Five-Card Draw Probabilities

Draw poker players have to ask themselves about two different kinds of probabilities. The methods are the same, but the contexts are different. The first is "What are my chances of being dealt a [whatever kind] of hand? More commonly, you may need to ask yourself something like, "Given that I hold a pair, what are my chances of improving my hand on a three-card draw?" The first question is sometimes called "absolute probability," like the case of flipping a coin or throwing dice. The second is often called "contingent probability," and asks for a calculation "given" some event that has already occurred.

Probabilities of Receiving Hands in the Deal

The Full House. What is the probability of being dealt a full house? The full house is trips and a pair. The trips can be any one of 13 ranks, using 3 out of four suits. The pair can be any one of 12 remaining ranks and 2 out of four suits. All five cards in the hand are thus defined. This probability is then

combin(13;1)*combin(4;3)*combin(12;1)*combin(4;2), divided by combin(52,5).

The result is 3,744 instances, or a chance of 0.14% in an initial deal.

The Flush. What is the probability of being dealt a flush? The flush is five cards of the same suit. The calculation is

combin(13,5)*combin(4,1), divided by combin(52,5).

As this single configuration imposes constraints on all five cards, no other terms are required. All five cards in the hand are defined.

The calculation comes to 5148 instances, or 0.198% in an initial deal. Included in this calculation are the 40 flushes that also happen to be straights.

The Straight. What is the probability of being dealt a straight? The first term would be [combin(10;1)*combin(4,1)]. This means that the first card in the sequence can be any of ten possible ranks. (In a suit of 13 cards, only 10 can initiate a straight). The suit is immaterial, any one of four possible. The next question relates to cards 2 through 5 in the straight. They must be of a specific rank (or combin(1;1) for the rank term, but they can be of any suit (or combin(4;1) for the suit term). In other words, once the starting card has been defined, there will be four cards in the deck that can satisfy the requirement for each of the next four cards in sequence. Thus the complete calculation is

combin(10,1)*combin(4,1)**5, divided by combin(52,5).

(Raising the combin(4,1) term to the fifth power is the same as multiplying out each combin(4,1) term for each of the five cards.) The number of instances is 984, including 40 that also are flushes.

The probability of being dealt any straight (including the flushes) is 0.394%.

A Pair. What is the probability of being dealt at least a pair, any pair? There are two ways to approach the question:

" Build up from the bottom: Calculate the number of pairs that could pop up, disregarding that the three "other" cards might have pairs, trips, or other benefits (including trips and quads of the "pair" as well), and adding back in the straights and flushes, or
" Come down from the top: Calculate the combinations that give you straight flushes, flushes, quads, full houses, straights, trips and pairs, cumulating them to reach your answer.

Both methods involve making separate calculations for different hands and then adding them up or separating them out to arrive at a single number. The second system is found most often in "cumulative probability tables," which tell you your chances of a specific outcome. Such a table is reproduced below:

The table shows that the chance of receiving a pair in a five-card draw is about 42.3% for just one pair. The chances of getting a pair or something better are about 49.9%. In other words, the chances of getting "nothing" except a hand that might turn into something later are 50.1%

Jacks or Better. Using the table, you know that 42.3% of hands will have just one pair. Four out of thirteen ranks will be jacks or better. So the chances of one pair of jacks or better are 4/13 times 42.3% or 13.02%. Add to this the cumulative possibility of a hand that is better than one pair (7.62%), and the answer is that "Jacks or Better" can be expected about 20.64% of the time, or once in every five hands on average.

Probabilities of Receiving Hands in the Deal.

Example One. What is the probability of drawing one card to a flush? (Assume, of course that you already have four suited cards.) The numerator is combin(9;1)*combin(1;1). There are only 9 ranks in the deck that will meet the requirement (as the other four are already in the hand), and in only one suit. (This could be determined just by thinking about it, but the formal procedure helps to avoid missing anything.) There are no "other" cards, as you are only looking for one card, and you are making a one-card draw. The denominator is combin(47;1), meaning a one card draw from a 47-card deck. So the chances are 9/47 or about 19%.

Example Two. I have a pair of nines and an ace. Is it better to save the ace and draw two cards or slough the ace and draw three cards?

The Two Card Draw: Here it is assumed that you cling to your pair of nines and also hold back your ace. The two card draw has four possibilities: a second pair, a third nine, a second ace, or a full house of nines and aces.

