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shermn27
06-28-2005, 10:34 PM
This is a question that has been bugging me for a while. I was playing in a home game (not sure which game I was playing) in which the low card in the hole and anything like it was wild. I am pretty sure it was a 7 card game. In any case, I made a straight flush and got all of my chips in, only to find out that I lost to 5 of a kind.

My question is, in a 7 card game (w/3 cards in the hole) and low in the hole and anything like it is wild, is it more likely to get a straight flush or 5 of a kind? It seemed to me that a straight flush would be more difficult to make. I am hoping some of you probability experts can help me with this problem or lead me in the right direction. Thanks.

AaronBrown
06-30-2005, 06:48 PM
No, there are many more straight flushes. It's easiest to see it if you consider it separately for different numbers of wild cards.

With a single wild card, there are 12 different fours of a kind you could get, and 160 different four-to-a-straight flush (that includes some with your wild card, which we shouldn't do, but the number to remove depends on what card it is, so call it about 150 combinations).

With two wild cards, there are 48 ways to get three of a kind and many ways to get three to a straight flush.

With three wild cards, there are 72 ways to get a pair and 184 ways to get 2 to a straight flush (again you have to say about 170).

With four wild cards you can get both.

So, regardless of the number of wild cards you have, a straight flush is at least as likely as five of a kind, so it must be more likely overall.

Siegmund
06-30-2005, 07:37 PM
The standard hand rankings are based on listing the hands from least to most frequent in 5 random cards with no joker.

When jokers are added to the deck, a paradox arises. It's no longer possible to construct a ranking system where each hand ranking is rarer than the next lower ranking. People choose to use a joker to make the best possible hand. (Consider video poker. With even one joker in the deck it's easier to make 3 of a kind than 2 pair even though trips has the higher payoff. If they reversed the payoff table, you would simply declare AAK2-Joker to be two pair instead of trips, and the higher-paying hand remains the more common one. In video poker / 5-card draw, it takes 3 jokers to make full houses more frequent than flushes.)

Aaron has showed that in your particular game with only 4 jokers, it remains harder to make five of a kind than a straight flush. However, for some sufficiently large number of jokers (51 is an upper bound, but it's probably a lot fewer) that won't be the case.

shermn27
06-30-2005, 08:22 PM
Thanks guys. I appreciate it. /images/graemlins/smile.gif

BruceZ
06-30-2005, 09:02 PM
[ QUOTE ]
When jokers are added to the deck, a paradox arises. It's no longer possible to construct a ranking system where each hand ranking is rarer than the next lower ranking.

[/ QUOTE ]

This isn't a problem when the joker is a bug rather than a wildcard. The bug may only be used as an ace, or to fill straights and flushes.

Siegmund
06-30-2005, 10:03 PM
[ QUOTE ]

This isn't a problem when the joker is a bug rather than a wildcard. The bug may only be used as an ace, or to fill straights and flushes.

[/ QUOTE ]

Adding sufficiently many bugs will still cause flushes to be more common than straights. (But yes, in real life, there is usually only one and that doesn't happen.)