Ask The Wizard #123
On a recent Travel Channel show about Las Vegas whales, they surprised me with the statement that casinos sometimes offer loss discounts to whales. In other words, the whale plays on credit, and are charged only a percentage of their total losses at the end of a visit. Does this make it possible to set up a positive-expectation game? Would betting systems start to make sense in this context?
Yes, this is true. It is not unusual for "whales" to get a 10% rebate on losses. In my opinion this is a very risky offer to make and a sharp player could easily abuse it and gain an advantage. The kind of player who would make an ideal recipient of this offer is one who grinds out a lot of play in a high house edge game. The kind of player who could best exploit this offer is one who plays a low house edge game, for a short time, and with a wide range of bet size. It sounds paradoxical but under this deal the player must lose to gain any benefit. Thus the player should set a high winning goal and relatively low losing goal. If we can ignore the house edge for the sake of example if the winning marker were $1,000,000 and the losing marker $100,000 then the probability of success is 1/11, as I show in a later question. The expected value after the 10% rebate is (1/11)*$1,000,000 + (10/11)*(0.9*-100,000) = +$9091. A good strategy to achieve a high winning goal quickly would be something like an anti-martingale, or anything where you bet more after you win.
On baccarat, are the odds perpetual (as in dice and roulette) or do the odds change as cards are dealt out of the shoe (as in blackjack)? I know that it is not at all probable, but is it mathematically possible for the Banker to win every single hand in the baccarat shoe?
In an effort to debunk betting systems I used to say that the past does not matter in gambling. However once in a while somebody would rebuke me by saying that the past does matter for card counters, which is true. So now I say that in games of independent trials, like roulette and craps, the past does not matter. As I show in my baccarat appendix 2 a shoe rich in small cards favors the player and a shoe rich in large cards favors the banker. Thus, in baccarat, there is an extremely slight disposition that the next outcome will be the opposite of the last. So, yes, the odds do change in baccarat as the cards play out, but only to a very small extent. For all practicable purposes the game is not countable. I do not know if the banker could win every hand but I speculate that the answer is yes.
If I flat bet until I either win $100 or lose $1000 what is my probability of reaching both goals?
Ignoring the house edge the probability of achieving the winning goal is the loss marker divided by the sum of the loss marker and the winning marker. In this case 1000/(1000+100) = 1000/1100 = 90.91%. However the house edge will reduce the probability according to the house edge of the game played and the amount bet per hand, the smaller the bet the lower the probability of achieving the winning goal.
Is there a way you can test my betting system outside of your $2000 challenge?
My fee to do a straight up test would still be $2000. That is the value of my time to do the test. It costs me almost nothing to offer $20,000 if you pass the challenge because it is mathematically nearly impossible that you will win.
In which order would you place these machines for playing or not playing? Thanks. Double Diamonds, Triple Diamonds, Triple Wild Cherry, Wild Cherry.
I would rank them all equally. Generic 3-reel slots tend to be set to about the same return for the same casino and coinage.
In one of your answers you state that the average number of rolls for a shooter in craps is 8.522551. How is that number obtained?
First, if the probability of an event is p then the expected number of trials for it to occur is 1/p. Let's call x the expected number of rolls per shooter. The probability that any given round will end in one roll (with a 2, 3, 7, 11, or 12) is 1/3. If the player rolls a 4 or 10 on the come out roll the expected number of additional rolls is 4, because the probability of rolling a 4 or 7 is (6+3)/36 = 1/4. Likewise If the player rolls a 5 or 9 on the come out roll the expected number of additional rolls is 3.6 and for a 6 or 8 is 36/11. Assuming a point was thrown the probability of it being a 4 or 10 is 3/12, a 5 or 9 is 4/12, and a 6 or 8 is 5/12. So the expected number of throws per round is 1+(2/3)*((3/12)*4 + (4/12)*3.6 + (5/12)*(36/11)) = 3.375758. Next, the probability that the player will seven out is (2/3)*((3/12)*(2/3) + (4/12)*(3/5) + (5/12)*(6/11)) = 0.39596. The probability that player will not seven out is 1 - 0.39596 = 0.60404. So...
x = 3.375758 + 0.60404*x
0.39596*x = 3.375758
x = 8.52551
What incentive does a dealer have to "tell on you" if he suspects you are card counting. Why would a dealer care if you are card counting or not? Doesn’t it mean bigger tips for him?
Good question. If the counter were tipping then the dealer has the choice of not telling and getting more tips or tattling to get on the good side of casino management. I think it in large part comes down to the attitude of the dealer, whether he is rooting for the player or the casino. Dealers who are loyal to their employer first will probably tell, and tipping may not help. Dealers share tips so the dealer you give your tip to may only get 1% of it. Tipping cynical dealers who resent tip sharing won’t buy you much cover. In my opinion dealers loyal to the casino are more likely to be women than men and Asians over any other race. One of my blackjack books goes into this in more detail but I can’t remember which one. The decision to tip is hotly debated in the counting community and many counters follow the Stanford Wong philosophy of only tipping if the cover it buys you is more than in value than the tip itself. This may explain the joke that the difference between a counter and a canoe is that a canoe sometimes tips. Other counters tip anyway whether they think it buys them cover or not because they believe in tipping.