17 Dec 2001
I haven't written about things mathematical in a while, so I think I will
talk about a little probability "paradox" that arises in poker. I've talked
about the value of bluffing in poker to maximize expected return, so now I
will look at particular probabilities themselves.
Now, I always forget the ordering of hands in poker, and when I do, I just
calculate the probabilities of getting each hand in 5-card stud; I remember
calculating for 7-card stud, and even best-chance for 5-card draw, and they
all result in the same ordering of hands. Logically, the least likely poker
hands have the highest precedence.
To begin with, some basic facts:
-using 52-card deck
-4 suits (hearts, clubs, diamonds, spades)
-13 face values per suit (2 - 10, Jack, Queen, King, Ace)
I'm going to ignore small details like royal flush vs. non-royal straight
flush. That's not of interest for this demonstration.
I'm not going to prove the following, because I don't really want to get into
the basics of combinatorics here, but the number of ways to choose k things
from a set of n things (and order doesn't matter) is n_C_k ("n choose
k") which is n!/(k! (n-k)!)
First of all, there are 52_C_5 = 2,598,960 different 5-card poker hands. This
will be the denominator of all the probabilities we are interested in,
because each individual hand is equally likely of being dealt (unless one is
playing against Maverick).
So, here are the possible hands:
Four of a kind (4 cards of same face value, 1 card not matching)
Three of a kind (3 cards of same face value, 2 cards not matching any others)
Two pair (2 cards with same face value, 2 cards with same face value
different from the first 2, and 1 card not matching the other 4 in face
One pair (2 cards with same face value, 3 cards not matching any others in
Full House (3 cards matching in face value, 2 cards matching in face value
but not matching first 3)
Straight flush (5 cards with same suit, and face values in consecutive order)
Flush (5 cards with same suit, but not consecutive face values)
Straight (5 cards with consecutive face values, but not all same suit)
High Card (the leftover category... none of the above)
So let's calculate some probability here!
Four of a kind:
First of all you have to choose the face value of the four matching
cards. There are 13 possible face values. Then, given the face value,
there's only one way to get those 4 with the same face value. Finally, there
are 48 remaining cards to choose 1 among. So here's the calculation of the
number of ways to get four of a kind:
13_C_1 * 4_C_4 * 48_C_1 = 13*1*48 = 624
Probability = 624/2598960 = .00024
Very unlikely - .02% - chance of getting this hand.
Three of a kind:
First of all, choose the face value of the three matching cards, and then out
of the four cards with that face value, choose 3. Finally, one must choose
two =other= face values for the remaining two cards, and for each of those
face values, there are 4 cards to choose 1 from.
number of ways to get three of a kind:
13_C_1*4_C_3*12_C_2*4_C_1*4_C_1 = 13*4*66*4*4 = 54912
Probability = 54912/2598960 = .0211
Not bad, one hundred times more likely than getting 4-of-a-kind.
First we have to choose the face values of the two pairs, then for each pair,
we have to choose 2 cards from the 4 with those face values. Finally, there
are 44 cards left over not matching in face value to the two pairs, and we
need one of those cards.
number of ways to get two pairs:
13_C_2*4_C_2*4_C_2*44_C_1 = 78*6*6*44 = 123552
Probability = 123552/2598960 = .0475
So this is about as twice as likely as getting three-of-a-kind.
First we choose the suit of the pair, then we get 2 of the four cards with
that face value. Then, for the three remaining cards, we have to choose
three different face values for these cards out of the 12 face values
remaining. Finally, for each of the remaining face values, one has to pick 1
card out of the 4 cards with those face values.
number of ways to get one pair:
13_C_1*4_C_2*12_C_3*4_C_1*4_C_1*4_C_1 = 13*6*220*4*4*4=1098240
probability = 1098240/2598960 = 0.4226
Wowza! This is getting into the "reasonable probability" range, as in, you
expect to see this hand crop up often enough you during poker games, as
opposed to its less likelier cousins...
First, we have to choose the face value for the three-of-a-kind, then pick 3
of the 4 cards with that face value. Then, we choose the face value for the
pair, and pick two of the four cards with that face value.
number of ways to get a full house:
13_C_1*4_C_3*12_C_1*4_C_2 = 13*4*12*6 = 3744
probability = 3744/2598960 = .00144
More likely that a four-of-a-kind, but less likely than three-of-a-kind (as
one would expect).
Now, I am going to just go with the convention that Ace can be low or high
(but not both), so the possible consecutive orderings start with: A,2,3,4,5
and go up to: 10,J,Q,K,A. So there are 10 possible consecutive runs. As
well, one needs to choose the suit of the run.
number of ways to get straight flush:
10*4_C_1 = 10*4 = 40
Proability = 40 / 2598960 = .0000153
This is even unlikelier than 4-of-a-kind, so this will have higher
First one has to pick the suit, and then pick 5 cards from that suit. Then
one must subtract all the straight flushes, since those will be counted in
number of ways to get a flush:
4_C_1*13_C_5 - 40 = 4*1287 - 40 = 5108
probability =5108/2598960 = .001965
More likely than full house, but less likely than three-of-a-kind.
