Poker Strategy

number of ways to hit the straight on fifth street: 8*47*46 = 17,296
number of ways to hit the straight on sixth street: 40*8*46 = 14,720
number of ways to hit the straight on the river: 40*39*8 = 12,480
17,296 + 14,720 + 12,480 = 44,496

Since the above caculation is a permutation, we must eliminate those cards that are being counted twice by by divding 44,496 by 3! (3*2*1=6). 44,496/6 = 7,416.

7,416/17,296 = .4287 which is rounded off to .43. This implies that there is a 43% chance that you will have a straight by the river.

on fifth street (method 1):

number of ways to choose the last two cards : 47 C 2 = 1,081

number of ways to choose the last two cards so that neither one is one of the eight straight cards: 39 C 2 = 741

Since there are 1,081 ways to miss the straight, there are 1,081 - 741 = 340 ways to hit the straight.

340/1,081 = .3145 which is rounded off to .31. This implies that there is a 31% chance that you will have a straight by the river.

on fifth street (method 2):

number of ways to choose the last two cards : 47 C 2 = 1,081

number of ways to hit the straight on sixth street: 8*46 = 368
number of ways to hit the straight on the river: 39*8 = 312
368 + 312 = 680

Since the above caculation is a permutation, we must eliminate those cards that are being counted twice by by divding 680 by 2! (2*1=2). 680/2 = 340.

340/1,081 = .3145 which is rounded off to .31. This implies that there is a 31% chance that you will have a straight by the river.

on sixth street:

number of ways to choose the last card : 46 C 1 = 46

number of ways to choose the last card so that it is one of the eight straight cards: 8 C 1 = 8

8/46 = .1739 which is rounded off to .17. This implies that there is a 17% chance that you will hit the straight on the river.

Hitting the Flush in Seven Card Stud

We calculated the odds of hitting a flush if a player's first four cards are suited giving him an "flush draw".

on fourth street (method 1):

number of ways to choose the last three cards: 48 C 3 = 17,296

number of ways to choose the last three cards so that neither one is one of the nine flush cards: 39 C 3 = 9,139

Since there are 9,139 ways to miss the flush, there are 17,296 - 9,139 = 8,157 ways to hit the flush.

8,157/17,296 = .4716 which is rounded off to .47. This implies that there is a 47% chance that you will have a flush by the river.

on fourth street (method 2):

number of ways to choose the last three cards: 48 C 3 = 17,296

number of ways to hit the flush on fifth street: 9*47*46 = 19,458
number of ways to hit the flush on sixth street: 39*9*46 = 16,146
number of ways to hit the flush on the river: 39*38*9 = 13,338
19,458 + 16,146 + 13,338 = 48,942

Since the above caculation is a permutation, we must eliminate those cards that are being counted twice by by divding 48,942 by 3! (3*2*1=6). 48,942/6 = 8,157.

8,157/17,296 = .4716 which is rounded off to .47. This implies that there is a 47% chance that you will have a flush by the river.

on fifth street (method 1):

number of ways to choose the last two cards : 47 C 2 = 1,081

number of ways to choose the last two cards so that neither one is one of the nine flush cards: 38 C 2 = 703

Since there are 1,081 ways to miss the flush, there are 1,081 - 703 = 378 ways to hit the flush.

378/1,081 = .3496 which is rounded off to .35. This implies that there is a 35% chance that you will have a flush by the river.

on fifth street (method 2):

number of ways to choose the last two cards : 47 C 2 = 1,081

number of ways to hit the flush on sixth street: 9*46 = 414
number of ways to hit the flush on the river: 38*9 = 342
414 + 342 = 756

Since the above caculation is a permutation, we must eliminate those cards that are being counted twice by by divding 756 by 2! (2*1=2). 756/2 = 378.

378/1,081 = .3496 which is rounded off to .35. This implies that there is a 35% chance that you will have a flush by the river.

on sixth street:

number of ways to choose the last card : 46 C 1 = 46

number of ways to choose the last card so that it is one of the eight flush cards: 9C 1 = 9

9/46 = .1956 which is rounded off to .20. This implies that there is a 20% chance that you will hit the flush on the river.

