Basketball probability question
Quote: acesideLet me switch back to probabilities. If a player shots a basketball continuously with a probability of making a goal at 1/100, how many shots will be needed by average to see two goals back to back, or gapped by only one miss?
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Thank you for this question as I am learning and interested in the answer of the math OGs here.
Let's see if I have learned anything. According to binominal calculation, a 2 out of 2 with 1/100 probability equals 0.01%.
Regarding gapped by only 1 miss looks to me like 2 out of 3 and eager to see the answer.
Quote: sevenQuote: acesideLet me switch back to probabilities. If a player shots a basketball continuously with a probability of making a goal at 1/100, how many shots will be needed by average to see two goals back to back, or gapped by only one miss?
link to original post
Thank you for this question as I am learning and interested in the answer of the math OGs here.
Let's see if I have learned anything. According to binominal calculation, a 2 out of 2 with 1/100 probability equals 0.01%.
Regarding gapped by only 1 miss looks to me like 2 out of 3 and eager to see the answer.
link to original post
Here is a table of expected length of a series of shots before you either make two shots in a row or 2 out of 3. I did it for a range from 1% to 100% probability of a made shot. I generated it via Monte Carlo simulation.
| Prob | Ave # Shots |
|---|---|
| 0.01 | 5125.23 |
| 0.02 | 1313.49 |
| 0.03 | 597.20 |
| 0.04 | 343.86 |
| 0.05 | 225.05 |
| 0.06 | 159.81 |
| 0.07 | 120.07 |
| 0.08 | 93.89 |
| 0.09 | 75.79 |
| 0.10 | 62.61 |
| 0.10 | 62.62 |
| 0.20 | 18.89 |
| 0.30 | 9.87 |
| 0.40 | 6.41 |
| 0.50 | 4.67 |
| 0.60 | 3.65 |
| 0.70 | 3.00 |
| 0.80 | 2.55 |
| 0.90 | 2.23 |
| 1.00 | 2.00 |
Quote: acesideLet me switch back to probabilities. If a player shots a basketball continuously with a probability of making a goal at 1/100, how many shots will be needed by average to see two goals back to back, or gapped by only one miss?
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For two in a row: 10,200
For two in a row with one miss allowed between them: 10,199,000 / 199, or about 5,125.
Quote: ThatDonGuyQuote: acesideLet me switch back to probabilities. If a player shots a basketball continuously with a probability of making a goal at 1/100, how many shots will be needed by average to see two goals back to back, or gapped by only one miss?
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For two in a row: 10,200
For two in a row with one miss allowed between them: 10,199,000 / 199, or about 5,125.
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Please tell me if I am right or wrong.
When you say 2 in a row with a miss allowed, do you mean hit - miss - hit?
and miss - hit - hit does not count?
thanks
Quote: ThatDonGuyQuote: acesideLet me switch back to probabilities. If a player shots a basketball continuously with a probability of making a goal at 1/100, how many shots will be needed by average to see two goals back to back, or gapped by only one miss?
link to original post
For two in a row: 10,200
For two in a row with one miss allowed between them: 10,199,000 / 199, or about 5,125.
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For two shots in a row, the average number of shots needed is 10100, and I know how to calculate this. But I still haven’t figured out how to calculate the second question. I will put the calculation into an excel spreadsheet to calculate it out. I will try it at home.
Quote: aceside“2 in a row with a miss allowed” means either …mmmmgmg or … mmmmmmgg counts.
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sorry now it gets more confusing at least for me.
each player = participant gets 3 tries which means to hit 2 in a row looks like
hit - hit no need for 3rd try
miss - hit - hit
hit - miss - hit
Quote: acesideFor my question, I do not set a restriction of 3 shots as a set. I just let the player shoot continuously until he gets 2 goals back to back or gapped by one miss.
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Thanks for the clarification. I was confused because my original question was 2 out of 3 for each player/participant.
Quote: acesideFor two shots in a row, the average number of shots needed is 10100, and I know how to calculate this. But I still haven’t figured out how to calculate the second question. I will put the calculation into an excel spreadsheet to calculate it out. I will try it at home.
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There's probably some Poisson method for this, but I use a Markov process:
Let E(a,b) be the expected number of shots remaining with a made and b missed in the current "run" (i.e. if you miss when b = 1, everything "resets").
Let p be the probability of making any particular shot.
The solution is E(0,0), since (0,0) is the starting point.
E(2,0) = E(2,1) = 0, since you are done.
E(1,1) = 1 + p E(2,1) + (1 - p) E(0,0) = 1 + (1 - p) E(0,0)
E(1,0) = 1 + p E(2,0) + (1 - p) E(1,1) = 1 + (1 - p) (1 + (1 - p) E(0,0)) = 2 - p + (1 - p)^2 E(0,0)
Note that E(0,1) is impossible, as if you miss a free throw at (0,0), you remain at (0,0)
E(0,0) = 1 + p E(1,0) + (1 - p) E(0,0)
(1 - (1 - p)) E(0,0) = 1+ p E(1,0)
p E(0,0) = 1 + p E(1,0)
Divide both sides by p:
E(0,0) = 1/p + E(1,0) = 1/p + 2 + p + (1 - p)^2 E(0,0)
(1 - (1 - p)^2) E(0,0) = 1/p + 2 - p
E(0,0) = (1/p + 2 - p) / (2p - p^2)
In this case, p = 1 / 100, so the solution is
(1 / (1/100) + 2 - 1/100) / (2/100 - 1/10,000)
= (10,199/100) / (199/10,000)
= 10,199,000 / 199
Quote: ThatDonGuyQuote: acesideFor two shots in a row, the average number of shots needed is 10100, and I know how to calculate this. But I still haven’t figured out how to calculate the second question. I will put the calculation into an excel spreadsheet to calculate it out. I will try it at home.
