2
$\begingroup$

You are in a tennis club with two other members. One of them, Roger (think Federer) is a really good tennis player, and you have a 0.1 probability of winning a match against him. The other player is Joe (think average). He is a pretty good tennis player, and you have a 0.5 probability of winning a match against him.

Now, you are given a choice of the following two sequences of matches:Roger, Joe, RogerorJoe, Roger, JoeThe rule is that you will collect a prize if and only if you win at least two games in a row among your set of three matches. (Note: winning two out of three is not enough; you have to win at least two consecutive matches.)

QUESTION 1) Which of these two schedules (RJR or JRJ) offers you the better chance of collecting a prize?

QUESTION 2) What is the expected number of wins for you in the sequence RJR? In the sequence JRJ?

$\endgroup$
  • $\begingroup$ What are your thoughts on these questions? $\endgroup$ – paw88789 Nov 21 '16 at 20:47
2
$\begingroup$

For the first question: given that you have to win two consecutive matches, you have to win the middle match. Given that Joe is much easier to beat, that suggests the RJR is the better sequence to try and win the prize.

Let's do the math though and see if this intuition is correct:

For RJR, you have to win against Joe (.5 chance), and then you also have to win at least one of two matches against Roger, which is 1- chance of losing both = 1-.9*.9= 1-.81=.19. So, the chance of winning the prize is .5*.19=.095

For JRJ, you have to win against Roger (.1), and win at least one against Joe (1-.5*.5=1-.25=.75), so the chance of winning a prize is .075

So yes, take the RJR sequence indeed!

And to see how the initial intuition was correct in general: Let $x$ be the chance of winning against A, and $y$ the chance of winning against B, and suppose $x>y$: then the chance of winning the prize for ABA is $y*(1-(1-x)^2) = y*(1-(1-2x+x^2)) = y*(2x-x^2)= xy(2-x)$, which by symmetry means the chance of winning the prize for BAB is $xy(2-y)$, and since $x>y$, we have that $xy(2-y)>xy(2-x)$, so yes, always try and have the easier opponent in the middle.

$\endgroup$
  • $\begingroup$ Thank you so much! And for QUESTION 2) What is the expected number of wins for you in the sequence RJR? In the sequence JRJ? i may be way over thinking things, but I am totally unsure how to even approach the question, any suggestions? $\endgroup$ – Nathan Lile Nov 24 '16 at 1:20
  • $\begingroup$ For the expected number of wins, multiply the number of wins times the probability of that many wins for each of the possible number of wins. So that is: Expected number of wins = P(1 win)* 1 + P(2 wins)*2 + P(3wins)*3. Now you 'just' have to figure out P(1), P(2), and P(3). P(3) is easy: it's the chance of winning each match so multiply all the relevant probabilities. The others take a little more thought : what are the possible ways to win exactly 1 match out of the three? How about exactly two? $\endgroup$ – Bram28 Nov 24 '16 at 1:24
  • $\begingroup$ Maybe a little easier is simply to do this: consider all possible 8 outcomes between the 3 matches: Win Win Win, Win Win Lose, Win Lose Win, WLL, LWW,LWL,LLW, and LLL. Figure out the probability of each of those 8 outcomes, multiply each of those by the number of wins in that particular outcome, and sum them all up: that is the expected number of wins. $\endgroup$ – Bram28 Nov 24 '16 at 1:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.