Hitting a Full house when rolled up with trips

on third street (method 1)

49C4 = 211,876
total number of ways to catch the last four cards

12C4 = 495
total number of ways to choose four unique ranks out of 12, for the last four cards

4C1 = 4
ways of choosing the singleton for each of the four ranks

495 x 4*4*4*4 = 126,720

126,720 / 211,876 = .5980
your chance of missing the full house is 60 percent

on third street (method 2)

you can end up with a full house or better in five ways:

1. catching the quad card, regardless of the other three cards
2. catching two pair
3. catching runner runner runner quads
4. catching a pair and two singletons
5. catching trips and a singleton

1. catching the quad card, regardless of the other three cards

ways to choose the quad card: 1C1 = 1
ways to choose the last three cards: 48C3 = 17,296
1 x 17,296 = 17,296

2. catching two pair

ways of choosing the ranks (A and B) of the two pair: 12C2 = 66
ways of choosing the two cards amongst rank A: 4C2 = 6
ways of choosing the two cards amongst rank B: 4C2 = 6
66 x 6*6 = 2,376

3. catching runner runner runner quads

ways of choosing a rank for the quads: 12C1 = 12
ways of choosing all four cards of that rank, resulting in quads: 4C4 = 1
12 x 1 = 12

4. catching a pair and two singletons

ways of choosing three ranks (A,B,C) for the last four cards: 12C3 = 220
ways of choosing two cards among Rank A for a pair: 4C2 = 6
ways of choosing one card among Rank B for one singleton: 4C1 = 4
ways of choosing one card among Rank C for another singelton: 4C1 = 4
220 x 6*4*4 = 21,120
the paired rank could be Rank A, B or C. Therefore, we multiply by three: 21,120 x 3 = 63,360

5. catching trips and a singleton

ways of choosing two ranks (A, B) for the remaining four cards: 12C2 = 66
ways of choosing three cards among Rank A for trips: 4C3 = 4
ways of choosing one card among Rank B for the singleton: 4C1 = 4
66 x 4*4 = 1,056
the tripped rank could be Rank A or B. Therefore, we mutiply by two: 1,056 x 2 = 2,112

17,296 + 2,376 + 12 + 63,360 + 2,112 = 85,156
adding up the numbers

85,156 / 211,876 = .4019
your chance of catching a full house or better is 40 percent

on fourth street (method 1)

total number of ways to catch the last three cards: 48C3 = 17,296

total number of ways to choose three unique ranks out of 11, for the last three cards: 11C3 = 165
ways of choosing the singleton for each of the three ranks: 4C1 = 4
165 x 4*4*4 = 10,560

10,560 / 17,296 = .6105
your chance of missing the full house is 61 percent

on fourth street (method 2)

You can end up with a full house or better in five ways:

1. catching the quad card, regardless of the other two cards
2. pairing the 4th street card once. The remaining two cards can be anything except the trip rank or the pair rank.
3. tripping up the 4th street card. The last card can be anything except the rank of either of the trips.
4. pairing the 4th street card three times, resulting in quads.
5. catching trips (of a new rank)
6. catching a pair and a singleton (neither one of which is the rank of the trip or the 4th street card).

1. catching the quad card, regardless of the other two cards

ways to choose the quad card: 1C1 = 1
ways to choose the last two cards: 47C2 = 1,081
1 x 1,081 = 1,081

2. pairing the 4th street card once. The remaining two cards can be anything except the trip rank or the pair rank.

ways of choosing one card among the rank of the 4th street card: 3C1 = 3
ways of choosing the remaining two cards so that neither one is of the trip rank or the pair rank (52-8=44): 44C2 = 946
3 x 946 = 2,838

3. tripping up the fourth street card. The last card can be anything except the rank of either of the trips.

ways of chossing two cards among the rank of the 4th street card: 3C2 = 3
ways of choosing the remaining card so that it is not of the rank of either trip (52-8=44): 44C1 = 44
3 x 44 = 132

