Number Of Attempts So That I Get X Free Throws In A Row

[fairyrak]

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

[jim360]

I don't think vascop has, because the question calls for a specific number of consecutive successes, and those are counted differently from just an arbitrary number of successes per attempts.

I believe the problem will require more sophisticated techniques, but I may be missing something.

[Jonathan-Joe]

(^ means to the power of: N^2 = N x N, N^3 = N x N x N and so on.)

Chance that one throw is successful : 0.717
Chance that three consecutive throws are all successful : 0.717^3 = 0.369
So chance that the run of three is not successful : 1 - 0.369 = 0.631

In a trial of seven throws there are five runs of three (each throw except the last two is the start of a run).
The chance that all five are unsuccessful : 0.631^5 = (slightly more than) 0.100 or 10%
So the chance that at least one run is successful is slightly less than 90%

In a trial of eight throws there are six runs of three.
The chance that all are unsuccessful : 0.631^6 = 0.063 or 6.3%
So the chance that at least one run is successful is about 93.7%

So eight throws are needed (but seven very nearly gives the 90% required).

Post a request if you want to see my general formula (the "x" and "y" of the OP) for the problem.

[jim360]

Please do post it.

[Jonathan-Joe]

p = probability of success on one throw
q = probability of success for n consecutive throws
A = number of throws

Chance that one throw is successful : p
Chance that n consecutive throws are all successful : p^n
So chance that the run of n is not successful : 1 - p^n

In a trial of A throws there are A - n + 1 runs of n throws
The chance that all are unsuccessful : (1 - p^n)^(A - n+1)
The chance that all are unsuccessful : 1 - q

(1 - p^n)^(A - n + 1) = 1 - q

The only way I know to solve this is to take logs

(A - n + 1)*log(1 - p^n) = log(1 - q)
A - n + 1 = log(1 - q)/log(1 - p^n)
A = log(1 - q)/log(1 - p^n) + n - 1

A scientific calculator (including that included with Windows &) will handle this.

[jim360]

I like the look of that solution -- it's behaving in the way I'd expect.

For the benefit of fairyrak, in his original notation the formula Jonathon-Joe has given would be written:

y = log(1-0.9)/log(1-0.717^x) + x -1

You can generalise this easily enough by changing 0.717 to whatever probability of success on one throw you'd like, and 0.9 to whatever probability of success of x consecutive free throws you wish.

[mibn2cweus]

27 then?

[jim360]

Assuming JJ's formula is correct, then to ensure a 90% probability of obtaining nine consecutive free throws, you would need to have fifty-three attempts.

[vascop]

Hi all
I'm not convinced yet that JJ's reasoning is correct - I'm still thinking about it.

[jim360]

While you're still thinking about it, vascop, would you be so kind as to check my working earlier, when I tried to calculate the probability for three straight throws given 4, 5, or 6 attempts? In principle, if JJ's formula is correct, then it should reproduce the results backwards, that is, that requiring a q of 0.577 would give A=5.

[jim360]

Should just add that I found the "martingale solution", but it's to a slightly different question: what is the expected number of trials needed before you obtain a chain of x successive free throws? The answer is:

[1/(1-p)]*{(1/p^x)-1}

for p = 0.717, x=3 this gives y = 6.05, so that you would expect to have a string of three successive free throws after six attempts or so. And the expected time to obtain nine consecutive throws is 67 attempts. But expectation doesn't mean 90% probability -- so perhaps, after all, this is a problem that needs to be calculated in a different way. I think some doubts remain over JJ's solution but it looks the right sort of approach.

http://mathproblems123.files.wordpress.com/2010/09/martingaleextras.pdf

[fairyrak]

thanks for your effort so far, yes I want the number of throws for the general case instead of than the expected value. Although thanks finding the expected value too, it's interesting. That formula also behaves well with some stats that I had, but I'm still analyzing too, I'll be checking in some hours

[fairyrak]

i rely less on the formula for now, there might be some error in one of the formulas or a mathematical explanation or another thing because
for
p=0.732
with a percentage of success wanted in x jumps in a row at least once=96%
x (the number of free throws in a row that i have to get)=26

it gives this result

y (the solution according to jonathan joe) =10748.8551
the expected value according to jim= 12429.302


also i made simulated stats with a program and got 0 succesfull chains of 26 or more times in 10700 tries

what do you think? was it just luck? or something happened like a very strong deviation caused by such a high x of 26 or more free throws in a row? and i entered 96% thinking that it was a high enough %

note: and i got 2 chains of 25 in the simulation

[dr b]

You'd have to simulate 10700 free throws, say, 1000 times. In 900 of those trials you would expect to have a run of 26 made free throws. Remember that trying 10700 free throws only gives a 90% chance of 26 in a row.

[vascop]

@jim
I have checked your calculations and I get:
3/3 37%
3/4 47%
3/5 43%
3/6 30%

Using JJs formula I can't get 3,4,5,6 from these percentages.
The reason I think JJ's formula might be wrong is because he wtites in his post:
"In a trial of A throws there are A - n + 1 runs of n throws "
I don't think this is correct. For example, for A=5 and n=3 this gives 3 which we know is incorrect and should be 8 I think, as you worked out in your post:
SSSSS
SSSSF
SSSFS
SFSSS
FSSSS
SSSFF
FSSSF
FFSSS

[dr b]

but if you take 5 free throws, your runs of 3 are free throws:
(1,2,3)
(2,3,4)
(3,4,5)

no?

[factor-fiction]

....although you can get each of these in different ways
e.g. (1,2,3) can come from a SSSFF, SSSSF, SSSFS, SSSSS
which i think is what vascop is saying

[vascop]

What about
(1,2,3,4) and (1,2,3,4,5) etc?
These are all situations where you have 3 consecutive successes.

[jim360]

"
3/3 37%
3/4 47%
3/5 43%
3/6 30%
"

Seems odd that three straight free throws is less likely having had six attempts than five, which is also less than four... Sure about those last two?

I was running simulations checking the 9/53 case and it doesn't seem to match up. Counting any set of 53 trials as successful if there is at least one string of nine free throws or greater, then the average success rate seems to be of order 60%, which is somewhat less than 90. Which goes against JJ's formula again. Back to the drawing board...

(method for simulation follows:

colA: = randbetween(1,1000)
collB: = IF(cellA

[vascop]

Sorry Factor, I posted at same time as you. That's exactly what I am saying.

[vascop]

Jim
Don't put greater than signs in your posts as it terminates the message.