A basketball player who shoots 80% from the free-throw line goes to the charity stripe with just 1.7 seconds remaining in a basketball game.
He has two shots, and his team is trailing by two points.

What is the probability that he will make only one of the two, and why?

(Assume that he is trying, of course, for his team to win the game.)

(In reply to re(2): Simple by Brian Wainscott)

Actually, Roger was right, although he did not explain why. Whether the player hits his first shot or not, he gets a second shot, so the first shot can do nothing but help him and he will therefore try his best. Result: 80%.

If he hits his first shot, he can try to make the second shot, which will put the game into overtime and give his team an opportunity to win, OR he can miss the shot and hope his team gets the rebound and makes a basket in the remaining 1.7 seconds of the game. Since the odds are in favor of the opposing team getting the rebound (since they are given better rebounding position by the rules of the game), the shooter will elect to try his best on the second shot as well, which he will miss 20% of the time. Odds of making one shot so far is .8*.2=16%

20% of the time the shooter will miss the first shot. In this scenario, hitting the second free throw will not help win the game – they will most likely not get the ball back in the final 1.7 seconds, and the opposing team will win by one point. The only winning strategy here is to deliberately miss the second shot, and hope his team can rebound and make a shot in the remaining 1.7 seconds to tie the game and force overtime (or kick it out for a 3-pointer, to win in regulation). Therefore the odds of missing the first free throw and then making the second one is 0%.

The total odds of making only one free throw, given the player’s desire to win the game, is 16%.

Posted by Bryan on 2003-08-15 11:52:31