4. pairing the 4th street card three times, resulting in quads.

we must choose all three cards among the remaining three cards of the 4th street card rank: 3c3 = 1

5. catching trips (of a new rank)

ways of choosing a rank for the trips: 11C1 = 11
ways of choosing three out of four among the selected rank: 4C3 = 4
11 x 4 = 44

6. catching a pair and a singleton (neither one of which is the rank of the trip or the 4th street card).

ways of choosing two ranks (A, B) among the eleven ranks (we must exclude the rank of the trip and the 4th street card): 11C2 = 55
ways of choosing a pair among the four card of Rank A: 4C2 = 6
ways of choosing the singleton among the four card of Rank B: 4C1 = 4
55 x 6 x 4 = 1,320

1,320 x 2 = 2,640
We may also choose Rank B for the pair and Rank A for the singleton, resulting in the same number of choices.

1,081 + 2,838 + 132 + 1 + 44 + 2,640 = 6,736
adding up all the numbers

6,736 / 17,296 = .3894
your chance of catching a full house or better on 4th street is 39 percent

on fifth street (method 1)

total number of ways to catch the last two cards: 47C2 = 1,081

total number of ways to choose two ranks out of the ten remaining ranks: 10C2 = 45
ways of choosing the singleton for each of the two ranks: 4C1 = 4
45 x 4*4 = 720

720 / 1,081 = .6660
your chance of missing the full house is 67 percent

on fifth street (method 2)

you can end up with a full house or better in five ways:

1. catching the quad card, regardless of the last card
2. pairing the 4th street card AND pairing the 5th street card
3. pairing the 4th street card and catching a singleton
4. tripping the 4th street card
5. pairing the 5th street card and catching a singleton
6. tripping the 5th street card
7. catching a pair (of a new rank)

1. catching the quad card, regardless of the last card

ways of catching the quad card: 1C1 = 1
ways of catching the last card: 46C1 = 46
1 x 46 = 46

2. pairing the 4th street card AND pairing the 5th street card

ways of choosing one card among the three remaining cards of the 4th street rank: 3C1 = 3
ways of choosing one card among the three remaining cards of the 5th street rank: 3C1 = 3
3 x 3 = 9

3. pairing the 4th street card and catching a singleton

ways of choosing one card among the three remaining cards of the 4th street rank: 3C1 = 3
ways of chossing the last card (it cannot be the rank of the trip, the 4th street card or the 5th street card: 52-12=40): 40C1 = 40
3 x 40 = 120

4. tripping the 4th street card

ways of choosing two cards among the rank of the 4th street card: 3C2 = 3

5. pairing the 5th street card and catching a singleton

ways of choosing one card among the three remaining cards of the 5th street rank: 3C1 = 3
ways of chossing the last card (it cannot be the rank of the trip, the 4th street card or the 5th street card: 52-12=40): 40C1 = 40
3 x 40 = 120

6. tripping the 5th street card

ways of choosing two cards among the rank of the 5th street card: 3C2 = 3

7. catching a pair (of a new rank)

ways of selecting a rank among the 10 ramaining ranks (it cannot be the rank of the trip, the 4th street card or the 5th street card): 10C1 = 10
ways of choosing two cards among the chosen rank: 4C2 = 6
10 x 6 = 60

adding up the numbers: 46 + 9 + 120 + 3 + 120 + 3 + 60 = 361

361 / 1,081 = .3339
your chance of catching a full house on fifth street is 33 percent

on sixth street

total number of ways to catch the last card: 46C1 = 46

ways of catching the last card so that it is not one of the ten cards that will result in a full house or quads: 36C1 = 36

36 / 46 = .7826
your chance of missing a full house or better on the river is 78 percent

ways of catching the last card so that it is one of the ten cards needed for a full house or quads: 10C1 = 10

10 / 46 = .2173
your chance of catching a full house or better on the river is 22 percent

Hitting a Full house with two-pair

on fourth street:

There are a total of 48C3 = 17,296 possible combinations for the last three cards. In order to NOT improve to a full house or better, we must eliminate all combinations that contain any of the four cards that will make a full house or better. 44C3 = 13,244.

