Dealer mistake for advantage

While at a local casino I noticed the dealer at three card poker would consistently make the same payout error. Whenever the player folded and the dealer didn't qualify, instead of the player losing the ante, he would pay the player 1 to 1 on it. I'm curious as to whether or not someone would know what the new edge is with this change and if it's enough to flip the advantage. It's regular three card, ante bonus's are 1 to 1 for straight, 3 to 1 for trips, and 4 to 1 for str flush.

Quote: Awz

While at a local casino I noticed the dealer at three card poker would consistently make the same payout error. Whenever the player folded and the dealer didn't qualify, instead of the player losing the ante, he would pay the player 1 to 1 on it. I'm curious as to whether or not someone would know what the new edge is with this change and if it's enough to flip the advantage. It's regular three card, ante bonus's are 1 to 1 for straight, 3 to 1 for trips, and 4 to 1 for str flush.
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You probably have a dozen members buying plane tickets to wherever you are……

This happened with only this one dealer, correct? How long did you observe this? What $$ were you betting? How much $$$ did you win?

Over multiple hours. I was only betting table minimun at the time and made a few hundred. Im not going to give out the location or the dealer for privacy reasons obviously but i would still like to know exactly how much advantage this gives the player.

Quote: Awz

While at a local casino I noticed the dealer at three card poker would consistently make the same payout error. Whenever the player folded and the dealer didn't qualify, instead of the player losing the ante, he would pay the player 1 to 1 on it. I'm curious as to whether or not someone would know what the new edge is with this change and if it's enough to flip the advantage. It's regular three card, ante bonus's are 1 to 1 for straight, 3 to 1 for trips, and 4 to 1 for str flush.
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This reflects a tremendous advantage.

If we look at Wizard's Three-Card Poker page here:

https://wizardofodds.com/games/three-card-poker/

What we immediately find is that the player will fold 32.5792% of hands, pursuant to optimal strategy, thereby incurring an expected loss of .325792 units.

I suspect that there would be mathematically optimal strategy changes such that players should actually fold more often. However, I am going to ignore that possibility as I can't think of an easy way to do it. Instead, what we will do is determine how often the dealer does not qualify.

Fortunately, this topic has already been addressed on these very forums, in which Tringlomane found:

https://wizardofvegas.com/forum/questions-and-answers/math/13595-3-card-poker-percentage-of-hands-dealer-qualifies/

Quote:

Total qualifying combos: 15,380

Total combos: 52*51*50/6 = 22,100

Dealer qualifies: 15,380/22,100 = 0.6959 = 69.59%

Ergo, the dealer does not qualify 30.41% of the time.

You will notice that this is very close to the probability of the player folding. The reason for that is because the dealer qualifies with a Queen, but in certain cases, a player with a queen should fold.

The combined probability of the player folding (based on optimal strategy) and the dealer not qualifying is ROUGHLY:

.3041 * .325792 = 0.0990733472

However, this would be an expected swing of double that number of units because, not only are you not losing the one unit when you fold, but you are also getting paid an additional unit. Thus:

0.0990733472 * 2 = 0.1981466944

Thus, it is swing of roughly .198 units for every hand played.

The WoO page suggests that the normal expected return of the ante is -0.033730, so you end up with -0.033730 + 0.1981466944 = 0.1644166944

Which reflects an overall player advantage of greater than 16%. This return Table also takes play into account.

NOTES AND CLARIFICATIONS

1.) As I mentioned, this is all pursuant to the normal strategy for playing the game. However, that normal strategy would have you losing one unit, when you fold, 100% of the time, as is supposed to happen.

In this case, if you fold and the dealer does not qualify, then you will have not only NOT LOST one unit, you also get one unit in profits.

For that reason, a modified strategy would have you fold more often.

Based on the WoO page, we find that a player loses with a Flush (or less) 0.223805 or 22.3805% of the time. The player wins with a flush (or less) 0.213906 or 21.3906% of the time.

Of course, this wouldn't change making the Play bet and losing into a win, but what it does do is make some folds win more often where they wouldn't win at all normally. At a guess, you would fold a high-card queen (with no pair or better) more often than not. The reason why is because if the dealer qualifies and beats you (which he will more often than not) you would lose two units. However, by folding, you lose one unit roughly 70% of the time and profit a unit roughly 30% of the time whereas you should lose one unit 100% of the time. The normal strategy exists the way it does because folding is typically an automatic loss of a unit; that is no longer the case. In fact, folding goes from a guaranteed -1 to an expectation of -.4. You might actually fold some high card kings with no pair, or better, I don't know.

2.) Technically, you not having a playable hand normally makes it slightly more likely that the dealer qualifies. The reason why is because, absent very bad queens, you've removed three cards from the deck that would normally help in the dealer not qualifying. That means there are fewer low cards for the dealer to get. If you fold very bad queens, then you have only removed two cards that would cause the dealer not to qualify and one that would cause the dealer to qualify.

That's another reason that you would fold more queens with no pair or better, if not all of them. I could probably figure it out but it would take a long time.