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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
17:17 Thu 29th Aug 2013
So it does, sorry abotu that:
if cell A greater than 717, then enter 1, otherwise 0. 1 represents a successful free throw.
then if the sum of nine consecutive cells in col B is 9, print 1, otherwise zero.
Then sum col C, if this sum is greater than zero then count 1 to the successful strings of 53 attempts, otherise ignore, and count total attempts. when doing this for 100 sets of 53, usually between 50 and 63 such attempts are successful.
if cell A greater than 717, then enter 1, otherwise 0. 1 represents a successful free throw.
then if the sum of nine consecutive cells in col B is 9, print 1, otherwise zero.
Then sum col C, if this sum is greater than zero then count 1 to the successful strings of 53 attempts, otherise ignore, and count total attempts. when doing this for 100 sets of 53, usually between 50 and 63 such attempts are successful.
I think the desired approach is this:
Take a run of 2 successes as an example. If you take 3 FTs there are 8 possible outcomes, 3 of which contain a run of 2. Suppose p(success) = .7, then the total probability of getting a run of 2 is p(sss) + p(fss) + p(ssf) = .637.
We need to find N = number of FTs to get that probability up to (say) .9
So in N tosses there are 2^N outcomes.
The probability of N successes in a row = .7^N
The probability of N-1 successes in a row = .7^(N-1)
The probability of N-2 successes in a row = .7^(N-2)
etc.
All of these would count as a run of at least 2 (as long as N>4)
Find N so that those probabilities add up to .9
As N goes up 2^N gets huge, of course, but conceptually I think that's the approach. With a 0,1 outcome something about the binomial distribution might help get an answer.
I also simulated JJ's equation and was getting empirical probabilities around 60% rather than 90%.
Take a run of 2 successes as an example. If you take 3 FTs there are 8 possible outcomes, 3 of which contain a run of 2. Suppose p(success) = .7, then the total probability of getting a run of 2 is p(sss) + p(fss) + p(ssf) = .637.
We need to find N = number of FTs to get that probability up to (say) .9
So in N tosses there are 2^N outcomes.
The probability of N successes in a row = .7^N
The probability of N-1 successes in a row = .7^(N-1)
The probability of N-2 successes in a row = .7^(N-2)
etc.
All of these would count as a run of at least 2 (as long as N>4)
Find N so that those probabilities add up to .9
As N goes up 2^N gets huge, of course, but conceptually I think that's the approach. With a 0,1 outcome something about the binomial distribution might help get an answer.
I also simulated JJ's equation and was getting empirical probabilities around 60% rather than 90%.
I think the problem is twofold: firstly there's an issue of counting the umber of possible ways in which you can get x successive free throws in y attempts, but this one is relatively easy to solve, I think -- lots of 2^(y-x+any excess). But the problem is, or appears to be, that each individual trial can count differently. Suppose you have a target of 3/5, then there are as established eight ways to achieve this, but these are all weighted differently, and ordering can matter and chance what counts.
For 100/200, you could count 101 separate cases, but some of these cases can turn out to be equivalent to each other, as, say, a string of 200 heads should only count once but can end up counting 101 times. In JJ's method, meanwhile, even if you fail in a particular string of 100, the circumstances may be such that you are likely to succeed in the next string -- if, say, you have FSSSSSSS then that would count as a failure, for a target of 9, but obviously there is a 71.7% chance that you will succeed at the next attempt.
So it's very confusing, and as it seems that empirical data doesn't support the JJ formula then something else is needed.
For 100/200, you could count 101 separate cases, but some of these cases can turn out to be equivalent to each other, as, say, a string of 200 heads should only count once but can end up counting 101 times. In JJ's method, meanwhile, even if you fail in a particular string of 100, the circumstances may be such that you are likely to succeed in the next string -- if, say, you have FSSSSSSS then that would count as a failure, for a target of 9, but obviously there is a 71.7% chance that you will succeed at the next attempt.
So it's very confusing, and as it seems that empirical data doesn't support the JJ formula then something else is needed.
Jim,
My spreadsheet, which I used to check your calculations, does the calculation for the case of at least 2 consecutive successes in 4 throws by multiplying the number of times strictly 2 consecutive successes can occur in 4 throws and the number of strictly 3 consecutive successes in 4 throws and then the number of strictly 4 consecutive guesses in 4 throws, and then multiplies each by the probability and adds them all together.
I can explain this more precisely if you like.
My spreadsheet, which I used to check your calculations, does the calculation for the case of at least 2 consecutive successes in 4 throws by multiplying the number of times strictly 2 consecutive successes can occur in 4 throws and the number of strictly 3 consecutive successes in 4 throws and then the number of strictly 4 consecutive guesses in 4 throws, and then multiplies each by the probability and adds them all together.
I can explain this more precisely if you like.
Sorry correction to my last post:
Jim,
My spreadsheet, which I used to check your calculations, does the calculation for the case of at least 2 consecutive successes in 4 throws by calculating the number of times strictly 2 consecutive successes can occur in 4 throws and the number of strictly 3 consecutive successes in 4 throws and then the number of strictly 4 consecutive guesses in 4 throws, and then multiplies each by the probability and adds them all together.
