# After the Flop

### QQ not having A or K hit by the river

Hypothesis: own hand dealt QQ. Event: A or K not hit by the river.

First, the probability for A or K to be hit by the river:

Favorable combinations for this event (final board configuration):

(Axyzt), in number of 4*C(42,4)

(AAxyz), in number of 6*C(42,3)

(AAAxy), in number of 4*C(42,2)

(AAAAx), in number of 1*42

(Kxyzt), in number of 4*C(42,4)

(KKxyz), in number of 6*C(42,3)

(KKKxy), in number of 4*C(42,2)

(KKKKx), in number of 1*42

(AKxyz), in number of 16*C(42,3)

(AKKxy), in number of 4*6*C(42,2)

(AKKKx), in number of 4*4*42

(AKKKK), in number of 4

(AAKxy), in number of 6*4*C(42,2)

(AAKKx), in number of 6*6*42

(AAKKK), in number of 6*4

(AAAKx), in number of 4*4*42

(AAAKK), in number of 4*6

(AAAAK), in number of 4

(where x, y, z and are different from A and K)

Totally, we have 1268092 favorable combinations for the event “A or K hit by river”.

The total number of possible combinations for the final board: C(50,5) = 2118760.

The probability: 1268092/2118760 = 59.850%

Then, the probability for the event “A or K not hit by the river” is 1 – 1268092/2118760 = **40.149% = 1 / 2.490**

### QQ versus AK heads up, A or K hitting by river

Hypothesis: own hand dealt QQ, one opponent dealt AK. Event: A or K hit by river.

We have 3 aces and 3 kings left.

The favorable combinations of the final board for the event to measure:

(Axyzt), in number of 3*C(42,4)

(AAxyz), in number of 3*C(42,3)

(AAAxy), in number of C(42,2)

(Kxyzt), in number of 3*C(42,4)

(KKxyz), in number of 3*C(42,3)

(KKKxy), in number of C(42,2)

(AKxyz), in number of 9*C(42,3)

(AKKxy), in number of 3*3*C(42,2)

(AKKKx), in number of 3*42

(AAKxy), in number of 3*3*C(42,2)

(AAKKx), in number of 3*3*42

(AAKKK), in number of 3*1

(AAAKx), in number of 3*42

(AAAKK), in number of 1*3

(x, y, z and t are different from A and K).

Totally, we have 861639 favorable combinations, from C(48,5) = 1712304 possible.

Probability: 861639/1712304 = **50.520% = 1 / 1.987**

### Flop being all on kind

Hypothesis: no cards seen.

For each card (as value), we have C(4,3)=4 favorable combinations for the event. Totally, 13*4 = 52 favorable combinations, from C(52,3) = 22100 possible.

Probability: 52/C(52,3) = **0.235% = 1 / 425**

### Four flush improving

Hypothesis: 4 same symbols in own hand plus flop. Event: at least one symbol hit by river. Let S be the same symbol of that four cards.

Favorable 2-card combinations of turn plus river:

(Sx), with x different from S, in number of 9*(47-9) and

(SS), in number of C(9,2) = 9*4.

Totally, we have 378 favorable combinations from C(47,2) = 1081 possible.

Probability: 378/1081 = **34.967% = 1 / 2.859**

### Open ended straight flush improving

Hypothesis: 4 cards from a straight flush (not starting with 2 or 10) in own hand plus flop. Event: one or two of the 2 cards left to be hit by river.

Let C and D be the needed cards (unique cards – value + symbol).

Favorable combinations of turn and river:

(Cx), with x different from C and D, in number of 47-2

(Dx), with x different from C and D, in number of 47-2

(CD), in number of 1

Totally, we have 91 favorable combinations from C(47,2) = 1081 possible.

Probability: 91/1081 = **8.418% = 1 / 11.879**

### Open ended straight improving

Hypothesis: 4 cards from a flush (not starting with 2 or 10) in own hand plus flop. Event: one or two of the 2 cards left (as value i.e. 8 outs) to be hit by river.

Let C and D be the needed cards (as value).

