Variance in Craps

Anyone have the variance of the different bets in craps, such as the pass line, free odds, place (4,5 & 6)?

Thanks

For a wager that pays x:1 with a probability of winning p, the house edge and standard deviation (per unit bet and per 'root decision') are:

ev = (x+1)*p - 1
std = (x+1)*Sqrt[p*(1-p)]

For even money bets, like the pass line, x =1 and p ~ .5, so std ~ 1.
If you place the four, the standard deviation is about 1.3 per unit bet.

Quote: DrEntropy

For a wager that pays x:1 with a probability of winning p, the house edge and standard deviation (per unit bet and per 'root decision') are:

ev = (x+1)*p - 1
std = (x+1)*Sqrt[p*(1-p)]

For even money bets, like the pass line, x =1 and p ~ .5, so std ~ 1.
If you place the four, the standard deviation is about 1.3 per unit bet.

Those are correct, as far as they go. They give you the information, per unit bet, for ONE BET. However, as you make multiple bets, the Standard Deviation and the ev increase at different rates, the ev increasing at the same rate as the number of bets, but the SD increasing with the square root of the number of bets.

So, for a $5 passline bet, with ev of -$.0707 and SD of $4.9995, for 100 bets the ev is -$7.07 but the standard deviation is $$49.995, increasing ten times (sqrt 100). Since the standard deviation rises more slowly than the expected loss, if you play long enough the SD will catch up to and pass the expected loss. This is why the longer you play the less likely you are to be ahead.

Bets with a high standard deviation tend also to have a high house edge. The bets on 2 and 12 have SD of over five times the amount of the bet, and the house edge is 13.88%.

Odds bets behind pass/come/don't pass/don't come bets add variance (SD's ranging from 1.0954 to 1.414 times the amount of the bet for rightside odds) without adding to the expected loss, since there is no house edge. They do, however, increase the "risk of ruin" simply because you are betting more.

These calculations are only good when the amount of the bet does not change. Actually, much of the variance in the way people actually play comes from varying the bet amount. For this, you need to do simulations to figure the variance.
Cheers,
Alan Shank

I suspect that either DrEntropy or Alan could help clear up the senior moment I am having with variance.

I think I understand the ev and std formulae that are given by DrEntropy and Alan’s discussion of the combined ev and std of a series of wagers. However, what if the multiple wagers are simultaneous?

I think that Alan’s calculation would apply correctly for a series of uniform single-number wagers on perhaps 38 consecutive spins of the roulette wheel. But if I bet all 38 numbers equally for the same spin, there is absolute certainty as to what the outcome will be. The ev is the same as for 38 sequential bets, but the std for the collection of simultaneous bets drops to zero (I think). I don’t think the formula works there, does it?

In a similar vein, we can analyze the standard deviation of a series of equal pass line bets in craps. Each of them should have the same ev and std values as a come bet, but if I place a pass bet followed by a come bet on each roll, does Alan’s formula apply while the bets are all dependent on the same roll(s) of the dice?

Obviously I don't have this clearly in my head, and I would appreciate some help on this. Thanks.

Quote: Doc

Obviously I don't have this clearly in my head, and I would appreciate some help on this. Thanks.

Short answer is that the Sqrt[n] formula doesn't apply to events that are not independent, which includes the two examples you cite. Eventually things get complicated enough that a simulation is the best and easiest way to compute the standard deviation...

Cheers!

Thanks! Not much into computer programming these days, so I think I will be doing my "simulations" in short runs at physical craps tables. Less scientific validity, greater entertainment value.

Quote: Doc

Thanks! Not much into computer programming these days, so I think I will be doing my "simulations" in short runs at physical craps tables. Less scientific validity, greater entertainment value.

When the bets affect each other, like hedging a pass bet with a field bet or Any Craps, etc., sometimes you can analyze them as a unit, as long as they are resolved "in step". For example:

$5 pass and $1 Any Craps (one Any Craps bet per pass decision)

For the SD, we do this:

So, the ev is -$.1818 and the SD is $4.75. Note that the SD for a $5 pass bet is $4.9995; the hedge reduces variance, as it's intended to.

However, if you make more than one Craps bet per pass decision, the number of craps bets varies, and you can't calculate it directly. Well, you could grind it out, considering the weighted probabilities of different numbers of Craps bets being resolved per pass decision, but that's laborious.

When you get into different bets resolving separately, and, especially, varying bet amounts, you really need to simulate.

The best craps simulator I have ever seen is WinCraps, available at:
http://www.cloudcitysoftware.com .

A steal at $19.95, you can download a demo for free.
Cheers,
Alan Shank

Quote: goatcabin

The best craps simulator I have ever seen is WinCraps, available at:
http://www.cloudcitysoftware.com .

A steal at $19.95, you can download a demo for free.
Cheers,
Alan Shank

I also own this one, and second this recommendation.

Quote: DrEntropy

I also own this one, and second this recommendation.

Hear, hear!

BTW, can entropy ever be reversed? >:-)
Cheers,
Alan Shank

Quote: goatcabin

The best craps simulator I have ever seen is WinCraps, available at:
http://www.cloudcitysoftware.com .

A steal at $19.95, you can download a demo for free.

I appreciate the statistics lessons I am getting on this forum. It has been too long since I studied the subject, and the brain has atrophied.

I would likely rush out and buy WinCraps on whim, but fortunately I can save my $19.95 because it is Windows software while I am strictly a Mac user. 8-)>