A and B plays chess everyday . The probability that A beats B in any game is twice the probability that B beats A . The probability that B beats A is three times the probability that the game ends in a draw . What is the probability that the game ends in a draw ?

My working :
P(A wins ) = 2(1- P(A wins ))
P(A wins ) = 2/3

P(B wins) = 3(1-P(A wins)-P(B wins))
P(B wins) = 3(1-2/3-P(B wins ))
P(B wins )=1/4

P(draw)=1-2/3-1/4=1/12 --- my answer

My answer is wrong . I wonder where my mistake is .....

mathaddict,

P(B wins) is not equal to 1- P(A wins). This is because you have to also consider the possibility of a draw.

Thus P(B wins) = 1- P(A wins ) - P(draw).
So work out the exercise again, carefully...

Good Luck

__________________

THanks Isomorphism ,

My second attempt :

P(A wins ) = 2[1-P(B wins)-P(draw)]
P(A wins) = 2-2P(B wins)-2P(draw) ------ 1st equation

P(B wins)= 3P(draw) ----- 2nd equation

P(draw) = 1-P(A wins)-P(B wins ) ---- 3rd equation

But after simplifying , i got P(draw)=1/4
My answer is wrong according to the book . Did i get the equations right in the first place ? Thanks for any help ..

Let Pr(B beats A) = Pr(A loses) = p.

Given: Pr(A beats B) = Pr(A wins) = 2p.

Given: Pr(A and B draw) = p/3.

p + 2p + (p/3) = 1.

Solve for p.

I get Pr(Draw) = 1/10.

__________________
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Thanks Mr F , but the answer given is 1/9

Let be the probability of a draw, probability of B winning and the probability of A winning.

Then:





and of course:



or:



so:



as Mr Fantastic has.

Now the ony way out I can see is if the question asked for the odds of a draw since the odds are the ratio of favourable to unfavourable outcomes the odds are .

.

__________________
Yes, I am an agent of Satan, but my duties are largely confined to the legal department.

Which shows why it's important to understand the difference between the probability of an event as represented by:

1. A number between zero and 1 inclusive .

2. Odds of the form a : b

3. Percentage

etc.

__________________
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