Gambling Guide
Ask the Wizard #15
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.
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/(450001/2*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%.
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 ai 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%.
a0 = 0
a1 = p*a2
a2 = p*a3 + (1-p)*a1
a3 = p*a4 + (1-p)*a2
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a1197 = p*a1198 + (1-p)*a1196
a1198 = p*a1199 + (1-p)*a1197
a1199 = p*a1200 + (1-p)*a1198
a1200 = 1
Divide the left side into two parts:
p*a1 + (1-p)*a1 = p*a2
p*a2 + (1-p)*a2 = p*a3 + (1-p)*a1
p*a3 + (1-p)*a3 = p*a4 + (1-p)*a2
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p*a1197 + (1-p)*a1197 = p*a1198 + (1-p)*a1196
p*a1198 + (1-p)*a1198 = p*a1199 + (1-p)*a1197
p*a1199 + (1-p)*a1199 = p*a1200 + (1-p)*a1198
Rearange with (1-p) terms on the left side and p terms on the right:
(1-p)*(a1) = p*(a2 - a1)
(1-p)*(a2 - a1) = p*(a3 - a2)
(1-p)*(a3 - a2) = p*(a4 - a3)
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(1-p)*(a1197 - a1196) = p*(a1198 - a1197)
(1-p)*(a1198 - a1197) = p*(a1199 - a1198)
Next multiply both sides by 1/p:
(1-p)/p*(a1) = (a2 - a1)
(1-p)/p*(a2 - a1) = (a3 - a2)
(1-p)/p*(a3 - a2) = (a4 - a3)
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(1-p)/p*(a1197 - a1196) = (a1198 - a1197)
(1-p)/p*(a1198 - a1197) = (a1199 - a1198)
Next telescope sums:
(a2 - a1) = (1-p)/p*(a1)
(a3 - a2) = ((1-p)/p)2*(a1)
(a4 - a3) = ((1-p)/p)3*(a1)
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(a1199 - a1198) = ((1-p)/p)1198*(a1)
(a1200 - a1199) = ((1-p)/p)1199*(a1)
Next add the above equations:
(a1200 - a1) = a1 * (((1-p)/p) + ((1-p)/p)2 + ((1-p)/p)3 + ... + ((1-p)/p)1199)
1 = a1 * (1 + ((1-p)/p) + ((1-p)/p)2 + ((1-p)/p)3 + ... + ((1-p)/p)1199)
a1 = 1 / (1 + ((1-p)/p) + ((1-p)/p)2 + ((1-p)/p)3 + ... + ((1-p)/p)1199)
a1 = ((1-p)/p - 1) / (((1-p)/p)1200 - 1)
Now that we know a1 we can find a1000:
(a2 - a1) = (1-p)/p*(a1)
(a3 - a2) = ((1-p)/p)2*(a1)
(a4 - a3) = ((1-p)/p)3*(a1)
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(a999 - a18) = ((1-p)/p)9998*(a1)
(a1000 - a19) = ((1-p)/p)9999*(a1)
Add the above equations together:
(a1000 - a1) = a1 * (((1-p)/p) + ((1-p)/p)2 + ((1-p)/p)3 + ... + ((1-p)/p)999)
a1000 = a1 * (((1-p)/p)1000 - 1)) / ((1-p)/p - 1))
a1000 = [ ((1-p)/p - 1) / (((1-p)/p)1200 - 1) ] * [ (((1-p)/p)1000 - 1) / ((1-p)/p - 1) ]
a1000 = (((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.
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.
