September 27th, 2011 at 5:18:42 PM
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Here is a question asked of me recently. Somebody I know hit six single-line royals in 5,000 hands once. He estimates he has played 25,000,000 hands of video poker in his lifetime. What are the odds of having such a streak at least once in 25M hands?

Of course, we could answer it by simulation, but that always leaves me so ... unsatisfied.

If we don't mind doing matrix multiplication 25 million times over, we could follow the kind of solution for a coin flipping problem I addressed in my June 4, 2010 Ask the Wizard column.

Any thoughts how to solve this? ME (MathExtremist), miplet, CM (CrystalMath), here is your chance to shine!

Of course, we could answer it by simulation, but that always leaves me so ... unsatisfied.

If we don't mind doing matrix multiplication 25 million times over, we could follow the kind of solution for a coin flipping problem I addressed in my June 4, 2010 Ask the Wizard column.

Any thoughts how to solve this? ME (MathExtremist), miplet, CM (CrystalMath), here is your chance to shine!

It's not whether you win or lose; it's whether or not you had a good bet.

September 27th, 2011 at 5:38:44 PM
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.0003425

September 27th, 2011 at 6:05:50 PM
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Quote:buzzpaff.0003425

Care to show your work?

It's not whether you win or lose; it's whether or not you had a good bet.

September 27th, 2011 at 6:15:50 PM
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Quote:WizardCare to show your work?

I'm pretty sure his work involved typing a random number. :) I have no clue how do it.

“Man Babes” #AxelFabulous

September 27th, 2011 at 6:22:49 PM
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NO ! Does speed count? I was first, you know. And so far no one has challenged my answer.

No fair miplet, how did you know ? LOL

No fair miplet, how did you know ? LOL

September 27th, 2011 at 7:23:11 PM
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at least n royals in N hands is

F(n,N) = p*F(n-1,N-1) + (1-p)*F(n,N-1),

where p is 0.00000154

At least one such occurrence in 25M trials is P=1-(1-F(n,N))^25,000,000

I calculate F(6,5000) as 2.867*10^-16, and P = 8.327*10^-9

Seems too simple ... What am I missing?

F(n,N) = p*F(n-1,N-1) + (1-p)*F(n,N-1),

where p is 0.00000154

At least one such occurrence in 25M trials is P=1-(1-F(n,N))^25,000,000

I calculate F(6,5000) as 2.867*10^-16, and P = 8.327*10^-9

Seems too simple ... What am I missing?

"When two people always agree one of them is unnecessary"

September 27th, 2011 at 8:12:31 PM
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Quote:weaselmanwhere p is 0.00000154...Seems too simple ... What am I missing?

0.00000154 = 1 in 649,350. You seem to be looking at the royal probability on the deal. Let's use 1 in 40,000 for the probability on the draw.

That will get you closer, but there is still the issue that the 25,000,000 period of time overlap.

It's not whether you win or lose; it's whether or not you had a good bet.

September 27th, 2011 at 8:32:47 PM
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Based on a quick estimate, my best guess at this time is 1 in 19,000.

It's not whether you win or lose; it's whether or not you had a good bet.

September 28th, 2011 at 7:37:22 AM
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I ran a simulation overnight. Here are the results:

Wins=3

Losses=19125

Recall that each "loss" is 25M hands, so that is almost half a trillion hands.

So, based on this small sample size I get a probability of 1 in 6376. I'm still trying to find a mathematical solution, but no epiphany so far.

Wins=3

Losses=19125

Recall that each "loss" is 25M hands, so that is almost half a trillion hands.

So, based on this small sample size I get a probability of 1 in 6376. I'm still trying to find a mathematical solution, but no epiphany so far.

It's not whether you win or lose; it's whether or not you had a good bet.

September 28th, 2011 at 8:02:31 AM
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Heck, I hit 4 single-line royals in probably less than 1000 hands! (Different machines, but all inside of 4 days)

and I haven't hit one since : ( lol

and I haven't hit one since : ( lol

Gambling calls to me...like this ~> http://www.youtube.com/watch?v=4Nap37mNSmQ