Favorable combinations of turn and river:

(Cx), with x different from C and D, in number of 4*(47-8)

(Dx), with x different from C and D, in number of 4*(47-8)

(CD), in number of 8

(CC), in number of C(4,2)

(DD), in number of C(4,2).

Totally, we have 332 favorable combinations from C(47,2) = 1081 possible.

Probability: 332/1081 = **30.712% = 1 / 3.256 **

### Two pair making a full house

Hypothesis: two pair in own hand plus flop. Event: Making a full house by river

Let’s denote by (TTDDC) the cards from own hand and flop, T, D and C mutually different (as value).

Favorable combinations of turn plus river for a full house to make:

(Tx), x different from T, D and C, in number of 2*(47-2-2-3)

(Dx), x different from T, D and C, in number of 2*(47-2-2-3)

(TT), in number of 1

(TD), in number of 2*2

(DD), in number of 1

(CC), in number of C(3,2)

(TC), in number of 2*3

(DC), in number of 2*3.

Totally, we have 181 favorable combinations from C(47,2) = 1081 possible.

Probability: 181/1081 = **16.743% = 1 / 5.972 **

### Trips improving to full house or better

Hypothesis**: **three same cards (as value) in own hand plus flop Event: making a full house or quads by river.

Let’s denote by (TTTCD) the cards (T, C and D mutually different – as value).

Favorable combinations of turn plus river:

(Cx), x different from T, C and D, in number of 3*(47-1-3-3)

(Dx), x different from T, C and D, in number of 3*(47-1-3-3)

(CT), in number of 3*1

(DT), in number of 3*1

(DC), in number of 3*3

(xx), x different from T, C and D, in number of 10*C(4,2)

(CC), in number of C(3,2)

(DD), in number of C(3,2)

(Tx), x different from T, C and D, in number of 47-1-3-3.

Totally, we have 361 favorable combinations, from C(47,2) = 1081 possible.

Probability: 361/1081 = **33.395% = 1 / 2.994**

# When you hold AK suited

Let S be the symbol of your suite.

### Q, 10, J (royal flush)

Favorable combinations: 1 (Q 10 J)

Possible combinations: C(50,3) = 19600.

Probability: **1/19600 = 0.0051%**

### Improves to 4 of a kind

Favorable combinations:

(AAA) – 1

(KKK) – 1

Totally, 2 favorable combinations from 19600 possible.

Probability: 2/19600 = **1/9800 = 0.0102%**

### Improves to full house

Favorable combinations:

(AAK) – in number of C(3,2)*3 = 9

(KKA) – in number of C(3,2)*3 = 9

Totally, 18 favorable combinations.

Probability: 18/19600 = **1/1089 = 0.00091%**

### Flopped flush

Favorable combinations: (SSS), without (10 J Q of S), in number of C(11,3) – 1 = 164

Probability: 164/19600 = **0.836% = 1/119.5**

### Pair of A or K w/ four flush

Favorable combinations:

(ASS), in number of 3*C(11,2) = 165

(KSS), in number of 3*C(11,2) = 165.

Totally, 330 favorable combinations.

Probability: 330/19600 = **1.683% = 1/59**

### Two of your suite with board paired 2-Q

Favorable combinations: (xxS), with one S-card in the paired cards, x being 2-Q.

The number is 11*11*C(3,2) = 363.

Probability: 363/19600 = **1.852% = 1/54**

### Two of your suite w/ no royal

(With only two of your suite a royal could not possibly be flopped. You need three for this. If we change the event to “Three of your suite w/ no royal”:

Favorable combinations: (SSS), no royal, in number of C(11,3) – 1 = 164. Probability: 164/19600 **= 0.836% = 1/119.5**

This is identical to situation “Flopped flush”.

### Two of your suite

Favorable combinations: (SSx), with x different from S, in number of C(11,2)*(50-11) = 2145.

Probability: 2145/19600 = **10.943% = 1/9.13**

### Any Q, J, 10

Favorable combinations: (QJ 10), in number of 4*4*4 = 64.

