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Say a casino has N slot machines all with the same expected value and standard deviation. Why is it when the standard deviation gets really large you end up with more machines under the expected value than over? I understand that the average of the samples will be approximately the expected value of the slot. But the samples means don't appear to be normally distributed. This seems not to be in line with the central limit theorem. I am wondering what bit I have confused. Thanks!

ndm
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    Where did you hear that you'll have more under than over? It may have to do with the fact that if the distribution is not symmetric then the median will not necessarily equal to mean (and hence the probability that a result is less than the expected value is no longer 0.5). – ConMan May 29 '17 at 23:12
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    I read it in a research paper (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwi9xOmYo5bUAhVDOSYKHak4DnIQFggnMAA&url=https%3A%2F%2Fgamblinghelp.nsw.gov.au%2Fwp-content%2Fuploads%2FA-20-Game-Survey-of-Volatility-in-NSW-final.pdf&usg=AFQjCNHdbW8fZqfBT97DmkOGxUffilpSCQ&sig2=vPb06bGFZE7CvXY6Rnn4pg), and verified with some simulations. You are correct, the median almost never equals the mean on slot machines. – ndm May 29 '17 at 23:40
  • Be careful. Not all distributions have a variance. Some, not even an expectation! Look at the assumptions behind the CLT carefully... – Malcolm May 30 '17 at 00:06
  • The usual view is the other way around: For a skewed distribution, the SD is often high because the extreme points contribute large squared deviations $(X_i - \bar X)^2$ to the variance. // Slot machines should have SD's, but the rate of convergence to normal in the CLT is often slow for skewed distributions. For example, a the mean of a dozen (symmetrical) std UNIF random variables is nearly normal, but not so for a mean of a dozen (highly skewed) exponential random variables. – BruceET May 31 '17 at 08:24

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Having looked at the paper you linked, it comes down to a word that even they use - skewness. While some random distributions, like the normal distribution, are nice and symmetrical, many of them aren't. Often, especially when we look at things relating to money or other quantities, the distribution will have a big peak in small values, and a long tail of rare but large values. When that happens, the mean, or expected value, of the distribution will tend to be larger than the median, or middle value.

For example, consider a 6-sided die where 5 of the sides have one pip, and the remaining side has 16. The expected value of a single roll is $\frac{1}{6}(1 + 1 + 1 + 1 + 1 + 16) = 3.5$, just like a normal d6. But the probability that a single roll will be less than that value is $\frac{5}{6}$, because of the way the sides are distributed.

Similarly, poker machines are often designed so that the vast majority of their payouts are minimal, with a couple of very big, but very rare, jackpots. So if you measure "proportion of payouts less than the average payout", you're going to get a number bigger than 50%.

ConMan
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  • I was aware the mean did not equal the median, I just was unaware this was required for the CLT to be valid. – ndm May 30 '17 at 00:44
  • The only thing the CLT says is that if you have a large number of rvs with certain properties, then the distribution of the mean of those rvs approaches a normal distribution. If the original rvs are skewed in a particular direction, then the distribution of their mean will probably still be skewed that way too, just less so. – ConMan May 30 '17 at 01:36
  • Good answer (+1). Here, the precise barrier to using the CLT is the slow convergence for a badly skewed distribution (as in my simulation). – BruceET May 31 '17 at 09:19
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Continuing my Comment with a concrete example:

Suppose a slot machine pays \$1, \$10, \$100, and \$1000 with probabilities proportional to .1, .01, .001, and .0001 respectively. (Approximate probabilities 0.90009001, 0.09000900, 0.00900090, and 0.00090009 would accomplish that.) The average amount won in one play is \$3.60 and the SD is about \$31.42.

Simulating the average winnings per session after 100,000 sessions of 1000 plays each of such a slot machine, I get the following histogram for the average winnings. The distribution for one play is so extremely skewed that not even the average of 1000 plays is anywhere near normal.

enter image description here

BruceET
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