wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> Basketball odds
        
(Message started by: Altamira_64 on Jul 6^th, 2013, 3:32am)

Title: Basketball odds
Post by Altamira_64 on Jul 6^th, 2013, 3:32am

A basketball player has a 50% free throw shooting average at his first two throws, which can be interpreted to mean that the probability of his hitting any single free throw (of the first two) is 0.5 or 5/10. Right thereafter, the probability to succeed to any new throw is equal to the true total average, of all throws from the first one up to the previous (of the one he is going to shoot now).
What are his chances
1. to succeed to exactly 50 of all 100 throws he will shoot.
2. to succeed to at least 50 of all 100 throws he will shoot.

Title: Re: Basketball odds
Post by jollytall on Jul 11^th, 2013, 9:35pm

Let's look at, what is the chances that he will score altogether only 1. After 2 shoots he has already one. So the probability that he will miss the third is 1/2. So either he will have 1 or 2 with equal chances.
Now that at the third one he will miss again, it is already 2/3 probability, i.e. the outcme has a 1/2 * 2/3 = 1/3 probability. For symmetry reasons having 3 after 4 shoots is also 1/3. Consequently having 2 after 4 is also 1/3.
With full induction we assume that 1..n-1 out of n has always 1/(n-1) probability.
k out of (n+1) can occur two different ways. Either he had already k out of n [1/(n-1) probability] and fails OR k-1 [with the same probability] and succeeds.
Having m out of n the probability to score is m/n, to fail is 1-m/n.
Now the probability of k out of n+1 is:
1/(n-1)*(k-1)/n + 1/(n-1)*(1-k/n) = 1/n.

So Q1: 1/99, Q2: 50/99

Title: Re: Basketball odds
Post by rmsgrey on Jul 12^th, 2013, 7:40am

I read it as 1/4 of the time he misses both the opening shots, and 1/4 of the time he sinks both of them, while the other 1/2 the time, he gets one in.

I've not checked jollytall's working beyond that, but if it holds up, I get answers:

1/198
199/396

Title: Re: Basketball odds
Post by Grimbal on Jul 12^th, 2013, 9:49am

I checked what it looks like on a spreadsheet and I also found that the odds for "0 success in n" and "n success in n" is 1/4 each, and the remaining 1/2 is spread evenly between the "k success in n" for 0<k<n.
This leads to the same result as rmsgrey.

Title: Re: Basketball odds
Post by jollytall on Jul 12^th, 2013, 9:25pm

I missed the words "can be interpreted". I took it as that after the first 2, he has exactly 1 in.
Obviously, if the first two are also random, then 0 and 2 has 1/4 chance each. If the first 2 are the same, i.e. 0 or 2, then all the rest are the same, i.e. 0 or 100 is at the end.

My solution was for the middle 1/2 cases only. If I take the chance to get to 1 out of 2 (i.e. 0/2), then I get the same.

Q1: 1/2 * 1/99 = 1/198
Q2: 1/4 + 1/2 * 50/99 = 199/396