[bonanova]

The ₅₂C₅ = 2 598 960 poker hands without jokers have

values inversely related to their frequency of occurrence:

SF = . . . .40 - straight flush

4K = . . . 624 - 4 of a kind

FH = . . 3 744 - full house

FL = . . 5 108 - flush

ST = . .10 200 - straight

3K = . .54 912 - 3 of a kind

2P = . 123 552 - 2 pair

1P = 1 098 240 - 1 pair

HC = 1 302 540 - high card

Jokers are wild cards: they can be counted as duplicating

any other card in the deck.

If a joker is added to the deck, does the order of

precedence of the hands change?

What if two jokers are added?

With a wild card in the deck, a 10^th hand: 5 of a kind,

becomes possible. Whatever its frequency, its presence

in the list does not affect the order of the other 9 hands.

Thus in answering the puzzle, ignore the relative

frequency of 5OAK, and consider only the relative

values of the original 9 hands.

[Guest]

Here's an interesting situation! A paradox infact!!

To see if the order changes or not, we add two possible scenarios:

1) when no joker is included in the 5 cards and

2) when one of the 5 cards is a joker

Now, when a joker is included and the other 4 cards are any of the 52 normal cards, in case of 2 pairs vs 3 of a kind, there is a conflict.

Both need the other 4 cards to be 1 pair and two any other cards. Now, if the joker be allocated as to make a 3K as it is "higher in order", the total number of ways of getting 3K become more than 2P even when both case 1 and 2 are added; 228 072 possibilities for 3K vs 123 552 for 2P.

This means that with a joker, you should then make a 2P instead of a 3K as now 2P is of a "higher order" than 3K!!

But if you make a 2P instead, 3K possibiities remain smaller than 2P, meaning that you should then make 3K and not 2P with the joker!

I guess this is the reason why jokers are not included in poker.

Any ways, if you still were to work further with 2 jokers, the only change is in positions of 3K and 2P. For looking at 2 jokers, we now have 3 cases

1) when no joker is included in the 5 cards

2) when 1 joker is included

3) when 2 jokers are included

[plainglazed]

...the number of possible combinations of high card only will not change with the introduction of wild cards but the total combinations obviously does. At some number of wild cards added, it will be more likely to have a pair then just a high card. My math's not strong enough to prove but it looks like a straight addition to the frequency of number of pairs with one wild card added makes it close to to the fixed frequency of high card only and adding two wild cards would make the pair more frequent than the high card only.

[plainglazed]

Had to learn some combinatorics, starting with what that word meant. As stated above, if you add wild cards, the number of possible hands that are high card only does not change - 1,302,540 possibles. If you add one joker, the number of possible hands that are pairs increases by this amount (₄C₁₃ - 41)(₄C₁⁴ - 4) = 169,848. That is to say the combinations of 4 different ranks out of the 13 minus the 41 combinations where, with the addition of the wild card, a straight is made and in any of the four suits minus the four flushes. So adding 169,848 to bn's natural pair combinations 1,098,240 gets 1,268,088 which is still less than the 1,302,540 high card only combinations. Now if you add a second joker, in effect you just double the 169,848 above since if both jokers occur in the hand, you have at least three of a kind (or the paradoxical two pair). This gives 1,437,936 which brings the frequency of having a pair greater than having high card only.

[bonanova]

Kudos to plainglazed for finding one-third of the answer.

Note to DeeGee: one always declares the higher ranking of two possible hands.

See the 2nd spoiler in this post. You would never call 4 Aces 2 pairs, for example.

Here are three clues to get us through the entire puzzle.

increases the number of hands to ₅₃C₅ [one joker] or ₅₄C₅ [two jokers].

But that does not matter for the purpose of this puzzle.

It's only the relative frequencies of the nine hands that are of concern, not their probability.

Leaves the frequencies of exactly two of the nine hands unchanged.

The frequencies of the other seven hands increase.

That allows the frequencies of precisely two of the nine hands to be

leapfrogged by their next higher ranked [originally less frequent] hands.

A moment's reflection discloses which two hands cannot be

created by the addition of a wild card.

Careful accounting of the effect of wild card also tells whether

one or two jokers are needed to increase the frequency of

a next-higher ranked hand, to achieve a switch in rank.

[see plainglazed's spoiler in the above post.]

