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In your tennis club there are two other members. One of them is a very good player. His name is Roger (think Federer) and you have a 0.1 probability of winning a match against him. The other player is Joe (think average). He is a decent tennis player and you have a 0.4 probability of winning a match against him. In a tennis tournament, you have two choices for the sequence of matches: Roger, Joe, Roger or Joe, Roger, Joe To get a prize in the tournament you need to win at least two matches in a row among your set of three matches. (Note: this does not mean winning two out of three matches; you have to win at least two consecutive matches.)

4a. Which of these two schedules (RJR or JRJ) gives you the better chance at a prize?

4b. What is the expected number of wins for you in the sequence RJR? In the sequence JRJ?

My solution:

For Part A, I just multiplied the probabilities of each comination of 3 that includes two consecutive wins (w,w,l), (l,w,w), (w,w,w). For RJR, this came out to be 0.076. For JRJ I got 0.064.

For Part B, I'm a little stuck. I started calculating the expected number of winning games like so:

For RJR:

0 games won * (0.9*0.6*0.9) = 0

1 game won * (0.1*0.6*0.9 + 0.9*0.4*0.9 + 0.9*0.6+0.1) = 0.432

2 games won * (0.1*0.4*0.9 + 0.9*0.4*0.1 + 0.1*0.6+0.1) = 0.156

3 games won * (0.1*0.4*0.1) = 0.012

0+0.432+0.156+0.012 = 0.6

And the same for the JRJ order. Is this correct though?

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  • $\begingroup$ The procedure is ok. I haven't checked the arithmetic. $\endgroup$
    – true blue anil
    Commented Apr 14, 2016 at 6:15
  • $\begingroup$ In, for example, your calculation for $3$ games won, you write $(0.1)(0.4)(0.1)=0.12$. Of course this is technically not right, you multiplied by $3$ to get the $0.12$. The procedure was correct but the description was not. $\endgroup$
    – André Nicolas
    Commented Apr 14, 2016 at 6:26

1 Answer 1

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Your strategy looks right. Let us check your RJR calculation by using a different method, which I hope is accessible.

Let random variable $X_1$ be equal to $1$ if you win the first game, and $0$ if you lose the first game. Define $X_2$ and $X_3$ analogously. Then the number $Y$ of games won is given by $Y=X_1+X_2+X_3$. By the linearity of expectation we have $E(Y)=E(X_1)+E(X_2)+E(X_3)$.

We have $E(X_1)=E(X_3)=0.1$ and $E(X_2)=0.4$, so $E(Y)=0.6$.

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  • $\begingroup$ Thank you! That makes sense $\endgroup$
    – dibdub
    Commented Apr 14, 2016 at 6:20
  • $\begingroup$ You are welcome. I hesitated to introduce random variables, since that may be premature. But since your analysis was correct I thought I would add information that will be useful, if not now then later. $\endgroup$
    – André Nicolas
    Commented Apr 14, 2016 at 6:28

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