The Three Card Poker problem

28 December 2017

By John Grochowski

QUESTION: In Three Card Poker, how should I divide my bets between ante-play and Pair Plus?

Should I bet equal amounts on them? Should I divide my bets so my Pair Plus is twice my ante? The latter is kind of my working theory, since whenever I bet after seeing my cards, that would bring my ante-play combination up to the same amount as the Pair Plus.

ANSWER: Given the current state of the game, I'd put my full bet on ante-play and skip Pair Plus entirely.

The house edge on ante-play is 3.37% of your ante or 2.01% of total action, once the play bet is taken into account.

On Pair Plus, the most common pay table available today has a house edge of 7.28%. The differences in the house edges is so great that I'd focus on ante-play and skip Pair Plus.

That assumes Pair Plus pays 40-1 on a straight flush, 30-1 on three of a kind, 6-1 on a straight, 3-1 on a flush and 1-1 on a pair.

The original pay table differed in one important respect. Flushes paid 4-1 instead of 3-1. With that difference, the house edge drops to 2.32%.

When that pay table was common, if I found myself at a $5 minimum table, I'd bet $5 on Pair Plus and ante $5 on ante-play. That gave me a shot at the bigger payoffs in Pair Plus while putting the larger total wager, including the play bet, on the portion with the 2.01% edge on total action.

As with all systems, sometimes it worked, more often it didn't, but the occasional big payoff was more than welcome.

Alas, there are no more 40-1 or 30-1 payoffs for me in Three Card Poker, because I won't play Pair Plus with the current common pay table.


QUESTION: Please explain something to me. I've been told that on a three-reel slot, you could have 10 of the same symbol on the first reel, 10 of the same on the second reel, and 9 of the same on the third, with just one loser symbol.

I've been told the game could be programmed so the loser symbol would come up 99% of the time.

How can that be with a game that's supposed to be random?

ANSWER: For the same of example, let's call the winning symbols 7s and the one loser symbol a lemon.

The first two reels are going to land on 7s every time.

For the third symbol, imagine the random number generator is working with a set of 1,000 numbers. It generates numbers randomly, and any number is as likely to come up as any other.

The game programmer maps those numbers onto a virtual reel. The game might be designed so that every time the RNG generates 43, the reel stops on the third 7, and every time it generates 21, the reel stops on the fifth 7, and so on.

Symbols can be allotted different proportions of random numbers. The virtual reel could be designed so that each 7 gets only one number while the lemon gets 991 numbers.

That would mean random results — the 1,000 numbers would be generated randomly and there would be no way of knowing what came next. But the 1,000 numbers would bring only nine 7s and 991 lemons to the pay line.

So even with your starting conditions — 10 7s each on reels 1 and 2 and nine 7s on reel 3 — random results would lead to losing spins 99.1% of the time.