Question

Players X and Y roll a pair of dice alternately. The person who gets 11 first wins. If X starts, what is his chance of winning?

Answer 1

First find the probability of getting 11.
Probability that X wins = Probability that X gets 11 first time + (prob that X doesnt get 11,Y doesnt get 11,X gets 11) + (prob that X gets 11 in his third trial) + ...

Its an infinite series which converges.

Answer 2

we may get 11 by = (5,6) or (6,5)
p(11) = 2/36
p(not 11)= 34/36
p(x wins) = p(x gets 11) + p(x doesnt get 11)*p(y doesnt get 11)*p(x gets 11) +
p(x doesnt get 11)*p(y doesnt get 11)*p(x doesnt get 11)*p(y doesnt get 11)*p(x gets 11) .... and so on
=>2/36 + (34/36)*(34/36)*(2/36) + ........
=>(2/36)(1 + 34/36 + (34/36)^2 ........... )
hope you can simplify further,,

and hope my soln was clear..

Answer 3

Answer is 18/35..not wat the series will give

Answer 4

how to solve it but??

Answer 5

sorry a correction there..not
1 + 34/36 + (34/36)^2 ...........
but is
1 + (34/36)^2 + (34/36)^4 ...........

and its an infinite GP whose common difference is a fraction
sum=a/(1-r)
a= first term
r= common difference
still doesnt give 18/35 though..hmm..

Answer 6

ohh wait,,it does give 18/35..

Answer 7

:D

Answer 8

how come... iam not getting it..

Answer 9

you evaluated the sum ?

Answer 10

its giving 324/35

Answer 11

not possible

Answer 12

note that the whole sum is multiplied by (2/36)

Answer 13

hey i got it...

Answer 14

yes i forgot to multiply 2/36..
thanks @shubhamsrg

Answer 15

nevermind,glad to help :)