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Number Of Attempts So That I Get X Free Throws In A Row
hi, I have this problem of maths, my probability is always 71.7% for every free throw . Assume I never get nervous or anything, I always have 71.7%
I'll do many free throws in a year, so whats the minimum number (y) of free throws that i need to try in order to make sure that in those number (y) of free throws I have at least 90% to success in (x) jumps in a row at least once?
I need a formula that gives me y based on x and on a probability of 0.717
I'll do many free throws in a year, so whats the minimum number (y) of free throws that i need to try in order to make sure that in those number (y) of free throws I have at least 90% to success in (x) jumps in a row at least once?
I need a formula that gives me y based on x and on a probability of 0.717
Answers
This is an interesting discussion of run probabilitie
16:17 Thu 29th Aug 2013
So you want to have a 90% chance of x consecutive successes... what is the expected number of free throws needed to generate such a chain of successes at least once, 90% of the time?
I've heard of similar problems, e.g what is the minimum amount of time it takes a monkey to type a certain message. I couldn't work out how to do that, have a feeling it involved "martingale" theory or something, but will give it some thought.
I've heard of similar problems, e.g what is the minimum amount of time it takes a monkey to type a certain message. I couldn't work out how to do that, have a feeling it involved "martingale" theory or something, but will give it some thought.
I think this can be solved using the binomial distribution. However the calculation is not easy to do the way the poster wants it. It's fairly easy to calculate the probability of k successes in n attempts:
(0.717)^k times(0.283)^(n-k) times (n!/((k!)(n-k)!))
but if you want to set the result to 0.9 and find n when k/n=0.9, its a difficult calculation to do without a computer.
(0.717)^k times(0.283)^(n-k) times (n!/((k!)(n-k)!))
but if you want to set the result to 0.9 and find n when k/n=0.9, its a difficult calculation to do without a computer.
The probability of making throw one is .717
The probability for making both one & two is .717^2 (read squared)
The probability for making three consecutive throws is .717^3
The probability for making nine consecutive throws is .717^9 ~5%.
0.9/0.05 = 18
On average nine consecutive throws will be successful in ~18 attempts.
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The probability for making both one & two is .717^2 (read squared)
The probability for making three consecutive throws is .717^3
The probability for making nine consecutive throws is .717^9 ~5%.
0.9/0.05 = 18
On average nine consecutive throws will be successful in ~18 attempts.
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Yes, I think 18 is far too low. There has to be a more efficient method than the one I used, but the problem for, say, 3 consecutive three throws amounts to working out firstly the number of ways this can occur in any number of throws, then multiplying each case by the probability of its occurring and summing these up. The analysis for 3 straight throws in 5, then, gives (S = success, F = failure) the following eight sequences that work:
SSSSS
SSSSF
SSSFS
SFSSS
FSSSS
SSSFF
FSSSF
SSSFF
These can be grouped into three groups with probabilities 0.717^5, 0.717^4*0.283, 0.717^3*0.283^2, and the total probability is about 0.58 or 58%.
Similar work for three consecutive throws give six attempts gives 19/64 possible sequences, with a total probability of 67%, and for four attempts the result is about 47% probability. Also, 0.717^3 ~0.36, so we have a sequence going something like:
3/3 = 37%
3/4 = 47%
3/5 =58%
3/6 = 67%
From this I project that you would need eight or nine free throws to have a 90% chance of a string of (at least) three consecutive throws. The method described above can be applied to any target number of consecutive free throws x, but is clearly very tedious and is surely not the best way to tackle the general problem. However it does mean that it's likely that mibn's "for 9 you need 18" answer is likely to be an underestimate.
SSSSS
SSSSF
SSSFS
SFSSS
FSSSS
SSSFF
FSSSF
SSSFF
These can be grouped into three groups with probabilities 0.717^5, 0.717^4*0.283, 0.717^3*0.283^2, and the total probability is about 0.58 or 58%.
Similar work for three consecutive throws give six attempts gives 19/64 possible sequences, with a total probability of 67%, and for four attempts the result is about 47% probability. Also, 0.717^3 ~0.36, so we have a sequence going something like:
3/3 = 37%
3/4 = 47%
3/5 =58%
3/6 = 67%
From this I project that you would need eight or nine free throws to have a 90% chance of a string of (at least) three consecutive throws. The method described above can be applied to any target number of consecutive free throws x, but is clearly very tedious and is surely not the best way to tackle the general problem. However it does mean that it's likely that mibn's "for 9 you need 18" answer is likely to be an underestimate.
yes, I meant free throws in a row (successful ones)
for example to have 90% of making 40 successful ones in a row I might have to throw like 30000 (just to name an example it with pure human guessing)
I'm not interested in the probability of making 40 in a row, I know that it's 0.717^40, what I'm interested is in the number y of attempts that i should do to have 90% of making x in a row (successful ones)
for example to have 90% of making 40 successful ones in a row I might have to throw like 30000 (just to name an example it with pure human guessing)
I'm not interested in the probability of making 40 in a row, I know that it's 0.717^40, what I'm interested is in the number y of attempts that i should do to have 90% of making x in a row (successful ones)
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