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.

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.

Quote: miplet

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.

You are not taking into account that given the premise that the player hand is composed of three suited cards including the Jack and two lower cards, not all of the listed combinations are equally likely.

Well, I still haven't looked at mkl's and miplets solutions in detail. I think one possible error on mkl's part (even allowing that straight flushes are included) is that (I think) his solution includes permutations, not just combinations. That is, he would be counting A56 and A65 as different flush hands. Did I misunderstand?


Edit: Comment withdrawn. I think I am the one in error.

Quote: mkl654321

You are not taking into account that given the premise that the player hand is composed of three suited cards including the Jack and two lower cards, not all of the listed combinations are equally likely.

Huh??? Which of those hands listed has a differant probability of being a dealer hand? I didn't include J or 6 or 9 as they were included in the OP's hand.

Quote: Doc

Well, I still haven't looked at mkl's and miplets solutions in detail. I think one possible error on mkl's part (even allowing that straight flushes are included) is that (I think) his solution includes permutations, not just combinations. That is, he would be counting A56 and A65 as different flush hands. Did I misunderstand?


Edit: Comment withdrawn. I think I am the one in error.

Nope, you are right. I just realized that mkl is using permutations and only counting hands that the first card is an A,K, or Q.

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.

Found one minor error, but it doesn't explain the full discrepancy between mkl and miplet.

If we use mkl's technique, for the 3rd card it should be 8/47 not 8/48=1/6.


Edit: I was typing at the same time is miplet. If the permutations vs. combinations comment is correct, then perhaps combined with the above 48 vs. 47 error it might explain a bunch of the difference. The remainder might be that mkl allows AKQ as a dealer hand. Don't know whether that explains everything or not. .... Still not completely sure that mkl is using permutations -- I'm waffling on this one.

Quote: miplet

Huh??? Which of those hands listed has a differant probability of being a dealer hand? I didn't include J or 6 or 9 as they were included in the OP's hand.

Any listed card that is a Jack or lower could already be in the player's hand; a Q/K/A could not be. Thus, once the condition of a higher flush is established, the remaining two cards are somewhat more likely to be another higher card than one of the J-or-lower cards.

Quote: Doc

Found one minor error, but it doesn't explain the full discrepancy between mkl and miplet.

If we use mkl's technique, for the 3rd card it should be 8/47 not 8/48=1/6.


Edit: I was typing at the same time is miplet. If the permutations vs. combinations comment is correct, then perhaps combined with the above 48 vs. 47 error it might explain a bunch of the difference. The remainder might be that mkl allows AKQ as a dealer hand. Don't know whether that explains everything or not. .... Still not completely sure that mkl is using permutations -- I'm waffling on this one.

Actually, my arithmetical booboo aside, I wasn't using permutations OR combinations. I was using the method of multiplying the probabilities of three conditions, all of which have to be true:

One card is a Q/K/A of the desired suit
The second card is also of that suit
And so is the third card

But upon writing this, I see the flaw. The multiplicative (?) method shouls be like this:

Case 1: draw high flush card (calculation as already discussed)
Case 2: draw low flush card, then calculate probability of completing flush with one or two high cards
Case 3: draw low flush card, then another low flush card, then calculate probability of drawing A/K/Q

So it seems that we can get that flush one of three ways: High/Any/Any, Low/High/Any, or Low/Low/High. So we need to sum all these probabilities:

High/any/any: (3/49)(8/48)(7/47)
Low/High/any: (10/49)(3/48)(8/47)
Low/Low/High: (10/49)(9/48) (3/47)

Adding the three products together would give you a cumulative probability, which I would suspect would be quite close to another poster's answer.

Quote: matthew333

DEALER

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

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

Quite simply,

There are 49 x 48 x 47 / 6 combinations remaining in the deck = 18,424.

The available combinations of having 3 spades is 10 x 9 x 8 / 6 = 120.

There are 34 combinations of Ace high (2 straight flushes) flushes
There are 28 combinations of King high (no straight flushes) flushes
There are 21 combinations of Queen high (no straight flushes) flushes
There are 15 combinations of 10 high flushes (no straight flushes)
There are 10 combinations of 8 high flushes (no straight flushes)
There are 6 combinations of 7 high flushes (no straight flushes)
There are 2 combinations of 5 high flushes (1 straight flush)
There are no combinations of 4 high flushes (1 straight flush)

There are therefore 116 flushes and 4 straight flush combinations. 83 will beat the player. Therefore the odds that the dealer will beat you with a FLUSH (not a straight flush) in the same suit given the player's card is 83 / 18,424 = 0.4505 percent... not that too far-fetched.

EDIT: Miplet is off by one because he included A32 which is a straight flush in my books.

Hi everyone! Looks like I came to the right place. Something I was wondering about. Not sure how to formulate a proper equation.

I sit down at the playing table and start off by being dealt a spade.
Now I'm thinking there are 51 cards left, and only 12 of them could be a spade.
So I get dealt my second card, another spade. Now, the dealer is holding 50 cards, and only 11 of them are spades.
So I get dealt my third card, another spade! Yes! Now dealer is holding 49 cards, only 10 of which could be a spade.

The dealer gets dealt his/it's first card. A spade! Dealer's now left with 48 cards, 9 of which are spades.
Next card...dam! Another spade. No worries. Still in my favor. 47 cards left, and only 8 of them are spades.
A winning hand aside, what are the odds that the dealer will come up with another spade?

Quote: matthew333

Hi everyone! Looks like I came to the right place. Something I was wondering about. Not sure how to formulate a proper equation.

I sit down at the playing table and start off by being dealt a spade.
Now I'm thinking there are 51 cards left, and only 12 of them could be a spade.
So I get dealt my second card, another spade. Now, the dealer is holding 50 cards, and only 11 of them are spades.
So I get dealt my third card, another spade! Yes! Now dealer is holding 49 cards, only 10 of which could be a spade.

The dealer gets dealt his/it's first card. A spade! Dealer's now left with 48 cards, 9 of which are spades.
Next card...dam! Another spade. No worries. Still in my favor. 47 cards left, and only 8 of them are spades.
A winning hand aside, what are the odds that the dealer will come up with another spade?

You said it yourself: 8 in 47, or 39 to 8 against. A little better than a 5 to 1 shot.