We must further eliminate out of those 13,244 all combinations which contain three running card of the same rank. We may choose thee rank in 11 ways and we may choose the specific three cards of that rank in 4C3 = 4 ways. 11*4 = 44. 13,244 - 44 = 13,200.

Therefore the chance of ending up unimproved is 13,200 / 17,296 = .7632 and the chance of improving to a full house or better is (17,296-13,200) / 17,296 = .2368 or 23.68%.

on fifth street:

There are a total of 47C2 = 1,081 possible combinations for the last two cards. In order to NOT improve to a full house or better, we must eliminate all combinations that contain any of the four cards that will make a full house or better. 43C2 = 903.

We must further eliminate out of those 903 all combinations which contain two running cards of the fifth street rank. We may choose those two cards in 3C2 = 3 ways. 903 - 3 = 900.

Therefore the chance of ending up unimproved is 900 / 1,081 = .8326 and the chance of improving to a full house or better is (1,081-900) / 1,081 = .1674 or 16.74%.

on sixth street (two-pair):

There are a total of 46 possibilities for the last card. Four of them will yield a full house. 4/46 = .0869 making it an 8.69% chance of catching a full house.

on sixth street (three-pair):

If you are going in to the river with three-pair, then your chances of filling up increase. You now have six cards that will make a full house. 6/46 = .1304 making it a 13.04% chance of making a full house.

The importance of live cards in SCS:

The importance of having live cards in SCS is illustrated by the following example: You made two-pair right away on fourth street and the pot is four-way action from third street on. By the time sixth street comes around, you have paid close attention and noted that none of the cards you need for a full house have shown up.

The total exposed cards are: seven third-street upcards, three fourth street cards, three fifth street cards and three sixth street cards 7+3+3+3=16. We therefore only consider 46-16=30 cards when calculating your chances of filling up. 4/30=.1333 giving you a 13.33% chance of filling up.

If you have three-pair and they are all live, then your chances are 6/30=.2 giving you an astonishing 20% chance of filling up, which is almost as big as the percentage of filling up accorded to trips (21.73%).

Comparison Chart

trips vs draws
	third street	fourth street	fifth street	sixth street
chances of improving trips to full house or better	40.19 %	38.94 %	33.39 %	21.73 %
chances of hitting the flush	18.01 % (3-card)	47.16 %	34.96 %	19.56 %
chances of hitting the straight	21.47 % (3-card)	42.87 %	31.45 %	17.39 %
chances of hitting the straight and winning vs. bigger straight-draw	16.86 %	24.49 %	21.55 %	14.36 %
chances of hitting the flush and winning vs. trips	10.77 %	28.79 %	23.28 %	15.30 %
chances of hitting the straight and winning vs. trips only	12.84 %	26.17 %	20.94 %	13.62 %
chances of hitting the straight and winning vs. flush-draw only	17.60 %	22.65 %	20.45 %	13.98 %
chances of hitting the straight and winning vs. trips AND flush-draw	10.52 %	13.83 %	13.62 %	10.94 %
chances of hitting the straight and winning vs. trips AND flush-draw AND bigger straight	8.26 %	7.90 %	9.34 %	9.04 %

two-pair vs draws
	fourth street	fifth street	sixth street
chances of improving two-pair to full house or better	23.68 %	16.74 %	08.70 %
chances of hitting the flush	47.16 %	34.96 %	19.56 %
chances of hitting the straight	42.87 %	31.45 %	17.39 %
chances of hitting the flush and winning vs. two-pair	36.11 %	29.20 %	17.85 %
chances of hitting the straight and winning vs. two-pair only	32.82 %	26.27 %	15.87 %
chances of hitting the straight and winning vs. flush-draw only	22.65 %	20.45 %	13.98 %
chances of hitting the straight and winning vs. two-pair AND flush-draw	17.34 %	17.08 %	12.77 %

SCS Three-Card Starting Hand Stats

Discussion: odds of being dealt a pair or better in the 3-card starting hand

There are 52C3 = 22,100 ways of being dealt the first three cards. In order not to have a pair or better, we must choose three cards from three distinct ranks 13C3 = 286. Since each rank can be chosen in four ways, we multiply by 4: 286*4*4*4 = 18,304. Therefore the odds of having a pair or better are (22,100-18,304)/22,100 = .1718. You will have a pair or better 17% of the time.

