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Ask the Wizard #80
You got it! Actually the system boasted an advantage of 7.94%. I'll up that and go to 8.00%. So here is "Wizards 8.0% advantage system." Here is how to play it.
- This system can be played on any even money game, including roulette, but craps is strongly suggested due to the lower house edge.
- Player makes only even money bets. In roulette any even money bet will do and the player may change the bet at will (as always the past does not matter).
- Player must be comfortable with a betting range of 1 to 1000 units.
- The first bet is 1 unit.
- After each bet the player will determine 8.1% (the extra 0.1% is a margin of safety) of his total past wagers. If his net win is less than this figure he will bet the lesser of the difference and 1000 units. If his net win is more then he will bet one unit.
- Repeat until 7500 bets are made.
In roulette I did a computer simulation of this experiment 10,000 times and the player made his 8.0% 4236 times and failed 5764 times. So the first time with live play it would not be unlikely that the player would report a success story. In craps betting on the pass line using the same system resulted in 6648 wins and 3352 losses, for a success rate of 66.48%. Going back to roulette, if the spread is 1 to 10,000 units the numbers of wins was 8,036 and 1,964 losses. In all cases when the system doesn't hold up over 7,500 spins the loss is big, more than 8.0% on average.
Of course this system is just as worthless as every other. The point I hope I have made is that it is easy to easy to design a system that usually wins. However when you do lose you lose big. Over the long run the losses will be more than the wins and the player will have a lot less money in his pocket.
I do plan to add more on sports betting in the future. That is where I’m personally focusing most of my gambling energy. However I haven’t found a good angle to exploit yet in baseball or hockey but hopefully I’ll think of something.
Because we already know an ace has been removed for the shoe. That leaves the rest of the shoe slightly ten heavy. A greater the ratio of tens in the shoe the more likely the insurance bet is to win. The fewer the decks the greater this effect is. Insuring a 20 (except A/9) increases the house edge on the insurance bet because there are two less tens in the shoe.
You are correct that option B has the greater probability of success, although the goal and the capital are the same. The reason is the average amount bet in option B is less, thus your money is exposed to the house edge less, thus the probability of winning increases. The amount bet in option A is always $500. The average amount bet in option B is (12/37)*125 + (25/37)*(12/37)*(125+187.5)+ (25/37)*(25/37)*(125+187.5+187.5) = 337.29.
When I was on the Vegas Challenge, with a few minutes to go, I had about $8,000 and needed to get to at least $24,000. So I split my bankroll into four piles of $2000 each and bet each one on a 4-number combination, each of which would have paid $22,000. This way I was not necessarily exposing my entire stake to the house edge, which increased my probability of winning.
The probability of this occurring in which your other two cards are any two 10-point cards is 4*COMBIN(6,4)*COMBIN(6*16,2)*(4/6)*(3/5)*(1/2)/combin(312,6) 1 in 22,307,231. However there are other ways you get four aces in the same hand, for example the last card might be an 8 or 9. I would have to do a computer simulation to consider all the other combinations. However to make a rough guess I’d say the 7 million looks about right.
1 in 52 factorial, or 1 in 80,658,175,170,943,900,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.