Alex flips a fair coin 20 times. Bert spins a fair coin 21 times. Bert wins if he gets more heads than Alex, else Alex wins. Note that Alex wins if there is a tie. What is the probability that Bert wins?

(In reply to Assuming .5 solution by Jer)

I don't know enough about the TI-83 to say what is wrong with the formula (but see below for an idea), but the following Basic program agrees with Oskar's analytic solution:

10   for A=0 to 20
15    Pa=combi(20,A):print Pa:Pat=Pat+Pa
20    for B=A+1 to 21
40     Pb=combi(21,B)
50     W=W+Pa*Pb
60    next
70   next
80   print W/2^(20+21),Pat

(it finds the probability 0.5).

A variation on this with

20    for B=A to 21

leads to a total of 0.6223856712476845132, the same as you had found, but this incorrectly awards ties to B, and shows not that winning a tie is better than getting an extra spin, but rather it's better to get both.

Another analytic solution is as follows:

The situation described can be shown to be equivalent to A and B each flipping 20 times; whoever has more heads wins; in the case of a tie, B flips again--the non ties favor neither player, and the tiebreaker favors neither player.

Why are these two scenarios equivalent?  If B already has more heads before the tie breaker, even playing the "tiebreaker" then wouldn't change things; If A already has more, then playing the "tiebreaker" (as it is in the given scenario, but not the alternative) would only lead to a tie, which this scenario awards to A anyway, so again there's no difference.  And if there were a tie, then the tiebreaker is in fact done, and again the scenario matches the given one.

But Oskar's solution was much more succinct.

Posted by Charlie on 2004-05-20 22:12:41