Texas Hold’em Probabilities

In addition to the application of probabilities in poker some special considerations apply to Texas Hold'em. The calculations are fairly simple, in that prior to the flop the "deck" (that is, the set of unseen cards, including those in other players' hands) has 50 cards. The chances that an ace will turn up in the flop (or any other specific card of which you have one already), are 16.5% or about 5:1 against. The chances of a match for either the ace or the king (or whatever) are double that, or around 33%. This is 2:1 odds.

From there, the calculations can become a little more complex, but still doable within the conventional methodology used in poker. For example, if the hole cards are suited, then the chances of having at least one match in the flop are pretty good, about 63%. The chances of having two matches are lower, 26%, and three is just 5%.

The phase of the game after the flop is where the play becomes dramatic. Fortunately, the math is much simpler, as each card is dealt one at a time. Players will add up the number of "outs" for their hand, that is, the number of different cards in the deck, which, if dealt on turn or river, would complete the hand. It is possible, of course, to make these calculations if you should need two cards for an "out," but then you would have to ask yourself why you are in the game at this stage, since the odds of getting the two necessary cards are probably much worse than the pot odds.

So let us assume that only one card is required for completion of the hand. However, more than one rank might do the trick, as in drawing for a flush, or more than one suit would work, if drawing for a straight or completion of a house or quad. Suppose a player has 4 "outs" for a full house. This means that he or she has two pair, and two of each are still in the deck. The deck has 47 cards in it before the turn card is dealt. The chances of completing this hand on the turn are 4 / 47 or 8.5%. If the hand did not complete on the turn, the chances of completing it on the river card are 4 / 46 or 8.7%. The compound probability of receiving one of the "outs" on either the turn or the river has a little more work to it, but essentially, there are 184 positive scenarios out of a total of 1081 possible one, for an outcome of 17%. Added to this is the remote possibility that one of the pairs will be matched on the board by a pair in the turn and the river. Only two chances out of 1081 would qualify, for a probability of 0.2%. The following table may be helpful in calculating the odds of a winning hand on the turn and river rounds, given the number of "outs." Be sure to include every combination of rank and suit.

Example One

I have suited hole cards, King high. What are the chances of two more in the suit out of the flop?

This question expresses the hope that if the flop comes to create a four flush that the player can bring in some good action and have a decent chance for closing a flush on the turn or river card. Unlike stud, where some other player's cards are out of the deck, in Hold'em the deck stays pretty much at 50 minus the flop, then the turn and the river in succession. This is not a lot of information to go on. (At least in five-card draw, you know what 5 cards are, plus the number you draw. In stud, the deck reduces quickly with all the face-up cards. But in Hold'em the unseen cards never go below 45.) The ranks available for desirable cards are 11 - the 13 in the suit minus the two in the hole. Assume we're on the war path for spades.

• The calculation for a flop with just one spade will have combin(11;1)*combin(1;1)*combin(39;2) different possibilities. This is 8,151. The unseen cards include only 39 non-spades, of which two are required, and any one of eleven spades.
• The two-spade flop will have combin(11;2)*combin(1;1)*combin(39;1) different possibilities or 2,145.
• The three-spade flop will have combin(11,3)*combin(1;1) different possibilities or 165.
• The universe is combin(50;3) possible combinations of three cards from a fifty-card deck, or 19,600.

Thus, the chances for a three-spade flop are less than 1 percent. But the chances for a flop with at least two spades increases to 11.79%. The chance of hitting at least one spade, but maybe more than one, is better than 50% -- 53.37% to be exact.

Example Two

I have an ace and a low card in the hole. Everybody slow-played through the flop, which was garbage. What is my chance of matching up that ace on the turn or river? There are still three aces in the deck, presumably, and the deck has 47 cards. The probability that the turn will be an ace is 3/47 or 6.4%. The probability that either the turn or the river will be an ace are combin(1;1)*combin(3;2)*combin(44;1) / combin(47;2) or 12.2%. The chance that both will be aces is less than 3 in one thousand (combin(3;2)/combin(47;2)).

Example Three

I have a pair of queens in the hole, which I bet the daylights out of until the flop came. It was a 5-6-7. How likely is it that someone else will be able to put a straight together? In one variation or another, this is a common scenario, when the flop makes unimportant hole cards to be much more interesting and powerful pairs to seem sickly. Winning straight combinations are 3-4, 4-8, or 8-9 in any suit. An easy, but somewhat inaccurate approximation would be to imagine that each player has a "draw" of four cards - the two hole cards, plus the turn and the river. The odds of a 3-4 in a draw or deal of four out of a 47 card deck are: combin(1;1)*combin(4;1)*combin(1;1)*combin(4;1)* combin(45;2). Some might ask why the "other" card term is (45;2) instead of (39;2). The 39 option would rule out the eight threes and fours as well as the two queens and the flop, which are known cards already. The 45 option would rule out the same known cards, but only two of the cards being drawn for instead of all eight. As we do not care if the "other cards" contain threes and fours in addition to the two required, the higher number is more appropriate to use.

