Calculating probability of a Jack high flop

How does one calculate for a specific card-high flop? I.e. What is the likelihood that the following conditions exist on the flop? 1) that a single Jack hits, and 2) that an Ace, King and Queen does not?
thank you in advance for any help you might be able to provide.

If you don't take any hole cards, including your own, into account, there are 52 x 51 x 50 / 6 = 22,100 different flops.

Of these, there are four possible cards for the "single Jack". Assuming you don't want to see any card paired in the flop (since you said you are looking for a "card-high" flop), there are 36 cards for the "first" card (any of the 2s through 10s), and 32 for the "second" card (any of the 2s through 10s that don't match the "first" card), but since order does not matter, each pair appears twice, so divide by 2. The total number of "Jack-high flops" = 4 x (36 x 32) / 2 = 2304.

The probability of a Jack-high flop is 2304 / 22,100 = about 1 in 9.6.

Here is the formula in Excel: =combin(9,2)*4^3/combin(50,3) = 0.11755102.

This is saying, the number of ways to pick the other two ranks lower than a jack (there are nine of them), times the 4^3 ways to pick a suit from each one, divided by the number of ways to choose 3 cards out of 50.

Quote: Wizard

Here is the formula in Excel: =combin(9,2)*4^3/combin(50,3) = 0.11755102.

This is saying, the number of ways to pick the other two ranks lower than a jack (there are nine of them), times the 4^3 ways to pick a suit from each one, divided by the number of ways to choose 3 cards out of 50.

Since you do know your hole cards, that info should be included, but it is apparent that the calculations quickly get out of hand:

1) You have a Jack and a lower rank card.
2) You have a Jack and a higher rank card.
3) You have a pair of Jacks.

Quote: Wizard

Here is the formula in Excel: =combin(9,2)*4^3/combin(50,3) = 0.11755102.

Double edit:
The correct answer for a J-high flop with two other unpaired cards is 0.104253. This answer is arrived at by modifying the last term in the wizards equation to combin(52/3). It can also be straightforwardly calculated as: 3*(4*36*32)/(52*51*50) where 3 is the number of distinct sequences in which the 3 flop cards might be dealt.

Quote: gordonm888

Double edit:
The correct answer for a J-high flop with two other unpaired cards is 0.104253. This answer is arrived at by modifying the last term in the wizards equation to combin(52/3). It can also be straightforwardly calculated as: 3*(4*36*32)/(52*51*50) where 3 is the number of distinct sequences in which the 3 flop cards might be dealt.

I agree with the Wizard in the use of combin(50,3) because there are already two known cards removed from the deck. What he doesn't consider, though, is the composition of the initial hand.

If we assume that the player's hand is composed of two non-suited cards under J that cannot make a straight with a J, then I calculate the probability to be:
4*combin(7,2)*4^2/combin(50,3) = 6.86%

If the two cards in the players hand are suited or can make a straight with a J, then the probability will be lower.

Quote: CrystalMath

I agree with the Wizard in the use of combin(50,3) because there are already two known cards removed from the deck. What he doesn't consider, though, is the composition of the initial hand.

If we assume that the player's hand is composed of two non-suited cards under J that cannot make a straight with a J, then I calculate the probability to be:
4*combin(7,2)*4^2/combin(50,3) = 6.86%

If the two cards in the players hand are suited or can make a straight with a J, then the probability will be lower.

No, I disagree strongly. It is not even stipulated by the OP that you are a player in the game, you may be an observer. The OP's question is simply -what is the probability of a jack high flop? Obviously, all ten players in the game would know what their two cards were (whether they were discarded or not) but this does not change the odds of the meta-question "what are the odds of a jack-high flop?"

Unless you stipulate "What are the odds of a Jack-high flop when you hold (say) Q-T in your hand?" you have a 52 card deck from which to form the flop. When you address a mathematical question, you must identify "What is known?"

Addressing your example, if you calculate the odds of a Jack-high flop when the player has

- two lower unpaired ranks
- a pair, with lower ranks
- one lower and one higher rank
- two higher unpaired ranks
- a pair of higher ranks
- one jack and a lower rank
- one jack and a higher rank
- a pair of jacks

and then probabilistically weight these scenarios, the final answer will be the answer I gave: 0.10425.

Both CrystalMeth and Wizard are in error.

Quote: gordonm888

CrystalMeth

Freudian or insult?

Why even call it a flop, then? Why not ask what is the probability of getting a J-high hand in three cards dealt from a full deck?

Quote: CrystalMath

Freudian or insult?

Why even call it a flop, then? Why not ask what is the probability of getting a J-high hand in three cards dealt from a full deck?

Whoops, I apologize for the typo - I did not mean to insult you. You have a very clever screenname, please forgive my slip/

The OP asked his question without revealing why he wanted to know. Maybe he is thinking about issues associated with the strength and playability of a pocket pair of Queens (or Jacks)? Those hands are very powerful and straightforward to play if the flop is jack high (or lower) but are vexing when an A or K hits the board. Or maybe he is wondering about the strength and playing strategy when you have one or two overcards (A,K,Q) to a jack high flop. I don't know why he asked. But giving him the probability of a Jack High flop when he holds two cards with ranks less than a jack is not the optimal way to respond, IMHO.

And the Wizard's response - stated to 9 digits from someone who boasts he is the last word in gaming odds - was worse than non-optimal, it was simply erroneous. He is a very smart guy and his calculations are usually highly reliable, but accepting some peer review from the very smart mathematical folks on this forum and acknowledging his infrequent goofs when he makes them would be a healthy practice.

Quote: gordonm888

Whoops, I apologize for the typo - I did not mean to insult you. You have a very clever screenname, please forgive my slip/

:)

Quote:

The OP asked his question without revealing why he wanted to know. Maybe he is thinking about issues associated with the strength and playability of a pocket pair of Queens (or Jacks)? Those hands are very powerful and straightforward to play if the flop is jack high (or lower) but are vexing when an A or K hits the board. Or maybe he is wondering about the strength and playing strategy when you have one or two overcards (A,K,Q) to a jack high flop. I don't know why he asked. But giving him the probability of a Jack High flop when he holds two cards with ranks less than a jack is not the optimal way to respond, IMHO.

Fair enough, I don't know enough about the game to even guess what he's looking for.

Quote:

And the Wizard's response - stated to 9 digits from someone who boasts he is the last word in gaming odds - was worse than non-optimal, it was simply erroneous. He is a very smart guy and his calculations are usually highly reliable, but accepting some peer review from the very smart mathematical folks on this forum and acknowledging his infrequent goofs when he makes them would be a healthy practice.

We're all human. He is humble and does admit when he's wrong, in my experience.

Given that there is evidently no knowledge of the two hole cards, I would like to amend my answer to:

=combin(9,2)*4^3/combin(52,3) = 0.104253394.

Maybe it is a bad excuse, but I'm just used to there being only 50 possible cards in the flop.