4

Players A and B play a tennis match that consists of 5 SETS. The probability of A winning the first set is 1/2.

If he wins this set, his probability of winning the next set remains 1/2. If he loses, his probability of winning the next set becomes 1/4.

If he wins now, the probability of winning the next one goes back up to 1/2, otherwise, it stays at 1/4.

What is the probability that A wins the MATCH?

My attempt: I considered every possible configuration, like (WWW, WWLW, WLLWW....) and came to the answer 5/16, which is CORRECT.

I have two doubts:

  1. Is there a more elegant solution to this problem?
  2. Surprisingly, if we were to just find the possible 5 letter permutations containing W and L, the ones containing 3 Ws are simply 5C3 and the total number of permutations are 25.

Hence the probability of a random permutation to have 3 Ws is 5C3 / 25.

Which is also 5/16! Does this have anything to do with anything?

Note: This problem is not a duplicate, since whether the player wins/ loses, affects the probability of winning the next set.

TEC0001
  • 91
  • 7
  • 3
    Possible duplicate of p chance of winning tennis point -> what f(p) chance of winning game? – Rob Feb 18 '18 at 14:56
  • How so? In my problem, the probability of winning a set isn't constant. It changes depending on whether he loses/wins – TEC0001 Feb 18 '18 at 15:10
  • This question doesn't mention Wii Tennis but that doesn't really change what the answers are going to be. – Robert Soupe Feb 18 '18 at 23:06
  • I'm sorry, but I'm unable to understand how that problem is related to mine. In that problem, it clearly states that there is a FIXED chance p of winning a point.

    Also, it also assumes that the game continues to the 5th point even after the winner is decided, while in my problem, it stops when one player wins 3 sets.

     – TEC0001 Feb 19 '18 at 04:34
  • I would solve the problem by drawing the tree of the game and adding the probabilities of all game scenarios where $A$ wins; I don't think that your numerology is related to the problem. – kludg Feb 21 '18 at 05:41
  • That's how I solved it as well. i was wondering if there was an easier way and why a simple permutation gave me the same answer. Maybe it really has nothing to do with it. – TEC0001 Feb 21 '18 at 05:43

0 Answers0