What is the probability of....?

[olds442jetaway]

My sessions are running in the 5,000 to 5,500 hands played. That said, long term, in airport deuces with a 98.91 return, I should hit a Royal long term about every 8 sessions. Assuming I play to make it easier in the calculation, the exact number of hands every session where I should hit quad deuces by the end of that session, what is the probability that by the end of 8 sessions played, I will NOT have hit quad deuces? I don’t think that has happened to me yet, but I have come close. Thanks in advance for a reply.

[olds442jetaway]

PS.....I know I should know the formula and I remember it being a fairly easy one, I am just drawing a mental blank!

[olds442jetaway]

By chance is it .0039. If someone could see if I did the calculation correctly, thar would be great.

[New2vp]

By chance is it .0039. If someone could see if I did the calculation correctly, thar would be great.

I think you are off by an order of magnitude.

To give a more general answer, please let me restate your question as:

What is the probability of getting zero successes in 8 cycles (in single hand video poker)?

If the outcome is rare, as it is in your example, this can be fairly closely estimated really quickly as EXP(-8), using Excel's notation, which equals 0.00033546 to eight places, which translates to only about 1 time out of 2981.

I really don't need to know anything else about the problem to get this number. If you have a different number of cycles and a different rare hand, like getting no royals in 3 cycles, just place the 3 in the expression instead of the 8, and you will get 0.04978707.

You can get more exact figures by knowing the length of the cycle, but I doubt that this precision really helps much in any analysis that you might be interested in. For the game in question, the Wizard of Odds website shows what it calls combinations for quad deuces as n = 3,727,422,492 and a total of t = 19,933,230,517,200. So, the probability of hitting deuces while following optimal strategy is n/t = 0.000186995 and the cycle length is t/n = 5347.72502. Eight cycles rounds to 42,782 hands.

So, a more exact calculation for the probability in question in Excel is BINOMDIST(0,42782,n/t,0) = 0.00033520 to eight places, which is closer to only 1 time in 2983. You can see that the earlier quick estimate is accurate to six places, which should be close enough for many concerns.

Bob Dancer's current week's column talks about applications of the binomial distribution, so you might get additional explanation there. Calculations using Euler's number e = 2.71828... might be a little harder to understand, but it's really quick to calculate if you have access to Excel or calculators that have the e^() or e^x button.

[olds442jetaway]

Thanks so much for the work and answer. That helps to explain why I haven’t missed in eight straight sessions so far. I owe you a beer if you get up to Ct and Mohegan Sun

[BobDancer]

my blog this week found at gamblingwithanedge.com covers how to use the binomial theorem to analyze this type of problem

[Tedlark]

my blog this week found at gamblingwithanedge.com covers how to use the binomial theorem to analyze this type of problem

But can it core a apple?

Kidding, kidding, should be interesting listening.

[BobDancer]

Kidding, kidding, should be interesting listening.

Actually, it is a written piece so reading will probably be the better input device than listening.

[Tedlark]

Should be an interesting read then.

[advantage playe]

OH IT CAN CORE A APPLE !! thanks ralph cramdon !!