jrlvnv
jrlvnv
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June 13th, 2013 at 8:33:01 PM permalink
Take a American wheel with 38 numbers. I will give you 38 spins and the object is to hit EVERY number once. What are the chances of hitting all 38 numbers in 38 spins? Would love to see some of the math behind it. Thanks everyone
MathExtremist
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June 13th, 2013 at 8:40:29 PM permalink
38!/(38^38)
=0.000000000000000486120346
or 1 in 2,057,103,777,254,330
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
rdw4potus
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June 13th, 2013 at 8:44:53 PM permalink
somehow, probably due to rounding in Excel, I missed that by 10 in my answer in the 666 thread. Odd!
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
jrlvnv
jrlvnv
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June 13th, 2013 at 8:48:28 PM permalink
Thanks for the quick reply... One last question..... Maybe :) .... 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
ChesterDog
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June 13th, 2013 at 9:01:27 PM permalink
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.
MathExtremist
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June 13th, 2013 at 9:05:09 PM permalink
Quote: rdw4potus

somehow, 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...
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
jrlvnv
jrlvnv
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June 13th, 2013 at 9:51:16 PM permalink
Quote: ChesterDog

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.



So if I understand your math..... Any number not being able to repeat in the next 20 spins will happen about 1 in 500?
rdw4potus
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June 13th, 2013 at 9:55:39 PM permalink
What 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?
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
sodawater
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June 13th, 2013 at 10:26:17 PM permalink
Quote: MathExtremist

38!/(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)?
jrlvnv
jrlvnv
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June 13th, 2013 at 10:30:36 PM permalink
Quote: rdw4potus

What 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.
ChesterDog
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June 13th, 2013 at 10:38:27 PM permalink
Quote: jrlvnv

So 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?"
DJTeddyBear
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June 14th, 2013 at 7:08:55 AM permalink
Quote: jrlvnv

One 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
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
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