It's my first time so I'm just gonna go with...Thread Title

I was playing 3-card poker at an on-line casino. I was rather pleased when I drew a flush.
WAS. Until the dealer drew a HIGHER flush with the SAME suit. I just thought it might be interesting if someone could figure out the odds on THAT.

Quote: matthew333

I was playing 3-card poker at an on-line casino. I was rather pleased when I drew a flush.
WAS. Until the dealer drew a HIGHER flush with the SAME suit. I just thought it might be interesting if someone could figure out the odds on THAT.

The answer would depend on how high YOUR flush was.

DEALER

FLUSH
K (spades) 3 (spades) 7 (spades)

PLAYER
FLUSH
6 (spades) 9 (spades) J (spades)

I sometimes enjoy taking a stab at these problems and then having someone else show me the correct way to solve it, so here goes....

Let's word the problem this way: What is the probability that you will get a Jack high flush that loses to a higher flush in the same suit?

For a jack high, I think there are

4 * (combin(9,2)-1)=140

possible flush hands, with the (-1) to discount the possibility of a straight flush. For each of those hands, I think there are

combin(9,2) + combin(8,2) + combin(7,2) -1 = 84

higher flush hands in the same suit available for the dealer. This gives a total of

140 * 84 = 11,760

qualifying combinations for the two hands. For a random deal, there are

combin(52,3) * combin(49,3) = 407,170,400

hands for the dealer and one player. The probability of your getting a Jack high flush and losing to a higher flush in the same suit becomes

11760 / 407170400 = 0.0000288822

I think. If you want to consider only spades, divide by four (again, I think). If you want to consider all possible flushes losing to higher flushes (not just Jack-highs), then have fun working your fingers off on your own calculator!

Now you mathematicians, show me how to do it correctly!

This might be simple:

Since the premise is that the player has a J-high flush, all we have to do is figure out the possibility of the dealer having a Q-high or better flush in the same suit.

There are ten cards remaining in that suit out of 49 total remaining.

By definition, one card must be the A, K, or Q of that suit. So, start with 3/49.

The next card must be one of the 9 remaining in that suit. So, 9/48.

The third card must be one of the 8 remaining in that suit. So, 8/48, or 1/6.

Multiply: (3/49)(9/48)(1/6) = 27/15,112, or something like 1 in 559

So this occurence, though unusual, would happen once every dozen or so playing sessions. Unless, of course, my math is wrong.

mkl: We addressed different problems. I included the probability of the player getting a jack-high flush in the first place. If we take it as a given that the player already has that hand (which is what I think your solution assumes), then I think your solution is close to correct, except you allow the possibility of losing to the dealer's AKQ straight flush rather than a dealer flush. A picky difference, but that was the OP question, and I think losing to a straight flush may not be so emotionally frustrating as to an ordinary flush just barely higher than the one you hold. It would cost just as much money, though.

Gentlemen! Thank you very much for your feedback. I should have mentioned that 3 card poker is played with
'one standard deck of 52 cards that ia shuffled before each game' I got to see a scenario where I was dealt 3 cards (all spades) then the dealer was dealt 3 cards of the same suit (higher of course) I was thinking that I had a better chance of being hit by lightning on my way to cash in my winning lottery ticket than seeing this solution.

Quote: Doc

mkl: We addressed different problems. I included the probability of the player getting a jack-high flush in the first place. If we take it as a given that the player already has that hand (which is what I think your solution assumes), then I think your solution is close to correct, except you allow the possibility of losing to the dealer's AKQ straight flush rather than a dealer flush. A picky difference, but that was the OP question, and I think losing to a straight flush may not be so emotionally frustrating as to an ordinary flush just barely higher than the one you hold. It would cost just as much money, though.

You could take that 3/49 figure and change it to 4/49 for a ten-high flush being beaten, 5/49 for a nine-high flush, etc. I would consider the straight flush(es) to be equivalent to the flushes, given that the OP's question didn't exclude them.

Quote: mkl654321

This might be simple:

Since the premise is that the player has a J-high flush, all we have to do is figure out the possibility of the dealer having a Q-high or better flush in the same suit.

There are ten cards remaining in that suit out of 49 total remaining.

By definition, one card must be the A, K, or Q of that suit. So, start with 3/49.

The next card must be one of the 9 remaining in that suit. So, 9/48.

The third card must be one of the 8 remaining in that suit. So, 8/48, or 1/6.

Multiply: (3/49)(9/48)(1/6) = 27/15,112, or something like 1 in 559

So this occurence, though unusual, would happen once every dozen or so playing sessions. Unless, of course, my math is wrong.

I believe your math is right, given the player having a J-high flush. But we should do the math on the Jack High flush probability as well. It's only 1 in 559 times IF the player gets his hand. It would be every 12 sessions if you got a Jack high flush every hand, which you most likely would not.

So I think you just have to set up these conditions:
We want to draw 6 cards out of a deck of 52. All 6 must be the same suit, and the first three must contain no card higher than a J, and the last three must contain at least 1 A,K, or Q. While the last three (dealers cards) may be a straight flush, the (first three (players) may not.

My wife and I were playing 3 card when her str8 flush was beaten by the dealers str8 flush. Still a nice pay on the bonus's though.