View Full Version : basketball probability question
jojobinks
04-04-2005, 09:42 PM
you're a 40% shooter from 3 point range. over the span of 500 shots, what is the chance that you'll have a streak of 10 misses at some point?
BruceZ
04-05-2005, 01:07 AM
[ QUOTE ]
you're a 40% shooter from 3 point range. over the span of 500 shots, what is the chance that you'll have a streak of 10 misses at some point?
[/ QUOTE ]
The probability of missing 10 shots in a row at some time in N shots P(N) obeys a simple recurrence relation:
P(N) = 0 for N < 10
P(10) = (.6)^10
P(11) = P(10) + (0.4)*(.6)^10
P(N) = P(N-1) + [1-P(N-11)]*(0.4)*(0.6)^10 for N >= 12.
That is, the probability of it happening in the first N shots is the sum of the probability of it happening in the first N-1 plus the probability of it happening on the Nth shot after not happening earlier, but to happen on the Nth shot, it must NOT happen in the first N-11 shots, and then there must always be 1 basket followed by 10 misses.
Soving this in Excel, P(500) = 70.55%.
Edit: Originally did this with 60% shooter (and got 3.0%).
gaming_mouse
04-05-2005, 02:11 AM
[ QUOTE ]
[ QUOTE ]
you're a 40% shooter from 3 point range. over the span of 500 shots, what is the chance that you'll have a streak of 10 misses at some point?
[/ QUOTE ]
The probability of missing 10 shots in a row at some time in N shots P(N) obeys a simple recurrence relation:
P(N) = 0 for N < 10
P(10) = (.6)^10
P(11) = P(10) + (0.4)*(.6)^10
P(N) = P(N-1) + [1-P(N-11)]*(0.4)*(0.6)^10 for N >= 12.
That is, the probability of it happening in the first N shots is the sum of the probability of it happening in the first N-1 plus the probability of it happening on the Nth shot after not happening earlier, but to happen on the Nth shot, it must NOT happen in the first N-11 shots, and then there must always be 1 basket followed by 10 misses.
Soving this in Excel, P(500) = 70.55%.
Edit: Originally did this with 60% shooter (and got 3.0%).
[/ QUOTE ]
Bruce,
Is there a way to solve such recurrences in closed form?
elitegimp
04-05-2005, 12:35 PM
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
you're a 40% shooter from 3 point range. over the span of 500 shots, what is the chance that you'll have a streak of 10 misses at some point?
[/ QUOTE ]
The probability of missing 10 shots in a row at some time in N shots P(N) obeys a simple recurrence relation:
P(N) = 0 for N < 10
P(10) = (.6)^10
P(11) = P(10) + (0.4)*(.6)^10
P(N) = P(N-1) + [1-P(N-11)]*(0.4)*(0.6)^10 for N >= 12.
That is, the probability of it happening in the first N shots is the sum of the probability of it happening in the first N-1 plus the probability of it happening on the Nth shot after not happening earlier, but to happen on the Nth shot, it must NOT happen in the first N-11 shots, and then there must always be 1 basket followed by 10 misses.
Soving this in Excel, P(500) = 70.55%.
Edit: Originally did this with 60% shooter (and got 3.0%).
[/ QUOTE ]
Bruce,
Is there a way to solve such recurrences in closed form?
[/ QUOTE ]
I'm not Bruce, but you can set up a Markov Chain with 11 stages:
Stage 1 = first shot or made the last shot
Stage 2 = missed last shot, made the one before it
Stage 3 = missed 2 in a row
...
Stage 11 = missed 10 in a row
for 1<=i<=10
P(going from stage i to stage i+1) = 0.6 (you miss 60% of the time, so you've now missed i in a row instead of i-1)
P(going from stage i to stage 1) = 0.4 (you make the shot 40% of the time, so you've now missed 0 in a row)
stage 11 is a sink, so P(going from 11 to 11) = 1
(once you've missed 10 in a row, you meet the criteria)
So let A be a matrix with A(i,j) = P(going from stage i to stage j). Then A^n is a matrix with the property that A^n(i,j) is the probability of being in stage j n events after starting in stage i.
Hence we want [A^500](1,11) = 0.7055
Dunno if that's what you were asking for or not /images/graemlins/smile.gif
Edit: And when I typed in the numbers backwards (make 60% shots), I got 0.0305
Also, here's some pseudo-code:
A = 11 x 11 matrix
for i=1,10 do
A(i,1) = 0.6
A(i,i+1) = 0.4
end for
B = A^500
print B(1,11)
gaming_mouse
04-05-2005, 04:57 PM
gimp,
that's exactly what i wanted.
Markov chains are cool.
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