Fact: There are 3744 ways that a full house can be formed if one is holding a five card hand in poker (where, of course, the position/order that the cards are held in the hand is irrelevant). Keep in mind that (KH, KS, KD, 7D, 7C) is a different Full House than (KH, KS, KD, 7D, 7H)

Fact: The probability that a full house would be dealt to a player in a five card hand would be 3744/(52!/(5!*47!)) = 3744/2,598,960.

In Texas Hold 'Em, one can use 7 cards to form a hand. Can somebody tell me how to figure the probability of getting a Full House in Texas Hold 'Em? Note: the 2 extra cards must be of different values than the two values used to form the Full House.

I think the formula is e = mc squared divided by pi. That's if you are playing with 5 or 6 players.

If you play against the experts on this forum, they can deal a full house to themselves from a shuffled deck without looking.

Guess magicians are naturally lucky
They always say poker requires skill and when they say "DEAL"...they never say "specifically" from where

Ok, first let's use basic intuition to get a idea of where the probability should be. It is certainly true that in there are 3,744 different full houses. And in a game where 5 cards are dealt, there are 2,598,960 starting hands.

How the number of starting hands is calculated is shown above by Munseys_Magic. But what is not shown is how to find the number of full houses. A simple way to view this problem is this:

In a 5 card hand, to have a full house, you must have three cards the same, and two cards the same of a different value. Your hand consists of 3 "x" cards, and 2 "y" cards. xxxyy

Let's use Aces to designate the three of the kind. There are 4 different three card combinations that could be formed with the four aces without respect to order. 4!/(3!) = 4

Next, the pair could be of any value, and there are 6 ways to form a two card combination from four cards. 4!/(2!*2!) = 6 Since there are twelve other values of cards that could complete the full house (besides the aces) we multiply 6 by 12 and get 72. There are 72 different pair combinations to complete our three of a kind (of which for the aces, there are four different combinations) So now, we multiply 4*72 and get 288. This gives us the number of full houses there are IF the three of a kind is aces. Now since there are 13 values of cards, which could potentially be the three of a kind, we multiply 288 by 13 to get the total number of full houses possible = 3,744.

Oh yeah, back to that basic inuition thing.....the probability of being dealt a FH in in a five card hand is 3,744/2,598,960 as mentioned before. It equals .14% . Now, we could look at hold'em as a 7 card hand. You are essentially being dealt 7 cards to choose the best five from. So, the fact is there are more cards being dealt out, that naturally implies more combinations of card as compared to a 5 card hand. Right off the bat, you should be able to deduce that your chances of getting a full house are largely increased. By how much we will soon find out.

So, just two more cards are being dealt. How does that affect the number of "hands" that are possible? It drastically changes this. In hold'em, the number of 7 card combinations is 52!/(7!*45!) = 133,784,560

So now we can look at this problem like this - you need your full house, but there are going to be two more cards dealt. xxx yy zz

We already know the amount of full houses possible : 3,744. But considering two more cards are being dealt, the full house could be scattered all over the place (i.e. one card in your hand, three on the flop, and one on the river....or two in your hand one on the flop, and one each on the turn and river...etc. you get the point.)

How many different ways can the full house be scattered? Well, there are 47 cards left right? There are 47!/(2!*45!) combinations those two cards can make = 1,081.

Multiply those two, and we get the number of different ways ANY full house can be scattered between the 7 cards. 1081*3744 = 4,047,264.

But there is one more problem to tackle. What is the remaing two cards actually make your hand better than a full house? We have to eliminate these scenarios. Let's go back and pretend that you have three aces in your full house. For simplicity, let's designate your "pair" as jacks.

If the remaining two cards dealt are both Jacks, you clearly have a four of a kind. So that's one scenario. If only one more jack is dealt (stay with me here), you are still in the clear since you will have aces over jacks. Remember, we are dealing with a full house consisting of three aces. We will account for the reverse situation soon. But, if an ace is dealt you will also improve your hand to four of a kind.

There is only one ace left, and there are 47 cards left unseen. The ace can mix with any of those other 46 to form a two card combo. (Now remember, we don't care about order specifically) So there are 46 two card scenarios with an ace in it, and there is also, the "jack jack" scenario. 47 total combinations that will make things worse for you if your full house has three Aces. Now, you could of course have three of a kind of any value so we take 47 and multiply it by 13 to get 611.

So of the 4,047,264 scenarios where a full house is within the 7 cards, 611 of those scenarios produce a hand which is actuallly better than a full house. So, we have to subtract them out. The probability of being dealt a full house in a 7 cards hand is thus:

(4,047,264-611)/133,784,560 =3.02%

Now let's not forget that this is an unconditional probability.


