Ask The Wizard #228
What is your opinion of the 5-Count strategy in craps?
For the benefit of other readers, the 5-Count is a method of slow-playing craps, as discussed in ’Golden Touch Dice Control Revolution’ by Frank Scoblete and Dominator. As the book states, it is a way of betting nothing on some rolls, reducing your expected loss on random shooters, while still getting the full comp value of table time.
The way the 5-Count works is you start counting rolls as soon as a new shooter throws any point number. When you get to five rolls after you start counting, the shooter is deemed worthy, and you start betting. However, you if the 5th roll is not a point number, it doesn’t count.
The book says you will only be betting 43% of the time, which I agree with. It is common for craps players to not bet, bet small, or bet the don’t pass on new shooters, as a way to qualify him. Once a shooter has made a point, or thrown lots of point numbers, the other players will gain confidence in him, and start betting with him. So, this kind of strategy seems natural. When casinos rate your average bet, they don’t lower the average for betting nothing some of the time. However, sometimes they will dock your time, especially if you are betting big.
An alternative strategy is to wait until the shooter makes a point. Under this strategy you will only be betting 40.6% of the time, less than the 43.5% with the 5-Count.
What is the correct strategy for Acey Deucey at a home poker game? The way we play is if the third card matches one of the first two, then the bet is a push.
The way you play, where a third-card match is a push, the odds swing in your favor when there are at least six ranks between the first two cards (a six-card spread). The way I played in Orange County, a third-card match resulted in a double loss. Under that rule, the odds are break-even with an eight-card spread. If a third-card match results in a 1x loss, then you need a seven-card spread for the odds to be in your favor.
How many traditional bingo cards would have to be in play to achieve a coverall in 40 numbers or less, statistically speaking?
The cards are randomly printed, so if you purchased enough, you would get repeats. So there is no number where you would be assured of winning. The probability of each card winning is 0.00000000243814, or 1 in 410,148,569. Suppose you would be happy with a probability of winning of p, the number of cards you purchased is n, and the probability of winning per card is c. Let’s solve for n:
P = 1-(1-c)n
1-p = (1-c)n
ln(1-p) = n×ln(1-c)
n=ln(1-p)/ln(1-c)
For example, to have a 90% chance of winning you would need to buy ln(1-.9)/ln(1-0.00000000243814) cards, which equals 944,401,974.
In Super Pan 9, using 8 decks, is a tie payout of 8 to 1 a player advantage bet? My crude analysis gives the player approximately a 2.5% edge.
My Super Pan 9 page shows the probability of a tie is 11.3314%. So if a tie paid 8 to 1, the expected return would be 9×0.113314 − 1 = 0.019826. Although a 1.98% player advantage is less than your figure, it is still a great bet. Where can I play it?
I count cards (14 count). I prefer to play alone at third base, and avoid really bad players. The continuous shuffle machines are impossible to play and count. Is there a current list of Las Vegas casinos using these machines?
I don’t blame you for wanting to avoid other players. They slow down the game, and if they smoke, they pose a health hazard. However, it shouldn’t matter whether they are good or bad. Just about all Vegas casinos have some mixture of continuous shufflers, automatic shufflers, and hand-shuffled games. For details on blackjack rules, including type of shuffle and penetration, there is no substitute for the Current Blackjack News, for which a paid membership is required.
I play at a casino that allows unlimited re-splitting of cards in a six-deck black jack shoe, except for no re-splitting aces. In general, it must be beneficial to the player, but was wondering by how much and if there are points at which you stop splitting. I have found no information on this rule.
Unless you are counting cards, you should keep re-splitting as much as possible, up to the maximum 24 hands in a six-deck shoe. The value of infinite resplitting compared to a maximum of three times is very little. It depends on whether or not double after split is allowed, but either way, the value is well under 0.01%.
Imagine a craps player who takes maximum odds, say 10x, on his pass line and come bets reducing the house edge to 0.18%. He avoids other bets that give the house a bigger edge. He is an "astute" right bettor in every way except this: he is determined to lose. Through bankroll management, and a determined effort to only leave the table a "loser," he hopes he can look back on his years of craps playing and say, "I wagered $1 million at the craps table over the years and gave back $50,000 to the house; because of my ’skill,’ I left 5% at the table." Is he deluding himself? Is he doomed, in spite of his efforts to leave the table a loser every time, to only give the house roughly 0.18% or $1800?
Yes! I’ve said many times that betting systems not only can’t beat a house edge game, they can’t even dent it. That includes denting it in the house’s favor. In other words, even if he tried to lose, he still only gives up 0.18% over the long-run, under your assumptions. Over a shorter time, he probably could do this, but not over "years." Some might argue that to deliberately lose, the player should do an anti-Martingale, where the player kept pressing his bets until he lost. However, a problem there is that a winning player will eventually reach the table maximum, which is rather low in craps. It just goes to show how futile betting systems are.