# Betting Systems - FAQ

Let a

_{i}denote the probability if the player has $i he will reach $1,200 before losing everything. Let p the probability of winning any given bet = 49.3212%.

a_{0} = 0

a_{1} = p*a_{2}

a_{2} = p*a_{3} + (1-p)*a_{1}

a_{3} = p*a_{4} + (1-p)*a_{2}

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a_{1197} = p*a_{1198} + (1-p)*a_{1196}

a_{1198} = p*a_{1199} + (1-p)*a_{1197}

a_{1199} = p*a_{1200} + (1-p)*a_{1198}

a_{1200} = 1

Divide the left side into two parts:

p*a_{1} + (1-p)*a_{1} = p*a_{2}

p*a_{2} + (1-p)*a_{2} = p*a_{3} + (1-p)*a_{1}

p*a_{3} + (1-p)*a_{3} = p*a_{4} + (1-p)*a_{2}

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p*a_{1197} + (1-p)*a_{1197} = p*a_{1198} + (1-p)*a_{1196}

p*a_{1198} + (1-p)*a_{1198} = p*a_{1199} + (1-p)*a_{1197}

p*a_{1199} + (1-p)*a_{1199} = p*a_{1200} + (1-p)*a_{1198}

Rearange with (1-p) terms on the left side and p terms on the right:

(1-p)*(a_{1}) = p*(a_{2} - a_{1})

(1-p)*(a_{2} - a_{1}) = p*(a_{3} - a_{2})

(1-p)*(a_{3} - a_{2}) = p*(a_{4} - a_{3})

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(1-p)*(a_{1197} - a_{1196}) = p*(a_{1198} - a_{1197})

(1-p)*(a_{1198} - a_{1197}) = p*(a_{1199} - a_{1198})

Next multiply both sides by 1/p:

(1-p)/p*(a_{1}) = (a_{2} - a_{1})

(1-p)/p*(a_{2} - a_{1}) = (a_{3} - a_{2})

(1-p)/p*(a_{3} - a_{2}) = (a_{4} - a_{3})

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(1-p)/p*(a_{1197} - a_{1196}) = (a_{1198} - a_{1197})

(1-p)/p*(a_{1198} - a_{1197}) = (a_{1199} - a_{1198})

Next telescope sums:

(a_{2} - a_{1}) = (1-p)/p*(a_{1})

(a_{3} - a_{2}) = ((1-p)/p)^{2}*(a_{1})

(a_{4} - a_{3}) = ((1-p)/p)^{3}*(a_{1})

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(a_{1199} - a_{1198}) = ((1-p)/p)^{1198}*(a_{1})

(a_{1200} - a_{1199}) = ((1-p)/p)^{1199}*(a_{1})

Next add the above equations:

(a_{1200} - a_{1}) = a_{1} * (((1-p)/p) + ((1-p)/p)^{2} + ((1-p)/p)^{3} + ... + ((1-p)/p)^{1199})

1 = a_{1} * (1 + ((1-p)/p) + ((1-p)/p)^{2} + ((1-p)/p)^{3} + ... + ((1-p)/p)^{1199})

a_{1} = 1 / (1 + ((1-p)/p) + ((1-p)/p)^{2} + ((1-p)/p)^{3} + ... + ((1-p)/p)^{1199})

a_{1} = ((1-p)/p - 1) / (((1-p)/p)^{1200} - 1)

Now that we know a_{1} we can find a_{1000}:

(a_{2} - a_{1}) = (1-p)/p*(a_{1})

(a_{3} - a_{2}) = ((1-p)/p)^{2}*(a_{1})

(a_{4} - a_{3}) = ((1-p)/p)^{3}*(a_{1})

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(a_{999} - a_{18}) = ((1-p)/p)^{9998}*(a_{1})

(a_{1000} - a_{19}) = ((1-p)/p)^{9999}*(a_{1})

Add the above equations together:

(a_{1000} - a_{1}) = a_{1} * (((1-p)/p) + ((1-p)/p)^{2} + ((1-p)/p)^{3} + ... + ((1-p)/p)^{999})

a_{1000} = a_{1} * (((1-p)/p)^{1000} - 1)) / ((1-p)/p - 1))

a_{1000} = [ ((1-p)/p - 1) / (((1-p)/p)^{1200} - 1) ] * [ (((1-p)/p)^{1000} - 1) / ((1-p)/p - 1) ]

a_{1000} = (((1-p)/p)^{1000} - 1) / (((1-p)/p)^{1200} - 1) =~ 0.004378132.

