Ask The Wizard #80
In your last column you said anybody could create a roulette system that showed a 6.5% profit over 7500 spins. Well, I’m anybody and am challenging you to give me one.
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.
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.
- 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.
Hello. I found your site very informative. Will you be providing any analysis on other sports like hockey betting and more on baseball in the future (such as o/u, selecting the best puck-line, etc)?
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.
Dear Wizard, Just a quick question, why does the house edge change for an insurance bet in blackjack depending on the number of decks used?
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.
Dear Sir, In a single zero roulette game, the PROBABILTY of winning increases if you place a portion of your money on fewer numbers for more spins versus covering more numbers per spin, an example: If you are willing to risk 500$ in order to win 250$ then you could: Option (A): place 250$ on any of two dozen and should you be a winner you will win 250$. The probabilty of that happening is 24/37=(.648648). Option (B): Place 125 on any one dozen and should you be a winner you will win 250$ and walk away. However, should you lose you can now bet187.5$ on the same dozen and should you be a winner you will win 375$ which will get you the 250$ and the 125$ you lost on the previous spin. Now should you lose on both spins you still have 187.5$ to play with and you can place 20.833333$ on any nine numbers and should you be a winner you will get 750$ which is equal to your 500 original capital plus 250$ in winning which was your goal. The probability of that happening meaning either hitting a dozen OR nine numbers at LEAST once in three spin is equal to[1-(25/37)x(25/37)x(28/37)]=0.65451. Hence, for the SAME capital and for the SAME payoff you are able to increase your PROBABILITY of success as in option (B) if you play fewer numbers with less money but for MAYBE more spins.(As you might win on the first spin) You can even improve your probability further if you play only six numberes at a time and try to win 250$. Any Explanation??!!!! Assuring you of my highest regards and awaiting the favor of your reply I remain.
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.
Dear wiz, I am a blackjack dealer here in Vegas and the other night dealing, I had 4 out of the 6 ace of spades in my hand. I had A-A-K-A-A-10, so good think is I busted, but quick calculations on the game, we figured getting 4 out of the six aces on one had is around 7mil to 1. Is this number a little high?
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.
When you open a new deck of cards they Ace to King of each suit. What are the odds of taking a shuffled deck of cards and reshuffeling it to the Ace to King state it was originally in?
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.