Last week I wrote about completing a royal flush cycle (one or more royals in each suit) within one trip. I was very lucky, to be sure, but just how lucky?
I have a hard time calculating how lucky something was after the fact. You can massage the numbers and come up with all sorts of probabilities. There is not one absolutely correct answer that everybody can agree on.
Let’s start with the 11-day trip. I used all of it, hitting the club royal on my last day there. Most players never have an 11-day out-of-town casino trip in their entire career! I’ve had 14-day trips to Cherokee. I would have been just as delighted to complete the cycle on one of those trips. But I never did.
It’s not the length of the trip, of course, that determines your chances, but rather the number of hands played. Near as I can tell, I played about 62,000 video poker hands. I know how many points I earned during the trip, but some of my play was on slots. I record my points earned each day, but not how that number of points is broken down between slots and video poker.
I know I’m going to be using the word “cycle” in a different way now, but a “royal cycle” is the number of hands, on average, to hit a royal flush in a particular game. In Jacks or Better the number is right around 40,000, and that’s the commonly used number for video poker royal cycles. But NSU has a longer cycle, 43,456 hands, because you play hands differently in this game.
Calling 62,000 hands 1.4 royal cycles is as close as I can get. The 43,456 number is fairly precise (I could tell you that number is 43,456.27, which would be more precise, but hardly more useful), while the 62,000 number is an educated guesstimate. Using the Binomial Theorem, connecting on exactly four royals in 1.4 cycles happens about 4% of the time, and connecting on four or more royals in 1.4 cycles happens about 5.6% of the time.
If I connect on exactly four royals, all four suits will be present only about 9% of the time. Were I fortunate enough to have connected on five-or-more royals (I wish!), it would have been easier to have all four suits represented. Not a lock, of course, but easier.
Now what do we do with the club royal being dealt? The “dealt-ness” of that royal was overkill. I would have also completed the cycle even if I had needed to draw one or more cards to get the club royal.
But the royal being dealt allows me to jack up my numbers when I tell people how rare this was. (It’s not something I normally do, but I’m discussing it here because there are always “How rare was it anyway?” questions.)
The dealt royal arrives approximately every 650,000 hands. But since at the time it hit, I needed the royal in clubs to complete the cycle, those only come around one-quarter as often — or about every 2,600,000 hands.
All these things had to happen on the same trip — namely playing 62,000 hands, collecting at least four royals, having every suit being accounted for in those four or more royals, and (this one is optional), one of these had to be dealt. To determine how likely all of this is, you need to multiply all of these probabilities together. I’ll let others do it, because I’m not at all convinced that figuring out how likely something was to happen — after you know it did happen — is a meaningful exercise at all.
There’s more on this trip. The deuces cycle in this game is about 5,356 hands. In 62,000 hands you have 11.6 of these cycles. I collected 12 sets of deuces — which is essentially spot on given the imprecision of the 62,000 number. The thing is, one of those sets of deuces was dealt.
Being dealt a specific quad happens every 54,167 hands, on average — so in 62,000 hands “it figures” I would have collected one or more. Mathematically, even though 62,000 is larger than 54,167, I was still a slight underdog to hit exactly one set of dealt deuces on the trip, although I was a sizeable favorite to collect one or more.
What this has to do with anything is that on the same trip I was dealt a royal and dealt deuces! (I was also dealt four aces with a deuce, which is another rare event that is called a 5-of-a-kind for $400 in $5 NSU Deuces Wild and it’s the kind of hand that makes you wish you were playing a different game!) Being dealt a royal is rare enough. But also being dealt deuces is even more rare!
It was, to be sure, a trip to remember!
Sometimes casinos restrict players who have too much success. I’m hoping that’s not the case here.
Have you ever had a big bonus
for a consecutive royal flush and
what did it pay
Not an AP myself but would like to hit $1 million coin on VP in a year to see the kinds of comps I would get at my local.
Bob does that and more on his 11 day trip.
An incredible difference in leagues. Welp, if I am ever of a financial status to afford that, it would pretty cool.
Better quit now. A brutal losing streaks always come right after the luckiest ones. I’ve told you before, and you’re heading straight into one. Don’t get caught.
Using your logic, Mike, since I have been losing consistently, I should play more as I am due to win?
I’m deserving as well, on a very long dry spell.
Mike, what is the basis for this statement? Losing streak and winning streak are not rigidly defined terms.
So, you believe that if you ahead money for a week, for example, you should not play the next week? Somehow, winning money the previous week will make you more likely to lose money the next week? If you don’t believe that each deal is independent and not based on previous hands, your statement is silly. If you do believe that past results affect future results, the games are unfair, not legal and you should never play again.
You are silly. The losing streak is due. Your positive variance will be evened out.
Good argument, Mike. You’ve made you case well………
The probability can be calculated but it’s a bit tricky. It requires De Morgan’s law and the inclusion-exclusion law. Here are the details. Let E_i be the event of no royals of suit i in n trials. We want Pr(not E_1 and not E_2 and not E_3 and not E_4). By De Morgan, this is 1 – Pr(E_1 or E_2 or E_3 or E_4), and by inclusion-exclusion, we get 1 – (4Pr(E_1) – 6Pr(E_1 and E_2) + 4 Pr(E_1 and E_2 and E_3) – P(E_1 and E_2 and E_3 and E_4)). Now let p be the probability of getting a royal of suit i. Our formula can now be written 1 – 4(1-p)^n + 6(1-2p)^n – 4(1-3p)^n + (1-4p)^n. With n=62000 and p=(1/4)(1/43456) as in the article, the formula reduces numerically to 0.00810012, or one chance is 123.455. Unlikely, but not ridiculously so.
Thanks to you and De Morgan I now completely understand how Bob gets so many Royals.
It probably won’t, but Stewart’s post should shut everybody up…..please, it’s a joke.
Stewart: My regards to De Morgan! While I’m surprised that the number you came up with was so low, I don’t usually argue with math professors about math problems, and I’m not going to start now!
My problem with any such number, though, is we (actually you) came up with it after the event was already over and we knew the outcome.
The only numbers you used were 62,000 hands, four royals of different suits, and a royal cycle of 43,456. Next time any of us complete this type of royal cycle, it will be with a different number of hands and likely on a different game.
We could have included any number of specific details to make the number higher. Such as:
1 The order of the royals was diamonds, spades, hearts, and clubs
2 One of the royals was dealt
3 This happened on the same bank of machines during one extended visit.
4 This was accompanied by certain additional interesting dealt hands
Any or all of these “extras” would have bumped the odds against this — and who’s to say exactly which ones shouldn’t be included?
Still, it was far more refreshing to think about what you wrote than mike’s “Quite while you’re ahead!” nonsense.
Hi Bob. I don’t blame you for being skeptical. Because of the alternating signs, the argument lacks good intuition. An alternative approach occurred to me, which may be more in line with your thinking. Let p be the probability of a royal (of any suit). The number of royals in n trials has the binomial(n,p) distribution. With n large and p small, this is well approximated by the Poisson(L) distribution with parameter L = np. The probability of k royals is then approximately f(k) = L^k e^(-L)/k! The next ingredient is, given k royals (4 or more), what is the probability that there is at least one royal of each suit, call it g(k). There is a formula for this, g(k) = 1 – (3/4)^k + 6(1/2)^k – 4(1/4)^k. For example, g(4) = 0 09375 and g(5) = 0.234375. To get the final result, we must sum the terms f(k)g(k) from 4 to infinity (4 to 100 suffices). With n = 62000 and p = 1/43456, we get 0.00810058, or one chance in 123.488. The result is similar to before, but it is easier to understand.