36#

#ul("Three rolls:")#
#P(1, 1, 3)=1/6*1/6*1/6=1/216#
#P(1, 2, 2)=1/6*1/6*1/6=1/216#
#P(1, 3, 1)=1/6*1/6*1/6=1/216#

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 "scoring 5 in two rolls" to "scoring 5 in any way".

Using conditional probability, we get:

#P("2 rolls"|"scored 5")=[P("2 rolls"nn"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)]#

#=(1/36)/(1/6+1/36+1/216+1/216+1/216)color(blue)(*216/216)#

#=6/(36+6+1+1+1)#

#=6/45=2/15#

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.