Question

A and B alternately throw a pair of dice. A wins if he throws 6 before B throws 7; and B wins if he throws 7 before Athrows 6. What are their respective chances of winning, if A throws the dice first?

• A. \(\frac{13}{16},\frac{31}{16}\)
• B. \(\frac{30}{61},\frac{31}{61}\)
• C. \(\frac{31}{61},\frac{41}{61}\)
• D. \(\frac{38}{61},\frac{28}{61}\)

Answer 1

The Correct Option is B

The probability that A throws \('6'\) \(=(\frac{5}{36})\) and

The probability of A not getting \('6'\) \(=1-\frac{5}{36}=\frac{31}{36}\)

The probability that B throws \('7'\) \(=\frac{6}{36}\)

The probability of A winning the game is given by,

= (In the first throw if A wins) or (In the 1st throw A looses, B also looses and in the 2nd throw A wins) or ( A and B looses in the 1st and 2nd throw's and A wins in the 3rd throw .....it goes on like that)

\(=\frac{5}{36}\)\(+\frac{31}{36}.\frac{30}{36}.\frac{5}{36}\)\(+\frac{31}{36}.\frac{30}{36}.\frac{31}{36}.\frac{30}{36}\)\(\frac{5}{36}\)\(+.......\)

It is in arithmetic progression,

\(=(\frac{5}{36})\)\((1-\frac{\frac{1}{31}}{36}\)\(.\frac{30}{36}\)\()\)

\(=(\frac{5}{36})\)\((\frac{1296}{1296-930})\)

\(=(\frac{5}{36})\)\((\frac{1296}{366})\)

\(=\frac{5×6}{61}\)\(=\frac{30}{61}\)

Probability of B winning the game is given by,

\(=1-\frac{30}{61}\)

\(=\frac{31}{61}\)

Hence, option B is the correct answer.The correct option is (B): \(\frac{30}{61},\frac{31}{61}\)