Posted by goatcabin
Mar 20, 2010

Two million players

Suppose you could get a couple of million people to visit their favorite casino and play craps for a couple of hours, betting nothing but the pass line. Each person would have enough bankroll to cover two hours' worth of betting, so no one would bust before two hours. Each person would report his/her net win or loss to me, and I would record and analyze the results. What would a graph of all the outcomes look like?

Since the player would never bust, and would always play the full two hours, the graph would be a symmetrical bell-shaped curve, its peak height at about -1 unit, its slope such that almost 70% of the area under the curve would be between about -8.5 units and about +7 units and 95% between about -16 units and +14.5 units.

The fact that the peak of the curve comes in the negative part of the graph is because of the house advantage of 1.414% built into the pass (and come) bet. It is actually the mathematical average of all the individual net outcomes, and is called the "expected value" or, in gambling, the "expected loss". The "house advantage" is usually expressed as a percentage, the "expected loss" as a number of units or a dollar value if we're talking about a specific bet amount.

This "expected loss" is always there; think of it as a "handicap" that you have to overcome in order to break even or better. If your luck is exactly average, then you will lose about one unit in two hours' pass play, and something like one quarter of those couple of million players would be expected to end up between -2 and +2 units. But others would experience widely different results, perhaps winning or losing up to 30 units, or even a bit more. Unfortunately, for any given number of units we expect more people to lose that many than to win that many. That's because of the built-in "handicap" everybody faces.

The fact that different people's results from a couple of hours' pass play will be so varied is due to what is called "variance". Without it, there would be no point to gambling. If you knew you were going to lose 1.414% of your total bet amount, you probably wouldn't play, right? But it's not only possible, it's expected that "your results will vary". When you make bets in craps, or any other game of chance, the house advantage is what you pay to buy variance. Variance is what gives you the chance to overcome that handicap and break even or win money from the casino.

Most books on craps compare different types of bets on the basis of their house advantages. We all know the numbers: 1.414%, 1.515%, 4.0%, all the way up to 16.67% for the "Any Seven". The higher the house advantage, the bigger the "handicap" you have to overcome to break even or come out ahead. However, the probability of overcoming that handicap for any given number of bets depends heavily on the degree of variance built into the bet. Let's compare a couple of bets:

pass line boxcars
units won 1 30
probability .4929 .0278
units lost 1 1
probability .5071 .9722

As a general rule, the larger the difference between the probabilities of winning and losing a bet, and the larger the difference between the payoff and the bet amount, the greater the variance. Remember that we are buying variance with the house advantage, and the more variance we buy, the more we have to pay. That is why the proposition bets and hardways carry a higher house advantage than the line and place bets. Higher variance means higher risk, for the player but also for the casino.

To see how this affects one's chances of overcoming the HA handicap, let's examine 72 bets on the passline vs. the boxcars:

expectation: pass 35W 37L = - 2 units
12 2W 70L = -10 units

one better pass 36W 36L = breakeven
12 3W 69L = +20 units

Because of the high variance, for the bet on the 12 all it takes is beating the expectation by one outcome to go from losing 10 units to winning 20 units! Of course, there's a flip side to that: variance works both ways.

To continue, the probability of overcoming the house advantage for any given number of bets is a function of both the expected loss and the variance, expressed as what we call the "standard deviation" (SD). I will not go into the calculation of the SD in this post, but if you get WinCraps from www.cloudcitysoftware.com, it gives you the SD for each bet, expressed as a percentage of the bet amount (I use a decimal fraction instead).

pass bet boxcars
expected loss .0141 .1389
standard deviation .9999 5.0944

In other words, the SD of a passline bet is essentially the amount of the bet, while the SD of the bet on the 12 (or any single combination paying 30:1) is over five times the amount of the bet. That's for one bet, but while the expected loss increases with the number of bets, the SD increases more slowly, with the square root of the number of bets.

For any number of bets (n):

pass bet boxcars
expected loss .0141 times n .1389 times n
standard deviation .9999 times sqrt n 5.0944 times sqrt n

Remember that about 70% of the area under that curve was between -8.5 and +7 units? That is one standard deviation worse and better than the mean outcome. The 95% of the area between -16 and +14.5 units represents two standard deviations worse/better than the mean.

For any bet, the probabilities of coming out one SD (or 2, or any multiple) better (or worse) than expectation are THE SAME. This means that the same amount of good luck is worth five times as much on a 30:1 bet as it is on a line bet (pass, come, don't pass, don't come).

Let's look at those 72-bet comparisons again:

pass bet boxcars
expected loss 1.018 10.0
standard deviation 8.484 43.2

The expected losses are 72 times those for a single bet, but the SDs are only 8.485 (the square root of 72) times those for a single bet.

Let's try to draw a picture of what this means:


The '0' indicates breaking even, the 'E' the expected value (loss) and the '1', '2' and '3' indicate outcomes 1, 2 and 3 standard deviations better, and worse, than the expected loss.

For any given number of bets, the probability of overcoming the expected loss is a function of the magnitudes of the expected loss and the standard deviation. For 72 pass bets, the expected loss is about 12% of the standard deviation, so one doesn't have to be extremely lucky to break even or better.

However, since the expected loss increases faster than the SD, the more bets you make, the larger the expected loss relative to the SD, and the luckier you have to be to overcome the handicap.

For a pass bet, the expected loss is the greater than the SD after 5001 bets, see below:

-.01414 * 5001 = -70.714
square root of 5001 = 70.717 * .9999 = 70.710

After 5001 pass bets, a player has to come out a whole standard deviation better than expectation just to break even, and mathematics tells us that the probability of doing that, or better, is under .16. Still, that means that if our two million friends all made 5001 pass bets, we would expect well over 300,000 of them to be even or better.

You often hear people say, or see written, "in the long run, you can't beat the house edge", but that's NOT TRUE. What is true is that, the longer you play, the less likely you are to be even or ahead. After our 200-roll experiment, we would expect over 950,000 of our two millions friends to be even or better.

Remember that more variance usually costs more, in terms of house advantage. However, there is a way to get more variance for free: the odds bet behind the line bets. More on that later.
Good luck,
Alan Shank


teddys Mar 20, 2010

Good stuff, GC. Hopefully people will come here instead of looking into the "1.41% is a hoax" thread.

kenarman Mar 20, 2010

Thanks Allan for explaining better than I have seen before variance and it's relationship to the odds. If Mrs. Kenar was only more mathematically inclined I could now justify my craps play over her slot play. he he

DrEntropy Oct 16, 2010


Just stumbled into this blog post. I also have been thinking along these same lines for some time and think that the concept of paying for variance is pretty solid. I like to compare games by comparing, for 1 hour of play at typical speeds, the cost for one hour and the standard deviation per one hour.


jkluv7 Jan 21, 2013

Alan -

My brain just melted reading all of this. I LOVED taking it all in !!

Thank you !

mustangsally Sep 24, 2015


here is 1 comment

Alan Shank says

"What would a graph of all the outcomes look like?

Since the player would never bust, and would always play the full two hours, the graph would be a symmetrical bell-shaped curve, its peak height at about -1 unit,..."

Oh oh

looks to me that over 72 bets resolved there is a top to this bell curve with a slight slope (for rain drainage) that one could easily build a "goatcabin" on from point -1 to 0

use your imagination for what the cabin would look like

here is what I show (of course the x-axis should go out to -72 and +72 but we round to be close)

I opted not to use Excel (on vacation)

I like the "goatcabin" location

high above the rest