Post Flop Odds Math

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