Hello,

If there has already been recent threads about 5-card draw probabilities or easily available and useful websites, I apologize in advance.

Suppose in basic 5-card draw, we have 99xyz and drew 3.

1. What are the odds of improving to two pairs?
2. What are the odds of improving to trips?
3. What are the odds of improving to a full house?
4. What are the odds of improving to quads?

And finally, would "two pairs or better" mean the sum of these percentages, or is that calculated in some other sort of way?

Suppose you held onto a kicker, say 99Axy and drew 2 instead. What are your odds of improving to two pairs or better?

Finally, my last question is if you hold AKxzy, would it be better to throw away the King and draw 4 or to keep the King and draw 3 to improve to one pair or better? I know in video poker, we would often hold onto a King especially if it is suited because of the bonuses, but that doesn't necessarily mean it improves the probability of improving your hand.

Thanks.

I'm sure these odds or probabilities can be found at several
sites; however, in the case of playing in a poker game where
there has already been some folding, certain cards, notably
aces and kings are more likely than if you were just playing
heads up (since almost all players will play KK or AA in a
five- or six-handed game). Assuming each unseen card has
equal probability of being drawn, here are the probabilities.


DRAWING THREE TO A PAIR
====================

Combinations are out of C(47,3) = 16 215. C(n,r) denotes
the number combinations of n objects taken r at a time =
n!/[ (n-r)! r! ].

1. Two pairs: 9C(4,2)(44-3)+3C(3,2)(44-2) = 2 592
or probability of about 0.1598519889 or odds of about
5.255787037 to 1 against.

2. Trips: 2[C(9,2)x4x4+9x4x3x3+C(3,2)x3x3] = 1 854
or probability of about 0.11433857540 or odds of about
7.7459546926 to 1 against.

3. Full house: 2[9C(4,2)+3C(3,2)]+9C(4,3)+3C(3,3) = 165
or probability of about 0.01017576318 or odds of about
97.27272727 to 1 against.

4. Quads: C(2,2)(48-3) = 45
or probability of about 0.002775208141 or odds of about
359.3333333 to 1 against.

Total combinations above: 2 592 + 1 854 + 165 + 45 = 4 656

One can then calculate the "two pairs or better" chances by
summing the above, but if the above isn't computed, it's
easier to find the probability of not improving and subtract
from one:

5. No improvement: C(9,3)x4x4x4+C(9,2)x4x4xC(3,1)x3+
9x4xC(3,2)x3x3+C(3,3)x3x3x3 = 11 559 or the probability is
about 0.7128584644; therefore, the probability of improving
to two pairs or better is 0.2871415356. As a check, adding
the combinations: 4 656 + 11 559 = 16215 = C(47,3).


DRAWING TWO TO A PAIR AND A KICKER
==============================

As mentioned in the first paragraph, in the context of a
game, if there are several folds, it will be slightly more likely
as compared to heads up play to catch an ace.

Combinations are now out of C(47,2) = 1 081.

1. Two pairs: 3[44-2]+9C(4,2)+2C(3,2) = 186
or probability of about 0.1720629047 or odds of about
4.811827957 to 1 against.

2. Trips: 2[44-2] = 84
or probability of about 0.07770582794 or odds of about
11.86904762 to 1 against.

3. Full house: 2x3+C(3,2) = 9
or probability of about 0.008325644218 or odds of about
119.1111111 to 1 against.

4. Quads: 1
or probability of about 0.0009250693802 or odds of exactly
1080 to 1 against.

Again, one can simply add the above combinations for
improving to two pairs or better: 186 + 84 + 9 + 1 = 280.
Alternatively, without the calculations above, one can
calculate the probability of not improving:

5. No improvement: C(9,2)x4x4+9x4x2x3+3x3 = 801
or probability of about 0.7409805735; therefore, the
probability of improvement is about 0.2590194265 or about
2.860714286 to 1 against.


DRAWING TO AK
============

Again, as mentioned in the first paragraph, if playing in a
game with some folds, aces and kings are bit more "live" as
compared to heads up play. You normally prefer to draw
three to AK instead of four to an ace since a pair of kings
has a decent chance of winning the pot even though
technically, drawing four to an ace increases the chances
of making a pair; however, if you were the SB or BB, AK
could be the best hand predraw so discarding the king could
cost you the pot.

