# Best Poker Hand probability when dealt 3OAK with 5 opponents

So I've gone through and found all possible number of hands and probabilities for each hand (royal flush, straight flush, straight, flush, full house, four of a kind, three of a kind, 2 pair, 1 pair). Had to show work obviously for these, so I was unable to just pull them from the millions of websites I could have found them on.

Now, I'm given a situation where I'm playing against 5 other people and I'm dealt a 3 of a kind. I need to find the probability that one or more of my 5 opponents were dealt a higher hand than I. Basically, what's the probability that I have the best hand. Assuming that all players' hands are independent.

Is this a simple math calculation using my already known probabilities, or is it a little more tedious?

My probabilities are:

Royal Flush 0.000154%

Straight Flush 0.00139%

Flush 0.197%

Straight 0.393%

1 Pair 42.3%

2 Pair 4.75%

3-of-a-Kind 2.11%

4-of-a-Kind 0.024%

Full House 0.144%

Nothing (high card hand) 50.1%

The only thing I can currently think of, since the hands are independent, is adding the probabilities of each hand that's better than a 3OAK, and raising it to the 5th power since there are 5 opponents.

• Assuming one deck is being used, the problem is that your knowledge of the cards in your own hand changes the probabilities of the other hands; for example, if you have three-of-a-kind K, you know for certain that no one else has a pair of K, a royal flush, a full house with K up, etc. If for whatever reason you are playing with many decks, you can assume your cards don't change the probabilities too much, and then proceed finding the probability that at least one person has a better hand. I also don't know what kind of poker this is, which probably changes the answer, too. – anonymouse Apr 13 '16 at 22:13
• Unfortunately all the information I am given for this assignment is that it's "Poker". No specific game.. And the problem itself says to assume that all players' hands are independent to make the simplifying assumption, which I take as the same thing as each player is dealt from an independent deck? – taskle Apr 13 '16 at 22:19

## 1 Answer

First, if you want to figure out the possibility that one of the 5 hands and the table beat you, figure out the probability that none of the other hands at the table beat you, and subtract that number from one.

Now, if you want to be very persnickety about this, the information that you have a 3 of a kind slightly increased the probability that someone else at the table has a 4 of a kind and slightly lowers the probability that they have a straight or a flush. It isn't so much that it is hard to figure out all of these probabilities, but it is laborious and likely above and beyond what is expected.

• So if I figure that each independent hand dealt to each of the other 5 opponents has a 0.759544% probability at getting a better hand than 3-of-a-kind, can I simply multiply these probabilities by 5 to get the probability that at least one beat me? Or does that calculate the probability that all 5 beat me? Doing that would yield 3.79772%. Or since they're each independent, is it simply 0.759544% that at least one beat my hand? I know I'm probably overthinking this. – taskle Apr 13 '16 at 23:25
• each opponent has a 97.2% chance of getting something worse than a 3 of a kind. (no pair + pair + 2 pair). There is also a 1.1% chance that another hand is also a 3 of a kind, but with a higher ranking 3. 3 of a kind beats all others = (96.1)^5. 82% chance your hand is the best hand. 18% chance that it is not. – Doug M Apr 13 '16 at 23:32