-C

Whoops, I did forget something.

Instead of going through it all, instead of subtracting 611, subtract 7301.
It's late, my bad.

MAY I recomend a book by the "wizzard of odds"? Lots easier than guessing.

Kudos Craig.

Andrei

If AAAKK comes and your sitting with AK you have a dozen+ full houses not to mention three four of a kind and so on. I would get Scarne on Cards. I have it but it's a long answer.

Any 3 Aces with any 2 Ks and the 3 ks with any 2 Aces - so I don’t know is that 18 boats in one hand?

C'mon people, let's get something going here. Don't make me give out the real answer for free.

Craig,

Thank you, thank you, thank you!!

By the way, if one subtracts 7,301, it still comes out to be 3.02% because we're dealing with, comparitively speaking, such large numbers.

Thanks again!

Jim

Dannydoyle implies with his post that my answer is wrong. He is indeed correct - the answer as given is incorrect.

The main thing about poker, and every other game for that matter, is that memorization of probabilities is useful to a degree - but, the more important thing is to truly understand WHY a probability is as it is.

What is wrong with my answer? Think about it.

Craig - I don't want to run the wrong track.

Is "Multiply those two, and we get the number of different ways ANY full house can be scattered between the 7 cards. 1081*3744 = 4,047,264." where the problem is?

In other words, are pretty much all steps correct up to that final phase?

Andrei

Here's a reply that I got from a math site from Drexel University. It actually gives the probabilities for all hands!
*********************************************************************************

There are C(52,7) total ways of getting 7 cards out of a standard
deck. That's 133,784,560 possible 7 card hands.

Now, let's go through the number of ways to get each particular hand.

Royal Flush: With five cards, there are only 4 possible royal flushes
(one of each suit). With 7 cards, we have to consider the 6th and 7th
cards. After the 5 cards for the royal flush, there are 47 cards left
over. There are C(47,2) = 1081 ways of picking the last two cards, or
4324 total Royal flushes possible.

Prob(Royal flush) = 4*1081 / C(52,7) = 0.00323%

Straight Flush: There are 9 possible straights possible for a straight
flush that is not a royal flush, one each for the lowest card in the
straight being anything from ace to 9. There are 4 suits. The last
two cards can be any card except the card making a higher straight
flush. For example, if you had A-2-3-4-5 of spades, you don't want to
let one of the other cards be a 6 spades, or you will be counting that
possibility twice. That leaves 46 cards for the last two, and C(46,2)
= 1035. 9*4*1035 = 37260 straight flushes possible.

Prob(Straight flush) = 4*9*C(46,2) / C(52,7) = 0.0279%

Four of a kind: There are 13 different unique 4-of a kinds (one for
each rank). There are C(48,3) ways of picking the remaining three
cards. So, there are 13*C(48,3) = 224,848 possible Four of a Kinds.

Prob(Four of a kind) = 13*C(48,3) / C(52,7) = 0.168%

Full Houses: This gets a bit complicated. There are 3 separate cases
we must consider.

Case 1: We could have two sets of triples and an extra card. There are
C(13,2) ways of picking which two ranks are the triples. For each
triple, there are C(4,3) ways of picking which 3 of the 4 cards of a
rank are in the triple. Finally, there are 44 extra cards to be used
as the 7th card. That's a total of C(13,2)*C(4,3)*C(4,3)*44 = 54,912
ways of getting this type of full house.

Case 2: 1 set of triples, 2 different unique pairs. There are 13 ways
to pick the rank of the triple (any of one the ranks). There are
C(12,2) ways to pick the rank of the pairs. For the triples, there are
C(4,3) ways of picking the suits. For each pair, there are C(4,2) ways
of picking the suits of the pair. That's a total of
13*C(12,2)*C(4,3)*C(4,2)*C(4,2) = 123,552 ways of getting this type of
full house.

Case 3: 1 set of triples, 1 pair, 2 blanks (don't match either the
triple, the pair, or each other). There are 13 ways to pick the rank
of the triples. There are 12 ways to pick the rank of the pair. And
there are C(11,2) ways to pick the ranks of the blanks. There are
C(4,3) ways of picking the suits of the triples. There are C(4,2) ways
of picking the suits of the pair. And for each blank, there are 4
suits available. So, there's a total of
13*12*C(11,2)*C(4,3)*C(4,2)*4*4 = 3,294,720 ways of getting this type
of full house.

In all, there are 3,473,184 ways of getting a full house.

Prob(Full house) = 3,473,184 / C(52,7) = 2.60%

Flush: For a flush, we again have 3 cases. We'll take each one at a time.

