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Best of Donald Catlin
Blackjack Variance5 October 2007
I recently received an email from one of my readers, Paul, who was curious about the effect on variance when playing multiple hands in Blackjack. Here is part of Paul's letter to me.
Well, Paul, let me address the question about expectation first since it is the easier of the two and we will need the answer anyway to address the other question. If f is a random variable (defined as a function whose domain is a sample space) then the expected value of f, written E(f), is defined as
E(f) = f(x1)p(x1) + f(x2)p(x2) + ... + f(xm)p(xm) (1)
where p(x) represents the probability of the outcome x. Note I am assuming a finite sample space. If we have two random variables, say f and g, then the sum f + g is defined by (f + g)(x) = f(x) + g(x) where x is any outcome. From this and (1) it is clear that
E(f + g) = E(f) + E(g) (2)
If f and g are the payoff functions for two different hands of blackjack then f + g is the payoff function for two hands. From (2), therefore
E(f + g) = E(f) +E(g) = 2E(f) (3)
because E(f) = E(g). Since we are assuming that E(f) = -.005, it follows from (3) that
E(f + g) = 2(-.005) = -.01 (4)
so your calculation, Paul, is correct. Notice that (3) suggests that
E(af) = aE(f) (5)
for any real number a and in fact this is the case; I'll leave that to you.
The standard deviation calculation would be correct if two separate hands of Blackjack were independent but unfortunately this is not the case. Certainly the cards in one hand can influence the cards in another hand and both hands are very influenced by the dealer's hand. So let's begin at the beginning.
If f1 and f2 are two random variables let us denote E(f1) by m1 (m for mean) and E(f2) by m2. We then define the covariance of f1 and f2 by
cov(f1, f2) = E((f1 - m1)(f2 - m2)) (6)
By the variance of a random variable f we mean the following:
var(f) = cov(f, f) (7)
and the standard deviation of f is the square root of the variance, that is
s.d. = sqr(var(f)) (8)
If f has mean m then from (3), (5), (6) and (7) we have the following calculation:
var(f) = E((f - m)(f - m))
= E(f2 - 2mf + m2)
= E(f2) - 2mE(f) + E(m2)
= E(f2) - 2m2 + m2
var(f) = E(f2) - m2 (9)
Formula (9) is very useful in general and will be a very helpful tool for us.
If f represents a payoff function for Blackjack we wish to calculate the variance of f. Let me turn to my old (and sadly deceased) friend Peter Griffin for help with this. To begin with the expected return m = -.005 so the square of this number is 0.000025, a number so small in comparison with the first term in (9) that we can ignore it, which is exactly what Peter did in his famous book The Theory of Blackjack (Huntington Press 1996, fifth edition). Notice that in (9) we need not worry about the sign of f since the quantity is squared. What possible values for f can occur? Here are the main players: 0 (a tie), plus or minus 1 (a win or loss), +1.5 (a Blackjack), or plus or minus 2 (a split or double). Now to be sure, if resplitting pairs or doubling after splitting is allowed f might take on values greater than 2 or less than minus 2 but these are very rare events. Peter chose to ignore these and I will do the same.
Where do we get the probabilities? These can be obtained from a simulation and that is exactly what Peter did. On page 167 of his text he essentially gives the following table.
If we sum the numbers in the fourth column we are in effect carrying out the calculation in (1) with f2 replacing f. The total is 1.2625. The square root of this is 1.124 although most writers just use 1.1.
Finally, Paul, we can address your question. Because of the linear nature of E (see equality (2)) the covariance is bilinear, that is
cov(f + g, h) = cov(f, h) + cov(g, h) (10)
and a similar relation holds for the second variable. If f and g represent the payoff functions for two hands of Blackjack then they are identically distributed but, as noted above, not independent. Then
var(f + g) = cov(f + g, f + g)
= cov(f, f + g) + cov(g, f + g)
= cov(f, f) + cov(f, g) + cov(g, f) + cov(g, g)
= var(f) + var(g) + 2cov(f, g)
since from (6) the covariance function is obviously symmetric. Now we have already noted that the variance for a payoff function is approximately 1.26 so from the above calculation we have
var(f + g) = 2(1.26) + 2cov(f, g) (11)
Here is where Peter comes to the rescue. On page 142 of his book, in remark E, he writes "From simulated hands I estimate the covariance of two Blackjack hands played at the same table to be 0.50." So there we have it
var(f + g) = 3.52 (12)
The standard deviation is 1.876.
The argument I have just made can easily be extended to n hands and the result is
var(n hands) = 1.26n + 0.5n(n - 1) (13)
and this is stated in remark E, page 142, of Peter's book. Paul, I hope this clears things up for you. See you next month.
Don Catlin can be reached at email@example.com
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