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admin answers:

Depends on the game actually.

In five card stud, the chance of a straight flush that is NOT a royal flush is one out of 72,193. (36 possible straight flushes out of 2,598,960 total hand combinations ).

In five card stud, the chances of a straight flush that includes the royal flush combination is one out of 64,974 (40 out of 2.598,960).

In Texas Holdem (or 7 card stud) this changes a bit, since you have 7 cards, not 5.

In this game, the chances of a straight flush that is NOT a royal flush is one out of 3,590.57. (37,260 out of 133,784,560)

And the chances of a straight flush including the royal flush is one in 3,217.21. (41,584 out of 133,784,560).

Quite a difference in the Texas Hold’em game!

And, of course, there are a lot of people that say these hands seem to occur much more frequently when you play online poker! LOL.

admin answers:

Because the rank of hands is determined by how many possibilities for each hand exist in a 52-card deck.

There are 5,108 possible flushes, and 10,200 possible straights. A flush is harder to get, and that’s why it ranks higher.

Following is a list showing each hand, and the number of possibilities. No wild cards are taken into consideration.

Royal Flush: 4
Other Straight Flush: 36
Four of a Kind: 624
Full House: 3,744
Flush: 5,108
Straight: 10,200
Three of a Kind: 54,912
Two Pair: 123,552
One Pair: 1,098,240
Nothing: 1,302,540

admin answers:

I believe you’re referring to the side bets, which can vary by casino. The link below shows the payouts and odds of some of the popular side bets at Pai Gow Poker.

Have fun.

admin answers:

Ok, you have three spades….you need two more and there are 10 left in the deck. Your chance of hitting two spades are 10/49 X 10/48…or about 25 to one.

You have three cards in sequence. Assuming both ends are open…say you got 456…..so you need a 7 or a three first….8/49

then you need a 8 or a 2….8/48…..about 36 to one to hit the straight with two cards to come.

I am not a poke rexpert, but i think this makes sense. I will welcome feedback and critissism on this

admin answers:

Here’s how I would approach this problem:

First, you have a four of a kind. One of three things happened:
1. The community cards hold a four of a kind
2. The community cards hold a 3 of a kind and you hold the fourth card
3. The community cards hold a pair and you hold the other pair

In the case of #1, nothing interesting can happen. It is impossible for another person to beat the four of a kind on the board, so this is uninteresting.

In the case of #2, there are four remaining cards – the other two community cards and your opponent’s two cards. So the following is possible:
2A. These four cards create a higher four of a kind that yours.
2B. These remaining four cards can be combined with one of the three of a kind to create a straight flush.

In the case of #3, there are five remaining cards – the three other community cards and your opponent’s two cards. In this case we get the following possibilities:
3A. These five cards contain a four of a kind that beats your four of a kind.
3B. These five cards form a straight flush.
3C. Four of these five cards combine with one of the showing pair of community cards to create a straight flush.

Now, these are the only ways you can get beat. This is a complicated question, and so I’m assuming you have some ability to calculate probabilities already. The key is to calulate the probability of each event above happening and combining them together:
P(2)*[P(2A)+P(2B)]+P(3)*[P(3A)+P(3B)+P(3C)]

Note: The above is the probability of this event happening. You might also be interested in the conditional probability of you getting beat give that you alreay have your four of a kind. In that case, you just exclude the P(2) and P(3), to get
P(2A)+P(2B)+P(3A)+P(3B)+P(3C)

I didn’t actually find the answer, but I’m sure it is VERY small!