RoboThrow

[superprismatic]

RoboThrow is a robot which shoots free throw basketball shots.

Once it is set up to shoot, RoboThrow always makes its first

shot and misses its second shot. The probability of RoboThrow

making any subsequent shot depends on its previous shots by

this simple rule: If RoboThrow has made X previous shots and

has missed Y previous shots, it will make the next shot with

probability X/(X+Y) or, equivalently, it will miss it with

probability Y/(X+Y).

What is the probability that RoboThrow will make 75 of its

first 100 shots?

[plasmid]

I made a spreadsheet to solve it at first, but after seeing what emerged, I was determined to solve it with equations. This might be a bit wordy, but:

Suppose you want to compute the probability that you will have made m baskets after taking n shots, denoted P(m,n). If you know P(m-1,n-1) and P(m,n-1), then you can calculate P(m,n) as

P(m,n) = P(m-1,n-1) * probability of making the next basket ...

...... + P(m,n-1) * probability of missing the next basket

...... = P(m-1,n-1) * (m-1)/(n-1) + P(m,n-1) * ((n-1)-(m))/(n-1)

...... = P(m-1,n-1) * (m-1)/(n-1) + P(m,n-1) * (n-m-1)/(n-1)

If you happen to know that P(m-1,n-1) = P(m,n-1), then you can just call that value P(n-1) and simplify the equation to be

...... = P(n-1) * [(m-1)/(n-1) + (n-m-1)/(n-1)]

...... = P(n-1) * [(m-1)+(n-m-1)] / (n-1)

...... = P(n-1) * (n-2) / (n-1)

P(m,n) = P(n-1) * (n-2)/(n-1)

Now to handle two special cases.

First, if m=1, then you know that P(m-1,n-1) must be zero because the machine has to make the first shot. In that case,

P(m,n) = P(m,n-1) * (n-m-1)/(n-1)

P(1,n) = P(1,n-1) * (n-2)/(n-1)

Note that it ends up being the same as the equation derived earlier.

Second, if m=(n-1) (that is, if the machine needs to make every shot after missing the second one) then you will know that P(m,n-1) = P(m,m) = 0 because the machine has to miss the second shot. In this case,

P(m,n) = P(m-1,n-1) * (m-1)/(n-1)

P(m,n) = P(m-1,n-1) * (n-2)/(n-1)

Again, note that it ends up being the same as the equation derived earlier.

The important conclusion from these equations is that P(m,n) is independent of the value of m. That is, the probability of making m shots out of n attempts is the same no matter what m is.

Recall that I said that the above would be true if you happen to know that P(m-1,n-1) = P(m,n-1). We know that the first shot will hit and the second shot will miss, so the third shot has a 50% chance of hitting and a 50% chance of missing, so P(1,3) = 0.5 and P(2,3) = 0.5.

Using the first equation for P(m,n) above, we know that P(2,4) = 0.5 * 2/3 = 0.333. Using the second and third equations to solve for P(1,4) and P(3,4) also gives 0.333.

Continuing to apply these equations for a long time gives P(75,100) = 1/99

[superprismatic]

I made a spreadsheet to solve it at first, but after seeing what emerged, I was determined to solve it with equations. This might be a bit wordy, but:

Suppose you want to compute the probability that you will have made m baskets after taking n shots, denoted P(m,n). If you know P(m-1,n-1) and P(m,n-1), then you can calculate P(m,n) as

P(m,n) = P(m-1,n-1) * probability of making the next basket ...

...... + P(m,n-1) * probability of missing the next basket

...... = P(m-1,n-1) * (m-1)/(n-1) + P(m,n-1) * ((n-1)-(m))/(n-1)

...... = P(m-1,n-1) * (m-1)/(n-1) + P(m,n-1) * (n-m-1)/(n-1)

If you happen to know that P(m-1,n-1) = P(m,n-1), then you can just call that value P(n-1) and simplify the equation to be

...... = P(n-1) * [(m-1)/(n-1) + (n-m-1)/(n-1)]

...... = P(n-1) * [(m-1)+(n-m-1)] / (n-1)

...... = P(n-1) * (n-2) / (n-1)

P(m,n) = P(n-1) * (n-2)/(n-1)

Now to handle two special cases.

First, if m=1, then you know that P(m-1,n-1) must be zero because the machine has to make the first shot. In that case,

P(m,n) = P(m,n-1) * (n-m-1)/(n-1)

P(1,n) = P(1,n-1) * (n-2)/(n-1)

Note that it ends up being the same as the equation derived earlier.

Second, if m=(n-1) (that is, if the machine needs to make every shot after missing the second one) then you will know that P(m,n-1) = P(m,m) = 0 because the machine has to miss the second shot. In this case,

P(m,n) = P(m-1,n-1) * (m-1)/(n-1)

P(m,n) = P(m-1,n-1) * (n-2)/(n-1)

Again, note that it ends up being the same as the equation derived earlier.

The important conclusion from these equations is that P(m,n) is independent of the value of m. That is, the probability of making m shots out of n attempts is the same no matter what m is.

Recall that I said that the above would be true if you happen to know that P(m-1,n-1) = P(m,n-1). We know that the first shot will hit and the second shot will miss, so the third shot has a 50% chance of hitting and a 50% chance of missing, so P(1,3) = 0.5 and P(2,3) = 0.5.

Using the first equation for P(m,n) above, we know that P(2,4) = 0.5 * 2/3 = 0.333. Using the second and third equations to solve for P(1,4) and P(3,4) also gives 0.333.

Continuing to apply these equations for a long time gives P(75,100) = 1/99

Nice work, plasmid!