Case 1: All 7 cards in one suit. There are C(13,7) ways of picking 7
different cards in one suit. But we want to get rid of possibilities
for straights. There are 8 ways all 7 cards could form a 7-card
straight. There are 47 ways to form 6 card straights, and there are
162 ways to form 5 card straights from the 7 cards. That's a total of
217 ways. So, there are 4*(C(13,7) - 217) = 1499 possible flushes of
this type.

Case 2: 6 cards of one suit, and a blank. There are C(13,6) ways of
picking 6 different cards in one suit. There are 9 ways to get a 6
card straight here and 62 ways of getting a 5 card straight. And 39
ways the blank could come up. That's a total of 4*(C(13,6) - 71)*39 =
256,620 possible flushes of this type.

Case 3: 5 cards of one suit, and a blank. There are C(13,5) ways of
picking 5 different cards in one suit. We remove the 10 possible
straights to get C(13,5) - 10. The last two cards cannot give us a
hand better than a flush. So, we compute the probability directly.
C(39,2) ways of picking the last two cards. 4*(C(13,5) - 10)*C(39,2) =
3,785,028

Adding them together, there are 4,043,147 ways of getting a flush.

Prob(Flush) = 4043147/C(52,7) = 3.02%

Straights: We calculate this in multiple parts as well.

Case 1: 7 distinct cards (no paired cards). Above, we calculated 217
ways to get a 5 card straight from 7 cards. All we have to do now is
discount flushes. There are 4 ways to choose all 7 cards in any of the
4 suits. For 6 cards in the same suit, there are 4 suits, and C(7,6)
ways of picking the 6 cards, and 3 choices for the suit of the
remaining card, or 4*C(7,6)*3 = 84 ways of a flush in this situation.
For 5 cards in the same suit, there are 4 suits, C(7,5) ways of
picking the 5 cards, and 3 choices for suit for each of the remaining
two cards, for a total of 4*C(7,5)*3*3 = 756. We subtract 756, 4, and
84 from 4^7, the total number of ways the cards can be suited, 4^7 -
756 - 4 - 84 = 15540. So, with 7 distinctly ranked cards with a
straight but without a flush, there are 217*15540 = 3,372,180 ways of
making it happen.

Case 2: 6 distinct cards (last card pairs one of the others). There
are 9 ways of have 6 card straights. We can also have a 5 card
straight with the 6th distinct rank separate. If the straight begins
with A or 10, there are 7 choices for the 6th card. Otherwise, there
are 6 choices. That's a total of 2*7 + 6*8 = 62 ways. In all, there
are 71 ways of producing a 5 card straight from 6 distinct cards.

There are 6 choices for which distinct card is paired. There are
C(4,2) = 6 ways to pick which suits make up the pair. The remaining
cards can be chosen from any suit, so there are 4^5 ways to pick the
suiting. We now get rid of flushes. There are 4 ways to pick all the 5
non-paired cards all in the same suit. In addition, we can't have 4 of
them match the suit of one of the paired cards. There are C(5,4) ways
of picking which 4, 2 choices for which paired card is picked for
suit, and 3 choices for the suit of the 5th card (that is a different
suit) for a total of 4 + C(5,4)*2*3 = 34. So, we subtract this from
the total ways to suit the 5 cards: 4^5 - 34 = 990. That makes for
71*36*990 = 2,530,440 ways of a straight of this type.

Case 3: 5 distinct cards (must include two pairs or a three of a
kind). There are 10 ways to make a straight with 5 distinct cards.
First, let's consider the three of a kind. There are 5 different
possible three of a kinds. There are C(5,4) for which suits comprise
the trips. The suits of the remaining 4 cards can be any of the 4
suits, for a total of 4^4 = 256. But we should eliminate the three
possibilities that match the suits with any in the trips (to prevent
flushes). So, with trips, that's a total of 10*5*4*(256-3) = 50,600.

For the case with two pairs, there are C(5,2) ways of picking which
two cards are the two pair. For each pair, there are C(4,2) = 6 ways
of picking suits. 6*6 = 36. But now, we have to split the 36 up into
cases to get rid of flushes.

6 of them have each pair matching two suits. 24 of them have one suit
matched, and 6 of them have no matching suits.

There are 4^3 = 64 ways of suiting the three remaining cards in the
hand. Of these, if two suits are matched, we don't want either
possibility of flush to come up. So, in this case, we have 62 ways of
suiting the remaining cards. If only one suit matches in the two
pairs, there's only 1 way of matching the suit, and 63 safe ways of
suiting the remaining cards. If both two pairs have different suits,
all 64 ways of suiting are safe. So, for a 5 card straight with trips
or with two pair, that's a total of: 50600 + 10*C(5,2)*[6*62 + 24*63 +
6*64] = 277,400 ways of getting a straight.

