Dicey Probability

Satvik and Agnishom are playing a game with 2 2 standard dice. Both the dice are rolled together and the total is counted.

Satvik says that a total of 2 2 will be rolled first.

Agnishom, whereas, says that two Consecutive totals of 7 s 7's will be rolled first.

They keep rolling the dice till one of them wins!

The probability that Satvik wins the game can be expressed as a b \dfrac{a}{b} where a a and b b are co-prime positive integers.

Find the value of a + b a+b .


The answer is 20.

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3 solutions

Satyen Nabar
Nov 14, 2014

Let the probability that Satvik wins the game be P.

There are 5 possibilities.

1) If the first roll is a 2 (probability 1/36), Satvik wins immediately.

2) If the first roll is a 7, and the 2nd roll is a 2, (1/6 *1/36 = 1/216) Satvik wins immediately

3) If the first two rolls are both 7, (1/6 *1/6= 1/36) Satvik loses.

4) If the first roll is a 7 and the 2nd roll is neither a 7 nor a 2, (1/6 * 29/36 = 29/216), Satvik is back to square one and can win with probability P.

5) if the first roll is neither a 7 nor a 2, (29/36), Satvik is again back to square one and can win with probability P.

Probability P is the weighted mean of all the above possibilities,

P = 1/36 + 1/216 + (29/216) P + (29/ 36) P

Solving P = 7/13

Answer is 7 + 13 = 20

Learnt something :) Nice approach!

SHIV GUPTA - 6 years, 6 months ago

Great solution. Learnt something new.

Rajdeep Dhingra - 6 years ago

Excellent solution. Loved it!

Shanthanu Rai - 5 years, 1 month ago
Ratish Dalvi
Feb 3, 2015

Even I obtained the same summation as you did but how did you evaluate the summation?

Indraneel Mukhopadhyaya - 5 years, 6 months ago
Sumit Goel
Dec 31, 2014

Let a(n) be the probability that no one wins in n moves and last move is not a 7.

Let b(n) be the probability that no one wins in n moves and last move is a 7.

P(Satvik wins)= 1/36+(a(1)+b(1))/36+...(a(n)+b(n))/36.......

P(Agnishom wins)= 1/36+a(1)/36+a(2)/36....a(n)/36.............

Sum of both probabilities=1......(i)

b(n)=a(n-1)/6.....(ii)

Using (ii)

P(Satvik wins)=(7/6) (1/36) (1+a(1)+a(2)......)

Let 1+a(1)+a(2)......=z

Using (i)

z/36+7z/216=1

z=216/13

Hence Probability Satvik wins=7/13

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