Quick probability question

[tommiedee]

here is my request, what are the odds of getting dealt 4 aces and the kicker?  I was playing at Harrahs AC last weekend and was dealt it on a 3 line Super Play machine where you play an extra dollar on each line for max bet of $18.  DDB is the game and it pays same as RF 4000 credits.  I hit them for $12000.  I sent a pic to publisher hopefully it gets in newsletter.  

[olds442jetaway]

Thanks again guys. The beer is on you know who if you get out this way.

[Minn. Fatz]

here is my request, what are the odds of getting dealt 4 aces and the kicker?

That would be 12 possible hands out of 2,598,960 or 1 in 216,580.

[joechgo]

[QUOTE=Minn. Fatz] Not to step on Vman or anyone else's toes here but...

It depends on what kind of strategy you're playing. Assuming a near-optimal strategy, the wonderful Wizard of Odds gives a 76 percent probability of no better than one pair on a given hand of 9/6 JB. So your probability of not getting 2P or better over any 36 consecutive hands would be:

.76 ^ 36 = .0000512

or one in 19,530 trials. Somewhat worse odds than that kid winning the British Open by age 25 then...

Haha, yeah I saw this but hadn't got around to it. Mainly because with the volume olds plays, the likelihood of seeing it at least once is nearly a certainty and I wanted to do the "streak calculation".

Probability of not getting 2 pair or better for 36 or more consecutive hands at least once playing the following amount of hands of 9/6 JoB:

36 hands: 0.0000512
1,000 hands: 0.011842
10,000 hands: 0.11543
50,000 hands: 0.45926
100,000 hands: 0.7077146
250,000 hands: 0.9538403
500,000 hands: 0.9978701
1,000,000 hands: 0.9999955 [/QUOTE]Hey VMan - can you explain the math used for these streak calculations?

[Vman96]

Hey VMan - can you explain the math used for these streak calculations?

Sorry I was in Tunica for the weekend.

It's a recursive formula.


It's a extension of this example from the Wizard of Odds:

Look for this question in the thread below:

"If a coin is flipped 100 times, what is the probability of getting a streak of at least 7 heads in a row at least once?"

But instead of 8 terms in the example question, there will be 37. And instead of using a 50% probability you'll be using 0.76002 and 0.23998 instead.

http://wizardofodds.com/ask-the-wizard/ ... ity/coins/

In our case the formula looks like this

... means continuing the pattern

f(nth hand) = f(n-1)*0.23998 + f(n-2)*0.23998*0.76002 + f(n-3)*0.23998*0.76002^2 + f(n-4)*0.23998*0.76002^3 + ... + f(n-35)*0.23998*0.76002^34 + f(n-36)*0.23998*0.76002^35 + 0.76002^36

And with this formula, you can throw it into a spreadsheet. Starting with 0 hands and force the number of hands less than your query (36 in our example) to have a probability of 0 since we need a minimum of 36 hands for the event to be possible.

Although I was lazier initially and used this calculator instead. But writing this up, I used the formula for 1000 hands above in a spreadsheet and got the same answer for 1000 hands: 0.0118422223


http://www.pulcinientertainment.com/inf ... enter.html