Kobe bryant bet

[low_limit_soldier]

I recently made a bet with a friend during an NBA playoff game between the Pistons and the Lakers. Kobe Bryant had just been fouled and was at the freethrow line. The bet was whether or not Kobe would make one of his freethrows. I bet that he would make AT LEAST ONE of the freethrows and my friend said he'd give me -115 odds and I took the action. my bet was $115, so if i won, i made $100 from him. If he won, he got my $115. I knew Kobe had a good freethrow percentage (I checked later and it was 85.2%) I also figured that since he had 2 shots at the freethrow line and he only had to make one, I was getting the better end of the deal even with -115 odds. Also, even if Kobe missed both freethrows, If any one of the lakers, including Kobe, made a freethrow before a piston did, I still won the bet. Now this really made me think that this bet had a positive expectation, but I am no math whiz, and I want to combine this information to find out what my win precentage will beif i keep making this bet with him. So here is a summary of the bet.....

Kobe Bryant's Freethrow bet
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- My odds = (-115) (bet $115 to win $100)
- Bryant's freethrow percentage = 85.2%
- only has to make 1 of the 2 shots for me to win
- Even if Kobe misses both freethrows, i still win if any Laker makes a freethrow before a Piston does.


If someone is able to figure this out I would greatly appreciate it, all the math hurt my head. Thanks for your time.
-lowlimitsoldier

[jdl22]

You made an insanely +EV bet.

Let's ignore the .2% and say he is an 85% free throw shooter. Let's also make the somewhat more dodgy assertion that his free throws are independent events. This is roughly correct anyway and probably is if it's the middle of the game and not the end.

The probability that he will miss both is .15*.15 = .0225. Hence the probability of him making at least 1 is .9775. To check this note that the probability of him making both is .85*.85 = .7225, making one missing one can happen two ways so the probability is 2*.85*.25 = .255. Hence the probability making at least one is .7225 + .255 = .9775 as we thought.

So the odds for you to win based only on Kobe's shooting is
9775:225 or 391:9 which is slightly better than 43:1. For a fair bet you should have paid 4300 dollars to win 100.