Given enough time, the odds are likely to catch up to the player in any game of luck and the bankroll will keep going down gradually. However, if you were to bet larger amounts your odds would be much better. The following are the odds of winning 20% before losing 100% at various units of bet size.

$5: 0.336507

$10: 0.564184

$25: 0.731927

$50: 0.785049

$100: 0.809914

For more on the math of this kind of problem, please see my MathProblems.info site, problem 116.

- 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.

Expected wins = 2391/2 = 1195.5

Actual wins above expectations = 107.5

Standard deviation = sqrt(2391*(1/2)*(1/2)) = 24.45

Standard deviations away from expectations = (107.5 + 0.5)/24.45 = 4.4174

Probability of 4.4174 standard deviations or more = normsdist(-4.4174) = 0.000005 = 1 in 200,000

As an example, consider Game 1 in which the probability of winning $1 is 49% and losing $1 is 51%. In Game 2 if the player’s bankroll is evenly divisible by 3 he has a 9% chance of winning $1 and 91% of losing $1. In Game 2 if the player’s bankroll is not divisible by 3 he has a 74% chance of winning $1 and 26% of losing $1.

Game 1 clearly has an expected value of 49%*1 + 51%*-1 = -2%.

In Game 2 you can not simply take a weighed average of the two possibilities. This is because the game quickly gets off of a bankroll remainder of 1 with a win, and often alternates between remainders of 0 and 2. In other words the bankroll will disproportionately play the game with a 9% chance of winning. Overall playing Game 2 only the expected value is -1.74%.

However by alternating two games of Game 1 and two games of Game 2 we break the alternating pattern of Game 2. This results in playing the 75% chance game more and the 9% less. There are an endless number of ways to mix the two games. A 2 and 2 strategy of playing two rounds of Game 1 and two of Game 2, then repeating, results in an expected value of 0.48%.

I should emphasize this has zero practical value in the casino. No casino game changes the rules based on the modulo of the player’s bankroll. However I predict it is only a matter of time before some quack comes out with Parrondo betting system, alternating between roulette and craps, which of course will be just as worthless as every other betting system.

My question is, is there any way to calculate an optimum playing range of wins/losses? That if a player reaches X number of losses it is highly unlikely that they will recover and should just quit? Likewise, if a player wins X amount, then the player has achieved respectable winnings considering the probabilities of the game and should quit while ahead.

My question isn’t about winning long term with systems, as we know that’s impossible. But might systems have a usefulness in ’tailoring’ the losing experience? For example, player A prefers that each trip to the casino he will either win or lose a moderate amount of money (of course he’ll lose slightly more often than win). Player B prefers a chance to make a little money 4 out of 5 trips, and lose lots of money 1 in 5 trips.

Both will lose money in the long run, but is there a betting system that might help each accomplish his goal?

My history professor at NMSU told our class that the only way to win at Blackjack was to bet a little at a time and walk away will small profits..$25. This logic doesn’t seem to work in my books...I know it’s false. My question is...let’s say I have $1,000,000 to gamble with in my lifetime. Do I have "better odds" by betting the whole million at one hand of Blackjack vs. small hands or are the odds always the same regardless? You have a great website and keep up the good work. Thanks a lot for your help!

Since I cannot control the statistics, my question involves something that I can control, session length (and bankroll). Since a million or a billion number of hands is composed of so many “sessions” of, for example, 300 − 1,000 hands, does it not make sense to play until you either a) reach a pre-established target win amount, or b) play until you recover from a losing streak and end the session at break even?