I can explain this more precisely if you like.
Jim,
My spreadsheet, which I used to check your calculations, does the calculation for the case of at least 2 consecutive successes in 4 throws by calculating the number of times strictly 2 consecutive successes can occur in 4 throws and the number of strictly 3 consecutive successes in 4 throws and then the number of strictly 4 consecutive guesses in 4 throws, and then multiplies each by the probability and adds them all together.
I can explain this more precisely if you like.
Even so, you have a result that implied that it was more likely to get at least three straight throws when you only have four attempts, than if you had five. That is logically absurd, so I wonder if you entered the two numbers for 3/5 and 3/6 wrong. The other two match what I got (3/3 = 37%, 3/4 = 38%).
supposing that x=3, we're searching for N such that:
.9 = (sum from 1 to 2^N)(probability of getting at least 3 in a row in a sequence of N tries).
i.e., the sum goes over all 2^N possible sequences that can occur in N attempts.
As N gets large the number of such sequences (with a run of say 3) rises but the probability of each sequence falls.
.9 = (sum from 1 to 2^N)(probability of getting at least 3 in a row in a sequence of N tries).
i.e., the sum goes over all 2^N possible sequences that can occur in N attempts.
As N gets large the number of such sequences (with a run of say 3) rises but the probability of each sequence falls.
The probability of a run is discussed here, so someone should be able to put that all together:
http://
http://
As a simple example, which I brute forced rather than trying to get it algebraically, if you need a run of 2 with 90% probability, and your success prob on any FT is 0.7, then if N = 5, prob is 0.86, if N = 6, prob is 0.91.
If N=6 there are 2^6 = 64 possible sequences, of which 43 contain at least two successes in a row, and the total prob. of those 43 sequences is 0.912.
not a useful approach for higher values of run-length, however.
If N=6 there are 2^6 = 64 possible sequences, of which 43 contain at least two successes in a row, and the total prob. of those 43 sequences is 0.912.
not a useful approach for higher values of run-length, however.
Gosh darn it, I was just discovering that sort of approach for myself. I realised that if you are only looking for at least one such run, then as soon as you get it all subsequent throws don't matter. Hence the probability can be expressed as a sum of simple terms that look like p^x or p^x*(1-p), where p is the probability of success and x is the length of run you are interested in. Thus, the probability that a run of x occurs up to y = 2x is given by the terms:
y = x: p^x
y = x+1: p^x + (1-p)*p^x
y = x+2: p^x + 2*(1-p)*p^x
y = x+a: p^x + a*(1-p)*p^x, where x + a is less than y:
y = 2x : p^x + x*(1-p)*p^x
y = 2x+1: p^x + x*(1-p)*p^x + (1-p^x)*(1-p)*p^x
The last term here is a conditional probability, based on the logic that if you are hunting for a run of length x, then it can't have happened earlier (else you'd have stopped looking). Eventually, I suppose, the sequence above can be shown to be equivalent to the formula contained in Wolfram, once you have gathered like terms properly.
At this point I'm inclined to let the matter rest. I have an approach that works; the final result is contained in the wolfram page. I'll probably find myself coming back to this to finish it off, but I've nominally solved the problem.
y = x: p^x
y = x+1: p^x + (1-p)*p^x
y = x+2: p^x + 2*(1-p)*p^x
y = x+a: p^x + a*(1-p)*p^x, where x + a is less than y:
y = 2x : p^x + x*(1-p)*p^x
y = 2x+1: p^x + x*(1-p)*p^x + (1-p^x)*(1-p)*p^x
The last term here is a conditional probability, based on the logic that if you are hunting for a run of length x, then it can't have happened earlier (else you'd have stopped looking). Eventually, I suppose, the sequence above can be shown to be equivalent to the formula contained in Wolfram, once you have gathered like terms properly.
At this point I'm inclined to let the matter rest. I have an approach that works; the final result is contained in the wolfram page. I'll probably find myself coming back to this to finish it off, but I've nominally solved the problem.
This is an interesting discussion of run probabilities:
http://
http://
I'll retract "interesting" and replace it with "fascinating." It works! If you have p = 70% and want a 90% chance of a run of 7 or more, N must be 79! If p is 71.7%, N falls to 69.
I'm not sure a nifty formula of the sort N = whatever is possible, but as the mathematician in the link says, "Computers were invented for a reason, people."
I'm not sure a nifty formula of the sort N = whatever is possible, but as the mathematician in the link says, "Computers were invented for a reason, people."
Thanks dr b-- I'm gratified to see that this looks not unlike the solution I was working towards in my earlier post at 4.03pm today, only taken all the way to the end.
Anyway, there is your answer fairyrak, in the link dr b provides. A simple-sounding problem, with a remarkably complicated solution!
Anyway, there is your answer fairyrak, in the link dr b provides. A simple-sounding problem, with a remarkably complicated solution!
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