I have come across a variation of Oasis Poker, played in the Baltics and Russia, called Royal Poker. Here are the rules in English. There have been various scenarios added for all sorts of added bets, and from what I’ve seen at the casinos, it turns players into suckers. They buy cards trying to turn even the most pathetic hands into monsters etc. Do you think you can come up with an algorithm to simulate some results and come up with the house edge and strategy for this game? Thanks!

Martin from Tallinn

I’m getting asked about this more and more frequently. Unfortunately, the number of combinations in this game would be an unholy gigantic figure. A brute-force looping program might take thousands of years to finish. However, a good programmer can find short cuts. Weighing the costs and benefits, I don’t find this project to be a good use of my time. If I lived in Russia or the Baltics, I would likely feel differently.

For the benefit of other readers, Royal Poker is like Caribbean Stud Poker, with the following added options. It is my understanding that the player may invoke any or all of these options, except he may not invoke options 2 and 3 both.

1. If the player can make two paying hands, which both beat the dealer, and neither hand is entirely within the other, then both are paid. I am not sure whether the ante and/or raise are paid twice. For example, if the player had six cards and could make a straight and a flush, then the player could be paid for both hands.
2. The player may switch one to five cards for the price of the Ante.
3. The player may buy a sixth card for the price of the Ante.
4. The player may buy "insurance" before the dealer turns over his four face-down cards. The insurance bet pays even money if the dealer does not qualify.
5. The player may force the dealer to switch his highest card for the next one in the deck for the price of the Ante wager.
6. There is an "AA Bonus" side bet, which pays 7 to 1 if the player’s first five cards are a pair of aces or higher.

I can say that the AA Bonus side bet and Insurance Option should never be taken, and thus are not worth anything. The house edge on the AA Bonus is 12.99%. The following table shows the house edge on insurance to range from 8.57% to 33.57%, depending on the dealer’s up card.

### Insurance in Russian Poker

 Dealer’s Up Card Combinations Probability Exp. Value A 132804 0.335714 -0.328571 K 132804 0.335714 -0.328571 Q 108528 0.457143 -0.085714 J 108732 0.456122 -0.087755 10 108936 0.455102 -0.089796 9 109140 0.454082 -0.091837 8 109140 0.454082 -0.091837 7 109140 0.454082 -0.091837 6 109140 0.454082 -0.091837 5 109140 0.454082 -0.091837 4 108936 0.455102 -0.089796 3 108732 0.456122 -0.087755 2 108528 0.457143 -0.085714

I also know from my page on Oasis Poker that just the option to switch cards lowers the house edge from 5.22% to 1.04%. I tend to think the rule about being double-paid, getting to keep the sixth card (as opposed to switching), and forcing the dealer to switch a card will get the game to a worthwhile player advantage, if you knew the proper strategy. Sorry to pull a Fermat on you, but that is that best I can do at this time.

P.S. I have heard some casinos add a rule that if the player wins, then the ante bet only pushes. This would work significantly in the casino’s favor, I think wiping out any advantage.

I recently entered a raffle where there are 7,033 prizes and they say the odds of winning a prize are 1 in 13. I bought 5 tickets. What are my actual odds of winning something? Also, there are 40 big prizes. What are my odds of winning a big prize?

anonymous from Mesa, AZ

For the sake of simplicity, let’s ignore the fact that the more tickets you buy the lower the value of each ticket becomes because you compete with yourself. That said, the probability of losing all five tickets is (12/13)5 = 67.02%. So the probability of winning at least one prize is 32.98%. There are 7033×13=91,429 total tickets in the drum before you buy any. 91,429-40=91,389 are not big prizes. The probability of not winning any big prizes with five tickets is (91,389/91429)5 = 99.78%. So the probability of winning at least one big prize is 0.22%, or 1 in 458.

According to you, the longer we play, the closer our loss gets to the negative expected value that is the house edge. Does it follow then that if we were perfectly logical players, we would always stake our entire bankroll on one single wager to avoid this gradual approximating function? This is the advice Bluejay gives at vegasclick.com.

Thus says Bluejay, "...if you know that the longer you play, the more likely you are to lose, then that means that the shorter you play, the better your chances of winning. And the shortest term you can have is just one play. And so statistically, that’s your best bet: making just one even-money bet, putting all your money on the line at once..."

Does the Wizard of Odds agree with this reasoning?

Peter from Sydney

Yes, absolutely! If your goal is to win or lose \$x, if limited to even money games, then you maximize your odds by making just one even money bet. This was once the situation, although not limited to even money bets, on a never-aired episode of “The Casino.” There I was consulted on how to maximize the odds of achieving a win of \$4,000, given ordinary games and a starting bankroll of \$1,000. I had them bet \$100 on the pass line and then \$900 on the odds in craps. Unfortunately, we lost. Had we won that bet, I would have had them bet enough to get to the \$4,000 goal.

However, if fun enters into the equation, you will get more by making smaller bets for a longer period of time. If you simply want to minimize your expected loss, then don’t play at all.

In a recent article, it was revealed that Ty Lawson, the starting point guard at UNC said, "The only time I lost was in Reno; that’s when everybody on the team lost," he said. "It’s the only place I lost. The other five or six times I did gamble, I won at least \$500.”

Ben from Austin, TX

If we ignore the house edge (which is very low in craps if played properly), the probability of winning \$500, as opposed to losing \$1,000, is 2/3. The probability of 4 out of 5 winning sessions would be 5×(2/3)4×(1/3) = 32.9%.