supergrass
supergrass
Joined: Apr 6, 2014
  • Threads: 13
  • Posts: 58
January 14th, 2016 at 7:14:51 AM permalink
Suppose on a 5c machine I have hit a particular prize (somewhat like a jackpot) 13 times after 1000 spins.
Suppose on a 2c machine I have hit it 5 times after 1000 spins.
How confident can I say the two machines have different settings for the particular prize?
Last edited by: supergrass on Jan 14, 2016
Romes
Romes
Joined: Jul 22, 2014
  • Threads: 27
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January 14th, 2016 at 7:52:08 AM permalink
Simple answer: 0
Playing it correctly means you've already won.
Bondy3
Bondy3
Joined: Jan 4, 2013
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January 17th, 2016 at 8:33:43 PM permalink
assuming the probability of a jackpot is 1%, the standard deviation is for 1000 spins is 3.6, so that means that the 5c machine is less than 1 standard deviation over and the 2c machine would be 1.5 standard deviation under, both are within statistically accepted ranges.

with such a small sample size its not possible to detect with confidence that the machines are on different settings

variance of a set of spins = npq

n=number of spins
p=probability of jackpot
q=probability of non-jackpot
supergrass
supergrass
Joined: Apr 6, 2014
  • Threads: 13
  • Posts: 58
January 19th, 2016 at 9:49:48 AM permalink
If I knew the 5c machine jackpot chance is 1.3% and the 2c machine only hit 5 times then I can say the 2c machine is different with over 98.5% confidence. Calculation:

1000 spins, mean is 13 hits, result is 5 times, so 8 times less than expectation.
The standard deviation is 3.6 hits, so 2.22 SD below expectation.
So only 1.5% chance it happened due to bad luck.

It seems to me even with 5 hits, the sample size is big enough to get over 95% confidence interval.
However the difficulty is I don't know the true 5c jackpot chance. I only know based on my sample test the chance is 1.3%.

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