Probability - blockers

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.

HMMM almost a 100 views and no responses. I wonder if they're trying to keep it a secret Lol.

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.
link to original post

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

Wow that's amazing. Really appreciate it.

Thanks again