The probability of team A winning any game is 1/2. Suppose A plays B in a tournament (and there are no ties). The first team to win two games in a row or three games wins the tournament. Find the expected number E of games in the tournament.

AA       (1/2)(1/2)=(1/4)
BAA      (1/2)(1/2)(1/2)=(1/8)
ABAA     (1/2)(1/2)(1/2)(1/2)=(1/16)
BABAA    (1/2)(1/2)(1/2)(1/2)(1/2)=(1/32)

So E(X)= 2( (2)(1/4) + (3)(1/8) + (4)(1/16) + (5)(1/32) ) = 2.5645 or 41/16

But the answer in my book is 23/8 or approximately 2.9

What am I doing wrong?

  • $\begingroup$ Is there an upper limit on the number of games in total? $\endgroup$ – numbermaniac Oct 3 '17 at 4:30
  • $\begingroup$ @numbermaniac It just asks what I have stated. Nothing more is asked. $\endgroup$ – pyuntae Oct 3 '17 at 4:35
  • 2
    $\begingroup$ @numbermaniac: after five games one side will have won at least three, so that is an upper limit. $\endgroup$ – Ross Millikan Oct 3 '17 at 4:45
  • $\begingroup$ @RossMillikan ah yes, I only saw the "two games in a row" and missed that bit somehow. Thanks. $\endgroup$ – numbermaniac Oct 3 '17 at 4:52
  • $\begingroup$ Please do not modify your post after you received answers. $\endgroup$ – Did Oct 3 '17 at 13:35

Note that your probabilities do not sum to $\frac 12$ so you have done something wrong. You have missed ABABA

Now you didn't add the numerator correctly. Putting everything inside the multiplication by $2$ over $16$ we get $4\cdot 2 + 2\cdot 3 +4+5=8+6+4+5=23$

  • $\begingroup$ I got mixed up and kept thinking that was a tie. $\endgroup$ – pyuntae Oct 3 '17 at 4:59

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