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

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Edit: Comment withdrawn.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Edit: Comment withdrawn.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.