Please forgive me if question is too complex.

So we have 6 ordinary decks(52 cards each) and 10 jokers. How do you calculate the probability for following 4 card poker hands?

If I am right the denumerator is COMBIN(322;4)=439633040

So I am looking the numerator for each hand. For example if we count how many four of kind hands with out jokers, I think that is 13*COMBIN(20;4)=62985

But its the jokers that puzzle me.

How you calculate how many hands when joker is wild (it can be any card)?

four of a kind
four card straight flush
four card flush
four card straight
three of a kind


Hopefully some one is able to count this.

13*12*C(24,4)*24 = 39,783,744

13*12 possibilities for the ranks (counting order so as to distinguish between which is the quads and which is the extra card).
C(24,4) ways to pick 4 from a given rank, and 24 ways to pick from the other rank.

With jokers that becomes:
C(13,2) * 24 * { [C(34,4) - C(10,4)] + [C(34,4) - C(10,4) - 24*C(10,3)] }
= C(13,2) * 24 * {2*[C(34,4) - C(10,4)] - 24*C(10,3)} = 167,454,144

C(34,4) ways to draw 4 from either the quads rank or the jokers. But we don't want 4 jokers because that would be 5 of a kind, so we subtract C(10,4) which is the #ways to pick 4 jokers.

But notice that if there are 3 jokers, then only the higher rank can be the quads. That's why I have C(13,2) instead of 13*12 (think about this until it makes sense). In my unsimplified line, the first set of [ ] includes the case of 3 jokers. The second one does not, ie it subtracts the #ways to get 3 jokers which is 24*C(10,3). So we have C(13,2)*6*[quads including 3 jokers] + C(13,2)*24*[quads w/o 3 jokers]. Just as there are C(13,2) ways to choose 2 ranks and let the higher rank be quads, there are C(13,2) ways for the lower rank to be the quads, so that's the 2nd thing being added (but it requires there to be fewer than 3 jokers, otherwise the higher rank would be the quads).

Finally, I just factored out the C(13,2)*24 and added the brackets together which results in my 2nd line.

As for the other hands, I'll get to them another time or someone else can take a stab.

EDIT #2 -- oh you said 4-card hands. Ok I'll leave this up now that I wrote it all. In a separate post I'll answer your actual question.

Only four cards total in hand.

Four-card with no jokers in the deck: 13*C(24,4) = 138,138

Four card with jokers: 13*C(34,4) - 12 = 602,876

We subtract 12 because 13*C(34,4) counted the 4-joker case 13 times.

Yes, of course. I understand this now. Thank you.

It would be great if you or someone had time to calculate the others too.

Shouldn't you subtract 12*C(10,4)? which gives 600368

I summed the cases AAAA, AAA*, AA**, A***, **** to 600368 : 13*Sum(i=0 to 3){C(24,4-i)*C(10,i)} +C(10,4)

For the other hands Id like to clarify that if you get 5 of spades, 6 of spades, joker and joker, that counts as straight flush not straight. If you get 2 of spades, 2 of spades, 2 of spades and joker, that counts as four of a kind not flush.

Rank of hands

Quads
Four card straight flush
Four card flush
Four card straight
Three of kind

So even if the four card flush was easier to get, we will put it above the straight.

Four card straight:

No jokers =11*(24*24*24*24-24)=3649272

One joker in hand =34*(24*24*24-24)*10=4692000

Two jokers in hand ?


Am I even close here?

Doh, yes. Because there are 10 jokers not 4.

@IceKing -- your "no jokers" straight looks correct. I'll check your one-joker math tomorrow.

11(24^4 - 4*6^4) = 3,592,512

10*34*(24^3 - 4*6^3) = 4,406,400

34 is the ways to make a straight using a joker.

One reason: 34 = 10*3 + 4
Broadway can happen 4 ways. The other 10 straights can only happen 3 ways because e.g. 567+joker can only count as the 8-high, not the 7-high.

Another reason: 34 = 11*2 + 12
There are two gutshot patterns with 11 ways to occur, and the three-card connector can occur 12 ways.

Not sure if it's a coincidence that 34 is a fibonacci number. The fibonacci sequence is related to other string/streak type counting problems. Maybe this problem is somehow analogous to one of those.

So how about the original question. 6 decks, 10 jokers, how many four card straights?

I think we are missing four card straight with two jokers, and then we will have our answer. Am I right?

