- #1
Millie
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I found this question in a book:
Two palyers A and B alternatively roll a pair of unbiased die. A wins if on a throw he obtain exactly 6 points, before B gets 7 points, B wining in the opposing event. If A begins the game prove that the probability of A winning is 30/61 and that the expected number of trials need for A's win is approximately 6.
I was trying to solve the problem using conditional probability and expectation. To get 30/61 I conditioned on the sum on the dice on the 1st roll and then on the sum on the dice on the second roll.
However I am not sure whether we can calculate the required expectation through conditioning. I know that the exact answer is 371/61=6.0819. Through conditioning I only can 5.57 which however can also be approximated to 6.
Also if X denotes "the number of trials needed for A to win"
And Yi be the sum on the ith roll
Conditioning on the first roll,
E[X]= E[X/Y1=6)P(Y1=6) + E[x/Y1 is not equal to six]P[Y1 is not =6)
= 1x5/36 + E[X/Y1 not =6)x31/36
Then conditioninf on the 2nd roll.
E[X/Y1 not = 6]= E[X/Y1 not=6, Y2 not = 9)P(Y2 not equal 9) + E[X/Y1 not=6, Y2=9)P(Y2=9)
That's kind of where I am stuck for what is the expected number of trials for A to win given that B wins on the second trial??
Or should I try to condition on the 3rd or last event?
Or should I approach the problem without using condition.
Man I am so confused now!
thanks!
Two palyers A and B alternatively roll a pair of unbiased die. A wins if on a throw he obtain exactly 6 points, before B gets 7 points, B wining in the opposing event. If A begins the game prove that the probability of A winning is 30/61 and that the expected number of trials need for A's win is approximately 6.
I was trying to solve the problem using conditional probability and expectation. To get 30/61 I conditioned on the sum on the dice on the 1st roll and then on the sum on the dice on the second roll.
However I am not sure whether we can calculate the required expectation through conditioning. I know that the exact answer is 371/61=6.0819. Through conditioning I only can 5.57 which however can also be approximated to 6.
Also if X denotes "the number of trials needed for A to win"
And Yi be the sum on the ith roll
Conditioning on the first roll,
E[X]= E[X/Y1=6)P(Y1=6) + E[x/Y1 is not equal to six]P[Y1 is not =6)
= 1x5/36 + E[X/Y1 not =6)x31/36
Then conditioninf on the 2nd roll.
E[X/Y1 not = 6]= E[X/Y1 not=6, Y2 not = 9)P(Y2 not equal 9) + E[X/Y1 not=6, Y2=9)P(Y2=9)
That's kind of where I am stuck for what is the expected number of trials for A to win given that B wins on the second trial??
Or should I try to condition on the 3rd or last event?
Or should I approach the problem without using condition.
Man I am so confused now!
thanks!