I've made these calculations before, but with AK(offsuit)
and no card in the ten to queen range, there are 7 772
combinations of improvement to a pair or better. Then,
there are the following "adjustments" to the combinations:

one card in Q/J/T range: -16
two cards in Q/J/T range: -28
AK "onsuit" without third card of suit: +164
AK "onsuit" with third card of suit: +119

Then, there is another "minor adjustment" of +1 if you toss
out a "suited ten" from AKT(suited); presumably, you would
often draw two to AKQ(suited) and AKJ(suited) as a pair of
jacks often wins in a draw game scenario.

Thus, the probability of improving on AK depends on the
cards in the ten to queen range and being "onsuit" and if my
calculations are correct, the average adjustment is very
roughly about +23.1 so the probability is about 0.4807 in
improving to a pair or better on average.

In the best case, the probability of improvement to a pair or
better is (7 772 + 164)/16 215 or about 0.4894233734 or
almost 49%.


DRAWING FOUR TO AN ACE
====================

Calculations are bit more difficult as there are now C(47,4) =
178 365 combinations. Making a flush now depends on how
many cards of the same suit are discarded (could be zero,
one or two) and making a straight depends on how many
cards in the ten to king and deuce to five range are discarded.

Even in the worst cases, drawing four to an ace will improve
to a pair or better more than 1/2 of the time; however, that
is not to say that it is clearly better in the context of a draw
poker game: e.g., if you are the big blind and are heads up
with a small blind that limps, drawing three can give you a
decent chance of bluffing whereas by drawing four, bluffing
won't nearly be as effective against typical opponents.

Here's a nearly "worst case" hand: AKT52

Combinations out of 178 365.

1. Pair:
AA: 3[C(8,3)x64+C(8,2)x16x12+8x4xC(4,2)x9+C(4,3)x27] =
32 388
unseen rank: 8C(4,2)[C(7,2)x16+7x4x12+C(4,2)x9] = 34 848
seen rank: 4C(3,2)[C(8,2)x16+8x4x9+C(3,2)x9] = 9 156
total for one pair: 76 392

2. Two pairs:
Aces up: 3[8C(4,2)(44-4)+4C(3,2)(44-3)] = 7 236
Other: C(8,2)x6x6+8x6x4x3+C(4,2)x3x3 = 1 638
total for two pairs: 8 874

3. Trips:
C(3,2)[C(8,2)x16+8x4x12+C(4,2)x9] + 8C(4,3)[44-4] +
4C(3,3)[44-3] = 4 102

4. Straight:
2x4x4x3x3 = 288

5. Flush: C(10,4) = 210

6. Full house: C(3,2)[8x6+4x3]+3[8x4+4x1] = 288

7. Quads: C(3,3)[48-4]+8 = 52

Total combinations for the above: 90 206

Thus, the probability of making 22+ is 90 206/178 365 or about
0.5057382334 or almost 50.6%, but clearly above 1/2. Thus,
drawing four to an ace will be better than drawing three to AK
if the objective is to maximize the chance of making 22 or
better; however, as mentioned earlier, drawing four is usually
inferior for other reasons in an actual draw poker game. As a
clear example: you are the big blind with the above hand and
the cutoff limps and his range is predominantly TT-QQ. Now,
you simply maximize your chances of making QQ+ which turns
out to be drawing three to AK.

(9*6*41 + 3*3*42) / C(47,3)

=~ 16.0% or 5.3-to-1

There are 9 ranks that are not x,y,z that can pair 6 ways, times 41 ways to choose the last card, plus the 3 ranks x,y,z that can each pair 3 ways, times 42 ways to choose the last card.


2*[36*32/2 + 36*9 + 9*6/2] / C(47,3)

=~ 11.4% or 7.8-to-1

There are 2 nines; times 36*32/2 ways to get 2 other cards that are not x,y,z, plus 36*9 ways to get the other 2 cards where 1 is from x,y,z; pus 9*6/2 ways to get both remaining cards from x,y,z.