Probability: 64/19600 = **0.326% = 1/306**

### QQ or less, one or less of your suite

The favorable combinations are:

(xxS), x different from S (as symbol), x different from A and K (as value) – in number of 11*C(3,2)*11 and

(xxy), y different from S (as symbol), x different from A and K (as value) – in number of 11*C(4,2)*(50-11).

We have to subtract now the common combinations, which are the trips (xxx), no one of them having the symbol S. These are in number of 11*(C(4,3) – 1).

So, the exact number of favorable combinations is

11*C(3,2)*11 + 11*C(4,2)*(50-11) – 11*(C(4,3) – 1) = 3267.

Probability: 3267/19600 = **16.668% = 1/6**

### Flop two pair

Favorable combinations:

(AKx), x different from A and K, in number of 3*3*(50-3-3)

(Axx), x different from A and K, in number of 3*11*C(4,2)

(Kxx), x different from A and K, in number of 3*11*C(4,2)

Totally, we have 892 favorable combinations.

Probability: 892/19600 = **4.551 = 1/22**

### QQQ-222

Favorable combinations: QQQ-222, in number of 11*C(4,3) = 44.

Probability: 44/19600 = **0.224 = 1/445**

### A or K

Favorable combinations:

(Axy), in number of 3*C(44,2)

(Kxy), in number of 3*C(44,2)

(AAx), in number of 3*44

(AAA), in number of 1

(KKx), in number of 3*44

(KKK), in number of 1

(AKx), in number of 9*44

(AAK), in number of C(3,2)*3

(KKA), in number of C(3,2)*3

Totally, we have 6356 favorable combinations.

Probability: 6356/19600 = **32.428% = 1/3**

### Four to a flush

Favorable combinations: (SSx), x different from S, in number of C(11,2)*(50-11) = 2145.

Probability: 2145/19600 = **10.943% = 1/9**

### AA or KK making trips

Favorable combinations:

(AAx), x different from A and K, in number of C(3,2)*44

(KKx), x different from A and K, in number of C(3,2)*44

Totally, we have 264 favorable combinations and the probability is 264/19600 = **1.346% = 1/74**

** **

# When you hold KK

### Flopping four of a kind

Favorable combinations: (KKx), x different from K, in number of 1*(50-2) = 48

Probability: 48/C(50,3) = 48/19600 = **0.244% = 1/408**

### KAA

Favorable combinations: (KAA), in number of 2*C(4,2) = 12

Probability: 12/19600 = **0.061% = 1/1633**

### AK any card

Favorable combinations: (AKx), x different from A and K, in number of 4*2*(50-4-2) = 352

Probability: 352/19600 = **1.795% = 1/55.7**

### Making any full house

Favorable combinations:

(xxx), x different from K, in number of 12*C(4,3) = 48

(Kxx), x different from K, in number of 2*12*C(4,2) = 144

Totally, we have 192 favorable combinations for a full house.

Probability: 192/19600 = **0.979% = 1/102**

### Trips

(Kxy), x, y different from K, x different from y – in number of 2*C(48,2) minus the number of (Kxx) combinations = 2256 – 144 (see “making any full house” section) = 2112.

Probability: 2112/19600 = **10.775% = 1/9.3**

### AAA

Favorable combinations: (AAA), in number of C(4,3) = 4.

Probability: 4/19600 = **0.020% = 1/4900**

### AA7

Favorable combinations: (AA7), in number of C(4,2)*4 = 24

Probability: 24/19600 = **0.122% = 1/816.6**

### Any pair w/ no K or AA

Favorable combinations: (xxy), x different from K and A, y different from K – in number of 11*C(4,2)*(50-2-2) = 2904.

Probability: 2904/19600 = **14.816% = 1/6.74**

### A and any other cards besides a K

Favorable combinations: (Axy), x, y different from A and K – in number of 4*C(50-4-2,2) = 4*C(44,2) = 3784.

Probability: 3784/19600 = **19.306% = 1/5.2**

** **

# When you hold QJ off suit

### Flop 4 Q’s or J’s

Favorable combinations:

(QQQ) – 1

(JJJ) – 1

Totally, 2 favorable combinations from C(50,3) = 19600 possible.

