Hello,
Could someone calculate the probability of seeing 8 royal flushes in a cycle of 5 play NSUD over about 24 hours of play. I am figuring about 500 hands per hour x 5 (2500 hands per hour)
at 24 hours totaling 60,000 hands.
Thanks
Stats Question
RXLowrys wrote:
Could someone calculate the probability of seeing 8 royal flushes in
a cycle of 5 play NSUD over about 24 hours of play. I am figuring
about 500 hands per hour x 5 (2500 hands per hour)
at 24 hours totaling 60,000 hands.
You've posed 2 forms of the question on two vp forums: vpFREE and
acvpp. One is more approachable than the other. I suspect the vpFREE
posting represents the session that you experienced, you write:
"Could someone calculate the probability of seeing 8 royal flushes in
a cycle of 5 play NSUD over about 24 hours of play. I am figuring
about 500 hands per hour x 5 (2500 hands per hour)
at 24 hours totaling 60,000 hands."
This is a question that I'm not prepared to address for reasons that
I'll get to shortly. Separately, you pose a simplified version of
this question on acvpp that is much more approachable:
"Could someone calculate the statistical probability of getting 8
royal flushes in approx. 60,000 hands of NSUD?"
The answer to this latter question is .033%.
To be honest, I find the question of the probability of getting 8, or
more, RF's here more interesting (since all these cases represent
exceptional cases). More on that later. What follows is an
explanation of how this answer was calculated (along with a fair
amount of explanatory detail for those not reasonably versed in
statistics). Indulge according to inclination ...
- H.
ยทยทยท
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I don't want to overwhelm you (or anyone else) with statistical
jargon, but the question bears some explanation. In either case,
probabilities associated with a normal distribution of an event can be
applied to the problem. I remark on this because it's frequently
noted that video poker results aren't normally distributed, i.e.
doesn't conform to the familiar "bell shaped curve" (although over the
long-term they do approach a normal distribution).
But this is because each of the various hand possibilities have their
own individual probabilities and distinct payouts. If a question
deals with the probability of a sessions profit or loss exceeding "x",
then the probability of the possible outcomes is skewed by these
differences between hands and the distribution isn't normal.
However, when it comes to the probabilities of any single hand type
(e.g. RF), the situation is fairly simple. The probability is clearly
defined for the outcome of any hand and, because successive hands are
independent of each other (unrelated), the probabilities over a series
of hands (or session) can be readily calculated. Furthermore, the
probabilities involved conform to a normal distribution. I'll note
that the shape of that distribution (degree of flatness) can be
defined in terms of the variance statistic. But we needn't concern
ourselves with that in performing the calculations in this case.
But when you pose a question involving multiplay, things get tricky.
This is because not all hands are independent of each other. Within
any five hands of a given play, each hand is related to another by
virtue of the common cards on the deal. This introduces a second
variable into the hand distribution, referred to as covariance. As it
bears out in this problem, you'll note that if you've hit a RF on one
hand of a play that it's more likely that you'll hit another RF on
that play. As you can see, this introduces a twist to the
consideration of 5 successive hands in single play. To be honest, I'm
not familiar with how to incorporate this second element into this
question, so I'm sticking with the simpler case.
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Ok, I might have handled that last section of my reply by saying,
"Duh, don't know about multiplay here". Hopefully you benefited from
the longer explanation 
The simpler question can be expressed in generic form as the
probability of r successes in n trials, where r is the occurrence of a
RF and n is the number of hands. (I didn't pick "r" and "n" out of
the air -- I'm resorting to a probability textbook; 25 years has
dimmed the stat I picked up in school 
There's an equation to solve this type of problem (which is referred
to as a binomial probability question, the "binomial" referring to the
fact that the problem involves a success or failure of hitting a RF on
each hand).
P(r) = C(n,r) * p^r * ((1-p)^(n-r))
The small p is the probability of hitting a RF on any hand. The C
refers to the number of ways in which we can have r RF's occur over
the course of n hands (combinations).
The formula for C is:
. . . . . n!
C(n,r) = --------
. . . . . r! (n-r)!
(those leading dots are to keep Yahoo! from messing with alignment)
Just to be thorough, n! = the product of all numbers less than or
equal to n. (e.g. 4! = 1*2*3*4 = 24). The prospect of calculating
60000! (n! in this case) may seem daunting, but with a little basic
division arithmetic (cancellation) you'll see that this formula is
readily solved in many cases.
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Let's cut to the chase:
n = 60,000
r = 8
p = 1/43456.3 (the probability of a NSUD RF)
C(60000,8) = 60000! / (8! * 59992!)
. . . . . .= (60000*59999*...*59993)/(8*7*...*1)
Remember that bit about "cancellation"? ... there's a bit more to go
here, but we're talking a big number, as you might expect. Head to a
calculator and you get:
P(r) = C(n,r) * p^r * ((1-p)^(n-r))
. . .= .033%
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Let's return to the multiplay question. I don't have an answer for
you, but I expect that the solution to your question would be a tad
higher than this value.
More than likely the probability of hitting only one RF would be
smaller in the multiplay case than the single play given that if you
hit at least one RF on any given play, the probability of at least one
additional RF on the remaining hands would be higher than if the 4
hands weren't related to the first by the original deal, right?
But hey, I'm on really loose ground here ... I'll leave this question
to someone else.
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As I've indicated, I'd be more keenly interested in the probability of
hitting 8 or more RF's in your session (again, treating it as single
play rather than multiplay). I'm not going to solve this, but you've
got the tools now if you're of a mind to do so yourself ...
That probability will equal 1 - p(hitting < 8 RF's). (The two
probabilities have to add to 1, right?)
p (< 8 RF's) = p(0 RF) + p(1 RF) + p(2 RF) + ... + p(7 RF).
Each of these can be solved in the same manner that we solved p(8 RF)
above. Enjoy
- Harry
Harry Porter wrote:
"Could someone calculate the statistical probability of getting 8
royal flushes in approx. 60,000 hands of NSUD?"The answer to this latter question is .033%.
This response, roughly 3 out of 10,000, troubled me as likely being a
little large. Turns out I slipped in placing a parenthesis when
calculating the value. The correct value, FWIW, is .0082%.
- H.