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Combinations in poker (combinatorics)

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You’ll have to understand combos first and how blocking works. Combinations are pretty self-explanatory and I believe that you can understand them by using, for an example, your favorite hand, pocket aces: We have 6 combos of AA, and for reference d=diamonds, c=clubs, h=hearts and s=spades:

AcAd AcAs AcAh AdAs AdAh AsAh

There we go! Wasn’t that hard. So, basically, combos are all the possible AxAy that you, or your opponent can have, preflop.

What happens when we’re holding the Ac? Our opponent can only now make these combos: AdAs AdAh AsAh

Thus, we are blocking 3 of his AA combos. The chance that he will have AA is now 3/6, 50% less than before. This is more powerful than you can ever imagine.

Now, let’s talk about unpaired hands, as you probably guessed, they are different than pairs. You have four Aces and four Kings in one deck, and you can make exactly: 4×4=16 combos

Each of the four Aces you pair with each of the four Kings and thus you can make 16 total combinations of AK.

What happens when we are holding the Ac? How many AK combos can our opponent have?

Well, now there’s three Aces in the deck and four Kings left. So, the total number of AK combos that our opponent can have is: 3×4=12 combos This means that our opponent can now have only 12 out of the 16 combos of AK. The ace that we have reduces the likelihood that he is holding AK by 4 combos out of 16 total(25%).

Now, suited is a lot simpler. If you want to figure out the amount of suited combos that you can have, you just figure out how many suits there are, right?

AKs: AcKc AdKd AhKh AsKs 4 combos.

Every suited hand has 4 combos.

Now, if we’re holding an Ac, there’s only 3 combos left. 75%. There is a bit of application to combos like blocking, that you just got a sample of, but basically if you’re holding AK there’s 50% less chance that your opponent has either AA or KK, and 9/16 combos left of AK that he can have (3Ax3K), so almost 50% less chance for AK. This also reduces AQs/KQs combos etc.

Now, combinations also have a lot of applications post-flop. You can count the hands that your opponent can have. Top Pair: The flop is A37r, your opponent can have AK, for example. The total combos of AK that your opponent can have is 3Ax4K, because there’s already an Ace down there, so there’s only 3 left in the deck.

Two Pair: The flop is A37r and you suspect your opponent can have A3 in his range. The total number of combos for A3o will be 3×3=9 combos. There’s three Aces left in the deck and three Nines. Thus, multiplied by each other, 9 total two pair combos. Two pair will always have nine combos. Trips: Flop is AA2 and you think that your opponent has AJ in his range. This means there are two Aces left and four Jacks. Thus, 2×4=8 combos of trips. Trips will always have eight combos.

Set: Flop is 864r and you’re thinking that your opponent can have 44 in his range. This means there are three Fours left in the deck. With these, you can make 3 combos. Sets always have 3 combos.

Straight: Flop is AKT and you think that your opponent has QJ. There is no blocking here. This means that QJ will have a total of 4×4=16 combos. Also, if you believe that your opponent opened from EP and he doesn’t open QJo from there, then he can only have 4 combos of QJs. Ditto.

Quads: Flop is 882 and you suspect that your opponent can have dreaded 88’s. There are only two 8’s left in the deck, so there’s only 1 combination of quads, because the order in which your opponent has the 8’s doesn’t matter(for example 8c8d or 8d8c is the same thing).

Flush: This depends more on your opponent’s pre-flop range, for example if you think that your opponent has any suited ace in his pre-flop raising range, then on a board that is 864sss your opponent will have As2s, As3s, As5s, As7s, As9s, AsTs, AsJs, AsQs, AsKs, a total of 9 combos.

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