In a dice game whoever brings 6 first will be declared winner. There are two players A and B. Given that A starts first what is the winning probability of both the respective players?
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A is likely to win 2:1 since he will have his turn first...at least I think that's the answer
... I got no idea what the rules for the dice game are, What does "bring a 6" mean?
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I would want to say that the probability of winning is 1/2 for each player
It would be if they both go at the same time, but since the first player went first he has a better chance of rolling a 6 first
A wins B loses
A loses B wins
but i still say the question is a bit too vague for me to be confident
bringing a 6 means...whoever throws a dice and gets six
whoever throws the dice and gets a 6..wins the game, given that A starts first,what is the probability of winning of each of the two players?
Let x equal probability that no one gets a 6 (i.e. 5/6)
Let A equal probability of A getting a 6 on any throw (i.e. 1/6)
Let B equal probability of B getting a 6 on any throw (i.e. 1/6)
A can win on first or 3rd, 5th, 7th, etc if no one won before.
So Probability of A winning = A + xxA + xxxxA + ...
Similarly, B can win on 2nd or 4th, 6th, etc if no one won before.
So, probability of B winning = xB + xxxB + xxxxxB + ...
what is xxA why we need to multiply like that?
xxA is prob of nobody winning on 1st and 2nd and A throwing a 6 on the 3rd throw. So probability would be for xxA = (5/6)(5/6)(1/6). Since there is 5/6 of a probability that nobody wins in 1st toss, and 5/6 that nobody wins on 2nd toss and 1/6 probability that A wins on 3rd toss.
So a can win by Winning on 1st toss and prob of this is A=1/6
Or by winning on 3rd toss and prob of this is xxA = (5/6)^2 A
Or by winning on 5th toss and prob of this is xxxxA = (5/6)^4 A
Adding all these possibilities where A can win is A + xxA + xxxxA + ...
Use the geometric series formula to compute this sum: