Saturday, April 7, 2007
Progress
The next things I'm going to try to compute are the number of pair hands with flush draws, pair hands with open-ended straight draws, no-pair hands with flush draws and no-pair hands with straight draws. Maybe I'll try to compute the number of hands in different no-pair categories and numbers for gutshot draws too, but I think there are very few situations in 5-card draw where it makes sense to draw to inside straights.
categories of two pairs
I calculated earlier that there are 1584 threes-up hands:
n(threes up hands) = 1584
n(fours up hands) = 1584 * 2 = 3168
n(fives up hands) = 1584 * 3 = 4752
n(sixes up hands) = 6336
n(sevens up hands) = 7920
n(eights up hands) = 9504
n(nines up hands) = 11088
n(tens up hands) = 12672
n(jacks up hands) = 14256
n(queens up hands) = 15840
n(kings up hands) = 17424
n(aces up hands) = 19008
n(threes up hands) = 1584
n(fours up hands) = 1584 * 2 = 3168
n(fives up hands) = 1584 * 3 = 4752
n(sixes up hands) = 6336
n(sevens up hands) = 7920
n(eights up hands) = 9504
n(nines up hands) = 11088
n(tens up hands) = 12672
n(jacks up hands) = 14256
n(queens up hands) = 15840
n(kings up hands) = 17424
n(aces up hands) = 19008
flush categories
There are no ordinary 5-high or 6-high flushes because they would be straight flushes. I'm going to calculate the number of ordinary flushes in each category. One of the rank combos for each category (two for aces) will be a straight combo, so it wouldn't count as an ordinary flush.
n(seven-high flushes) = suit combos * (rank combos - straight combos) = 4 suits * ((5*4*3*2/4*3*2*1)-1) seven-high flushes/suit = 16 seven-high flushes
I'll explain the computation for rank combos in a little more detail. I pretend like I've looked at the 7 and I somehow know that I have a 7-high flush. There are five ranks below a 7. Once I've taken the 7 out of my hand, the first of the remaining cards can be any of the 5 cards that would give me a chance to make a 7-high flush. The next card can be any of the four remaining cards and so on. How these cards are arranged doesn't matter, so I divide the arranged rank combos by the number of possible arrangements (4*3*2*1).
n(eight-high flushes) = 4 suits * ((6*5*4*3/4*3*2*1)-1) eight-high flushes/suit = 56 eight-high flushes
n(nine-high flushes) = 136
n(ten-high flushes) = 276
n(jack-high flushes) = 500
n(queen-high flushes) = 836
n(king-high flushes) = 1316
n(ace-high flushes) = 4 suits * ((12*11*10*9/4*3*2*1)-2) ace-high flushes/suit = 1972
There are two straight combos for aces because the ace is interpreted as the highest ranking card of a flush, but it would be smallest ranking card of a five-high straight. We already counted the 5-high straight flush as a straight-flush, so we can't count a2345 suited as an ace-high flush.
n(seven-high flushes) = suit combos * (rank combos - straight combos) = 4 suits * ((5*4*3*2/4*3*2*1)-1) seven-high flushes/suit = 16 seven-high flushes
I'll explain the computation for rank combos in a little more detail. I pretend like I've looked at the 7 and I somehow know that I have a 7-high flush. There are five ranks below a 7. Once I've taken the 7 out of my hand, the first of the remaining cards can be any of the 5 cards that would give me a chance to make a 7-high flush. The next card can be any of the four remaining cards and so on. How these cards are arranged doesn't matter, so I divide the arranged rank combos by the number of possible arrangements (4*3*2*1).
n(eight-high flushes) = 4 suits * ((6*5*4*3/4*3*2*1)-1) eight-high flushes/suit = 56 eight-high flushes
n(nine-high flushes) = 136
n(ten-high flushes) = 276
n(jack-high flushes) = 500
n(queen-high flushes) = 836
n(king-high flushes) = 1316
n(ace-high flushes) = 4 suits * ((12*11*10*9/4*3*2*1)-2) ace-high flushes/suit = 1972
There are two straight combos for aces because the ace is interpreted as the highest ranking card of a flush, but it would be smallest ranking card of a five-high straight. We already counted the 5-high straight flush as a straight-flush, so we can't count a2345 suited as an ace-high flush.
Summary
Here are the number of hands in each category:
Straight flush: 40
Quads: 624
Full house hands: 3 744
Flushes: 5 108
Straights: 10 200
Trip hands: 54 912
Two pair hands: 123 552
Pair hands: 1 098 240
No pair hands: 1 302 540
You can find the odds of being dealt any of these hands by dividing the number by 2 598 960
Straight flush: 40
Quads: 624
Full house hands: 3 744
Flushes: 5 108
Straights: 10 200
Trip hands: 54 912
Two pair hands: 123 552
Pair hands: 1 098 240
No pair hands: 1 302 540
You can find the odds of being dealt any of these hands by dividing the number by 2 598 960
Odds of no pairs
A "no pair hand" can be defined as a hand that doesn't fit into any other category. The number of no pair hands can be found by subtracting the number of hands in the other categories from the total number of hands. If you do this you get the following result:
n(no pair hands) = n(total hands) - n(hands that beat ace-high) = (2 598 960 - 1 296 420) no pair hands = 1 302 540 no pair hands
NB: You could also calculate the number of no pair hands by calculating kicker combos where all five cards have to have different ranks, and subtract the number of straights and flushes from this. 4 out of the 1024 possible suit combinations for hands with five cards of different ranks will be flushes.
n(no pair hands) = n(total hands) - n(hands that beat ace-high) = (2 598 960 - 1 296 420) no pair hands = 1 302 540 no pair hands
NB: You could also calculate the number of no pair hands by calculating kicker combos where all five cards have to have different ranks, and subtract the number of straights and flushes from this. 4 out of the 1024 possible suit combinations for hands with five cards of different ranks will be flushes.
Odds of two pairs
Two pair hands are defined by the rank combos of the pairs, the suit combos of the biggest and the smallest pair, and the kicker combos. The first pair can be bigger or smaller than the second pair. We're not interested in the arrangement of the pairs, so we divide the arranged rank combos by two.
n(two pair hands) = rank combos * suit combos(big pair) * suit combos(small pair) * kicker combos = (13*12/2) pair combos * 6 suit combos/pair combo * 6 suit combos/suit combo * 44 two pair hands/suit combo = 123 552 two pair hands
Aces up is two pair where the biggest pair is a pair of aces, kings up is two pair where the biggest pair is a pair of kings and so on. These categories might not be enough to decide if one hand beats another, since two players might have two pair in the same category.
n(threes up hands) = small pair combos * suit combos * kicker combos = 1 small pair * 36 suit combos/small pair * 44 threes up/suit combo = 1584 threes up
I just lumped together the suit combos here, because there are always 6 suit combos for each pair and these combos are independent from each other. 1584 is actually the number of possible hands for any combination of two specific pairs (1584 "fours and deuces hands", 1584 "aces and kings hands" and so on). The only pair that is smaller than threes, are deuces. Fours up can be either "fours and threes" or "fours and deuces", so there are twice as many fours-up hands as trees-up hands.
n(two pair hands) = rank combos * suit combos(big pair) * suit combos(small pair) * kicker combos = (13*12/2) pair combos * 6 suit combos/pair combo * 6 suit combos/suit combo * 44 two pair hands/suit combo = 123 552 two pair hands
Aces up is two pair where the biggest pair is a pair of aces, kings up is two pair where the biggest pair is a pair of kings and so on. These categories might not be enough to decide if one hand beats another, since two players might have two pair in the same category.
n(threes up hands) = small pair combos * suit combos * kicker combos = 1 small pair * 36 suit combos/small pair * 44 threes up/suit combo = 1584 threes up
I just lumped together the suit combos here, because there are always 6 suit combos for each pair and these combos are independent from each other. 1584 is actually the number of possible hands for any combination of two specific pairs (1584 "fours and deuces hands", 1584 "aces and kings hands" and so on). The only pair that is smaller than threes, are deuces. Fours up can be either "fours and threes" or "fours and deuces", so there are twice as many fours-up hands as trees-up hands.
Odds of a full house
A full house hand consists of trips of a certain rank, and a pair of a different rank. I have already stated the parameters that define trips and pairs. When you have a full house, there are no kickers so you ignore kicker combos. Trip aces and a pair is called aces full, trip kings with a pair is called kings full and so on.
n(aces full hands) = suit combos(trips) * pair combos * suit combos(pair) = 4 suits * 12 pairs/suit * 6 aces full hands/pair = 288 aces full hands
There are also 288 kings full hands, 288 queen full hands and so on.
n(full house hands) = trip combos * suit combos(trips) * pair combos * suit combos(pair) = 13 trips * 4 suits/trip * 12 pairs/suit * 6 full house hands/pair = 3744 full house hands
NB: Once you have trips of a certain rank, you can't have an additional pair of the same rank. There are only 12 possible pairs for a certain category (jacks full for instance) of full house hands.
n(aces full hands) = suit combos(trips) * pair combos * suit combos(pair) = 4 suits * 12 pairs/suit * 6 aces full hands/pair = 288 aces full hands
There are also 288 kings full hands, 288 queen full hands and so on.
n(full house hands) = trip combos * suit combos(trips) * pair combos * suit combos(pair) = 13 trips * 4 suits/trip * 12 pairs/suit * 6 full house hands/pair = 3744 full house hands
NB: Once you have trips of a certain rank, you can't have an additional pair of the same rank. There are only 12 possible pairs for a certain category (jacks full for instance) of full house hands.
Wednesday, April 4, 2007
Odds of pairs
Pair hands are defined by the rank of the pair, the suit combos and the kicker combos. The number of arranged suit combos are 12. The first (paired) card can be any of the 4 suits, and the second can be any of the remaining 3 suits. The arrangement of the suits doesn't matter, so there are actually 6 suit combos. The first kicker can be any of the 48 cards with a different rank than the pair. The second kicker can be any of the 44 cards that doesn't pair the first kicker or equals the pair in rank. The third kicker can be any of the 40 cards that doesn't pair any of the other kickers or equals the rank of the pair. There are 3*2*1 = 6 different ways to arrange the kickers.
n(pair of ace hands) = suit combos * kicker combos = 6 suit combos * (48*44*40/6) pair of ace hands/suit combo = 84480 pair of ace hands
There are also 84480 pair of king hands, 84480 pair of queen hands and so on.
n(pair hands) = pair combos * suit combos * kicker combos = 13 pairs * 6 suits/pair * 14080 pair hands/suit = 1098240 pair hands
n(pair of ace hands) = suit combos * kicker combos = 6 suit combos * (48*44*40/6) pair of ace hands/suit combo = 84480 pair of ace hands
There are also 84480 pair of king hands, 84480 pair of queen hands and so on.
n(pair hands) = pair combos * suit combos * kicker combos = 13 pairs * 6 suits/pair * 14080 pair hands/suit = 1098240 pair hands
Odds of trips
I define trip hands by the suit combos, the rank of the trips and the kicker combos. The suit combos are most easily defined by the suit of the card you would need to make quads, so there are four suit combos. The kickers have to be of different ranks, both of them different from the rank of the trips. There are 48 cards that meet this requirement for the first kicker, and 44 for the second kicker. If you multiply these numbers, you get the number of arranged kicker combos. The arrangement doesn't matter, so the number of kicker combos are half as big as the number of arranged kicker combos.
n(trip ace hands) = suit combos * kicker combos = 4 suits * (48*44/2) trip ace hands/suit = 4224 trip ace hands
There are also 4224 trip king hands, 4224 trip queen hands and so on.
n(trip hands) = trip combos * suit combos * kicker combos = 13 trips * 4 suits/trip * 1056 trip hands/suit = 54912 trip hands
n(trip ace hands) = suit combos * kicker combos = 4 suits * (48*44/2) trip ace hands/suit = 4224 trip ace hands
There are also 4224 trip king hands, 4224 trip queen hands and so on.
n(trip hands) = trip combos * suit combos * kicker combos = 13 trips * 4 suits/trip * 1056 trip hands/suit = 54912 trip hands
Odds of quads
In ordinary 5-card draw, the kicker(sidecard) doesn't play. Two people can't have the same quads. I earlier calculated the number of possible hands you can get dealt, and all five cards was taken into account when I made this calculation. Therefore, I have to define quad hands by the kicker card as well as the rank of the quads.
It makes no sense to use an ace as a low card when you have quads. The ace gets the rank 14, and the smallest quads you can have is quad deuces. There are 13 different quads. When you have defined the rank of the quads, there are 48 different kickers you can have with these quads.
n(quad ace hands) = kicker combos = 48 quad ace hands
There are also 48 quad king hands, 48 quad queen hands and so on
n(quad hands) = quad combos * kicker combos = 13 quads * 48 quad hands/quad = 624 quad hands
It makes no sense to use an ace as a low card when you have quads. The ace gets the rank 14, and the smallest quads you can have is quad deuces. There are 13 different quads. When you have defined the rank of the quads, there are 48 different kickers you can have with these quads.
n(quad ace hands) = kicker combos = 48 quad ace hands
There are also 48 quad king hands, 48 quad queen hands and so on
n(quad hands) = quad combos * kicker combos = 13 quads * 48 quad hands/quad = 624 quad hands
Odds of a flush
A flush is determined by the suit and the rank combos (the number of possible hands from a deck where three suits have been removed). The number of rank combos are calculated similarly to the total number of 5-card hands. There are 13*12*11*10*9 arranged hands. We divide by the number of arrangements (5*4*3*2*1) to get 1287 rank combos.
n(flushes) = suit combos * rank combos = 4 suits * 1287 flushes/suit = 5148 flushes
This number includes straight flushes, so there are 5108 ordinary flushes. These flushes can be divided into ace-high, king-high and so on. I might come up with numbers for those categories later.
n(flushes) = suit combos * rank combos = 4 suits * 1287 flushes/suit = 5148 flushes
This number includes straight flushes, so there are 5108 ordinary flushes. These flushes can be divided into ace-high, king-high and so on. I might come up with numbers for those categories later.
Odds of a straight
The number of possible straights are determined by the number of possible straight combos (10 as mentioned in the straight flush article) and the number of suit combos(4 suits pr. card, 4^5 combos)
n(straights) = suit combos * straight combos = 4^5 suit combos * 10 straights/suit combo = 10240 straights
40 of these straights are actually straight flushes, so there are 10200 ordinary straights. 1020 of these are ace-high straights, and there are obviously 1020 straights in each category(k-high, q-high and so on).
NB: 4^5 = 4*4*4*4*4 = 1024
n(straights) = suit combos * straight combos = 4^5 suit combos * 10 straights/suit combo = 10240 straights
40 of these straights are actually straight flushes, so there are 10200 ordinary straights. 1020 of these are ace-high straights, and there are obviously 1020 straights in each category(k-high, q-high and so on).
NB: 4^5 = 4*4*4*4*4 = 1024
Odds of being dealt a straight flush
A straight flush can be determined by the suit of the flush and the highest card of the straight. The ace is usually the highest card in the deck, and can be given rank 14 (king is 13, queen is 12, jack is 11). You can choose to use the ace as a low card instead. The ace can be given the rank 1 if you so desire. The only time you'd want to do this, is if it can be used in small straight. 5432A is the smallest possible straight, and is called a 5-high straight. AKQJT is the biggest possible straight ("14-high") and is called an ace-high straight. There are 10 different ranks that can be the highest rank of a straight, so there are 10 different "straight combos" in each suit.
n(straight flushes) = suit combos * straight combos = 4 suits * 10 straight flushes pr. suit = 40 straight flushes
You can find the odds of being dealt a straight flush by dividing the number of straight flushes you can be dealt by the total number of hands you can be dealt (2 598 960).
n(straight flushes) = suit combos * straight combos = 4 suits * 10 straight flushes pr. suit = 40 straight flushes
You can find the odds of being dealt a straight flush by dividing the number of straight flushes you can be dealt by the total number of hands you can be dealt (2 598 960).
Calculating the total number of five-card hands
Hi,
I'm going to demonstrate how to compute the odds of being dealt specific starting hands in 5-card draw. The first thing that would be interesting to know is the total number of starting hands (with a regular deck containing 52 cards).
You have been dealt a 5-card hand. The cards are dealt face down on the table. The first card you look at can be any of the 52 cards from the deck. Once you've looked at the first card, the next card can be any of the remaining 51 cards from the deck and so on. After multiplying these numbers (52, 51, 50, 49 and 48), you have calculated the number of arranged hands you can be dealt.
Let's say that you've made a royal flush in spades. It doesn't matter if the first card you looked at was the ace or the ten of spades. It's a royal flush regardless of the sequence of the cards. If you know that you've been dealt the royal flush in spades (but have not looked at your hand), any of the five cards you have can be the ace. Once you've determined where the ace is, there are four cards that can be the king and so on. You have to divide the number of arranged hands by the number of ways to arrange five different cards (5*4*3*2*1) to get the total number of hands you can be dealt in five card draw.
Total number (n): n = (52*51*50*49*48)/(5*4*3*2*1) = 2 598 960
I'm going to demonstrate how to compute the odds of being dealt specific starting hands in 5-card draw. The first thing that would be interesting to know is the total number of starting hands (with a regular deck containing 52 cards).
You have been dealt a 5-card hand. The cards are dealt face down on the table. The first card you look at can be any of the 52 cards from the deck. Once you've looked at the first card, the next card can be any of the remaining 51 cards from the deck and so on. After multiplying these numbers (52, 51, 50, 49 and 48), you have calculated the number of arranged hands you can be dealt.
Let's say that you've made a royal flush in spades. It doesn't matter if the first card you looked at was the ace or the ten of spades. It's a royal flush regardless of the sequence of the cards. If you know that you've been dealt the royal flush in spades (but have not looked at your hand), any of the five cards you have can be the ace. Once you've determined where the ace is, there are four cards that can be the king and so on. You have to divide the number of arranged hands by the number of ways to arrange five different cards (5*4*3*2*1) to get the total number of hands you can be dealt in five card draw.
Total number (n): n = (52*51*50*49*48)/(5*4*3*2*1) = 2 598 960
Subscribe to:
Comments (Atom)