Note that adding a wild card to 1 pair does not produce 2 pairs,

but three of a kind; although adding a wild card to 2 pairs changes

it into a full house. Likewise, adding a joker to 3 of a kind does

not produce a full house, but 4 of a kind. And so forth.

changes, in the case of one or two jokers [but which?]

the frequencies of two adjacently ranked hands [but which hands?]

in a way that they achieve precisely equal probability!

Bonus points for identifying the two hands involved in this rather remarkable result.

[Guest]

Note to DeeGee: one always declares the higher ranking of two possible hands.

See the 2nd spoiler in this post. You would never call 4 Aces 2 pairs, for example.

The point DeeGee makes is that either 2P or 3K can be the created less frequently by the joker and thus dynamically changes the definition of the "higher hand".

It actually is a paradox if one were to do this on the spot. For example, if one were to take the results of a joker in the deck with the original hand structure, they would notice 3K is more frequent, and thus a lower hand. This would cause them to recalculate the frequencies, using the joker to build 2P rather than 3K this time, as it is the better hand. The output of that would yield 3K as being less frequent again. Thus this process would continue infinitely, hence the paradox.

I think your OP intended for calculations to always choose the higher hand from the original structure.

Hand  	Without Joker  	        Joker in Hand  	        With Joker  	        2 Jokers in Hand  	        With 2 Jokers

SF 	40 			164 			204 			64 				432

4K 	624 			2,509 			3,133 			3,796 				9,438

FH 	3,744 			2,808 			6,552 			0 				9,360

FL 	5,108 			2,696 			7,804 			1080 				11,580

ST 	10,200 			10,332 			20,532 			192 				31,056

3K 	54,912 			82,368 			137,280 		16,968 				236,616

2P 	123,552 		0 			123,552 		0 			        123,552

1P 	1,098,240 		169,848 		1,268,088 		0 				1,437,936

HC 	1,302,540 		0 			1,302,540 		0 			        1,302,540

Total 	2,598,960 		270,725 		2,869,685 		22,100 				3,162,510

With one Joker in the deck, we only have 2P < 3K.

With two Jokers, FH < 4K, 2P < 3K, and 1P < HC.

[plainglazed]

Would never have thunk it wihtout bn's hint and peeking at the work of vinays84 but I think with two jokers added -

natural four of a kind = 624

three of a kind, joker, X = (

13

C

1

)(

4

C

3

) 2 (

12

C

1

)(

4

C

1

) = 4992

pair, 2 jokers, X = (

13

C

1

)(

4

C

2

) 1 (

12

C

1

)(

4

C

1

) = 3744

or 624 + 4992 + 3744 = 9360

natural full house = 3744

two pair and joker = (

13

C

2

)(

4

C

2

)(

4

C

2

) 2 = 5616

or 3744 + 5616 = 9360

- the frequency of four of a kind and a full house are the same.

Combinatorics are fun. Am including the following link for anyone like me a couple of days ago who'd like to know more about the math in the spoiler above. Thank you wiki.

[Guest]

I'm guessing that those are the two hands adjacent with equal probability to which bonanova referred.

For 2 Jokers, my calculations showed 9,438 not 9,360.

So where do the extra 78 hands come from?

There are two extra cases to consider when 2 Jokers are involved:

4K in hand, Joker = ₁₃C₁*2 = 26

3K in hand, 2 Jokers = ₁₃C₁*₄C₃ = 52

26 + 52 = 78

Both these cases have a 1 joker not needed (used for the kicker).

Edited January 22, 2010 by vinays84

[superprismatic]

DeeGee is correct no matter how many wild cards there are.

DeeGee is right. There is no way to rank the hands as the OP requires. This was proven

in the paper Inconsistencies of "wild-card" poker (Chance 9(No. 3, 1996):17-22)

where John Emert and Dale Umbach, the authors, state "There is no possible ranking of

hands in wild-card poker that is based solely on frequency of occurrence". This has

also been echoed in several publications of The Mathematical Association of America.

Although I didn't check DeeGee's numbers, his argument is basically Emert & Umbach's:

With wild cards, there always exist types of hands A and B which could both be made from

the same dealt hand containing wild card(s) such that, if A is made to be of higher rank

than B, using the wild card(s) to always make A will make it easier to make than B;

But switching the ranks of A and B (making B of higher rank) will now turn the tables

where the wild card(s) will always be used to make B, thereby making it easier to make

B than A. So, just using the frequencies of occurrence to rank hands (assuming, of

course, that the higher rank hand will always be made), cannot be done. I wouldn't call

this a paradox because it's just the effect the ability to make two different hands from

the same dealt hand has on the situation. I'd prefer to call it surprising result.