basketball probability question

[jojobinks]

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]

[ 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]

[ 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]

[ 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 [img]/images/graemlins/smile.gif[/img]

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]

gimp,

that's exactly what i wanted.

Markov chains are cool.