Consider the following game, played with three standard six-sided dice. If the player ends with all three dice showing the same number, she wins. The player starts by rolling all three dice. After this first roll, the player can select any one, two, or all of the three dice and re-roll them. After this second roll, the player can again select any of the three dice and re-roll them one final time. For the following questions, assume that the player used the following optimal strategy: if all three dice match, the player stops and winds; if two dice match, the player re-rolls the die that does not match; and if no dice match, the player re-rolls them all.

Requied:
a. Find the probability that all three dice show the same number on the first roll.
b. Find the probability that exactly two of the three dice show the same number after the first roll.
c. Find the probability that the player wins, conditioned on exactly two of the three dice showing the same number after the first roll.

Answer:

a

 [tex]P(A) =  \frac{1}{36}[/tex]

b

[tex]P(B) = \frac{15}{36}[/tex]

c

[tex]P(U) = \frac{11}{36}[/tex]

Step-by-step explanation:

From the question we are told that

   The  number of dice is  [tex]n  =  3[/tex]

Generally  for all three dice to show the same number the second and the third dice must have the same outcome  as the outcome of the first dice.

This means that the number of outcome the first die can  have is  6  (6 sides), the number of outcome which the second and the third dice can have is  1 (they must match the first )

So the probability that all three dice show the same number on the first roll is mathematically represented as

      [tex]P(A) =  \frac{6}{6} * \frac{1}{6} *  \frac{1}{6}[/tex]

=>   [tex]P(A) =  \frac{1}{36}[/tex]

Generally for two of the three dice show the same number after the first roll then the number of outcome for  one of the  dice would be is  6 , the number of outcome for another one must be   1(i.e it must match the first ) , the number of outcome for the remaining one must be   5 (i.e it can show any of the remaining  5  sides which the first and second dice are not showing )

Now the number of ways of selecting this 2 dice that show the same number from the 3 dice is mathematically represented as

     [tex]N  =  ^3C_2[/tex]

Here C stands for combination

So

      [tex]N  =  ^3C_2  = 6[/tex]

So the probability that exactly two of the three dice show the same number after the first roll is mathematically represented as  

      [tex]P(B) = N   \frac{1}{6} *  \frac{5}{6}[/tex]

=>  [tex]P(B) = \frac{15}{36}[/tex]

Generally from the question we are told that if two dice match, the player re-rolls the die that does not match.

Now the probability that  the die that did not match the first  time will match the second time is

     [tex]P(E ) =  \frac{1}{6}[/tex]

Generally if that one die does not show the same number in the second round , the probability that it will match  in the third round is

    [tex]P(R) =  \frac{5}{6} *  \frac{1}{6} [/tex]

Generally the probability that he wins (i.e when all three are showing the same number ) and  exactly two is showing the same number is mathematically represented as

       [tex]P(K) =  P(E)* P(B) + P(R)* P(B)[/tex]

=>    [tex]P(K) =  \frac{1}{6} * \frac{15}{36}  +  \frac{5}{6} *  \frac{1}{6} * \frac{15}{36}[/tex]

=>  [tex]P(K)= \frac{165}{1296}[/tex]

So the probability that the player wins, conditioned on exactly two of the three dice showing the same number after the first roll is mathematically represented as  

    [tex]P(U) =  \frac{\frac{165}{1296}}{\frac{15}{36} }[/tex]

=>  [tex]P(U) = \frac{11}{36}[/tex]

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