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There's probably some Poisson method for this, but I use a Markov process:
Let E(a,b) be the expected number of shots remaining with a made and b missed in the current "run" (i.e. if you miss when b = 1, everything "resets").
Let p be the probability of making any particular shot.
The solution is E(0,0), since (0,0) is the starting point.
E(2,0) = E(2,1) = 0, since you are done.
E(1,1) = 1 + p E(2,1) + (1 - p) E(0,0) = 1 + (1 - p) E(0,0)
E(1,0) = 1 + p E(2,0) + (1 - p) E(1,1) = 1 + (1 - p) (1 + (1 - p) E(0,0)) = 2 - p + (1 - p)^2 E(0,0)
Note that E(0,1) is impossible, as if you miss a free throw at (0,0), you remain at (0,0)
E(0,0) = 1 + p E(1,0) + (1 - p) E(0,0)
(1 - (1 - p)) E(0,0) = 1+ p E(1,0)
p E(0,0) = 1 + p E(1,0)
Divide both sides by p:
E(0,0) = 1/p + E(1,0) = 1/p + 2 + p + (1 - p)^2 E(0,0)
(1 - (1 - p)^2) E(0,0) = 1/p + 2 - p
E(0,0) = (1/p + 2 - p) / (2p - p^2)
In this case, p = 1 / 100, so the solution is
(1 / (1/100) + 2 - 1/100) / (2/100 - 1/10,000)
= (10,199/100) / (199/10,000)
= 10,199,000 / 199
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Let me reword your solution in a different format. Let a basketball miss=0 and goal=1. The possible outcome strings are:
0 —> (00, 01) —> [(000, 001), (010, 011)];
1 —> (10, 11) —> [(100, 101), (stop, stop)].
From the first step, we have
n0 = 1 + (1-p) n00 + p n01 = 1+ (1-p) n0 + p n01;
From the second step second branch, we have
n01 = 1+ (1-p) n010 + p n011 = 1+ (1-p) n10;
From the fourth step first branch, we have
n10 = 1+ (1-p) n100 + p n101 = 1 + (1-p) n0.
To solve this set of three equations, we also need these conditions, n00=n0, n011=0, n010=n10, n100=n0, and n101=0. The solution is
n0 = (1+2p-p^2)/(2p^2-p^3).
If p=1/100, n0= 1,019,900/199.
But to be honest again, I envy you guys
Quote: acesideRecently I did some experiments on this basketball probability. I found my probability of making a successful goal is doubled if I use a size 6 basketball, which is 28.5”, a bit smaller than the standard size 7 of 29.5”. This means the probability is critically dependent on the size.
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interesting. Thanks for that, which means it must be a basketball size 7 in case of the challenge
Quote: sevenQuote: acesideRecently I did some experiments on this basketball probability. I found my probability of making a successful goal is doubled if I use a size 6 basketball, which is 28.5”, a bit smaller than the standard size 7 of 29.5”. This means the probability is critically dependent on the size.
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interesting. Thanks for that, which means it must be a basketball size 7 in case of the challenge
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Oh, and if this is a hobby of yours, please tell me which challenge is harder in terms of probability, a 2 out of 3 half court shot or a 30 second progressive shot where the player has to do the following?
a lay-up
a free throw
a 3 pointer
half-court shot
half-court shot only one try is allowed and the others as many as needed in case of a miss. all in 30 seconds
p^2 (3 - 2p).
Here, p is the probability of scoring one shot. If p=1/100, the probability is 0.000298.
For this new challenge of the four-type combination, let me assume these parameters, as listed in the table below.
Shooting type; probability of one shot; time needed for one shot; number of shots allowed.
1. A half-court; p, 6 sec, 1 attempt.
2. A 3-pointer; Sqrt(p), 5 sec, 2 attempts.
3. A free-throw; Sqrt[Sqrt(p)], 4 sec, 2 attempts.
4. A lay-up; Sqrt{Sqrt[Sqrt(p)]}, 3 sec, 2 attempts.
Also, the shooter must follow the above assigned shooting order. With these numbers, the final probability is calculated as
p x Sqrt(p)x[2- Sqrt(p)] x Sqrt[Sqrt(p)] x [2- Sqrt[Sqrt(p)] x Sqrt[Sqrt[Sqrt(p)] x [2- Sqrt[Sqrt[Sqrt(p)] = 0.000818.
Therefore, the 2 out of 3 half-court challenge is about 2.7 times harder.
Quote: acesideEarlier, ThatDonGuy calculated the probability of scoring at least 2 out of 3 half-court shots,
p^2 (3 - 2p).
Here, p is the probability of scoring one shot. If p=1/100, the probability is 0.000298.
For this new challenge of the four-type combination, let me assume these parameters, as listed in the table below.
Shooting type; probability of one shot; time needed for one shot; number of shots allowed.
1. A half-court; p, 6 sec, 1 attempt.
2. A 3-pointer; Sqrt(p), 5 sec, 2 attempts.
3. A free-throw; Sqrt[Sqrt(p)], 4 sec, 2 attempts.
4. A lay-up; Sqrt{Sqrt[Sqrt(p)]}, 3 sec, 2 attempts.
Also, the shooter must follow the above assigned shooting order. With these numbers, the final probability is calculated as
p x Sqrt(p)x[2- Sqrt(p)] x Sqrt[Sqrt(p)] x [2- Sqrt[Sqrt(p)] x Sqrt[Sqrt[Sqrt(p)] x [2- Sqrt[Sqrt[Sqrt(p)] = 0.000818.
Therefore, the 2 out of 3 half-court challenge is about 2.7 times harder.
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quick question
why did you change the shooting order?