Probability: 2/19600 = **0.010% = 1/9800**

### QQJ or JJQ

Favorable combinations:

(QQJ), in number of C(3,2)*3 = 9

(JJQ), in number of C(3,2)*3 = 9

Totally, 9 favorable combinations.

Probability: 18/19600 = **0.091% = 1/1089**

### AK 10

Favorable combinations: (AK 10), in number of 4*4*4 = 64.

Probability: 64/19600 = **0.326% = 1/30**

### K 10 9 or 89 10

Favorable combinations:

(K 10 9), in number of 4*4*4

(89 10), in number of 4*4*4

Totally, 128 favorable combinations.

Probability: 128/19600 = **0.653% = 1/153**

### Flopping any straight

The possible straights are: (89 10 JQ), (9 10 JQK) and (10 JQKA).

Favorable combinations for these straights:

(89 10), in number of 4*4*4

(9 10 K), in number of 4*4*4

(10 KA), in number of 4*4*4

Totally, we have 192 favorable combinations.

Probability: 192/19600 = **0.979% = 1/102**

### Open ended straight

The possible open-ended straights are: (9 10 JQ) and (10 JQK).

Favorable combinations:

(9 10 x), x different from 8 and K, in number of 4*4(50-2-4-4) = 640

(10 Kx), x different from 9 and A, in number of 4*4(50-2-4-4) = 640

Totally, 1280 favorable combinations.

Probability: 1280/19600 = **6.530% = 1/15.3**

### 3 suited cards of a suite you hold

Let S be one of the symbols you hold.

Favorable combinations: (SSS), in number of C(12,3) = 220.

Totally, we have 440 favorable combinations (for all two symbols).

Probability: 440/19600 = **2.244% = 1/45.54**

### Flopping trips (no full house)

Favorable combinations:

(xxx), x different from J and Q, in number of 11*C(4,3) = 44

(JJx), x different from J and Q, in number of C(3,2)*(50-3-3) = 3*44

(QQx), x different from J and Q, in number of C(3,2)*(50-3-3) = 3*44

Totally, we have 308 favorable combinations.

Probability: 308/19600 = **1.571% = 1/63.63**

### No A or K

Favorable combinations: (xyz), x, y and z different from A and K.

Let’s count the combinations holding A or K:

(Axy), in number of 4*C(50-4-4, 2) = 4*C(44,2)

(AAx), in number of C(4,2)*(50-4-4) = 6*44

(AAA), in number of C(4,3) = 4

(Kxy), in number of 4*C(50-4-4, 2) = 4*C(44,2)

(KKx), in number of C(4,2)*(50-4-4) = 6*44

(KKK), in number of C(4,3) = 4

(AKx), in number of 4*4*(50-4-4) = 16*44

(AKK), in number of 4*C(4,2) = 24

(KAA), in number of 4*C(4,2) = 24

(x, y different from A and K)

Totally, we have 8856 such combinations.

The number of favorable combinations will be C(50,3) – 8856 = 10744.

Probability: 10744/19600 = **54.816% = 1/1.82**

### Q’s up or J’s up

Favorable combinations:

For J’s up:

(Jxx), x different from J, Q, K, A, in number of 3*9*C(4,2) = 162

For Q’s up:

(Qxx), x different from J, Q, K, A, in number of 3*9*C(4,2) = 162

(QJx), x different from J, Q, in number of 3*3*(50-3-3) = 396.

There are no common combinations between them to subtract.

Totally, we have 720 favorable combinations and the probability is

720/19600 = **3.673% = 1/27.2 **

** **

# Odds of holding / not holding an Ace

The complementary event is A = ”At least one player holds an ace”. (“an ace” means “one or more aces”).

Let’s find the formula for P(A). The probability we are looking for will be then 1 – P(A).

For one player (a specific one), let’s find the probability of that player holding an ace:

Favorable combinations:

(Ax), x different from A, in number of 4*(52-4) = 192

(AA), in number of C(4,2) = 6.

