8

By standard poker hands I mean:

  • high card
  • one pair
  • two pair
  • three of a kind
  • straight
  • flush
  • full house
  • four of a kind
  • straight flush
  • EDIT: five of a kind

I'm curious if the ranking of hands should change with the presence of a wild card added as a 53rd card.


As was pointed out in an answer, a full wild card adds another possible hand. I am specifically looking for the chances of making any of the above hands in particular so that the hands can be arranged in the order of most likely to occur.

  • As i've said below, i don't believe the odds on any of these hands would change significantly enough compared to each other to be re-ordered. Your current list shows them from most to least likeliest. – Lewis Goddard Jan 10 '12 at 20:03
  • @LewisGoddard I expect that to be the case, however, a % chance of making the hand would give all the evidence necessary. – Nick Larsen Jan 10 '12 at 20:06
5

I'm no mathematician, but I think the joker switches the rankings of full houses and flushes.

With 53 cards, you have c(53,5) or 2,869,685 possible poker hands. I'm using this site http://en.wikipedia.org/wiki/List_of_poker_hands as a reference for normal deck hand numbers.

5 of a kind

This has to be exactly 4 cards of the same rank plus the joker. So that means there are exactly 13 five of a kind hands. 13/2,869,685 = % 0.000502

Straight flush

You would have all of the normal sized deck straight flushes, 40, plus all of those with one card replaced by the joker, 40 * c(5,1) = 200. 240/2,869,685 = % 0.000923

4 of a kind

For this one you use all the regular deck four of a kinds, 624, plus all the joker four of a kinds. One card is the joker then three of a kind, 13 * c(4,3), then the 5th card can't match the three of a kind so there are 53 - 5 = 48 cards left. 624 + 13*c(4,3)*48 = 3120/2,869,685 = % 0.12

Full house

Normal deck full houses, 3744, plus jokered full houses, c(4,2)*c(4,2)*13*12. 3744 + c(4,2)*c(4,2)*13*12 = 9360/2,869,685 = % 0.36

Flush

Normal deck 5 to a flush, c(13,5)*4, plus four to a flush c(13,4)*4, minus all straight flushes. c(13,5)*4 + c(13,4)*4 - 240 = 7768/2,869,685 = % 0.29

Straight

A normal sized deck has 10,200 straights so that will still be more common than flushes and full houses.

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  • Full house is too big. Some of the normal full house get bumped to quads. – paparazzo Apr 19 '17 at 18:34
1

Scenario 1:

If the wild card is played as itself (eg 9 of Diamonds, etc), it would effect the probabilities very little.

Say we take high card:

You have a K.
Your "natural" chances are 44/51 for a win, 3/51 for a draw, and 4/51 for a lose.
If the wild card is higher than a K, then you are left with odds of 44, 3, and 5 /52 respectively.
If it is a K then the odds are odds of 44, 4, and 4 /52.
If it is lower than a K then you are left with 45, 3, 4 /52.

If you know what the wild card is then you can figure it out, if not then in this case it is more likely lower, and i doubt it would effect hand rankings very much, except highest hand becomes five of a kind.

However...

Scenario 2:

The wild card is played as a "bug", where the card is (usually a joker) treated as an ace, unless it completes a five-card set.

As wikipedia says:

Under this rule, a hand such as K-K-Joker-5-2 is just a pair of kings (with an ace kicker), but any four same-suit cards with a bug make a flush, and a hand such as 7-Joker-5-4-3 makes a straight.

This would increase the chances of having an ace (obviously), and completely any five-card hand.

Scenario 3:

The card is "fully wild", and is defined as "a card that is fully wild can be designated by its holder as any card they choose with no restrictions."

Again, this adds an extra, highest hand, five-of-a-kind. I should only space out the odds of the other hands, not prompt any reorganisation.

Summary:

Apart from "five-of-a-kind", hand should not reorganise their rankings, either through rules of odd

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  • 1
    For every hand listed in the question, there is an objective, numerical answer. It seems clear that is what is being asked for. This does not answer the question. – Michael McGowan Jan 11 '12 at 15:44
0

A paradox arises with the probability of pairs and high card hands, if you're playing a game like hold'em, where you use seven cards to make your best five card hand. It's actually harder to make no pair than to make a pair, so it seems like a high card hand should rank higher than a pair, but if you change the rankings accordingly, then if you have the joker, and no straight, flush, etc., then you would make the joker represent the highest card that is not already there, because that's a better hand than top pair. This then changes the probabilities again, and a pair should once again outrank a high card hand, because it's less likely to be your "best five cards". So it makes sense to leave the rankings as they are.
You could probably make the same case about Omaha, because it's pretty hard to make no pair when you have nine cards to work with, but a pair still beats no pair because it makes more sense that way.
If you're using two or three wild cards, I wouldn't be surprised if it make a flush easier to make (lower ranking) than a straight, but I'm not sure of the exact numbers.

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0

A wild card would mess with the odds

Two pair would go from 4 outs to 5
5/4

A flush draw would go from 9 to 10
10/9
Not only that the wild card is killed by the boat
So flush still only has 9 outs versus 2 pair or trips (any hand to be afraid of)

Trips double the outs

Five of a kind is less common than a straight flush.

Two pair would be less common that trips as you would use the joker to makes trips not 2 pair.

Only let joker used as ace, straight, or flush is more common so it does not get caught up in flopping probabilities.

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