Adding all 3 cases together, we get 6,180,020 ways of getting a straight.

Prob(Straight) = 6,180,020/C(52,7) = 4.62%

Three of a kind: A hand that is three of a kind must have 5 distinct
ranks of cards (fewer and you have a full house or two pair or
something else). There are C(13,5) ways of picking the ranks, of which
we remove the 10 possible straights. That leaves 1277 possibilities.
We can pick any of the 5 ranks for the trips, and there are C(5,4)
ways of suiting the trips. Of the remaining 4 cards, we want to get
rid of the possibility of a flush. There are 4^4 = 256 ways of suiting
the 4 cards and 3 ways of completing the flush. So, there are 253 safe
ways of completing the flush.

That's a total of 1277*5*C(5,4)*253 = 6,461,620 ways of getting a
three of a kind.

Prob(Three of a kind) = 4.83%

Two pair:

Case 1: As above, a hand that is two pair can have 5 distinct ranks,
leaving 1277 qualifying hands 5 rank hands. As in the case above for
straights, there are 10 ways of picking which two pair and [6*62 +
24*63 + 6*64] ways of not also making a flush. That's a total of:
1277*10*[6*62 + 24*63 + 6*64] = 28,962,360 ways of getting two pair of
this type.

Case 2: Three pairs plus a blank card. There are C(13,3) ways of
picking the rank of the cards. For each pair, there are C(4,2) ways of
picking the suits, and 40 cards left over for the blank. That's a
total of C(13,3)*C(4,2)^3 * 40 = 2,471,040 ways of getting two pair of
this type.

In all, there are 31,433,400 ways of getting two pair.

Prob(Two Pair) = 23.50%

Pair: In a pair, there must be 6 distinct ranks in the hand. Above, we
saw there were C(13,6) - 71 = 1645 ways of doing this without
straights showing up. There are 6 choices for which rank is paired,
and C(4,2) ways of picking the suits of the pair. The suits of the
remaining cards can be any of the 4, leaving 4^5 choices. We want to
eliminate flushes now. We can't have all 5 cards in the same suit.
This can occur in 4 ways. We also cannot have 4 of them match suits
with either card in the pair. There are C(5,4) ways of choosing which
4 cards, 3 ways of picking the suit of the extra card, and 2 suits to
choose from in the pair, or C(5,4)*2*3 = 30. So, of the 4^5 ways of
suiting the 5 cards, 4^5 - 30 - 4 = 990 are safe. In total, there are
1645*6*6*990 = 58,627,800 ways of getting a pair.

Prob(Pair) = 43.82%

High card: For a high card situation, there must be 7 distinct ranks.
There can be no pairs. Getting rid of straights, as above, we are left
with 1499 sets of 7 cards without a straight. Now, we have to get rid
of flushes. There are 4 ways for all 7 cards to have the same suit.
For 6-flushes, there are C(7,6) ways of picking the 6 cards, 4
possible suits, and 3 possible suits for the last card. For 6-flushes,
C(7,6)*4*3 = 84. For 5-flushes, there are C(7,5) ways of picking the 5
cards, 4 possible suits, and 3 choices for each of the remaining two
cards for C(7,5)*4*3*3 = 756 ways to get a 5-flush. Of the 4^7
possible suitings of the 7 cards, 756 + 84+4 = 844 are bad. So, there
are a total of 1499*(4^7 - 844) = 23,294,460 possible ways to get a
high card.

Prob(High card) = 17.41%

If we add up all the possibilities we calculated for each type of
hand, we get 133,784,560, which is the appropriate number, giving us
an indication our calculations are correct.

Here's a chart summarizing the information:

Hand Possibilities Probability
Royal Flush 4,234 0.00324%
Straight Flush 37,260 0.0279%
4 of a kind 224,848 0.168%
Full House 3,473,184 2.60%
Flush 4,043,147 3.02%
Straight 6,180,020 4.62%
Three of a Kind 6,461,620 4.83%
Two Pair 31,433,400 23.50%
Pair 58,627,800 43.82%
High Card 23,294,460 17.41%

So, you see that in 7 card poker games, high card occurs less than 20%
of the time and both two pair and a single pair are much more likely
outcomes (a single pair is more than twice as likely as high card in 7
cards).

I don't realy care about these numbers... lol I just stack 3 of kind on bottom and I deal 2 of kind in hand, that's all...

Paul

Ah Paul, you are such an animal

Regards,

Paul H

Un huh

Bingo

Bingo! No not kind of full house, we are talking about poker.. I think.

What are the odds of getting BINGO on any given card?

Don't know, I think it is a certainty, but they stop playing before you get it.