# Ask the Wizard #15

I like your site very much. It is very informative. Thanks for putting out your thoughts. I noticed a betting strategy for craps suggested at Crappers Delight called "classic regression". In it he suggests, placing a 6 and 8, after a point is established. Then taking it down after one of them is hit. He said there are 10 combined ways to make the 6 and 8, but only 6 combined ways to make the 7. It sounds logical, but I've seen where you are able to show, that what appears logical on the surface is not so bright once it is analyzed. What are your thoughts on this strategy and what would the true odds be, if you did take the bets down after one hit?

Michael

This is similar to a question I got last week. Yes, it is true that there are ten ways to roll a 6 or 8, and six ways to roll a 7. However, one must not look at the probabilities alone, but weight them against the payoffs. The place bet on the 6 and 8 pays 7 to 6 odds when fair odds would pay 6 to 5. By making six unit place bets on the 6 and 8, and taking the other down if one wins, the probability of winning 7 units is 62.5% and the probability of losing 12 units is 37.5%. If the player must cover both the 6 and 8, then the place bet is the way to go. This rate of return isn't bad but could be better. For the player who puts a priority on minimizing the overall house edge, the best strategy is to make combinations of pass, don't pass, come, and don't come bets, and always take the maximum allowable odds.

How can I determine the odds of flat betting (no counting, no progressions , etc ) of being ahead in a negative game such as blackjack, w/o counting, with a 0.5% disadvantage after 45,000 or so hands? Is it even possible?

Kevin

This is a typical question one might encounter in an introductory statistics class. Because the sum of a large number of random variables always will approach a bell curve we can use the central limit theorem to get at the answer.

From my section on the house edge we find the standard deviation in blackjack to be 1.17. You won't understand this if you haven't studied statistics, but the probability of being at a loss in your example will be the Z statistic of 45000*0.005/(45000^{1/2}*1.17) =~ 0.91.

Any basic statistics book should have a standard normal table which will give the Z statistic of 0.8186. So the probability of being ahead in your example is about 18%.

I was curious -- I'm sure I can't get better than house odds - but wanted to test a modest gambling approach -- the quit while you're ahead scenario. Say I start with an even $1000. What percentage of time will I leave with $1,200, rather than leave with $0, assuming that I have to leave once I hit one or the other. Up 20% rather than be down 100% betting on the player in baccarat.

Brian from Denver, Colorado

Two key pieces of information you left out is how much you are betting and in which game. I'll assume that you are flat betting $1 at a time on the Player bet in baccarat. The probability that the player will win, given that there isn't a tie is 49.3212%.

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.

Why are basic strategy blackjack charts setup with the apparent theory that the dealer has a "10" card in the hole. When in reality I believe that the odds are 9-4 against a "10" card being anywhere. Am I missing something? Your website is very interesting. Thanks a lot.

Eddie from New Orleans, Louisiana

Assuming that the dealer has a 10 in the hole is just a memory device, it has nothing to do with the way the basic strategy was constructed. I can't stand in when I hear one player telling another, "You always assume the dealer has a ten in the hole." If this were true the player should hit a 19 against a 10, certainly that is an unreasonable play.