Roulette Probabilities for multiple spins

For a variety of reasons that are unimportant, I am doing additional math for my Poker For Roulette idea, to be based upon a four spin outcome. Because I suspect this question may create a debate and/or tangents, I am creating this new thread, rather than add it to my Poker For Roulette thread.


One of the results I'm seeing has me stumped. I'm half expecting my head to explode before the planet does in about 8 hours....

In my pea brain, the number of two pair results should equal the number of three of a kind results.

FYI: 0 and 00 are wild, so a four spin result that include a pair as well as a pair of wilds, is actually a wild four of a kind. Similarly, a three of a kind that has one number and three wilds or trips and one wild, is also a wild four of a kind.

Therefore, I'm thinking that the math for 2 pair is this: First spin is any of 36 numbers. Second spin is any of the remaining 35 numbers, Third spin is either of those numbers. Fourth spin is the same as the the third for trips, or the other number for two pair. I.E. 36 * 35 * 2 * 1 = 2,520. Then multiply by some factor to account for the various sequence combinations of those four spins.

Are you with me so far?

Here's my problems:

A - I don't know how to figure out the combination factor I mentioned. I.E. The "simple" math for the eludes me.

B - I created a program to cycle thru all 2 million four spin combinations. The results were 3,780 cases of two pair and 5,040 cases of three of a kind. I've been beating myself up for a couple days trying to find an error, but I'm coming up empty. If it's right, it suggests a combination factor of 1.5 and 2, respectively. How can that be?

Quite frankly, I originally expected more two pairs than three of a kinds, similar to the results I got for the 5 spin results. Now, when thinking about the 4 spin math, it seems to me that they should be an equal number. The results I got make no sense to me. And it's also making me question the math I had done for the 5 spin results.

Note: Typical Combin functions don't apply. Unlike real poker hands, there is no card removal issues to deal with. It's as if each card were drawn from independent decks.


So, can anyone explain these two pair and three of a kind calculations and results to me?

Thanks.

I'd break it down by number of wilds, rather than trying to first break it down by hand type. E.g. 4 wilds is always quads. Three wilds is also always quads. Two wilds is either trips or quads or a straight (if you do that) depending on what the numbers are. One wild is either pair, trips, or straight. So right there, you can't have two pair with a wild because if you have A/A/B/Wild, you have three of a kind instead of two pair.

I already did the wild breakdown, and eliminated the wild hands. Note, that the wilds are only partially wild, so there isn't much to eliminate.

Wilds are only wild for making quads or straights. I.E.:

A pair, with one wild, and some other number, is still just a pair.
A pair with any two wilds is quads.
Two numbers with matching wilds, is just a pair.
Two numbers with non-matching numbers is nothing.
One number with any three wilds is quads.
Trips, with a wild is quads.

I adjusted my program so there are NO wilds.

My pea-brained math says to now use the formula: 38 * 37 * 2 * 1 = 2,812 * some-combination-factor.

The results are 4,218 two pair and 5,624 trips - the same 1.5 and 2 multiplier factor as I got before.

I am not at home right now, but can work on it when I do in about 20 hours if you haven't figured it out by then

Quote: DJTeddyBear

Therefore, I'm thinking that the math for 2 pair is this: First spin is any of 36 numbers. Second spin is any of the remaining 35 numbers, Third spin is either of those numbers. Fourth spin is the same as the the third for trips, or the other number for two pair. I.E. 36 * 35 * 2 * 1 = 2,520. Then multiply by some factor to account for the various sequence combinations of those four spins.

Two pairs:
Ok let's count the possibilities for the pattern AABB.
It is 36 * 1 * 35 * 1.
You have 3 of those patters, AABB, ABAB, BAAB
Hence you get 3 * 36 * 35 = 3780

Three of a kind:
Pattern AAAB
36*1*1*35
You have 4 of those patterns, AAAB, AABA, ABAA, BAAA.
4*36*35 = 5040

MangoJ -

Thanks! It's so friggin' simple the way you mapped it out. And thanks for preventing my own head from exploding almost last minute before the planet explodes... (It's 5:30 here in NJ).


And now that you showed me how to do it, I recall doing similar illustrations when I did the math for the five spin system.

Good luck then. 5 spins are easy, but the wilds are not.
You should separate into the number of wilds (0-5) individually, and then sum those up in the end.

Suppose you like three of a kinds with one single wild (x):
Then a representative pattern is xAABC. This will be 2*36*1*35*34 for that specific pattern.
The number of patterns you should use the multinominal "5 choose 1,2,1,1" = 5! / (1! 2! 1! 1!) = 60
where n! is the factorial 1*2*3*...*(n-1)*n

Thanks for the additional advice.

For the record, I originally did the 5 spin stuff as you described, but the wilds got my head all screwed up, so I wrote a program to cycle thru all 79 million unique outcomes. Then I did the same for the 2 million four spin outcomes. Because I had come up with a mental picture that two pair and trips should be the same number, I REALLY started to doubt all the math I had done previously.

You put it all back into perspective for me, and showed me how to figure it out. I'm once again confident about my program. Thanks.