3 Card Poker Correct Play

The Wizard Of Odds site says to raise in three card poker if you have Q/6/4 or higher. Yet the site atsays "The minimum hand you must have before raising is at least an ace high with either a king or queen, otherwise one should fold." that's a huge difference in strategy. I know poor gambling advice abounds on the internet, but just how badly will people do if they follow this advice?

The site looks like its approached from more of a Poker perspective than a mathematical view.

Quote: Wizard

If you want to know why queen/6/4 is the borderline hand it is because if you raise on queen/6/3 you can expect to lose 1.00255 units, more than the 1 unit by folding. However if you raise on queen/6/4 the expected loss is .993378, less than the 1 unit by folding.

I have been asked several times about the strategy of raising on any queen or better, in other words mimicing the dealer. This is not a bad strategy but you will lose more with it than the optimal strategy above. The house edge playing the mimic the dealer strategy is 3.45%. Raising on everything, or playing blind, results in a house edge of 7.65%.

Maybe someone else can provide the rest of the answer with the proper maths.

I don't know how they came up with that advice. Several others have analyzed Three Card Poker, and all get the same results. The game analysis is a classic exercise in programming, but the number of combinations is not prohibitive to take hours to run.

I had my three card poker days, if you only raise with an Ace/Q+ you'll be folding 90% of the time. Alright that's exaggerating a bit, but you get the gist.

There are 22,100 combinations (52 x 51 x 50 / 6). From the Wizard's site, there are 48 Straight Flushes, 52 3 of a Kinds, 720 straights, 1,096 flushes, and 3,744 pairs. The dealer also has 16,440 other hands that are none of the above. Of those 16,440, 9,600 are Q high or better. So, the dealer (and you) will qualify on 15,260 of the 22,100 combinations.

Your choices when being dealt are to fold (expected value): -1 or to raise.

At this point, you will either win one unit (the dealer does not qualify), two units, the dealer qualifies and you beat them, or lose two units (the dealer qualifies and you lose). So, you have to solve for those three modes. You stay when your hand's expected value is > -1 and fold when it is below -1.

Let's start with an unsuited J-10-8, the highest Jack high hand. I am not taking into account card exclusion. The dealer will qualify on 15,260 combinations with a loss of 2 units. The dealer will not qualify on 6,840 combinations for a win of 2 units.

(15,260 x -2 + 6,840 x 1)/22,100 = -1.071493 units.

By staying on J high, you will lose on average -1.10588 units.

Let's now move to Q-6-4, the recommended stay hand. You will win two units when the dealer qualifies with Q-6-3 or less. This happens on 480 combinations (Q-2-3, Q-2-4, Q-2-5, Q-2-6, Q-3-4, Q-3-5, Q-3-6, Q-4-5).

(14,720 x -2 + 6,840 x 1 + 480 x 2 + 60 x 0)/22,100 = -.97919 units.

Move now to Q-6-3. you win two units when the dealer qualifies with Q-6-2 or less. This happens only on 420 cominations (Q-2-3, Q-2-4, Q-2-5, Q-2-6, Q-3-4, Q-3-5, Q-4-5).

(14,780 x -2 + 6840 X 1 + 480 x 2 + 60 x 0)/22,100 = -.99005 units.

I am not keeping in mind card exclusion (the fact that there is one less Queen in the deck which will change the combinations of everything) but you get the jist.