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 A Coin Game (Posted on 2004-05-20)
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?

 See The Solution Submitted by Brian Smith Rating: 3.5000 (4 votes)

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 re: Assuming .5 solution | Comment 5 of 12 |
(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 2015    Pa=combi(20,A):print Pa:Pat=Pat+Pa20    for B=A+1 to 2140     Pb=combi(21,B)50     W=W+Pa*Pb60    next70   next80   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

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