Ask The Wizard #172

It is correct that in general if your 16 has at least one 4 or 5 in it then the odds genrerally favor standing. This is obviously because a 4 or 5 will give you a 20 or 21, and there is at least one less of them in the remaining deck. 16 against a 10 is so borderline that this effect can make the difference in the play. In a two-deck game, I show following the rule of 45 lowers the house edge by 0.015%. In a six-deck game it only lower the house edge by 0.003%.

Thanks for the kind words. Yes, I gave Casino Life permission to use my material. I’m happy they gave me a good plug. No, they didn’t pay me. I also have never been paid for my appearances on the Travel Channel here in the United States. I do it for the fun and the publicity.

I hope you’re happy, I spent a week on this, off and on. Using the hi-low strategy you’ll need to spread to 4 or 5 units in a six-deck game. However, I devised a much simpler adaptation of the ace/five count. You can read all about it in my section on the Wizard Ace/Five Count.

Thanks for your comments. That sounds more or less like the usual policy in Vegas as well.

Thanks for the kind words. The three games where you can get the house edge under 0.5% are blackjack, craps, and video poker. Of the three I would recommend blackjack. I would start betting $10 a hand. By playing conservatively at the beginning you won’t go bust too soon and have to beg your friends for a loan. Let’s say you want to double your bankroll or go bust trying, then press your bets as you get close to the end of your trip. Be sure to memorize the basic strategy before you play and don’t accept anything more than a 0.4% house edge.

The probability of any one player having a flush is combin(11,2)/combin(49,2) = 55/1176 = 4.68%. Assuming independence between hands, which is not the case, the probability of 9 players not having a flush is (1 − 0.0468%)⁹ = 64.98%. So the probability of at least one player having a flush is 1-0.6498 = 35.02%. This is just a quick estimate. If I did a random simulation I think the probability would be just a little bit higher, because of the dependence between hands.

Much of the value of suited connectors comes from the relatively high probability of forming a straight, flush, or straight flush. These are premium hands, which usually win against any number of players. In a two-player game you only are rewarded with one person’s money, but in a 10-player game you can milk it for a much larger pot.

The basic formula for comps is that the casino will give you back a percentage of your theoretical loss. That percentage can vary by game, the higher the house edge the higher the percentage. I asked a former Vegas casino manager and he said the comp rebate is about 15%. Other pertinent pieces of the equation are 60 hands per hour in blackjack with an average house edge of 1%. So the value of comps you could expect would be (average bet) × (hours played) × 60 × 1% × 15%. Let’s assume 16 hours of play. You can then back out the average bet required. Let’s assume food and beverage has a value of $500. Then the average bet required would be 500/(16*60*0.01*0.15) = $347. A free room might be worth $1,000, so an average bet of $694 would be required. There is a whole spectrum of suites, roughly ranging in value from $1,000 to $10,000 a day, so an average bet of $1,389 to $13,889 would be required. Free golf might be worth $500, so back to $347 for that. I’ve heard of Vegas casinos comping shopping sprees at the Fashion Show mall, but they don’t sell cars there. If we assume $2,000 for airfare then $1,389. At high levels of play this may also be subject to skill level, the better you are the less you will get. They also might have some sympathy and give you more than you are entitled if you had a really bad run of luck. For rooms, you will have more bargaining power if you ask for one during a slow time when they have vacancies anyway.