1. ## probability

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

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

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) 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

3. ## Re :

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 ..

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.

5. ## Re :

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

6. Let $\displaystyle p_d$ be the probability of a draw, $\displaystyle p_b$ probability of B winning and $\displaystyle p_a$ the probability of A winning.

Then:

$\displaystyle 3 p_d=p_b$

$\displaystyle 2 p_b=p_a=6 p_d$

and of course:

$\displaystyle p_a+p_b+p_d=1$

or:

$\displaystyle 6 p_d+3p_d+p_p=1$

so:

$\displaystyle p_d=1/10$

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 $\displaystyle 1:9$.

.

7. Originally Posted by Constatine11
Let $\displaystyle p_d$ be the probability of a draw, $\displaystyle p_b$ probability of B winning and $\displaystyle p_a$ the probability of A winning.

Then:

$\displaystyle 3 p_d=p_b$

$\displaystyle 2 p_b=p_a=6 p_d$

and of course:

$\displaystyle p_a+p_b+p_d=1$

or:

$\displaystyle 6 p_d+3p_d+p_p=1$

so:

$\displaystyle p_d=1/10$

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 $\displaystyle 1:9$.

.
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.