On a video poker machine, does playing more lines increase the variance of the game? Does playing a higher denomination increase the variance of the game?
[Editor's Note: We started discussing this question and got bogged down almost immediately. Instead of begging off it, we offered it to a good friend of LVA, whom regular readers have seen on occasion in comments and expertise in the answers. Stewart Ethier is one of the world's top gambling mathematicians and a gambling historian/scholar par excellence. His answer is direct evidence that seemingly simple questions, like this one, can actually be quite complex.]
That's actually quite an interesting question, which I've thought about before. Here's my answer, explained as simply as I can.
First, a few remarks on terminology. Everyone knows the term expected (or mean) return. It is the long-term average return. For example, at 9/6 Jacks or Better with a one-unit (max-coin) bet, it's a weighted average of the 10 possible payouts (800, 50, 25, 9, 6, 4, 3, 2, 1, 0), weighted by the probabilities with which they occur.
800 (0.000024758268) + 50 (0.000109309090) + ... + 0 (0.545434669233) = 0.995439,
often written as 99.5439%. The variance is the expected squared deviation from the mean. In our example, it is
(800 - 0.995439)^2 (0.000024758268) + (50 - 0.995439)^2 (0.000109309090) + ... + (0 - 0.995439)^2 (0.545434669233) = 19.5147.
The variance determines the game's volatility by virtue of a result known as the central limit theorem. Alternatively, one can consider the standard deviation, which is the square root of the variance, the advantage being that it is measured in the same units as the payouts themselves (e.g., dollars instead of squared dollars). In our example, the standard deviation is 4.41754 units, while the variance is 19.5147 squared units.
Now, let's consider the easier second question first and let's work with standard deviation (the square root of the variance) instead.
If the random variable X is the payout from a one-unit bet, then bX is the payout from a b-unit bet, and SD(bX) = bSD(X). So the payout from a bet of size b has standard deviation proportional to b.
Example: In 9/6 Jacks or Better, the standard deviation of the payout from a one-unit (max-coin) bet is 4.41754. So for a $5 bet, the standard deviation is $22.09. For a $25 bet, it is $110.44.
Ultimately, the answer to the question depends on what is meant by the standard deviation of the game. If it is the standard deviation of the actual payout (e.g., $22.09), then the answer is yes: Playing a higher denomination does increase the variance of the game. If it is the standard deviation of the "payout per unit bet" (e.g., 4.41754), then the answer is no. For what it's worth, I prefer the latter interpretation.
For the first question, I'm assuming that the questioner meant a multi-play video poker machine (instead of multi-line). This is also an interesting question that has been discussed in at least two sources. Its answer again depends on interpretation.
Hannum and Cabot, in Practical Casino Math (2005, p. 162), wrote, "Without delving further into the mathematics, suffice it to say that for multi-play video poker, there is no change in the expected value (house edge), but the variance (volatility) increases." Ethier (that's me), in The Doctrine of Chances (2010, p. 568), wrote, "In n-play video poker, the variance of the 'payout per unit bet' is decreasing in n."
This apparent contradiction is the result of different interpretations. Let S_n be the sum of the payouts from the n plays, assuming one unit bet on each play. Hannum and Cabot were referring to Var(S_n)/n, which is increasing in n, while Ethier was referring to Var(S_n/n), which is decreasing in n. The justification for the former is that this would be the appropriate normalization if betting on n independent single-play machines. The justification for the latter is that, when comparing single-play and n-play machines, total amount bet should be the same, namely one unit. Not surprisingly, I prefer the second approach.
Notice that we are not providing a numerical example here. What is Var(S_n/n) with n = 10 for 9/6 Jacks or Better? Nobody knows! It has not been computed, to the best of our knowledge. We leave it as a challenging exercise for the interested reader.
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