Problem Solving

A and B play a game. A and B alternately roll a pair of dice (A starts the game). The first person to throw a score more than what the other has scored in the previous throw is said to have won the game. What is the probability that B wins the game in the second throw?

1. 575/1296
2. 1/2
3. 175/1296
4. 23/36

srcc25anu wrote:A and B play a game. A and B alternately roll a pair of dice (A starts the game). The first person to throw a score more than what the other has scored in the previous throw is said to have won the game. What is the probability that B wins the game in the second throw?

1. 575/1296
2. 1/2
3. 175/1296
4. 23/36

Probability that B wins the game in the 2nd throw (first throw for B, however 2nd overall for the game as A has the first throw) =

P(A=2)P(B>2)+P(A=3)P(B>3)+P(A=4)P(B>4)+P(A=5)P(B>5)+P(A=6)P(B>6)+P(A=7)P(B>7)+P(A=8)P(B>8)+P(A=9)P(B>9)+P(A=10)P(B>10)+P(A=11)P(B>11)

=1/36*35/36+2/36*33/36+3/36*30/36+4/36*26/36+5/36*21/36+6/36*15/36+5/36*10/36+4/36*6/36+3/36*3/36+2/36*1/36

= 1/36*36 * (35+66+90+104+105+90+50+24+9+2) = 575/1296. 1

A is the correct answer. thanks for the explanation.

hi Anshu here's the Quest. https://challenge.2iim.com/test_2/question_6.shtml Answer https://challenge.2iim.com/test_2/answer.shtml However, if B wins in the first throw - there's no second throw ... game is over with no solution I lost my 0.5 hours here

anshumishra wrote:

srcc25anu wrote:A and B play a game. A and B alternately roll a pair of dice (A starts the game). The first person to throw a score more than what the other has scored in the previous throw is said to have won the game. What is the probability that B wins the game in the second throw?

1. 575/1296
2. 1/2
3. 175/1296
4. 23/36

Probability that B wins the game in the 2nd throw (first throw for B, however 2nd overall for the game as A has the first throw) =

P(A=2)P(B>2)+P(A=3)P(B>3)+P(A=4)P(B>4)+P(A=5)P(B>5)+P(A=6)P(B>6)+P(A=7)P(B>7)+P(A=8)P(B>8)+P(A=9)P(B>9)+P(A=10)P(B>10)+P(A=11)P(B>11)

=1/36*35/36+2/36*33/36+3/36*30/36+4/36*26/36+5/36*21/36+6/36*15/36+5/36*10/36+4/36*6/36+3/36*3/36+2/36*1/36

= 1/36*36 * (35+66+90+104+105+90+50+24+9+2) = 575/1296. 1

Night reader wrote: hi Anshu here's the Quest. https://challenge.2iim.com/test_2/question_6.shtml Answer https://challenge.2iim.com/test_2/answer.shtml However, if B wins in the first throw - there's no second throw ... game is over with no solution I lost my 0.5 hours here

@night reader
I hear you. The question should be accompanied with the OA (at least if the poster knows). Sometimes the answer helps to understand the problems which are not clearly written, e.g. here I also had a doubt whether the question is asking about B's second throw, however after looking at the given options (1296 = 36^2), I guessed it was the first throw for B.