• Second Pair - The discards need to be considered in calculating the chances for a second pair. There will be one term that deals with 9 ranks having 4 cards in the deck, and another term with 2 ranks having only 3 cards in the deck (because of the discards). Two ranks are not represented, as they are the nines and the ace being retained. There is no "other card" element in drawing two cards for a pair. The calculation is combin(9;1)*combin(4;2) + combin(2;1)*combin(3;2), which equals 60.
• Second Ace - There is no need to consider the discards. The calculation is combin(1;1)*combin(3;1)*combin(44;1), or 132.
• Third Nine - There is no need to consider the discards. The calculation is combin(1;1)*combin(2;1)*combin(45;1) or 90.
• Full House - There is no need to consider the discards. The second ace is combin(1;1)*combin(3;1) and the third nine is combin(1;1)*combin(2;1). Their product is 6. These numbers, added together, and divided by combin(47;2) provide the outcome for the two card draw.

The combined chances are 288 out of 1081 or 26.64% that something good will happen.

Drawing Three Cards. The three card draw means that the ace has been sloughed to make way for three cards to come into the hand. This draw has five possibilities: four of a kind, three of a kind, a second pair, a full house of nines and a pair, and a full house of a trip and the pair of nines. The denominator is combin(47;3) or 16,215.

• Four of a kind requires the other two nines and a third card: combin(1;1)*combin(2;2)*combin(45;1) or 45.
• Three of a kind requires one of the other two nines and two "other" cards or a fresh draw of three of some other rank. The first piece is combin(1;1)*combin(2;1)*combin(45;2). This yields 1980 combinations. The second piece also divides into two parts. The one for ranks that have been discarded, and hence only have three matches in the deck, and the one for ranks that have not been discarded and have four matches in the deck. The calculation is combin(9;1)*combin(4;3) + combin(3;1)*combin(3;3). This is 45. So the total is 2025 combinations.
• A second pair is like the analysis of drawing a second pair in a two card draw, except that there is an "other" card in each piece of the calculation, as follows: One term deals with 9 ranks having 4 cards in the deck, and another term with 3 ranks having only 3 cards in the deck (because of the discards). One rank is not represented, the nines. The "other card" element varies with whether three or four cards are being sought for the pair. It starts with 47, then reduces by 2 for the nines, and by three or four for the other pair, depending on which term is looked at. The calculation is combin(9;1)*combin(4;2)*combin(41,1) + combin(2;1)*combin(3;2)*combin(42;1), which equals 2,466.
• A full house of nines and a pair would be in two parts, like the analysis of drawing an off pair in the two-card draw discussion. One part would be combin(1;1)*combin(2;1)*combin(9,1)*combin(4;2) + combin(1;1)*combin(2;1)*combin(3,1)*combin(3;2).
• A full house of three of some kind plus the pair of nines would be combin(9,1)*combin(4,3) + combin(3,1)*combin(3;3).

The overall chance of improvement in the hand is better with three cards in an absolute sense (28.4% versus 26.6%). Holding an ace and drawing two cards gives a 17.76% chance of achieving two pair (one of them being an ace in several instances). Drawing three cards drops the chance of an additional pair to 15.21%. The reason is that the probabilities for yet stronger hands have gone up in the three-draw hand. Three of a kind, for example, will occur 8.33% in a two-card draw, but will occur 12.49% of the time in a three card draw. The three-card draw also admits of four of a kind and a full house, with combined probabilities of approximately 1¼ percent.

The two-card draw is preferable if the objective is to have two pair, and the three card draw is preferable if the objective is to have three of a kind or better. Even discounting the very small probabilities for the more exotic hands, the three-card draw seems to be a better choice, as the overall chances for hand improvement are better.

This decision become less debatable as the rank of the "kicker" in the two-card draw goes down. The benefit of holding on to the ace, of course, is that it will beat all other pairs if it pairs up, and it will dominate in tie-breaking. But as the kicker drops in rank, the value of drawing to the held back card goes down quickly.

Example Three.

If I have a pair, but I want to bluff a trip by asking for just two cards on the draw, how much am I giving up in the way of a chance to improve my hand? Compare the results obtained in the previous example, as indicated in the tables. For one thing, you give up the small chance for some strong hands, like a full house. (Of course, it is still possible to obtain a full house in a two-card draw, but the chances are extremely small. How small? combin(1;1)*combin(2;1)*combin(1;1)*combin(3;1) divided by combin(47;2). This is 0.56%.) Focus on the chances for two pair and three of a kind. The results of the two tables are combined here for comparison:

The trade-off is essentially a drop of 4 percentage points for three of a kind and an increase of around two points for two pair. Whether this trade-off is worth it as the price of a bluff is something a "combin(n;k)" calculation can not determine.