If you look at your door card and see that it's an ace, you will have an ace or better (two aces) in the hole 11.5 percent of the time. There are 51C2 = 1,275 ways to choose the two hole cards and 48C2 = 1128 ways to choose them so that they do not contain any one of the three aces. (1,275-1,128)/1,275 = .1153.

To calculate the odds of having one ace ONLY in the hole, we deduct the combinations containing two aces in the hole. We can choose two aces out of three in 3C2 = 3 ways. Therefore the chance of having a pair of aces only is (1,275-1,128-3)/1,275 = .1129

If you have looked at your other opponents cards and you've noticed that none of them are showing an ace, then your odds of having an ace or better in the hole increase by two percent. Since there are eight exposed cards, the hole card selection can be made from only 44 cards. 44C2 = 946. Ways of not having any ace in the hole are 41C2 = 820. You will therefore have an ace or better 946-820=126 out of 946 times. 126/946=.1331

Discussion: odds of being dealt a three-flush

For any given suit, we must choose three out of thirteen cards in that suit in order to have a three-flush: 13C3 = 286. Since there are four suits in the deck, there are 286*4 = 1,144 ways of being dealt a three=flush out of a total of 22,100 (52C3). 1,144/22,100 = .0517. You will therefore be dealt a three-flush 5.2 percent of the time.

Discussion: odds of being dealt a three-straight

The three cards must be in the form of x + (x+1) + (x+2) where x can be any of the twelve ranks A-Q (A,2,3,4,5,6,7,8,9,T,J,Q). Each of three cards can be chosen in four ways. We therefore calculate 12*4*4*4 = 768 for the ways one can be dealt a consecutive three-straight. Since 768/22,100 = .0347, you will be dealt a three-straight 3.5 percent of the time.

another form of three-straight contains one gap so that if one catches the right card on fourth street, he will have an open-ended straight draw. This kind of three-straight must be in the form of x + (x+1) + (x+3) or x + (x+2) + (x+3). For each of these forms, the x can be any of the eleven ranks from Ace through Jack (A,2,3,4,5,6,7,8,9,T,J) and each card can be selected in four ways. 2*(11*4*4*4) = 1,408. Adding this number to the number of ways one can be dealt a consecutive three-straight we get 1,408 + 768 = 2,176 so that your chances of being dealt a one-gapped straight or better are 2,176/22,100 = .0985; that's 9.9 percent of the time.

Discussion: odds of being rolled up

For each rank, we can choose three out of four in three ways (4C3 = 4). We therefore calculate 13*4 = 52 for the ways to choose the first three cards so that they are trips. 52/22,100 = .002352941

Discussion: odds of being dealt a pair of aces

We can shoose the pair in 4C2 = 6 ways and we can shoose the "kicker" in 48 ways. 6*48 = 288; therefore we can choose a pair of aces (or any given pair) in 288/22,100 = .0130 ways. The percentage chance of being dealt aces is 1.3%

Three-card starting hands in SCS
Three-card hand	chances of being dealt such hand

a pair of the rank of the upcard	11.29%
a pocket pair	5.65%
a pair of aces	1.3%
a premium pair (J-A)	5.21%
a middle pair (7-T)	5.21%
a small pair (2-6)	6.51%
any pair	16.94%
a pair or better (trips)	17.18%
rolled up trips	0.24%
three low cards (for hi-lo stud)
three-flush	5.17%
three-straight (consecutive)	3.47%
three-straight (one-gap)	6.37%
three-straight (consecutive or one-gap)	9.85%
a draw hand (3-straight or 3-flush)	8.64%
a pair, trips, three-straight or three-flush	25.82%

Lowball