This calculation results in a sum of 15,890 combinations out of a possible 178,365, for a probability of 8.89%. As there are three scenarios that will accomplish the same result (3-4, 4-8, and 8-9), the chances are 26.6% that someone at the table will have a straight by the time the river card is turned. A couple of limitations should be placed on this number. First, if the straight is completed entirely from the last two cards, everyone will be able to play it, so no one will win unless there is a hole card that will complete a higher straight. On the other hand, if someone had the two cards to complete the straight in the hole, they probably would have folded already in light of the betting that went on over the pair of queens.

Thus, the most likely scenario is that a player has stayed in the game because one of the hole cards was not a 3 or 4 or 8 or 9, but rather something better; however, the player nevertheless has a straight-making card in the hole, perhaps an 8, let's suppose. Then, on the turn or river, the other straight-maker might appear - a four or a nine. This effectively cuts the chances of a straight back down to the range of 8% or 9% or so.

To reiterate, one-third of the possibilities (having the two hole cards fill the straight) probably would have resulted in folds. One-third of the possibilities (closing on the last two cards) would cause everyone to have the same straight. The pair of queens would be harmed only if a player were to have one card in the hole and the other on the turn or river, and the chances of that happening are less than 9%.

Example Four

I have a King-Jack unsuited in the hole. The flop came with Queen-Nine-Eight rainbow (three different suits). What are my chances to clinch a strong hand on the turn or river? Texas Hold'em uses the term "out" to describe a card that can make your hand strong if it is dealt as a turn or river card. Count the number of "outs" there are to make your hand a good one, and it will be simple to figure the probability. Any given live card has a 2.13% chance of turning up on the turn, and a 2.17% chance of turning up as the river card. The combined probability of seeing that card in either position is 4.26%. Thus, if you can identify 6 outs (for example, three other live kings for a pair and three live tens for a straight), then the chances of getting one of them in the turn is 6 * 2.13% or 12.77%. The chances of getting one of them in the river are 6 * 2.17% or 13.04%. The combined probability for one of three suits in either of two ranks out of a deal of two cards (6 outs for completing a pair of kings or catching one of three tens) is 25.53%. There are 6 different ways you can win, and any one of 46 other cards in the deck that will be OK as a second card, so that is a total of 276 combinations out of a universe of 1081, or 25.53%. This is just about the same thing as multiplying the combined probability for one "out" by six. Suppose you want to add Jacks to the probabilities. The chance of getting one of three Kings, Jacks or Tens in either the turn or the river card go up to 38.3%.

Here is how the calculation is accomplished.

For one out, it would be combin(1;1)*combin(1;1) / combin(46;1) for the turn. For the river card, the denominator is 45. For the compound probability, it would be combin(1;1)*combin(1;1)*combin(46;1) / combin(47;2) or 4.26%. If all three suits are live, then the odds of a specific rank will triple to 12.77% because the second element becomes "combin(3;1)." Add in a second rank in three suits (by changing the first element to "combin(2;1)", and the chances double to 25.53%. For three potential ranks, with three suits of each available, the chances go up to 38.30%.

Calculating the number of "outs" is not always that easy, since lots of alternative ways of looking at the flop are possible. For example, if the hole cards are suited, and if two more of the same suit are sitting in the flop, the number of outs would be 13 minus 4 or 9 to complete the flush -- all the other ranks in that suit that are yet unseen.

Calculating compound probabilities for "outs" is trickier, as in the case of receiving two needed cards for a straight or a flush in the turn and river cards. The method is the same as ever: count the number of combinations and divide by 1081, the total number of combinations of two cards dealt from a deck of 47. In the case of two cards for a straight, the possible choices are 16, that is, using both rank and suit elements, combin(1;1)*combin(4;1)*combin(1;1)*combin(4;1). By definition, all four suits of each must be live, or else they would already be in the hole or in the flop, and hence not needed. The result is 1.48%.

Hoping for two cards to complete a flush out of the turn and river cards is a better percentage because the number of possible combinations [combin(10,2)*combin(1;1)] is almost three times as large. Ten ranks are available, and two are required, and there are 45 different ways of combining two cards out of a universe of ten. The result is 45 out of 1081 or 4.16%.