Totally, 198 favorable combinations from C(52,2) = 1326 possible. The probability is 198/1326 = **99/661**

**Maximum 4 players could simultaneously hold an ace.**

-** The probability for 2 players to simultaneously hold an ace:**

Favorable combinations:

(Ax)(Ay), in number of 4*48*3*(52-3-2) = 27072

(AA)(Ay), in number of C(4,2)*2*(52-3-1) = 576

(Ax)(AA), in number of 576

(AA)(AA), in number of C(4,2)*1 = 6.

(x, y different from A)

Totally, we have 28230 favorable combinations from C(52,2)*C(50,2) possible.

The probability is **941/54145**

**- The probability for 3 players to simultaneously hold an ace:**

Favorable combinations:

(Ax)(Ay)(Az), in number of 4*48*3*47*2*46

(AA)(Ay)(Az), in number of 6*2*48*1*47

(Ax)(AA)(Az), in number of 6*2*48*1*47

(Ax)(Ay)(AA), in number of 6*2*48*1*47.

(x, y, z different from A)

Totally, we have 12*48*47*95 favorable combinations from C(52,2)*C(50,2)*C(48,2) possible.

The probability is **228/162435**

**- The probability for 3 players to simultaneously hold an ace:**

Favorable combinations:

(Ax)(Ay)(Az)(At), in number of 4*48*3*47*2*46*1*45.

(x, y, z, t different from A).

Possible combinations: C(52,2)*C(50,2)*(C48,2)*C(46,2).

The probability is **16/270725**

By applying now the probability calculus formula for non-independent events, we find:

P(A) = 99*n/661 – C(n,2)*941/54145 + C(n,3)*228/162435 – C(n,4)*16/270725 =

99*n/661 – n*(n-1)*941/108290 + n*(n-1)*(n-2)*38/162435 – n*(n-1)*(n-2)*(n-3)*2/812175,

where n is the number of players.

Therefore, the probability we are looking for is:

1 – 99*n/661 + n*(n-1)*941/108290 – n*(n-1)*(n-2)*38/162435 + n*(n-1)*(n-2)*(n-3)*2/812175.

By giving values to n, we find:

For n=2, the probability is **71.783%**

For n=3, the probability is **60.141%**

For n=4, the probability is **49.962%**

For n=5, the probability is **41.118%**

For n=6, the probability is **33.486%**

For n=7, the probability is **26.949%**

For n=8, the probability is **21.396%**

For n=9, the probability is **16.723%**

For n=10, the probability is **12.831%**

** **

# You have one Ace, probability of no one else having one:

We have 50 unknown cards.

The complementary event is A = ”At least one player holds an ace”.

**3 players could simultaneously hold an ace**. Let’s find the formula for P(A). The probability we are looking for will be then 1 – P(A).

For one player (a specific one), let’s find the probability of that player holding an ace:

Favorable combinations:

(Ax), x different from A, in number of 3*(50-3) = 141

(AA), in number of C(3,2) = 3.

Totally, 144 favorable combinations from C(50,2) = 1225 possible.

The probability is **144/1225**

**Maximum **

- **The probability for 2 players to simultaneously hold an ace:**

Favorable combinations:

(Ax)(Ay), in number of 3*47*2*46

(AA)(Ay), in number of C(3,2)*1*47

(Ax)(AA), in number of C(3,2)*1*47

(x , y different from A)

Totally, we have 13254 favorable combinations from C(50,2)*C(48,2) possible.

The probability is **47/4900**.

**- The probability for 3 players to simultaneously hold an ace:**

Favorable combinations:

(Ax)(Ay)(Az), in number of 3*47*2*46*1*45

(x, y, z different from A)

We have 3*47*2*46*1*45 favorable combinations from C(50,2)*C(48,2)*C(46,2) possible.

The probability is **1/2450**.

By applying now the probability calculus formula for non-independent events, we find:

P(A) = 144*n/1225 – C(n,2)*47/4900+ C(n,3)/2450 =

144*n/1225 – n*(n-1)*47/9800 + n*(n-1)*(n-2)/14700,

where n is the number of your opponents.

