## Draw Poker – Improving the Hand

We need a technical term. I have a source against which I can check my answers, so I want to compute the “odds against” improving a hand. The odds against something are computed as the number of bad outcomes divided by the number of good outcomes.

For example: we saw in the previous poker post that the probability of being dealt at least a pair is the good outcomes (“poker” in the previous post) divided by the total outcomes (“hands”):

OK, the probability of being dealt at least a pair is almost 50%.

The odds against being dealt at least a pair is bad outcomes divided by good outcomes:

That is, the odds against being dealt at least a pair are just barely greater than 1; so close in fact that we might as well call it “even odds”.

Now let’s suppose you’ve been dealt at least a pair – or that you’ve been dealt four cards toward a straight or flush. (Worthless as they stand, they have great potential.)

## Drawing one card

If we discard and receive one card, the calculations are pretty simple.

odds against improving two pair to a full house

Holding two pair, what are the odds against drawing a card for a full house?

We draw one card… there are 47 out there, and 4 good ones (2 each for each pair we hold), so there are 43 bad outcomes:

That is, the odds against improving two pair to a full house are 10.75 to 1. Jeff Rubens “Win at Poker” said 10.8; looks good to me.

odds against improving a 4-flush to a flush

Holding a 4 flush (four cards of the same suit), what are the odds against drawing the flush?

We draw one card, there are 47 out there, and 13-4 = 9 of them are good ones:

he said 4.2; good.

odds against improving a straight?

There are two possibilities: the straight is open-ended, so that either of two ranks will complete it; or it’s an inside straight (e.g. 2-3-5-6, needing a 4), so that only one card will complete it. By the way, the four-straights 1-2-3-4 and J-Q-K-A count as inside straights, because only one card will complete them.

For an open-ended straight, there are 8 good cards:

Rubens said 5 to 1.

Odds against improving an inside straight?

There are only 4 good cards, so…

(That’s the same as the odds against drawing a full house having two pair. Of course: in both cases there are 4 good cards.)

## holding three of a kind

Now life gets a little more interesting – we’re going to discard and receive two cards. Someday maybe I will consider the possibilities that we have either a three-straight or a three-flush and nothing else. For now I’m only going to look at one possibility: we are holding three of a kind.

There are 47 cards out there, and we choose 2, so there are…

$\binom{47}{2}$

= 1081

possible draws of two cards. We will need this number in order to compute the number of bad outcomes once we have the number of good ones.

There are two ways to improve three of a kind: by getting four of a kind, or by getting a full house.

improving to four of a kind?

There are 47 cards out there, 1 of which is the fourth card; there are 46 other cards, hence 46 different sets of 2 which contain the fourth card. The odds against getting 4 of a kind are…

(Sorry about the reduction of the fraction to lowest terms; the original denominator really was 46.)

Let me check that 46. There is one rank of which I hold three cards, and that doesn’t count as “other”; two ranks of 3 cards, for the two I discarded; and ten ranks of 4. That’s 6 + 40 = 46.

To phrase that cleanly: choose 1 out of 1… and then choose 1 out of 2 ranks and 1 out of 3 suits; or 1 out of 10 ranks and 1 out of 4 suits:

$\binom{1}{1} (\binom{2}{1} \binom{3}{1}+\binom{4}{1}\binom{10}{1})$

= 46.

getting a full house?

There are 12 ranks other than our three of a kind, and they’re not all the same.

2 ranks of the 12 have only 3 cards out there (because I discarded one each of two different ranks), so choose 1 out of 2 ranks, and then 2 out of 3 suits:

$\binom{2}{1} \binom{3}{2}$

= 6.

That is, there are 6 distinct full houses using one of the ranks I discarded.

Then there are 10 ranks with 4 cards out there. Choose 1 out of 10 ranks, and 2 out of 4 suits:

$\binom{10}{1} \binom{4}{2}$

= 60.

That is, there are 60 ways to draw a pair other than the two ranks I discarded.

Combined, there are 66 different full houses we could construct. The odds against are, therefore:

Now let’s add the possible four-of-a-kind hands; there were 46. The combined good outcomes (full house or four of a kind) are 66 + 46, so the the total good outcomes is… and the odds against improving are…

Rubens said 8.7; looking good.

