I’ve noticed that the CSM (Continuous Shuffler Machine) at the blackjack table does not shuffle ALL of the cards at the end of each hand. There are a few cards left in the shoe part of the machine (anywhere from 1 to 20 or so) that are not shuffled. Is there any way this can be used to advantage? For example, I was thinking that there is a lower (but still not zero) probability of having a card repeated two hands in a row. Sit out if there were a lot of high cards last hand . . . bet higher if there are a lot of low cards last hand. The CSM I saw used four decks so, on a full table, there are actually quite a few cards played each hand and you could potentially get a true value of plus/minus one if you made the simplifying assumption that none of those would repeat. Maybe enough to skew the odds?
Chuck from New York
You’re right, the discards are not mixed among all the cards but can not be placed close to the top of the shoe. I don’t know the exact size of this buffer but it is about 10-20 cards I think. As a card counter it would probably be safe to use a true count from just the last hand played and off the top of a shoe. When converting to the true count you will rarely get anything far from +/-1. If you’re any kind of counter at all I would forget about playing against a CSM, it isn’t worth the bother.
What basic strategy should you use for a CSM (Continuous Shuffler Machine)? Is it the same as a regular shoe for the same number of decks? It seems like the strategy could be a little different (maybe a 4 deck CSM should be played like a 3-deck shoe).
Chuck from New York
Yes, same as a regular shoe with the same number of decks. Most CSMs use five decks, for which you should use my 4-8 deck strategies.
I understand what the odds of being dealt a royal, straight flush are for any individual on a Caribbean Stud Poker or Let it Ride table are and how they are derived. But my question is this: as a 3rd party watching the game what are the odds of seeing any one of these hands being dealt to a player at the table on any given deal. I must believe it is dependant on the number of hands in play...is it merely the individual’s odds times the number of hands in play. ie. seeing a royal dealt on any particular hand with 4 players on a table means 4*odds of getting a royal? I’m slightly perplexed!
Amyn from Brantford, Canda
Your method is a good approximation. However by that logic, when flipping a coin, the probability of at least one person in 3 flipping a heads would be 3*50%=150%. Assuming independent events the probability of at least one success out of n trials, where the probability of each success is p, is 1-(1-p)n. In the case of the coin flipping example this would be 1-.53=0.875. In the case of four players of Caribbean Stud Poker the probability of at least one royal flush would be 1-(1-4/2598960)4 = 0.00000615629. However since all the cards are dealt out of the same deck the events are not independent. The math gets very complicated to determine the exact right answer and the approximation should be very close to the right answer.
The chart for Double Exposure indicates to split 10’s against a dealer 13-16. Does this mean that you continue to split additional 10’s? I’ve done this and wound up with 4 hands under 18 which all lost. Luckily I was only playing for fun on an Internet gaming site at the time.
Joe from San Diego
Yes, you should keep splitting as long as you keep getting tens. Playing one hand and losing does not disprove anything. Millions of hands must be played both ways and the results tabulated to truly know the best play.
For blackjack, which is the probability to obtain three suited seven in a 6-deck shoe?
Rodrigo from Costa Rica
I attempt to work this out in my blackjack appendix 8 but I’ll work through it more slowly here. We’ll ignore dealer blackjacks to keep things simple and assume the player always hits after two cards. The number of ways to arrange 3 cards in a 6-deck shoe is combin(312,3)=5,013,320. There are 24 sevens in the shoe. The number of ways to arrange 3 sevens out of 24 is combin(24,3)=2024. The probability is the number of winning combinations divided by total combinations, or 2024/5013320=0.0004, or about 1 in 2477.
Great site. I was playing video poker this weekend when the conversation turned to whether it is better to play one machine or try several machines looking for the one that is paying. After much discussion, the only thing I could really offer was that we were no more or less likely to hit the Royal Flush on any given machine at any given time. (You’ve told us enough times about the independence of trials that I almost understand.) Ok, with that said, here’s the question. If the Wizard had $200 and walked in to a bank of 10 full-pay machines, how would he play this bankroll? Would he put $200 in one machine? Or would he split the bankroll and play $50 in four machines? Or would he play $20 in each one? I think the mathematical answer is, that it doesn’t matter, but how would the Wizard play it?
Gil from Saint Petersburg
You’re right, the mathematical answer is that it doesn’t matter. I would choose the machine either randomly or based on environmental factors. My highest priority is that if there were any smokers in the vicinity I would sit as far from them as possible. Otherwise I would distance myself from any loud noises, including other players. If the machines were crowded I would pick an aisle machine, giving me a little more elbow room and one less neighbor.
What piece of information am I missing? If the odds of pulling a ten count card out of a deck is about 30.7% and the odds of pulling out an ace is 7.8% then it seems to me that the combined odds of this happening are about 2.4%. Why do blackjack simulators and blackjack authors state that the odds for a blackjack are 4.7% which happens to twice the calculated odds. What am I missing?
Jeffrey from Loveland
You are forgetting that there are two possible orders, either the ace or the ten can be first. Multiply by 2 and you’ll have your answer.
In Three Card Poker would it be prudent to increase your bet after say 5 or 6 losses? I know the Martingale system is bad news but since in Three Card Poker you have bonus payouts for better hands I thought it might be worth a chance. Please think about this for a minute before you answer.
John from Crestwood, Illinois
In the long run it doesn’t matter what you do. As I have said numerous times measured by long term results all betting systems are equally worthless. Systems in which you chase loses with bigger bets increase the probability of a modest short term win but at the cost of even bigger losses when your luck is at it’s worst.