=0.000000000000000486120346
or 1 in 2,057,103,777,254,330
Quote: jrlvnv.... What are the odds of a number not repeating in 20 spins? If you can provide "some" of the math... Awesome... Thanks though either way
My answer is 0.2073%. I got it by doing this product: (38/38)*(37/38)*(36/38)*(35/38)*...*(19/38), which can be expressed in Excel with:
=permut(38,20)/38^20.
Quote: rdw4potussomehow, probably due to rounding in Excel, I missed that by 10 in my answer in the 666 thread. Odd!
Much more likely due to different rounding in my version of Excel (which is ancient) vs. yours. The real answer, per Wolfram Alpha, ends in 334.438...
Quote: ChesterDogMy answer is 0.2073%. I got it by doing this product: (38/38)*(37/38)*(36/38)*(35/38)*...*(19/38), which can be expressed in Excel with:
=permut(38,20)/38^20.
So if I understand your math..... Any number not being able to repeat in the next 20 spins will happen about 1 in 500?
Quote: MathExtremist38!/(38^38)
=0.000000000000000486120346
or 1 in 2,057,103,777,254,330
So the chances that you will have 0 repeats of your upcard in 52 hands of 7-card-stud are 52!/(52^52)?
Quote: rdw4potusWhat exactly are you asking? Say an 11 just hit. Do you want to know the odds of 20 more spins without two more 11s, or the odds that those 20 spins will contain no repeat numbers at all?
I was asking what are the odds of the next 20 spins not having a single repeating number in the 20 number sequence. The answer that came out was .20 something.... Then I wanted to know if that .20 something meant 1/5th of 1% equaling 1 time in 500 it will happen where 20 spins does not produce a repeat.
Quote: jrlvnvSo if I understand your math..... Any number not being able to repeat in the next 20 spins will happen about 1 in 500?
Yes. .2% means .2 in 100, which is the same as 1 in 500.
However, I might have answered the wrong question. I answered this question, "What is the probability that the next 20 numbers will be all different?"
Quote: jrlvnvOne last question..... What are the odds of a number not repeating in 20 spins?
The prior answer to this question assumes that you don't want duplication of ANY number.
I think you want is to predict the odds of X sleepers in Y spins. In that case, you want the odds of any other number hitting for the number of spins. Here's the formula:
(( 38 - X ) / 38 ) ^ Y