OMC
OMC
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September 30th, 2024 at 5:32:43 PM permalink
The term blockers is all the rage in poker, so I'm wondering how the concept actually applies. I have two examples.

In 9 handed holdem game, if I'm dealt Jx, what is the probability that another player is dealt a J.

In a 9 handed holdem game, if I'm holding one heart and there's two hearts on the flop, what is the probability another player holds two hearts.

Tx in advance.
OMC
OMC
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October 1st, 2024 at 7:47:14 PM permalink
HMMM almost a 100 views and no responses. I wonder if they're trying to keep it a secret Lol.
DogHand
DogHand
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October 2nd, 2024 at 3:35:32 AM permalink
Quote: OMC

The term blockers is all the rage in poker, so I'm wondering how the concept actually applies. I have two examples.

In 9 handed holdem game, if I'm dealt Jx, what is the probability that another player is dealt a J.

In a 9 handed holdem game, if I'm holding one heart and there's two hearts on the flop, what is the probability another player holds two hearts.

Tx in advance.
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OMC,

I'll answer the first one.

With 9 players (you plus 8), the other players have a total of 16 hole cards. You want to know P(J): the probability that of those 16 cards, one (or more) is a Jack. Let's first compute P(no J), that is, the probability that they're holding no Jacks, after which we can easily find P(J) because P(J) = 1 - P(no J).

By the way, you didn't specify at what point in the hand you want to know the probabilities, so I will assume you want them BEFORE the dealer exposes any of the community cards. This means that the only cards you know are your two hole cards, a J and a non-J, so to you 52 - 2 = 50 cards are unknown.

Now let's compute P(no J).

The probability that your left-hand player has no J is (47/50)*(46/49). This is because for his first card the deck has 47 non-Jacks out of 50 cards, while for his second card the deck now has 46 non-Jacks out of 49 cards. Since BOTH of these must be true simultaneously for him to have no Jacks, we MULTIPLY their probabilities. Since (47/50)*(46/49) = 0.8824..., just over 88% of the time your left-hand player will not be holding a Jack.

Continuing to the next player, his no Jack probability is now (45/48)*(44/47), so now the no J probability for both of them combined is (47/50)*(46/49)*(45/48)*(44/47) = (47*46*45*44)/(50*49*48*47).

Continuing the rest of the way around the table, we finally get:

P(no J) = (47*46*45*44*43*42*41*40*39*38*37*36*35*34*33*32)/(50*49*48*47*46*45*44*43*42*41*40*39*38*37*36*35)

or, using Factorial notation:

P(no J) = (47!/31!)/(50!/34!)

I did this calculation in Excel using the formula

=(FACT(47)/FACT(31))/(FACT(50)/FACT(34))

and found:

P(no J) = (47!/31!)/(50!/34!) = 0.305306122

Finally, the P(J) = 1 - P(no J) = 0.694693878

So just over 30% of the time your opponents will all be Jack-free, and almost 70% of the time, your opponents will hold at least one of the remaining three Jacks.

Hope this helps!

Dog Hand
OMC
OMC
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October 2nd, 2024 at 6:54:59 AM permalink
Wow that's amazing. Really appreciate it.

Thanks again
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