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