[ 2*(9*6 + 3*3) + 9*4 + 3*1 ] / C(47,3)

=~ 1.0% or 97.3-to-1

We can get one of the 2 remaining 9s, with a pair from 9 ranks that are not x,y,z, times 6 ways to get a pair for each rank, OR a pair from one of the 3 ranks x,y,z times 3 ways to get a pair from each rank. Or we can draw 3 cards the same from 9 ranks that are not x,y,z times 4 ways to draw these cards from each rank. OR 3 cards that are the same from x,y, or z, with just 1 way to draw these cards from each rank.


1*45 / C(47,3)

=~ 0.28% or 359-to-1

There is 1 way to get 99, times 45 remaining cards.


Yes, excluding straights, flushes and straight flushes. [EDIT: Those hands not possible]. The above sum to about 28.7% or 2.5-to-1. You could also compute 1 minus the probability of getting 3 different ranks other than 9:

1 -
[ C(9,3)*4*4*4 +
C(9,2)*C(3,1)*4*4*3 +
C(9,1)*C(3,2)*4*3*3 +
C(3,3)*3*3*3 ] / C(47,3)

=~ 28.7% or 2.5-to-1


There are 9 ranks other than 9,x,y,z which have 4 cards each, and there are 3 ranks x,y,z which have 3 cards each. The above counts each possible combination of the 3 cards drawn cards coming from these 2 groups.

This agrees exactly with the sum if we don't round off, so this checks that everything above is correct.


Using the second method above, we get:

1 -
[ C(9,2)*4*4 +
C(9,1)*C(2,1)*4*3 +
C(2,2)*3*3 ] / C(47,2)

=~ 25.9% or 2.9-to-1

This time there are 9 ranks other than 9,A,x,y which have 4 cards each, and there are 2 ranks x,y which have 3 cards each. The above counts each possible combination of the 2 cards drawn cards coming from these 2 groups.

I also did each of these separately and summed, and the answer was exactly the same, so we know this is correct.


Drawing 4 is more likely to draw a pair or better by about 3%.

[EDIT - this ignores straights and flushes, but these make little difference except for AKs].

P(pair or better drawing 4) =

1 - P(no pair drawing 4) =

1 -
[ C(8,4)*4*4*4*4 +
C(8,3)*C(4,1)*4*4*4*3 +
C(8,2)*C(4,2)*4*4*3*3 +
C(8,1)*C(4,3)*4*3*3*3 +
C(4,4)*3*3*3*3 ] / C(47,4)

=~ 50.3%

There are 8 ranks other than A,K,x,y,z which have 4 cards each, and there are 4 ranks K,x,y,z which have 3 cards each. The above counts each possible combination of the 4 drawn cards coming from these 2 groups.


P(pair or better drawing 3) =

1 - P(no pair or better drawing 3) =

1 -
[ C(8,3)*4*4*4 +
C(8,2)*C(3,1)*4*4*3 +
C(8,1)*C(3,2)*4*3*3
C(3,3)*3*3*3 ] / C(47,3)

=~ 47.5%

There are 8 ranks other than A,K,x,y,z which have 4 cards each, and there are 3 ranks x,y,z which have 3 cards each. The above counts each possible combination of the 3 drawn cards coming from these 2 groups.

So we can see that it is more likely to draw a pair or better by drawing 4 cards rather than 3, by about 3%.

You forgot about flushes and straights.

At most this difference between drawing four and drawing three
is about 2.9% and depends on the suits of the discards (if the AK
is "onsuit") and cards in the range 2-5, T-Q. With the hand

AK876

drawing four will result in 495 combinations of flushes and
3x64+256-2 = 446 combinations of straights or 941 extra
combinations subtracted from the 88 657 combinations of
"nonpairing" hands giving 87 716 combinations. Then the
probability of making at least a pair is 1 - 87 716/C(47,4) or
about 0.5082219045.

Drawing three will result in a straight in 64 combinations so
these are subtracted from the 8 507 combinations of
"nonpairing" hands giving 8 443 combinations. Then the
probability of making at least a pair is 1 - 8443/C(47,3) or
about 0.4793092815.

Thus, for this hand, the difference is about 2.8912623%.


At the very far extreme, one could start with AK "onsuit":

AK432

and drawing four will now result in a flush in only 330
combinations and a straight in only another 299 or a total
of 629 combinations. Then the probability of making at
least a pair is 1 - (88 657-629)/C(47,4) or about
0.5064726824.

