4 Card Poker: Two Dealer Cards Exposed

In another thread I recently postulated that seeing one extra dealer card would result in decreasing the house advantage or even in creating a positive expected value for the player.

Several contributors to this site replied that the math was just too prohibitive to make an accurate analysis. Someone suggested the game could be beaten with three exposed dealer cards.

Since then I have been keeping a close eye on the 4 Card Poker tables at my local casino. I have noticed when the dealer's up-card and exposed bottom card are both of the same suit (and when there are relatively few cards of that suit in the players' hands) the dealer ends up turning over a flush the majority of the time. When the dealer's two known cards are suited connectors the likelihood of a flush or straight increases significantly.

Conversely, when the dealer's two exposed cards are of different suits and too far apart to make a straight possible (eg. King of Hearts; Four of Spades) the dealer usually ends up flipping over two pairs or less.

I would very much appreciate the opinions and input of other experienced gamblers on this subject -- especially on the topic of variations in the basic betting strategy capitalizing on this extra information.

Well if you see two cards that are the same suit, you can calculate the probability of a flush. First, you calculate that the other four cards are a flush of a different suit.

Those odds are 39/50*12/49*11/48*10/47 = .012108 (including a straight flush draw).

Then, you must calculate the odds that 2 or more cards are the same suit as the two first cards. There are six combinations to consider: the identity of the 3rd and 4th, 3rd and 5th, 3rd and 6th, 4th and 5th, 4th and 6th, and 5th and 6th card. Also assume that you do not know any other cards including your own.

The odds that that 3rd and 4th card are the same suit as the 1st and 2nd card is 11/50 * 10/49 = .044898. The odds that the 3rd and 5th card or the 4th and 5th card is the same as the 1st and 2nd card is 2 x 39/50 * 11/49 * 10/48 = .07296. The odds that the 3rd and 6th card, the 4th and 6th card, and the 5th and 6th card are 3 x 39/50 * 38/49 * 11/48 * 10/47 = .088482.

Add them up and the odds of the dealer having a flush (or straight flush) is .012108 + .20634 = 21.84%.

I guess my thought is that the best you can get is intuitive. Knowing the cards around you and the cards and the dealer up cards can help you make better decisions. For example, if you know that you and the two players around you don't have any of the same suit that the dealer has, the odds become 11/35 * 10/34 + 2 x 39/50 * 11/34 * 10/33 + 3 * 39/50 * 38/49 * 11/33 * 10/32 = 43.44%. At that point, you would probably think about folding everything under aces plus as well as only raising with 1 unit up to a straight (assuming that the player plays aces plus).

You would also make decisions as well if you know that two of the dealers cards are next in a straight to each other (between a 3 and Q) then the dealer has a 31.3% of making the straight. If the two cards are two cards next in rank and the same suit, the odds of making a straight or a flush jump to 45.5% and once again you change your strategy accordingly.

It's more obvious when you see a pair. But the odds of them getting the 3 of a kind to the pair is 15.51% while the odds of making the 2nd pair (or better) is 34%. At this point because the probability of having better than one pair is over 50% you would probably fold anything under Aces and only raise one unit up to two pair (and only if your high pair beats the pair that you see).

Take this even further ... say after having a peek at that second card, you have 2 of the same suit. Now the odds are even lower. If your friend next to you (in this social game) has 1 or 2 of that suit, the odds drop even more.

Quote: boymimbo

Add them up and the odds of the dealer having a flush (or straight flush) is .012108 + .20634 = 21.84%.

Hmmmm....I've always been a gambler who's relied on math and logic to back up each and every decision I make at the tables. This is why I have yet to bet a dollar on 4 Card.... Although I sincerely appreciate your meticulosity and respect your answer it just seems to me that dealers here turn over flushes at a rate much higher than one in five, given the first two cards are matching suits...

I also wonder whether or not it is a bad idea to bet the max of three units behind the ante when the dealer is holding two unpaired, inconsecutive cards of different suits; you have a mid-pair (say, eights) and both of the dealer's cards are lower than your pair (say, 7 of hearts, 2 of clubs). Instead of following the standard advice of betting one unit, why not wager three?

There is also the obvious advantage of folding a high pair when you can see that the dealer has a pair greater than your own (eg. you have QQ and the dealer, AA). All of the above-mentioned examples combined must translate into at the very least a reduction in the H.A.

Okay. Let's look at this more closely. First, let's acknowledge how we calculate the number of poker hands there are in a standard 52 deck. It is 52 x 51 x 50 x 49 x 48 / (5 x 4 x 3 x 2 x 1) = 2,598,960. This agrees with any calculation on the number of

In our sample below, there are 50 cards remaining. There are therefore 50 x 49 x 48 x 47 / (4 x 3 x 2) = 230,400. 11 of them match the same suit of the first two cards.

The number of hands with zero of the same suit is (39 x 38 x 37 x 36) / 24 = 82,251.
The number of hands with one of the same suit is (39 x 38 x 37 x 11) x 4 / 24 = 100,259.
The number of hands with two of the same suit is (39 x 38 x 11 x 10) x 6 / 24 = 40,755.
The number of hands with three of the same suit is (39 x 11 x 10 x 9) x 4 / 24 = 6,435.
The number of hands with four of the same suit is (11 x 10 x 9 x 8) / 24 = 330.

Therefore, the odds of completing a flush [or straight flush] is (40,755 + 6,435 + 330) / 230,400 = 20.634%.