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

Quote: kenarman

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.

Never seen that. Although I had trip Ace's once and was beat by a 789spade straight flush. At least the bonus paid like you said.

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.

Your math is wrong. ;+) You need to consider that either 1 or 2 of the cards are higher than the jack, with the 3rd lower, and just one higher than the jack on 2 lower.
2 higher:
You can pick 2 of 3 higher cards (A,k) (A,Q) (K,J) and one of the lower (2,3,4,5,7,8,10) 3*7=21
2 lower
You can pick 1 of 3 higher cards (A,K,Q) and 2 of the 7 lower combin(7,2)=21 3*21=63
For a total of 84 higher flushes. There are combin(49,3)=18424 total dealer hands.
84/18424=0.004559270516717 or 1 in 219 1/3.

This didn't happen in 3 card, but I once saw a dealers 4 of a kind (3's) get beat by a players 4 of a kind (kings) at a caribbean stud table once. I am sure that is pretty rare.

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.

At least some of the disagreement on calculations seems to be centered on whether it meets the OP criteria in the case that the dealer's hand is a straight flush rather than just a more ordinary flush. The other source of disagreement in calculations seems to be whether it is a given that the player has a jack-high flush or whether matthew was asking about the chances of being dealt that hand and losing to a higher flush.

Since the OP (quoted above) says "HIGHER flush", I'm going to stick with the position that a straight flush is a different hand and was not what matthew was asking about. His later report of the actual hands showed that the dealer's flush was not straight, though that may not make any difference to anyone. Also, the way I read the OP, he was asking the probability of both events occurring on a random deal. That is what I tried to calculate, and I am still waiting for the mathematicians to show me the correct solution.

Quote: Doc

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!

It's amazing the bizarre things that go through my mind as I try to sleep at night. Here is my stab at the solution to this expanded question: What is the probability of being dealt any (non-straight) flush and losing to the dealer's higher (non-straight) flush in the same suit?

My attempted solution: Figure the total number of cases with two hands all the same suit, discount by the number of hands where one or both has a straight flush, divide by the total number of possible dealt hands, and multiply by 0.5 to account for the chance that the dealer is the one with the higher hand.

P = 0.5*4*{combin(13,3)*combin(10,3)-[2*12*combin(10,3)-(8+8+7+6+5+4+3+2+1)]} / [combin(52,3)*combin(49,3)]

= 2*{286*120-[24*120-44]} / [22100*18424]

= 2*{34320-[2880-44]} / 407170400

= 62968 / 407170400

= 0.00015464778

I apologize for the string of integers summed -- I don't know how to type in the summation function here. That group (totaling 44) is intended to adjust for having double-deducted cases of both hands being straight flushes when counting the cases with one or more straight flushes in the previous term (the 2880).

So I'm now looking for a mathematician to tell me whether this is correct for losing with any flush to a higher flush in the same suit and also whether my earlier solution is correct when restricting to the case of the player's flush being jack high. Perhaps more properly, I would like to see the correct solution, since I highly suspect that I screwed it up somewhere along the way.

Quote: miplet

Your math is wrong. ;+) You need to consider that either 1 or 2 of the cards are higher than the jack, with the 3rd lower, and just one higher than the jack on 2 lower.

Actually, I don't. Given the premise that a jack high flush is beaten, it is sufficient to establish the condition that the dealer has either the Q/K/A of that suit---a 3/49 chance. Once that is established, all other cards of that suit are equal for the purposes of this calculation, whether they also happen to be higher than the Jack or not: A42 and AK2 btoh fulfill the condition of a higher flush.

Where you are going wrong is in attempting to calculate by use of combinations, not realizing, for one thing, that there are more possible combination matrices of "low" cards than just the one you list,

In most three card poker tables the automated shuffler deals all hands regardless of how many players are at the table. When that's the case, then everyone is calculating that a hand will beat one particular flush with a higher-rank flush of the same suit. Shoulnd't you factor in a 1 in 6 chance that the dealer gets that hand? After all, had another player gotten it, then the OP might not have lost.