Drawing three will result in a flush in 165 combinations and
a straight in another 63 combinations for a total of 228
combinations. Then the probability of making at least a pair
is 1 - (8 507-228)/C(47,3) or about 0.4894233734.

Thus, for this hand, the difference is only about 1.7049309%,
but of course, this is a "very extreme" case. Still, drawing
four is better than drawing three by more than 1.7% if the
objective is to make at least a pair.


Of course, the average will be around 2.7% since the hands
are two "extreme" examples.

I didn't forget, I purposely excluded them since her other questions were about hands that contained pairs only, so I took "1 pair or better" to mean "1 pair or more". (*) Of course it makes sense to include them. It makes very little difference on average as my difference was 2.8%, and you found that the average difference was 2.7% by including them. It is useful to know that this number can vary depending on the hand.

(*) Note that her question would "two pairs or better" mean the sum of these percentages? would seem to imply this interpretation is what she intended, and as I pointed out, it excludes straights, flushes, and straight flushes.

Fair enough, but this statement in the OP:

suggests that bonuses were paid for straights, flushes and
straight flushes and being "onsuit" does make a significant
difference when drawing three cards to AK (119/16 215 to
164/16 215 or as much as 1%). Of course, Hello Kitty did
not use "flush", "straight" or any word at all similar; so in
context, it's hard to know precisely what she(?) wanted.

Your interpretation is easily acceptable as explained.


The other technical "nitpick" is that when drawing four to an
ace, the resulting hand will often end up with worse than AK
about 35% of the time. Hello Kitty's post can be interpreted
in a five-card draw poker situation, so there will be many
situations where drawing four is "incorrect" despite the
chances of improving to a pair or better is higher as compared
to drawing three.

Hi,

Thanks for the thorough and detailed responses. They have a lot of stuff in it that is going to take me a while to look at, as I am not very familiar with the math involved.

Just to clarify, in the first two examples involving improving a pair, I didn't include straights and flushes because it is impossible to make a straight or flush. However, with the AKxzy example, straights and flushes would be included, although I realize any difference is still very minimal.

Thanks for the responses. I'll re-read them carefully.

Hi : There are this game in my town, can i beat this game?

You put ante bet to receive 5 cards you can change max 2 cards after that you must choose if you call the bet ( ante *2) , if you fold you loose your ante bet. The house open with minimun 8,8 or better.
The bonus table are : How many time pay the call ( 2 *ante)
Three ok a kind * 2
Straight *3
Flush *5
fullhouse *7
poker *20
Straight Flash *50
Royal Flush *100

How can i program to calculate ev or my advantage, how can determine the best strategy to play this game ?
Can i calculate how many units win if i call with minimun QQ or better ?
If i play with some friends and i know some dead cards, can i modify the ev ? suppose if i know my friends have (T,Q,K,A,9) big cards, the house have less possibility to open whit minimun 8,8 or better, i suppose that i can call with no good hand.

Can you respond my questions ?

Paolo

This is a house game from the sound of it? Then no, you can not beat it.

I get slightly different answers to pooch.

Note, I'm not taking into account that you didn't discard a pair (presumably) so it's slightly more likely you'll make 2pr etc. than if you discarded 3 unknown cards, but this effect should be minuscule.


1) 99YYX: 3.(45.3.41)/(47.46.45)
2) 999XY: 3.(2.45.41)/(47.46.45)
3) 99XXX: (45.3.2)/(47.46.45)
+ 999XX: 3.(2.45.3)/(47.46.45)
4) 9999X: 3.(2.1.45)/(47.46.45)


actual values are 17.1%, 11.4%, 1.11%, 0.277% respectively

You are not taking into account that he can draw the same ranks that he is throwing away (x,y, and/or z), and there are different numbers of those remaining. See my first post in this thread for correct calculations (small part of the tl;dr post).

Guys let me ask you something. When drawing 4 cards holding A high what are my chances of getting pair of aces or better?. I already read above that the chances of getting a pair or better is around 50% but got curious about drawing to pair of aces or better.

Holding Ak i fell like the chances of making pair of kings or better is a little bit better than 30%. I would guess 33% like in holdem. Am i right?