Calculating q for Probability of Scoring: A Math and Basketball Conundrum

[Panphobia]

Homework Statement

(This is just a question for fun, not homework or anything)
Barbara is a mathematician and a basketball player. She has found that the probability of scoring a point when shooting from a distance x is exactly (1 - x/q), where q is a real constant greater than 50.

During each practice run, she takes shots from distances x = 1, x = 2, ..., x = 50 and, according to her records, she has precisely a 2 % chance to score a total of exactly 20 points.

Find q and give your answer rounded to 10 decimal places.

The Attempt at a Solution

Would I need to iterate through all the q's and figure out all combinations of x's to get the answer? Or is there a smarter solution that I am just missing?

 

[D H]

This is a bad question. Is it 20 points or 20 baskets? If it's points rather than baskets, whose three foot rule is being used?

 

[Panphobia]

Scoring from any point x gives only one point. So it is baskets.

 

[lendav_rott]

Let the probability score be P and not score be P'.
The probability of Not scoring from any given x is
1 - (1 - x/q) , which is just x/q = P'
The probability of missing 30 points with the constant q is 98%

 

[Panphobia]

Yea I know that, I set up the question like that too. But I am not sure of how to figure out q without iterating through all combinations of x, Binomial[50,20] = 4.7129212e+13, which is not feasible.

 

[D H]

That 50 choose 20 is a *big*number is a big part of the problem. The other big part of the problem is that no two of the individual probabilities are the same.

 

[Panphobia]

Are you hinting something with *big*? Also when you say no two of the individual probabilities are the same, isn't that kind of intuitive? How does that help with getting an answer?

 

[lendav_rott]

Well, if she were to miss all of the shots the probability of it would be 50! / q⁵⁰

The problem is the distance is changing. Can't treat it like a simple dice problem anymore. Now we would have to calculate each and every combination out differently and there are...4.7something * 10^something different combinations to calculate :S

I will try and see what happens if she has 4(x= 1, 2, 3 and 4) shots with x/q chance to miss and has to score , say, 2 points with the probability of 70% of Not succeeding. Maybe there's a pattern to how the calculation performs. I don't know :/

 

[Panphobia]

Which is basically what I said, iterating. But for this, even using a computer, it would take a REALLY long time to compute. So there is obviously a better solution.

 

[Ray Vickson]

Homework Statement

(This is just a question for fun, not homework or anything)
Barbara is a mathematician and a basketball player. She has found that the probability of scoring a point when shooting from a distance x is exactly (1 - x/q), where q is a real constant greater than 50.

During each practice run, she takes shots from distances x = 1, x = 2, ..., x = 50 and, according to her records, she has precisely a 2 % chance to score a total of exactly 20 points.

Find q and give your answer rounded to 10 decimal places.

The Attempt at a Solution

Would I need to iterate through all the q's and figure out all combinations of x's to get the answer? Or is there a smarter solution that I am just missing?

Assuming the successive scores are independent (but distance-dependent, of course) and N = number of successful shots, we want $P\{ N = 20 \} = 0.02$, by adjusting the value of q. This can be done using a generating function and a computer algebra package: $P\{ N = 20\} =$ coefficient of $z^{20}$ in the expansion of
$$F(z) \equiv Ez^N = \prod_{i=1}^{50} \left[ \frac{i}{q} + \left( 1 - \frac{i}{q}\right) z \right]$$
I used Maple to find the coefficient for a given numerical value of q; each such computation is essentially instantaneous on an HP Probook. One can just keep trying various values of q (or let Maple try to solve the equation numerically). Here are the commands I used in the 'manual' approach, and with two nearby values of q:
f:=seq(i/q+(1-i/q)*z,i=1..50):
fqz:=seq(subs(q=52.649456,f),i=1..50): <---- q = 52.64956
Fqz:=mul(fqz,i=1..50):P:=evalf(coeff(Fqz,z,20));
P := .01999999536 <-------------------- P(N = 20}
fqz:=seq(subs(q=52.649457,f),i=1..50): <---- q = 52.649457
Fqz:=mul(fqz,i=1..50):P:=evalf(coeff(Fqz,z,20));
P := .02000000102 <-------------------- P(N = 20}

In the above,fqz = sequence of terms with a given numerical value of q; Fqz = F(z) for that q, and P = the (floating point) numerical value of the coefficient of z^20 in Fqz.

The P-values show that q = 52.64956 is a bit too small and q = 52.649457 is a bit too large. We could keep going like this to get a more accurate solution.

 

[Panphobia]

thanks! By the way, I only have a grade 12 probability understanding so far, would I need to more to successfully solve a problem like this on my own? By the way I solved the question in the way you said, in Mathematica.
answer: 52.6494571953

 

[Ray Vickson]

thanks! By the way, I only have a grade 12 probability understanding so far, would I need to more to successfully solve a problem like this on my own? By the way I solved the question in the way you said, in Mathematica.
answer: 52.6494571953

Another way to do it in Maple is to get an expression for P with q left as a symbolic constant. That gives a high-degree polynomial in 1/q, and one can then solve the equation P = P(q) = 0.02 numerically, using the Maple command fsolve(P=0.02,q=50 ..60) for example (which spits out the answer almost instantly). There are issues, however: the exact expression for P(q) has many terms with large numerators, like
$$-\frac{994127516915891993390185476709418888222447116029208791189437137551360000000000}{q^{47}}$$
and
$$\frac{323011458963304388485529194120664234646395732000697907575432055074324480000000}{q^{46}}$$
with both + and - signs. Numerics in such cases can be very tricky, due to round-off errors, so it is advisable to increase the numerical precision to ensure accurate calculations.

Like you get from Mathematica, Maple gives 52.6494571953091714497882428732 in 30-digit precision, which rounds down to 52.6494571953 in 12-digit (10 decimal) precision. This is a bit different from the result that would be obtained by using 12-digit precision throughout; this which demonstrates the need for higher precision, as already explained.