-
Flop Probability Math
This is for you number crunchers out there. This post shows how to calculate the probability of different flops occurring. As a starting point, there are 52 cards in the deck. This can produce 132,600 unique flops or permutations. Since the order of the cards on the flop doesn’t matter, you should know that 22,100 different flop combinations are possible.
For each example, the 1st card is drawing from 52 possible cards.
All Possible Flops = (52 * 51 * 50) = 132,600 permutations (“order matters”)
All possible Combos = 132,600 / 3! = 22,100 combinations (“any order”)
______________________________________________________________
Monotone = (52/52)(12/51)(11/50) = 5.17%
Rainbow = (52/52)(39/51)(26/50) = 39.76%
Two-Suited = (52/52)(12/51)(39/50)*3 = 55.05%
Trips Flop = (52/52)(3/51)(2/50) =0.23%
Paired Flop = (52/52)(3/51)(48/50) * 3 = 16.94%
Unpaired Flop = (52/52)(48/51)(44/50) = 82.82%
3 Straight Rainbow = (52/52)(3/51)(2/50) * 6 = 1.30%
_______________________________________________________________
How to interpret the math?
— monotone
= the 1st card can be any card, it chooses the suit
= the 2nd card is drawn from the 12 remaining of that suit
= the 3rd card is draw from the 11 remaining of that suit
— rainbow
= the 1st card can be any card, it chooses the suit
= the 2nd card is drawn from the 39 = 13 * 3 for the other 3 suits
= the 3rd card is draw from the 26 = 13 * 2 for the other 2 suits
— two suited
= the 1st card can be any card, it chooses the suit
= the 2nd card is drawn from the 12 remaining of that suit
= the 3rd card is draw from the 39 = 13*3 for the other 3 suits
— trips
= the 1st card can be any card, it chooses the rank
= the 2nd card is drawn from the 3 remaining of that rank
= the 3rd card is draw from the 2 remaining of that rank
— paired
= the 1st card can be any card, it chooses the rank
= the 2nd card is drawn from the 3 remaining of that rank
= the 3rd card can be any of the 48 remaining cards not of that rank
= there are 3 ways to display the paired flop (AAK, AKA, KAA)
— unpaired
= the 1st card can be any card (lets say Ace)
= the 2nd card can be any (K,Q,J,T,9,8,7,6,5,4,3,2) thus 12 * 4 = 48 possible
= the 3rd card must be a different rank than the others thus 11 * 4 = 44 possible
— 3 straight rainbow
= the 1st card can be any card (let’s say Ah)
= the 2nd card must be no-gap and of different suit thus any 2c,2s,2d or Kc,Ks,Kd will work with 3 cards each way. (Assume the flop is Ah 2c or the flop is Ah Kc)
= the 3rd card is a 1-gap and of different suit thus any 3s,3d,Qs,Qd. This represents 2 cards each way.
= there are 6 ways to display the 3-straight flop (A23, A32, 2A3, 23A, 3A2, 32A)
_______________________________________________________________
You can use this technique to calculate your chances of say:
— flopping a pair
— flopping a set
To flop 1 pair (let’s say you have AK)
— there are 3 remaining aces and 3 remaining kings that can hit either the 1st, 2nd, or 3rd card. Since you know two cards, there are 50 unknown cards in the deck. For this calculation, it’s easier to think weird and assume you don’t hit one pair, thus 1 – (44 / 50)(43 / 49)(42 / 48) = 1 – 0.675 = 32.5% of flopping 1 pair or better. On the 1st card, you have 44 cards that aren’t an ace or king.
To flop a set, (let’s say you have 55 🙂
— there are 2 remaining fives in the deck. Assume you miss to do the math and subtract that chance by 1 to find how often you hit it. Thus, the calculation is equal to 1 – (48 / 50)(47 / 49)(46 / 48) = 1 – 0.8824 = 11.76% chance of flopping a set or better.
-
1 Reply
-
Flopping a flush w/ AKss
— there are 13 cards of each suit (c, d, h, s). When you have a suited starting hand, there are 11 left and 3 of them need to hit the flop and avoiding the 39 non flush cards. Thus, (11/50)(10/49)(9/48) = 990 / 117600 = 0.00841 or 0.84%.
Flopping a flush draw w/ AKss
— on this flop, you have 11 spades and 2 of them have to be on the flop along w/ 1 of the 39 non flush cards. Thus (11/50)(10/49)(39/48) = 0.0364 or 3.64%. However, you need to think of all the different ways of sorting a unique 2-tone flop. The ways are (abc, acb, bac, bca, cab, cba) which is 6 however you need to make another adjustment. Assume that (a = b) because they are the same suit and c is the non-flush card. If you look at it this way, (aac, aca, aac, aca, caa, caa), you can see that we have replicates making only 3 unique ways of repping the 2-tone flop.
Thus, 3.64% * 3 = 10.9%
Hitting the backdoor flush w/ AKss
— this implies the flop is a rainbow and we will see all five cards
— the flop should have 1 of the 11 spades, and 2 of the 39 non-flush cards (c,d,h).
— the turn should be 1 of the remaining 10 spades
— the river should be 1 of the remaining 9 spades
— thus, the calculation is (11/50)(39/49)(38/48)(10/47)(9/46) = 0.00577 however lets think of how many ways to express the board with a rainbow flop and the backdoor hitting. Think of it as (abcaa, acbaa, bacaa, bcaaa, cabaa, cbaaa) which is 6 ways, thus (0.00577 * 6) = 0.0346 or 3.46%.
Doing similar math post-flop
— if you look at the math post-flop with the turn and river to be dealt, then (10/47)(9/46) = 4.16% is the chance that the turn and river will be the same suit.
Log in to reply.