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

[dr b]

145

[dr b]

I wrote a VBA function that implements the recursive formula given in the "mathematicians"' discussion in the above link. Plug in FT % (.717), desired prob (90%) and run length (9) and out pops the answer. I'm insufferably pleased with myself.

[jim360]

We hadn't noticed... :)

[mibn2cweus]

145 sounds about right to me. Since there is a ~1/20 (.717^9) chance of making 9 in a row at the get go, I figured at most (20 * .9) such attempts (162 throws) should get the job done, but then that doesn't factor in the overlapping of seperate attempts.

So give yourself another pat on the back from me too while you're there ;o)

[mibn2cweus]

I've been throwing some virtual baskets and here's the results I came up with -

First basket made on average in 1.3947 ~ (1/.717) attempts.
2*100*100 iterations

First two consecutive baskets made in 3.3288 attempts.
3*50*50, 3.336, 3.3008, 3.3496

First three consecutive baskets made in 6.1 attempts.
1*50*50

First four consecutive baskets made in 9.8984 attempts.
2*50*50, 9.9608, 9.836

First five consecutive baskets made in 15.15 attempts.
1*50*50

First six consecutive baskets made in 22.2838 attempts.
2*50*50, 22.7708, 21.7968

First seven consecutive baskets made in 31.8924 attempts.
1*50*50

First eight consecutive baskets made in 46.9 attempts.
2*50*50, 46.4328, 47.44

First nine consecutive baskets made in 67.38 attempts.
2*50*50, 67.1548, 67.614

First ten consecutive baskets made in 92.8152 attempts.
1*50*50


The program used to obtain the values listed -
Set initial values:
Shot accuracy variable used (.717)
Set length of run of consecutive successful baskets to be made.
Counter of consecutive baskets made (initially set to zero and reset to zero upon each unsuccessful attempt)
Counter of attempts made to achieve a successful run (initially set to zero at beginning of each new trial)
Set number of successfully completed runs to be averaged.

Add number of attempts required to complete each run and divide by number of successful runs completed.
Accumulate an equal number of such trials together and divide sum by number of trials to obtain an overall average.
In the examples given I typically averaged together 50 sets of the averages for 50 successful runs each for a total of 2500 successful runs at each level.

[mibn2cweus]

Just realised I threw over a million baskets, and made more than 717000 thousand of them!

Oh dear, look at the time . . . CUL8r

[jim360]

Yes, those numbers make sense for the expected time to get a run of a particular length, and those numbers match the formula I quoted earlier pretty well:

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

where x is the run length and p is the probability, E(y) is the expected value of y. Slightly different problem from the original question, but interesting all the same!

[mibn2cweus]

An amazingly simple formula confirming the somewhat ambiguous results laboriously thrashed out by my program. Then I reckon when we try to get from 50% to 90% assurance either way, all hell breaks loose . . .