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.)

yes the probabilty of him getting only one is 16%
getting both is 64%
getting neither is 4%
but logically he is trying to win the game so his options are
a) make both and get into overtime 64% chance to tie
b) getting the first and intentionally missing the second in hopes of rebounding for a two point taking the win - 80% chance of getting the first shot, and then he intentionally misses therefor only getting one of the two shots. Hopefully they rebound it then and win the game.
c) if he misses the first, he has to try for the second shot and still hope for a rebound. then his odds are 16% between both shots or 80% on the second shot.

regardless of how he shoots, his team must get the rebound and make a shot inorder to pull off a win, unless they can manage to steal the ball from a team that will probably hold it close to their body when they rebound it. Once they do get the ball they will have nearly no time to put the ball up in the air and win.

I suspect that the game is rather hopeless, but miracles have happened.

Also the shooter can miss the first and bounce the ball off the backboard to the three point line where one of his players would be ready to recieve it and shoot a 3-pointer to win the game. That would require either a hit and intentional miss 80% or two intentional misses 100% (unless you you assume that even in two intentional misses there is the 20% chance of making it which would mean that two intentional misses still have a 4% chance of making it)

Posted by Roger on 2003-08-16 05:05:40