Just wondering

Quote: mkl654321

This might be simple:

Since the premise is that the player has a J-high flush, ....

...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: I just now paid attention to your last paragraph. I think I disagree (if I understand). I think you conclude that there is ~ 1 chance in 559, given that the player has a jack-high flush. Do you think that the jack-high flush occurs often enough for you to see 559 of them in a dozen or so playing sessions? I suppose that depends on how many hands your sessions last.

Also, I'm not sure, but miplet may have been working along the line of not counting hands with straight flushes, as I think you did. I haven't followed through the numbers completely, and it seems to be a case of which way the question is defined anyway.

Quote: Nareed

In most three card poker tables the automated shuffler deals all hands regardless of how many players are at the table. When that's the case, then everyone is calculating that a hand will beat one particular flush with a higher-rank flush of the same suit. Shoulnd't you factor in a 1 in 6 chance that the dealer gets that hand? After all, had another player gotten it, then the OP might not have lost.

Just wondering

I don't think so. We've considered the probabilities that the one player and the dealer get their specific sets of three cards. We don't really care where the other 46 cards go. If the higher flush hand was given to another player at the table, that's just one more of the many instances that the dealer did not get the higher flush.

Quote: Doc

mkl: I just now paid attention to your last paragraph. I think I disagree (if I understand). I think you conclude that there is ~ 1 chance in 559, given that the player has a jack-high flush. Do you think that the jack-high flush occurs often enough for you to see 559 of them in a dozen or so playing sessions? I suppose that depends on how many hands your sessions last.

Also, I'm not sure, but miplet may have been working along the line of not counting hands with straight flushes, as I think you did. I haven't followed through the numbers completely, and it seems to be a case of which way the question is defined anyway.

A flush is a relatively easy hand to get: you'll be dealt one about 5% of the time. At least half of all flushes are Jack-high or better--I'd say 60 percent (I'm being lazy). So you need about 33 hands to get a J-high flush, and then you have a 1/559 chance of getting that hand beaten by a higher flush (or, AKQ straight flush).

So for the individual, it would take 19,000 or so hands for these two events to happen. For this to happen on a table, to SOMEBODY, would take about 3,000+ deals, assuming the table was full. I misinterpreted the question as how long would it take to happen at all, not how long it would take to happen to "me".

Quote: mkl654321

Actually, I don't. Given the premise that a jack high flush is beaten, it is sufficient to establish the condition that the dealer has either the Q/K/A of that suit---a 3/49 chance. Once that is established, all other cards of that suit are equal for the purposes of this calculation, whether they also happen to be higher than the Jack or not: A42 and AK2 btoh fulfill the condition of a higher flush.

Where you are going wrong is in attempting to calculate by use of combinations, not realizing, for one thing, that there are more possible combination matrices of "low" cards than just the one you list,

Here is my list of the 84 posible dealer hands:
AKT , AK8 , AK7 , AK5 , AK4 , AK3 , AK2 , AQT , AQ8 , AQ7
AQ5 , AQ4 , AQ3 , AQ2 , AT8 , AT7 , AT5 , AT4 , AT3 , AT2
A87 , A85 , A84 , A83 , A82 , A75 , A74 , A73 , A72 , A54
A53 , A52 , A43 , A42 , A32 , KQT , KQ8 , KQ7 , KQ5 , KQ4
KQ3 , KQ2 , KT8 , KT7 , KT5 , KT4 , KT3 , KT2 , K87 , K85
K84 , K83 , K82 , K75 , K74 , K73 , K72 , K54 , K53 , K52
K43 , K42 , K32 , QT8 , QT7 , QT5 , QT4 , QT3 , QT2 , Q87
Q85 , Q84 , Q83 , Q82 , Q75 , Q74 , Q73 , Q72 , Q54 , Q53
Q52 , Q43 , Q42 , Q32
The are combin(49,3)=18424 total dealer hands. 84/18424 = 3/658 or 1 in 219 1/3.