Therefore, the probability we are looking for is:

1 – 144*n/1225 + n*(n-1)* 47/9800 – n*(n-1)*(n-2)/14700.

By giving values to n, we find:

For n=1 (2 players), the probability is **88.244%**

For n=2 (3 players), the probability is **77.448%**

For n=3 (4 players), the probability is **67.571%**

For n=4 (5 players), the probability is **58.571%**

For n=5 (6 players), the probability is **50.408%**

For n=6 (7 players), the probability is **43.040%**

For n=7 (8 players), the probability is **36.428%**

For n=8 (9 players), the probability is **30.530%**

For n=9 (10 players), the probability is **25.306%**

** **

# Ace on flop, chances that someone has an ace down

We have 49 unknown cards. The event is A = “At least one player has one ace”.

**-The probability for one player (a specific one) to hold one ace:**

Favorable combinations: (Ax), x different from A, in number of 3*(49-3) = 138

Possible combinations: C(49,2) = 1176

Probability: 138/1176 = **23/196
**

Maximum three players could simultaneously hold one ace.

- **The probability for two players to simultaneously hold one ace:**

Favorable combinations: (Ax)(Ay), x, y different from A, in number of 3*46*2*45.

Possible combinations: C(49,2)*C(47,2)

Probability: 3*46*2*45/(C(49,2)*C(47,2)) = **45/4606
**

**- The probability for three players to simultaneously hold one ace:**

Favorable combinations: (Ax)(Ay)(Az), x, y, z different from A, in number of 3*46*2*45*1*44

Possible combinations: C(49,2)*C(47,2)*C(45,2)

Probability: 3*46*2*45*1*44/( C(49,2)*C(47,2)*C(45,2)) = **1/2303**

By applying now the probability calculus formula for non-independent events, we find:

P(A) = 23*n/196 – C(n,2)*45/4606 + C(n,3)/2303 =

23*n/196 – n*(n-1)*45/9212 + n*(n-1)*(n-2)/13818, where n is the number of players.

By giving values to n, we find:

For n=5, the probability is **49.337%**

For n=4, the probability is **41.250%**

For n=3, the probability is **32.316%**

** **

# Other important odds

### No pair improving to a pair on the flop

Let’s denote by (CD) the hole cards (C different from D as value).

Favorable combinations of the flop:

(Cxy), x, y different from C and D, x different from y – in number of 3*(C(50-3-3, 2) – 11*C(4,2))

(Dxy), x, y different from C and D, x different from y – in number of 3*(C(50-3-3, 2) – 11*C(4,2))

Totally, we have 5280 favorable combinations from C(50,3) = 19600 possible.

Probability: 5280/19600 = **26.938% = 1/3.7**

### Suited hole cards improving to a 4 flush

Hypothesis: (SS) hole cards (S symbol)

Favorable combinations of the flop: (SSx), x different from S, in number of C(11,2)*(50-11) = 2195

Probability: 2195/19600 = **11.198% = 1/8.92**

### One pair on flop improving by river

Hypothesis: hole and flop cards: (CCxyz), x, y, z different from C (as value)

Favorable combinations of turn and river:

(Cx), x different from C, in number of 2*(47-2) = 90

(CC), in number of 3

(xx), x different from C, in number of 12*C(4,2) = 72

Totally, 165 favorable combinations from C(47, 2) = 1081 possible.

Probability: 165/1081 = **15.263% = 1/6.55**

### Pocket pair improving to trips after flop

Pocket cards: (CC) Flop: (xyz), x, y, z different from C

Favorable combinations of turn and river: (Px), x different from P, in number of 2*(47-2) = 90, from C(47,2) = 1081 possible.

Probability: 90/1081 = **8.325% = 1/12**

### Two over cards improving to a pair

Let’s denote by (AB) the hole cards and by (CDE) the flop cards. Hypothesis: each of A and B higher than C, D or E (as value), A different from B, C, D, E mutually different.

