You play a game where you roll two dice and then sum up the faces. If on the fir
ID: 3051055 • Letter: Y
Question
You play a game where you roll two dice and then sum up the faces. If on the first roll you rol a 7 or 11 you win. If you roll a 2,3, or 12 then you lose on the first roll. In the other events then roll again until you roll that same number or a 7. If you roll the same number before a 7, then you win. Otherwise you lose. (a) What is the probability of winning the game given that you first roll was a 6? (b) What is the probability of winning the game by first rolling a 4 and then rolling a 4. (c) What is the probability of winning the game.Explanation / Answer
a)The sample space of rolling two dice is
(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)
(2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)
(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)
(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)
(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)
(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)
The probability of winning the game given that the first roll was a 6 is
Assume the first roll was a 6.
Let Pr(x) stand for the probability of event x happening on any given roll.
Probability of getting 6 on first roll=Pr(6) = 5/36, Pr(anything other than 6 and 7) = 1-5/36-6/36 = 25/36.
Pr(rolling a 6 before a 7)
= 5/36 + (25/36 * 5/36) + ((25/36)2 * 5/36) + ((25/36)3 * 5/36) + ...
= 5/36 * sum for i = 0 to infinity of (25/36)i
= 5/36 * (1/(1-25/36))
= 5/36 * 36/11 = 5/11.
b)The probability of winning the game by first rolling a 4 and then rolling a 4 is
Let Pr(x) stand for the probability of event x happening on any given roll. The answer is:
Pr(4) +
Pr(anything other than 4 and 7) * pr(4) +
Pr(anything other than 4 and 7)2 * pr(4) +
Pr(anything other than 4 and 7)3 * pr(4) +
Pr(anything other than 4 and 7)4 * pr(4) +
+ ...
Pr(4) = 3/36 = 1/12, pr(anything other than 4 and 7) = 1-3/36-6/36 = 27/36 = 3/4.
Pr(rolling a 4 before a 7)
= 1/12 + (3/4 * 1/12) + ((3/4)2 * 1/12) + ((3/4)3 * 1/12) + ...
= 1/12 * sum for i = 0 to infinity of (3/4)i
= 1/12 * (1/(1-3/4))
= 1/12 * 4 = 1/3.
The probability of winning the game by first rolling a 4 and then rolling a 4 is=Pr(4)*Pr(rolling a 4 before a 7)=(1/12)*(1/3)=1/36
c) The probalility of rolling 7 is Pr(7)= 6/36 = 1/6 and the probability of rolling an 11 is Pr(11)= 2/36 = 1/18
Similarly,Pr(rolling 4 before a 7)=1/3, Pr(4)=1/12,
Pr(5)=4/36=1/9,Pr(rolling 5 before a 7)=2/5,
Pr(6)=5/36,Pr(rolling 6 before a 7)=5/11
Pr(8)=5/36,Pr(rolling 8 before a 7)=5/11,
Pr(9)=1/9,Pr(rolling 9 before a 7)=2/5,
Pr(10)=1/12,Pr(rolling 10 before a 7)=1/3
Probability of winning the game=
Pr(7) + Pr(11) + Pr(4)*Pr(4 before a 7) + Pr(5)*Pr(5 before a 7) + Pr(6)*Pr(6 before a 7) + Pr(8)*Pr(8 before a 7) + Pr(9)*Pr(9 before a 7) + Pr(10)*Pr(10 before a 7)
= 6/36 + 1/18 + (1/12 * 1/3) + (1/9 * 2/5) + (5/36 * 5/11) + (5/36 * 5/11) + (1/9 * 2/5) + (1/12 * 1/3)
= 6/36 + 1/18 + 1/36 + 2/45 + 25/396 + 25/396 + 2/45 + 1/36
= 330/1980 + 110/1980 + 55/1980 + 88/1980 + 125/1980 + 125/1980 + 88/1980 + 55/1980
= 976/1980 = 244/495.
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