## holding a pair

Now life gets even more interesting: we discard and receive three cards. There are a lot of ways to improve one pair.

First, we need the total number of possible draws: Choose 3 out of 47:

$\binom{47}{3}$

= 16215.

getting 4 of a kind: 45

There are exactly two cards out there that would give me four of a kind, and I need both of them. But I draw three cards total, so there are 45 different good outcomes. (47 cards after the deal, then I draw the two I need, leaving 45 possibilities for the third replacement card.)

getting three of a kind: 1854

Adding to the pair requires that I get 1 of the 2 cards out there. That part’s easy.

Now choose two ranks for the remaining cards… but there are three possibilities, depending on whether they are both from 9 or both from 3 or one from each.

Huh?

I discarded three different ranks, so there are 3 ranks out there with only 3 cards left, and 9 ranks with all 4 cards left. In addition, I may draw two ranks from the “3”, or two ranks from the “9”, or I might draw one from each.

Three different computations for the two “other cards” I get.

First, both from the “3”. To get three of a kind, I must choose one of the 2 cards out there… then choose 2 out of 3 ranks… and for each of those two ranks, choose 1 of the 3 suits.

$\binom{2}{1} \binom{3}{2} \binom{3}{1}^2$

Second, both from the “9”. I must choose 1 of the 2 cards that give me three of a kind… then choose 2 out of 9 ranks… and within each of those two ranks, choose 1 of 4 suits.

$\binom{2}{1} \binom{9}{2} \binom{4}{1}^2$

Third, one from the “3” and one from the “9”. (I’ll take Gandalf’s Ring of Adamant, please; and the ring from the King of Angmar.)

Again, I must choose 1 of 2 to get three of a kind… 1 of 9 for one rank and 1 of 4 for the suit… and (not “or”, so keep multiplying) 1 of 3 for the other rank, and 1 of 3 for the suit.

$\binom{2}{1} \binom{9}{1} \binom{4}{1} \binom{3}{1}\binom{3}{1}$

Computing, I get

1152 + 54 + 648

= 1854.

That is, there are 1854 hands that improve one pair to three of a kind.

getting 2 pair: 2592

How many ways can I improve one pair to exactly two pair?

We have the same dichotomy: there are 9 ranks of 4 cards out there, and 3 ranks of 3 cards. But there are only two alternatives: draw the new pair from the “9”, or draw it from the “3”.

From 9 ranks, choose 1 for the pair… choose 2 suits out of 4… then choose another card from what’s left.

What’s left? There were 47 after I was dealt 5… I do not draw any of the pair I’m holding (that leaves 45)… and I draw a new pair (which takes two cards, and I do not draw either of the other two).

I think there are 41 cards left. We’ve seen how to compute that cleanly (improving three of a kind to four of a kind), but we could compute it directly:

Suppose I hold a pair of aces and discard 3, 5, 7; and I draw a pair of 2s. The fifth card cannot be ace or 2, and that’s the only restriction: any of the remaining 11 ranks are allowed. There are three ranks of 3 and eight ranks of 4, so 3*3 + 4*8 = 9 + 32 = 41.

Is it the same for the other case? Suppose I draw a pair of 3s (having discarded a 3). The fifth card cannot be ace or 3, but any of the remaining 11 ranks are allowed. There are two ranks with 3 (i.e. 5s and 7s) and 9 ranks of 4: 6 + 36 = 42.

Not the same.

From the “9”, then: choose 1 of 9 for the rank, choose 2 out of the 4 suits, and choose 1 of the remaining 41:

$\binom{9}{1} \binom{4}{2} \binom{41}{1}$

= 2214.

From the “3” ranks: choose 1 of 3 for the rank of the pair… choose 2 suits out of 3… and another card from the remaining 42:

$\binom{3}{1} \binom{3}{2} \binom{42}{1}$

= 378.

Combine them: 2214 + 378 = 2592.

Let me work out that 41 and 42 cleanly.

