[Guest]

This is less a riddle and more a complex problem for math geeks.

In poker, the hands you can get in any 5 card deal are:

high card (highest card value)

1 pair (any 2 cards same value)

2 pair

3 of a kind (any 3 cards same value)

straight (5 cards ascending order value)

flush (5 cards same suit)

full house (any 3 of one value and 2 of another)

4 of a kind (any 4 cards of same value)

straight flush (any 5 cards same suit ascending order)

Royal Flush (Ace, King, Queen, Jack, Ten of same suit)

So, the possible number of hands are 52*51*50*49*48 = 311,875,200 possible hands in poker!

The odds of getting a high hand are the number of ways to get higher than a high hand subtracted from the above.

The odds of a royal are 4 in 311,875,200 or 1 in 77,968,800

The odds of a 4 of a kind are 13 in 311,875,200 or 1 in 23,990,400; but, 3's beat 2's, etc.

Each player has equal odds of getting any one hand.

Assuming a 5 handed game, what are the odds that you are holding the winning hand with 5 different cards to each player?

[Guest]

As a followup, if someone gets this and wants to keep going, how do the odds change if there are 2 unique cards to each player and 3 common cards? (like Texas Hold 'em)

[Guest]

As a followup, if someone gets this and wants to keep going, how do the odds change if there are 2 unique cards to each player and 3 common cards? (like Texas Hold 'em)

Are we doing typical, like two people have a pair of aces, with one a kicker of king other kicker is queen. Win goes to King kicker?

Also, what do we do with a straight flush tie just in different suits?

[itachi-san]

Are we doing typical, like two people have a pair of aces, with one a kicker of king other kicker is queen. Win goes to King kicker?

Also, what do we do with a straight flush tie just in different suits?

Usually the ranking is Spade, Heart, Club, Diamond. But it can sometimes be reverse alphabetical: Spade, heart, diamond, club. Spade is always most valuable at any rate

[Brandonb]

The odds of holding the winning hand are 1in5. The types of hands don't matter, only the odds of winning. Everyone having an equal chance of any hand is just plain old 1in5 for the winner.

[Guest]

For polish, ingore kickers and count ties as a loss to make it easier, I want odds of winning and a tie isn't a win.

For Brandon, this is the exact solution I had in mind, but wanted to see if it really was the answer. The reason I doubt it is because the lower your hand, the better the odds of getting it. If every hand had equal chances of showing, 1 in 5 would be the answer, but the weighted probabilities (I think) changes things.

Itachi, ignore suit rankings. They don't matter in poker. For example, if you have a spade royal flush and mine's a club royal, we "tie" and split the pot. Since ties aren't a win, ignore in the probabilities.

[Brandonb]

For Brandon, this is the exact solution I had in mind, but wanted to see if it really was the answer. The reason I doubt it is because the lower your hand, the better the odds of getting it. If every hand had equal chances of showing, 1 in 5 would be the answer, but the weighted probabilities (I think) changes things.

Well that's your answer. The reason the odds are better is because- ie, getting a pair, there are 312 ways to get it. With a royal flush there are only 4 ways to get it. This applies for each individual playing.

The deal here is, the odds for one person obtaining any of those hands is: as you have already stated... These exact same odds apply to the guy next to him (the odds of getting any of those hands), and for the guy next to him and so on. Each guy has the same odds of picking and one of the various hands.

So, because each one has the same odds of obtaining each hand, then their relevance to each other is nothing more than 1:1:1:1:1. The odds of holding the winning hand is 1in5.

The same applies for hold-em, from the flop all the way down to the river card (so long as each player is blind). Though the odds change once someone can compare theirs to the community cards in the middle.

[Guest]

Yeah, I get what you are saying. Really, then, you have to know at least one hand (either yours or one other) to determine if your odds are favorable or not, right? For example, if I know I have a pair of twos, I can determine the odds of having the best hand, etc. or if I know that you have a pair of twos I can determine the odds of having a hand that beats it.

Is that right?

Also, if the answer IS 1:5, does the first player dealt always have a better chance of winning? All other players would have progressively one less shot at getting a given hand, right?

[Brandonb]

Yeah, I get what you are saying. Really, then, you have to know at least one hand (either yours or one other) to determine if your odds are favorable or not, right? For example, if I know I have a pair of twos, I can determine the odds of having the best hand, etc. or if I know that you have a pair of twos I can determine the odds of having a hand that beats it.

Is that right?

Also, if the answer IS 1:5, does the first player dealt always have a better chance of winning? All other players would have progressively one less shot at getting a given hand, right?

No, that only applies if you see your cards, if you are blind to the cards then there is no difference. No one has 1 card less of a chance because there is no way to determine if each card dealt is in any way valuable. However, once you see your own hand then the odds change from a blind 1:5, to the odds of your hand v.s. all other possibilities v.s. the number of cards dealt.

Basically, there are no odds when the cards are dealt. It's all luck at that point.

[Guest]

So, the possible number of hands are 52*51*50*49*48 = 311,875,200 possible hands in poker!

I may be wrong in my thinking here, but you're treating a poker hand as a permutation (in which the order your cards are dealt would change the value of your hand) when it should be a combination (where being dealt, say, A 2 3 4 5 of hearts is the same as being dealt 5 4 3 2 A of hearts).

Rather than 52 P 5, aren't there 52 C 5 combinations of cards?

52 C 5 = (52*51*50*49*48) / (5*4*3*2*1) = 2,598,960

Though I think this isn't entirely relevant to the question assuming no one had looked at their cards as I agree the chances of anyone winning under those circumstances would be 1 in 5.

[Guest]

I may be wrong in my thinking here, but you're treating a poker hand as a permutation (in which the order your cards are dealt would change the value of your hand) when it should be a combination (where being dealt, say, A 2 3 4 5 of hearts is the same as being dealt 5 4 3 2 A of hearts).

Rather than 52 P 5, aren't there 52 C 5 combinations of cards?

52 C 5 = (52*51*50*49*48) / (5*4*3*2*1) = 2,598,960

Though I think this isn't entirely relevant to the question assuming no one had looked at their cards as I agree the chances of anyone winning under those circumstances would be 1 in 5.

You are correct that it should be combinations and not permutations, and that 2.6 million is the correct amount. There are 52*51*50*49*48 hands if order matters (in other words, if AKQJ10 of Hearts is DIFFERENT than KJA10Q of Hearts, which is not the case in poker).

Also, the answer of 1/5 is correct before you look at your cards, and different the minute any single card is known.