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 TI83 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 againthe 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 20040520 22:12:41 