A and B roll a pair of dice A wins if he roll 6 and B wins if he rolls

5/36
6 can be rolled as 1,5 5,1 2,4 4,2 and 3,3
9 years agoHelpfull: Yes(5) No(1)
let winning probability be p and losing probability be q.
A B
p=3/36;q=30/36; p=6/36;q=30/36;
P(Win A)=p+(q^2*p)+(q^2*q^2*p)+.....
taking p as common
P(Win A)=p[1+q^2+(q^2)^2+....)
terms are in GP
hence sum =p[1/(1-q^2)]
therefore general formula for prob of win A is p[1/(1-q^2)]
Since here losing probabilities are different we take
P(Win A)=p/[1-(q1*q2)]
=>(3/36)/[1-(33/36)(30/36)]
Solving this we get 13/24
9 years agoHelpfull: Yes(1) No(3)
1/6
6/36 becomes 1/6
9 years agoHelpfull: Yes(0) No(0)
is it 2/3?
It is forming an infinte series...
9 years agoHelpfull: Yes(0) No(0)
two dice thrown sample space will be 36 and A's winning possibilties are 6
=> 6/36= 1/6.
what is the correct answer?
9 years agoHelpfull: Yes(0) No(0)
1/2 correct answer
9 years agoHelpfull: Yes(0) No(0)
There are 5 ways of obtaining 6 {(1,5), (5, 1), (2, 4), (4, 2), (3, 3)}
and Sample space while throwing 2 dice = 6×6 = 36
So P(E1)=5/36 and P(E1)=1−536=31/36

P(E2)=6/36 and P(E2)=1−16=5/6
9 years agoHelpfull: Yes(0) No(0)

A and B roll a pair of dice. A wins if he roll 6 and B wins if he rolls 7. They continue starting with A till one of them throws their respective number. What is the probability that A wins?