One last question, can you recommend a source for a software simulation system that can handle all rule variations, stop/loss provisions, extracting “sessions” of variable lengths, and variable hit/stand strategy based upon the size of the bet. I would love to give my approach a shot at the computer.

Thus says Bluejay, "...if you know that the longer you play, the more likely you are to lose, then that means that the shorter you play, the better your chances of winning. And the shortest term you can have is just one play. And so statistically, that’s your best bet: making just one even-money bet, putting all your money on the line at once..."

Does the Wizard of Odds agree with this reasoning?However, if fun enters into the equation, you will get more by making smaller bets for a longer period of time. If you simply want to minimize your expected loss, then don’t play at all.

Is Andy correct in that the best way to turn $30 into $1,000 is to put the whole $30 on a single number in roulette?

After much trial and error, I devised my "Hail Mary" roulette strategy, which will increase the odds of turning $30 into $1,000 to 2.8074%.

Wizard's "Hail Mary"strategy for roulette:

This strategy assumes that bets must be in increments of $1. In all bet calculations, round down.

Let:

b = Your bankroll

g = Your goal

- If 2*b >=g, then bet (g-b) on any even money bet.
- Otherwise, if 3*b >=g, then bet (g-b)/2 on any column.
- Otherwise, if 6*b >=g, then bet (g-b)/5 on any six line (six numbers).
- Otherwise, if 9*b >=g, then bet (g-b)/8 on any corner (four numbers).
- Otherwise, if 12*b >=g, then bet (g-b)/11 on any street (three numbers).
- Otherwise, if 18*b >=g, then bet (g-b)/17 on any split (two numbers).
- Otherwise, bet (g-b)/35 on any single number.

In other words, always try to reach the goal, with just one bet, if you can, without exceeding the goal. If there are multiple ways to accomplish this, then go with the one with the greatest probability of winning.

What about other games, you might ask? According to the Discovery Channel voice-over guy, "Everyone agrees that roulette is the best get rich quick scheme in the casino." Well, I don't. Even limiting ourselves to common games and rules, I find craps to be better. In particular, betting the don't pass and laying the odds.

Following my Hail Mary strategy for craps (explained below), the probability of turning $30 into $1,000 is 2.9244%. This assumes the player may lay 6x odds, regardless of the point (which is the case when 3x-4x-5x odds are allowed taking the odds). This probability of success is 0.117% higher than my Hail Mary strategy for roulette, and 0.2928% higher than the Andy Bloch strategy.

Andy might argue that my argument above relies on an assumption of a minimum bet of $1, which is hard to find in Vegas on a live dealer game. Expecting somebody might say that, I ran through both games under an assumption of a $5 minimum and betting in increments of $5. In that case, the probability of success using my Hail Mary strategy is 2.753% in roulette and is 2.891% in craps. In both cases, greater than the 2.632% under the Andy Bloch strategy.

In all fairness, the Discovery Channel would have never put the insane rant above on the air and was surely looking for something simple that the masses would understand. Andy was surely giving them something they wanted to hear. The basic premise of his advice is that if you want to reach a certain goal, then a hit-and-run strategy is much better than letting the house edge grind you down with multiple bets. That is definitely true and something I've been preaching for 17 years.

Wizard's "Hail Mary" strategy for craps.

This strategy assumes that bets must be in increments of $1 and wins will be rounded down to the nearest dollar. In calculating bets, never bet so much that you overshoot the goal. Also, never make a bet amount that will cause you to get rounded down.

Let:

b = Your bankroll

g = Your goal

- Bet max($1,min(b/7,(g-b)/6)) on the don't pass.
- If a point is rolled, and you have enough for a full odds bet, then lay the full odds. Otherwise, lay whatever you can.

So, I hope Andy and the Discovery Channel are happy. I've spent days running simulations to prove them wrong.

This question was raised and discussed on my forum at Wizard of Vegas.