First, one picks one of the ten consecutive runs, and then, for each card
in the run, there are 4 possible suits. Of course, one must subtract off the
number of straight flushes, which are counted in this method.
number of ways to get a straight:
10*4^5 - 40 = 10200
probability = 10200/2598960 = .003925
More likely than flush, less likely than three-of-a-kind.
This is the leftover one, so to get the number of ways, simply add up all the
others and subtract from the total number of poker hands:
number of ways to get high card:
2598960 - (10200 + 5108 + 40 + 3744 + 1098240 + 123552 + 54912+624)
probability = 1302540/2598960 = 0.5012
So the probability is a little greater than 1/2 that one will get "nothing".
Now we have the ordering of hands: (highest to lowest precedence)
Four of a kind
Three of a Kind
Ok, it's a little easier to just check online to see what the ordering of
hands are, and one can tell that, yes indeed, one has this ordering of hands
(with five-of-a-kind being highest precedence in a game with wild cards.)
What's that about wild cards? Well, many people don't like the fact that
over half the time one doesn't even get one pair. So, to make "more
interesting" hands more likely, they change some of the existing cards to
wild cards, or throw in extra cards (jokers) to act as wild cards. Wild
cards generally can take on any suit and any face value. The ranking of
hands remains the same, except for that extra 5-of-a-kind.
Let's throw in a single joker as a wild card, and let's see how that affects
probability of hands. Now there are two ways to make any kind of hand - a
"natural" way (no wild cards -- the numbers we got from above) and the
"unnatural" way (with the wild card - the extra we have to calculate now).
First we have to update our denominator:
Number of poker hands: 53_C_5 = 2869685
So, let's recalculate the probabilities. This is going to be a little messy,
but since I've added the joker, rather than turned one of the regular cards
into a wild card, it's not as messy as it could be.
You must get the wild card to get this hand, plus a "natural" four-of-a-kind.
ways of getting 5-of-a-kind:
13_C_1*4_C_4*1 = 13
probability = 13/2869685 = .00000453
One can get a 4-of-a-kind in a natural way (and not getting the joker), or by
having three-of-a-kind, the joker, and one other card.
ways of getting 4-of-a-kind:
624 + 13_C_1*4_C_3*1*48_C_1 = 624 + 13*4*48 = 624 + 2496 = 3120
probability = 3120/2869685 = .001087
The probability for 4-of-a-kind has bumped way up due to this wild card!
One can get it naturally, or by having two pair and the joker (you can't get
it with three-of-a-kind and the joker, because that would count as
four-of-a-kind, since it's ranked higher!)
ways of getting full house:
3744 + 13_C_2*4_C_2*4_C_2*1 = 3744 + 78*6*6 = 2808 + 3744 = 6552
probability = 6552/2869685 = .002283
One can get it naturally, or by having one pair and the joker.
ways of getting three-of-a-kind:
54912 + 13_C_1*4_C_2*1*12_C_2*4_C_1*4_C_1 = 54912 + 13*6*66*4*4 =137280
probability = 137280/2869685 = .04784
One can get two pair naturally, and that's it, because if you have one pair
and a joker, due to hand precedence, it would count as a three-of-a-kind.
ways to get two pair:
probability = 123552/2869685 = .043054
Hey! Wait a sec! I have two pair being less likely to occur than
three-of-a-kind with one wild card, and three-of-a-kind has higher
precedence! Well, that just won't do. Why don't we change the precedence
for this game, so that two pair has higher precedence over three-of-a-kind?
In that case, one can get three-of-a-kind only "naturally" (because if you
had one pair and a joker, you would pick the higher precedence two pair
hand): that's 54912 ways, and then to get two pair... hey, you would have at
=least= 123552 ways to get two pair via the natural way!
And that, my friends, is the trouble with wild cards. If you let even one
wild card in the game, you will always get a contradiction between hand
precedence and hand probability. The fact is, one wants to get the highest
hand possible, so one will always, when getting a wild card, fit it to make
the best hand possible.
Moreover, the more wild cards there are in the game, the more and more likely
flush hands are, which are pretty high-ranked hands. I will leave it to the
reader as an exercise to see what happens to the probabilities when =two=
jokers are thrown into the mix.
So how do casinos deal with wild cards? Generally, they reduce the payout
for hands, and don't worry about the precedence. If you're playing video
poker, then hand precedence only counts re: payout. And since they horribly
underpay you for your results, they're going to make money off of you, wild
cards or no wild cards.
But that's a subject for tomorrow.