A, B and C are three friends who often like to compete with one another.
(a) they are equally skillful in playing table tennis. A plays B first while C waits. the winner then plays C while the loser waits. This method of play goes on until one of them wins two consecutive sets. this person then gets a prize. find the probability that C gets the prize. ( ans: 2/7)
(b) they are given the same multiple-choice quiz to answer independently. the quiz has 10 questions, each provided with 4 possible answers. only one of which is correct. A knows eight of the correct answers, B knows seven, and C knows six. All of them then answer the remaining questions by sheer guess. No penalty is imposed on a wrong answer. What is the probability that C beats A and B?
(ans: 0.0259)

I can't figure out the method of solving this problem even after looking at the answers. Can someone help?

2. Re: probability problem about tournaments

For (b)

List out the combinations of successful guesses that allow C to beat A and B. We get

$\begin{matrix}A &B &C \\--- &--- &--- \\ 0 &0 &3 \\ 0 &0 &4 \\ 0 &1 &3 \\ 0 &1 &4 \\ 0 &2 &4 \\ 1 &0 &4 \\ 1 &1 &4 \\ 1 &2 &4 \end{matrix}$

A is a binomial distribution $Binomial(2,1/4)$
B is $Binomial(3,1/4)$
C is $Binomial(4,1/4)$

It's assumed the guessing is independent amongst the students so the probability of each row of the table is just the product of the 3 individual probabilities.

Then sum up all the row probabilities to get the final probability that C beats A and B

3. Re: probability problem about tournaments

a)

Look at this state diagram

Two things to note.

i) The left side, which I didn't draw, is identical in probability to the right side. So each side will contribute half of the overall probability.

ii) Each vertex has probability $\dfrac 1 2$ as all the players are equally skilled.

If you examine the diagram you'll see that C wins at $p=(1/2)^3$, and $p=(1/2)^6$ and will continue to win at $p=(1/2)^{3k}$

so the total probability of C winning is

\begin{align*} &P = 2 \sum \limits_{k=1}^\infty~\left(\dfrac 1 2 \right)^{3k} = \\ &2 \sum \limits_{k=1}^\infty~\left(\dfrac 1 8\right)^{k} = \\ &2\left(\dfrac {1}{1-\frac 1 8}-1\right) =\\ &2\left(\dfrac 8 7 - 1\right) = \\ &\dfrac 2 7 \end{align*}

4. Re: probability problem about tournaments

So two consective sets means winning A and B continuously?

5. Re: probability problem about tournaments

It means C wins twice in a row. Not continuously, but consecutively.