Probability problem 1

[susumandrai]

Ali, Benedict and Wiremu decide to have an arm-wrestling tournament amongst themselves. Two of them play first, and the winner of this game plays the third to decide the overall winner. The probability that Ali beats Benedict is 0.4, the probability that Benedict beats Wiremu is 0.7, and the probability that Wiremu beats Ali is 0.8. If any two are equally likely to play first, find who is most likely to be the overall winner and the probability that this person wins.

The textbook says answer is 0.506(6 recurring)

My answer came to 0.56. I made 3 different possibilities of starting a fight so 3 different sets of probability trees!
Please tell me how I should answer this question.

 

[susumandrai]

Here are my possible combinations:
Combination 1:

P(Ali)=0.08 P(Benedict)=0.42 P(Weiramu)=0.05

Combination 2:

P(Ali)=0.08 P(Benedict)=0.68 P(Weiramu)=0.24

Combination 3:

P(Ali)=0.34 P(Benedict)=0.42 P(Weiramu)=0.24


How do I come to .506(6 recurring) from here.....

 

[soroban]

Hello, susumandrai!

Ali, Benedict and Wiremu decide to have an arm-wrestling tournament amongst themselves.
Two of them play first, and the winner of this game plays the third to decide the overall winner.
The probability that Ali beats Benedict is 0.4, the probability that Benedict beats Wiremu is 0.7, and the probability that Wiremu beats Ali is 0.8.
If any two are equally likely to play first, find who is most likely to be the overall winner and the probability that this person wins.

The textbook says answer is 0.506(6 recurring)

Note: \(\displaystyle A>B\) means "A beats B".

We have: .\(\displaystyle \begin{array}{|c|c|} \hline P(A > B) \:=\:0.4 & P(B >A) \:=\:0.6 \\ \hline P(B > W) \:=\:0.7 & P(W > B) \:=\:0.3 \\ \hline P(W > A) \:=\:0.8 & P(A > W) \:=\:0.2 \\ \hline \end{array}\)


There are four ways that B can win the tournament.

(1) A and B play: \(\displaystyle P(A\text{ vs.} B) = \frac{1}{3}\)
. . B beats A: .\(\displaystyle P(B >A) = 0.6\)
. . Then B beats W: .\(\displaystyle P(B > W) = 0.7\)
. . \(\displaystyle P(B\text{ wins}) \:=\: \frac{1}{3}(0.6)(0.7) \:=\:0.14\)

(2) B and W play: .\(\displaystyle P(B\text{ vs. }W) = \frac{1}{3}\)
. . B beats W: .\(\displaystyle P(B > W) = 0.7\)
. . Then B beats A: .\(\displaystyle P(B > A) = 0.6\)
, , \(\displaystyle P(B\text{ wins}) \:=\:\frac{1}{3}(0.7)(0.6) \:=\:0.14\)

(3) A and W play: .\(\displaystyle P(A\text{ vs. }W) = \frac{1}{3}\)
. . A beats W: .\(\displaystyle P(A > W) = 0.2\)
. . Then B beats A: .\(\displaystyle P(B > A) = 0.6\)
. . \(\displaystyle P(B\text{ wins}) \:=\:\frac{1}{3}(0.2)(0.6) \:=\:0.04\)

(4) A and W play: .\(\displaystyle P(A\text{ vs. }W) = \frac{1}{3}\)
. . W beats A: .\(\displaystyle P(W > A) = 0.8[/t3ex]
[color=beige[. . [/color]Then B beats W: .\(\displaystyle P(B > W) = 0.7\)
. . \(\displaystyle P(B\text{ wins}) \:=\:\frac{1}{3}(0.8)(0.7) \:=\:\frac{0.56}{3}\)

Therefore: .\(\displaystyle P(B\text{ wins}) \:=\:0.14 + 0.14 + 0.4 + \frac{0.56}{3} \:=\;0.506666\text{ ...}\)\)

 

[susumandrai]

Hello, susumandrai!


Note: \(\displaystyle A>B\) means "A beats B".

We have: .\(\displaystyle \begin{array}{|c|c|} \hline P(A > B) \:=\:0.4 & P(B >A) \:=\:0.6 \\ \hline P(B > W) \:=\:0.7 & P(W > B) \:=\:0.3 \\ \hline P(W > A) \:=\:0.8 & P(A > W) \:=\:0.2 \\ \hline \end{array}\)


There are four ways that B can win the tournament.

(1) A and B play: \(\displaystyle P(A\text{ vs.} B) = \frac{1}{3}\)
. . B beats A: .\(\displaystyle P(B >A) = 0.6\)
. . Then B beats W: .\(\displaystyle P(B > W) = 0.7\)
. . \(\displaystyle P(B\text{ wins}) \:=\: \frac{1}{3}(0.6)(0.7) \:=\:0.14\)

(2) B and W play: .\(\displaystyle P(B\text{ vs. }W) = \frac{1}{3}\)
. . B beats W: .\(\displaystyle P(B > W) = 0.7\)
. . Then B beats A: .\(\displaystyle P(B > A) = 0.6\)
, , \(\displaystyle P(B\text{ wins}) \:=\:\frac{1}{3}(0.7)(0.6) \:=\:0.14\)

(3) A and W play: .\(\displaystyle P(A\text{ vs. }W) = \frac{1}{3}\)
. . A beats W: .\(\displaystyle P(A > W) = 0.2\)
. . Then B beats A: .\(\displaystyle P(B > A) = 0.6\)
. . \(\displaystyle P(B\text{ wins}) \:=\:\frac{1}{3}(0.2)(0.6) \:=\:0.04\)

(4) A and W play: .\(\displaystyle P(A\text{ vs. }W) = \frac{1}{3}\)
. . W beats A: .\(\displaystyle P(W > A) = 0.8[/t3ex]
[color=beige[. . [/color]Then B beats W: .\(\displaystyle P(B > W) = 0.7\)
. . \(\displaystyle P(B\text{ wins}) \:=\:\frac{1}{3}(0.8)(0.7) \:=\:\frac{0.56}{3}\)

Therefore: .\(\displaystyle P(B\text{ wins}) \:=\:0.14 + 0.14 + 0.4 + \frac{0.56}{3} \:=\;0.506666\text{ ...}\)\)

\(\displaystyle



Thank you so much Soroban
However I had a good look again at my answer with trees above and found this:
Probability Benedict winning = (1/3)x(.42+.68+.42) = 0.50666666...

Would you explain to me why after multiplying with 1/3 I get the correct answer!

Thanks again\)