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

• Panphobia
In summary, Barbara, a mathematician and basketball player, has found a probability for scoring a point when shooting from a distance x. She takes shots from distances 1-50 and has a 2% chance of scoring exactly 20 points. The task is to find the real constant q, which is greater than 50, and give the answer rounded to 10 decimal places. There are several ways to approach this problem, including using a generating function and a computer algebra package, or solving for q symbolically and using a numerical solver. The solution is q = 52.6494571953.
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.

## 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?

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?

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

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%

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.

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.

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?

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 * 10something 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 :/

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.

Panphobia said:

## 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.

## 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.

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.

Panphobia said:
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.

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.

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## What is the definition of probability of scoring?

The probability of scoring refers to the likelihood or chance of a particular event or outcome occurring in a game or experiment.

## How is probability of scoring calculated?

Probability of scoring is calculated by dividing the number of desired outcomes by the total number of possible outcomes.

## What factors influence the probability of scoring?

The probability of scoring can be influenced by various factors such as the skill level of the player, the difficulty of the task, and any external variables such as weather conditions or equipment.

## What is the role of sample size in determining the probability of scoring?

The larger the sample size, the more accurate the probability of scoring will be. This is because a larger sample size provides a more representative sample of the population and reduces the impact of random chance.

## How can probability of scoring be used in decision making?

Probability of scoring can be used in decision making to assess the likelihood of different outcomes and make informed choices based on the expected probabilities. It can also be used to evaluate the effectiveness of strategies or interventions in achieving a desired outcome.

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