Favorable combinations of flop and turn cards:

(Ax), x different from A, B, C, D, E, in number of 3*(47-3*5) = 96

(Bx), x different from A, B, C, D, E, in number of 3*(47-3*5) = 96

Totally, 192 favorable combinations from 1081 possible.

Probability: 192/1081 = **17.761% = 1/5.6**

### Two overs plus gutshot improving to a pair or better

Let’s denote by (AB) the hole cards and by (CDE) the flop cards. Let (ABCD) be the gutshot and F be the missing card from the straight. Hypothesis: each of A and B are higher than C, D or E, E is different from A, B, C or E.

Favorable combinations of turn and river card:

(Ax), x different from A, B, C, D, E, F, in number of 3*(47 – 5*3 – 4)

(Bx), x different from A, B, C, D, E, F, in number of 3*(47 – 5*3 – 4)

(AA), in number of 3

(AB), in number of 9

(AC), in number of 9

(AD), in number of 9

(AE), in number of 9

(AF), in number of 12

(BB), in number of 3

(BC), in number of 9

(BD), in number of 9

(BE), in number of 9

(BE), in number of 9

(BF), in number of 12

(Fx), x different from A, B, C, D, E, F, in number of 4*(47 – 5*3 – 4)

(FF), in number of 6

Totally, we have 372 favorable combinations from 1081 possible.

Probability: 372/1081 = **34.412% = 1/2.9**

### Gutshot straight draw hitting by river

Let’s denote by ABCD the gutshot cards and by E the missing card for straight.

Hypothesis: hole and flop cards: (ABCDx), x different from E

Favorable combinations of turn and river:

(Ex), x different from E, in number of 4*(47-4) = 172

(EE), in number of 6

Totally, 178 favorable combinations from 1081 possible.

Probability: 178/1081 = **16.466% = 1/6.07**

### Gutshot plus pair improving to two pair or better

Let’s denote by ABCD the gutshot cards and by E the missing card for straight.

Hypothesis: hole and flop cards: (AABCD)

Favorable combinations of turn and river:

(Bx), x different from B, in number of 3*(47-3)

(BB), in number of C(3,2)

(Cx), x different from C, in number of 3*(47-3)

(CC), in number of C(3,2)

(Dx), x different from D, in number of 3*(47-3)

(DD), in number of C(3,2)

(xx), x different from A, B, C and D, in number of 9*C(4,2)

(AA), in number of 1

(Ex), x different from E, in number of 4*(47-4)

(the (EE) combinations are included in (xx) category).

Totally, we have 726 favorable combinations from 1081 possible.

Probability: 726/1081 = **67.160% = 1/1.48**

### Backdoor flush

Hypothesis: hole and flop cards: (SSSxy), x, y different from S (symbol)

Favorable combinations of turn and river: (SS), in number of C(10,2) = 45, from 1081 possible.

Probability: 45/1081 = **4.162% = 1/24**

### Backdoor flush w/ over card improving to a pair or flush

Let (AB) be the hole cards and (SSS) the flop cards (S symbol). Hypothesis: A different from B, A and B different from S (as symbol), each of A and B higher than each of flop cards.

Favorable combinations of turn and river cards:

(Ax), x different from A and B, in number of 3*(47-3-3) = 123

(Bx), x different from A and B, in number of 3*(47-3-3) = 123

(SS), in number of C(10,2) = 45

Totally, 291 favorable combinations from 1081 possible.

Probability: 291/1081 = **26.919% = 1/3.71**

### Backdoor flush with gutshot improving by river

Let ABCD be the gutshot cards, E the missing card for straight and S a symbol.

Hypothesis: hole and flop cards: (ABCDx), with three of them having symbol S

Favorable combinations of turn and river:

(Ex), x different from E, in number of 4*(47-4) = 120

(EE), in number of C(4,2) = 6

(SS), in number of C(10,3) = 45

, from which we have to subtract the (E(S), S) combinations, in number of 9 (these were counted twice).

Totally, we have 289 favorable combinations from 1081 possible.

Probability: 289/1081 = **26.734% = 1/3.74**