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Ask The Wizard #67
I recently learned some information about dicesetting strategies in craps. Some believe that you can set the dice a certain way before the throw, and by keeping the roll of the dice to just one axis of rotation, you can have fewer possible sevens with certain dice sets. I wanted to know if there is any truth to this or is it just a fallacy.
I don’t believe in it. So far I have yet to see a name I respect endorse the method, nor any evidence that it works. While I don’t entirely rule out the possibility I am extremely skeptical of it. I may live in Nevada but when it comes to things like dice setting I’m from Missouri, "show me" it works.
On your Super Fun 21 strategy chart you show doubling your 10 against an ace if you have less that 4 cards. Why is this? Also, do you have a strategy based on simple card counting for this game? Thank you.
The reason you double with 2-4 cards and hit with 5 or more is because a hand of 6-cards automatically wins, and a 5 card total of 21 pays 2-1. However these incentives do not apply after doubling. Once the player already has four cards there is a stronger incentive to hit as opposed to double. I nice count strategy has been introduced in the green chip section at bj21.com.
In Mississippi, is the law the same for video poker as in Nevada, that each hand is a new game?
I assume you are asking whether each hand is dealt from a fresh deck, with no memory of past hands. This is how video poker is supposed to be played and I’m sure Mississippi is no exception.
Our $3 craps game pays $4.50 on a place bet of a 5 or 9, and $5.50 for a place bet on the 4 or 10. Could you tell me what the house advantage is on these bets? (I'm especially curious about the 5 or 9 since we are actually paying true odds for a place bet.)
There are 4 ways to roll a 5 (1+4, 2+3, 3+2, and 4+1) and 6 ways to roll a 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). So the probability of rolling a 5, given that a 5 or 7 was rolled, is 4/(4+6) = 0.4. The expected value of the place bet on 5 is (0.4*$4.50 + 0.6*-$3.00)/3 = 0. So you're right there is no house edge on the 5, or the 9. On the 4 and 10 the expected value is ((1/3)*$5.50 + (2/3)*-3)/3 = -0.0556. In other words the house edge is 5.56%. I take it this is only true of $3 bets, the casino rounding the winnings up from $4.20 to $4.50. As I reported in the last issue the player can gain an advantage in blackjack if the dealer pays $4.00 for a blackjack on a $2.50 bet.
At the Spielbank Stuttgart, one is given the option in blackjack to take even money immediately if one gets a black jack and the dealer shows a ten or an ace (to avoid a push). Is this a sucker bet? I won 390 euro thanks to your card. Many thanks!
Yes, this is a sucker bet, especially with a 10 showing. This is equivalent to taking insurance. The casinos here always offer "even money" in this situation and both dealer and other players will treat you like a fool if you decline it. However the expected value of a blackjack when the dealer has an ace showing is 1.04 units, better than the 1.00 units you’ll get by taking the even money. So unless you’re a card counter and know the remaining deck to be 10-rich then always decline even money. I’m glad my basic strategy helped you win.
Please explain "playing perfectly".
Playing perfectly means the same thing as using an optimal strategy. In other words making perfect use of the information available to maximize the expected outcome of the bet.
Are dice truly unbiased? It seems like the sides with the larger numbers which have more holes would be lighter than the sides with the smaller numbers and less holes. This seems to suggest that the heavier sides would more likely land face down with the larger numbers more likely landing face up. I can imagine a craps system that could try to exploit this principle, but I wonder if it would really work. What do you think?
With ordinary dice, the like those you get in a board game, this is true. However casino dice have inlaid spots. At the factory they drill holes for the spots then insert white colored spots into the holes, of the same density as the die itself. So the die is essentially a perfect cube. Even if they did use ordinary dice from a board game I doubt the bias would be nearly enough to overcome the house edge.