In both cases, we hold a pair and discard 3 other ranks; this splits the remaining cards into 3 ranks of 3 and 9 ranks of 4. We draw a pair, and then one other card.

If the pair comes from the “3”, then the other card is either: 1 of 2 for the rank, and 1 of 3 for the suit; or 1 of 9 for the rank, and 1 of 4 for the suit:

$\binom{2}{1} \binom{3}{1}+\binom{4}{1} \binom{9}{1}$

= 42.

If the pair comes from the “9”, then the fifth card is either: 1 of 3 for the rank amd 1 of 3 for the suit; or 1 of 8 for the rank and 1 of 4 for the suit:

$\binom{3}{1} \binom{3}{1}+\binom{4}{1} \binom{8}{1}$

= 41.

Good.

getting a full house: 165

There are two ways to do this. We could draw three identical cards; or we could draw another pair, and a card to improve the pair we held to three of a kind.

Either way, all three cards must be good, so there are no “other” cards to worry about.

draw 3 identical cards: 39

Suppose I draw 3 identical cards… some of which I’ve thrown away. That is, we must still distinguish whether the three come from one of the 3 ranks with 3 cards remaining, or from the 9 ranks with all 4 cards remaining.

One possibility: choose 1 out of 9 for the rank, then 3 out of 4 for the suits…

$\binom{9}{1} \binom{4}{3}$

= 36.

The other possibility: choose 1 out of 3 for the rank, and draw all 3 out of 3 suits:

$\binom{3}{1} \binom{3}{3}$

= 3.

The sum of the two alternatives is 39.

draw a pair and one of what we’re holding: 126

Choose 1 rank of 2 to improve the pair I’m holding. Then there are 3 ranks of 3, 9 ranks of 4. Choose a rank: either 1 of 9 and 2 of 4 suits…

$\binom{2}{1} \binom{9}{1} \binom{4}{2}$

= 108.

… or 1 of 3 and 2 of 3 suits:

$\binom{2}{1} \binom{3}{1} \binom{3}{2}$

= 18.

Since 108 + 18 = 122, that’s how many ways of drawing a pair and improving the original pair to three of a kind.

The sum of all possibilities is

39 + 108 + 18 = 165.

If we just consider the odds against improving a single pair to a full house, we get

Which leaves me wondering. I thought that Dr. Reid said the odds were 100,000 to 1 in an episode of Crminal MInds; now I wonder if he said 100 to 1. I hope so. But I have no idea what episode that was.

And now you know why I started these computations!

summary of improving one pair:

There are all together this many good outcomes… so the odds against are:

Rubens says 2.5 for the odds against improving one pair in every possible way. OK.

And that’s enough. I am not now going to consider the odds if we don’t have at least a pair or a four-straight or a four-flush.

### 6 Responses to “Draw Poker – Improving the Hand”

1. pool accessories Says:

Hi
I love Poker .. Now i become expert in this game..

thanks..

2. rip Says:

Well, I found that episode of Criminal Minds in which Prentiss keeps two cards and draws three to get a full house. It’s “… A Thousand words”, episode 20 of season 5. Best of all, Reid says the odds were 100 to 1 against – not 100,000 – and she corrects him: 97 to 1. Too bad he didn’t counter with our computed answer, ” 97.2727 to 1″.

3. http://tinyurl.com Says:

Thanks for the post for posting “Draw Poker –
Improving the Hand | Rip’s Applied Mathematics Blog” brukkon-puzzle-game . Imay absolutely wind up being coming back for much more reading through and commenting in the near future. With thanks, Sebastian

4. Frankq Says:

let’s say you have 3 suited cards and a low pair against ~3 other players with random hands in an un-raised pot … better to try to improve the pair or draw to the flush? If the decision is to stick with the pair, would anything change if the 3 suited cards are also connected (say QT9)?

5. yukonrcmpvets Says:

My wife and I played draw poker to see who is cleaning up the dishes. She got four of a kind, (8’s) and a two. First 5 cards. No draw for her. I got aces over two’s. Two pairs……………groan. I’m doing the dishes.

6. Reed Chaber Says:

Dr. Reid said 100 t0 1…that’s how I found this page. 🙂