The non-jokers can be connected, one-gapped or two-gapped.
There are 13 connectors + 12 one-gappers + 11 two-gappers. = 36 ways

36*100*(24^2 - 4*6^2) = 1,555,200

+ 3592512 + 4406400 = 9554112 four-card straights.

I just noticed the quads figure is still wrong because 3 or 4 jokers results in a straight flush, not quads.

13*[C(34,4) - C(10,4) - 24*C(10,3)] = 562,718

Edit: nvm, you ranked quads the highest. Disregard. Kapw's number is correct

Edit #2:
The number of 0-joker straight-flushes is 11*4*6^4 = 57024
The number of 0-joker quads is 13*C(24,4) = 138138

So the SF's should be ranked higher. (Plus optionally, you could make a new hand called quad-flush which would only have 13*C(6,4) = 195 combos without jokers.)

I was thinking to put on top of the list

4 jokers =C(10,4)=210
Perfect Quads - that is 4 exactly same cards =13*C(6,4)=195
4 Aces =C(34,4)-C(10,4)-4*C(6;4)=46106

Is this right for the four card straight flush, if you count that three joker hand is quads?

11(4*6^4) = 57024
10*34*(4*6^3) = 293760
36*100*(4*6^2) = 518400

518400+293760+57024=869184

Fmp.

Fyp

Thank you sir. You have helped me a lot.

How about the flush and trips?

Maybe Flush like this?

No jokers = C(13,4)*(4*6^4) - 11*(4*6^4)
+
One joker in hand = 10*C(13,3)*(4*6^3) - 10*34*(4*6^3)
+
Two jokers in hand = C(13,2)*C(10,2)*(4*6^2) - 36*C(10;2)*(4*6^2)

=13371696

Don't quote me on this just yet, but I did it as follows.

N(suited quads) = C(10,4) + C(10,3)*52*6 + C(10,2)*13*4*C(6,2) = 72750

N(flush) = 4*C(88,4) - N(straight flush) - N(suited quads) = 4*C(88,4) - 584064 - 72750 = 8730746

The ranks don't have to be different because flush beats trips and two-pair. So it's not C(13,4)*... but instead simply C(78,4). And you forgot to subtract the 195 perfect quads. There are 5648581 jokerless flushes. Haven't checked your other lines. Your math would have been elite if right though (cwidt?).

Yes of course, that makes sense. 78 spades + 10 jokers choose 4. Too obvious.

I started to think the perfect quads. Maybe 13*4*C(6,4)=780 is the right number?

Does 4*C(88,4) count 4 jokers cases four times?

Maybe flush should be
N(flush) = 4*C(88,4) - Four Jokers - Perfect Quads - Suited Quads - Straight Flush

= 4*C(88,4) - 4*C(10,4) - 13*4*C(6,4) - (C(10,3)*52*6 + C(10,2)*13*4*C(6,2)) - 584064

= 9327560 - 840 - 780 - 72540 - 584064 = 8669336

Yes, good catch. I subtracted the 4 jokers case as part of the suited quads, but only once. It needs to be subtracted four times (which means, another 3 times separately after having subtracted suited quads).

Four Jokers and Perfect Quads are subsets of Suited Quads. However, my Suited Quads calculation was incomplete because it didn't include the one-joker cases or perfect quads. Plus, like you said, my Perfect Quads count needs to me multiplied by 4, so instead of 195 it's 780.

N(suited quads) = 72750 + 780 + 13*4*10*C(6,3)

N(flush) = 4*C(88,4) - above - 584064 - 3*C(10,4)

But I badly need sleep so this whole post could be nonsense. GN

So flush is

=(4*C(88,4)-(72750+780+13*4*10*C(6,3))-584064-3*C(10,4))=8658936

Correct?

And correct? results:

Six decks, ten jokers, four card combinations

Four Jokers
=C(10,4)=210

Perfect Quads
=13*4*C(6,4)=780

Four Aces
=C(34,4)-210-780=45386

Four of A Kind
=13*C(34,4)-840-780-45386=553992

Straight Flush
=11*(4*6^4)+10*34*(4*6^3)+36*C(10,2)*(4*6^2)=584064

Flush
=(4*C(88,4)-(72750+780+13*4*10*C(6,3))-584064-3*C(10,4))=8658936

Straight
=36*C(10,2)*(24^2-4*6^2)+11*(24^4-4*6^4)+10*34*(24^3-4*6^3) =8698752

Three Of A Kind
=?

All combinations
=C(322,4)=439633040

Can anyone confirm the numbers above; heehaww?

How about the three of a kind?