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Probability problem 1

susumandrai

New member
Joined
Feb 7, 2012
Messages
47
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

New member
Joined
Feb 7, 2012
Messages
47
Here are my possible combinations:
Combination 1:
combi1.jpg
P(Ali)=0.08 P(Benedict)=0.42 P(Weiramu)=0.05

Combination 2:
combi2.jpg
P(Ali)=0.08 P(Benedict)=0.68 P(Weiramu)=0.24

Combination 3:
combi3.jpg
P(Ali)=0.34 P(Benedict)=0.42 P(Weiramu)=0.24


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

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
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

New member
Joined
Feb 7, 2012
Messages
47
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\)
 
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