Non-math geek and variance: thanks guys

Thanks to the dudes that did their best to guide me out of my math-fog when it comes to explanation of the term, 'variance.'

To Bootlegger for his creative 'Rocky Mountains/Great Plains' analogy and this sentence: 'The greater the variance, the greater the individual highs and lows will be. The smaller the variance, the smaller the highs and lows.'

To Panama Rick for his "99% of the time variance will be between
EV -3SD and EV +3SD."

And ETFan for his "Variance helps us quantify luck."

Now, including BJCobbson and Anonimuss, nobody mentioned my little individual, single box #5, where I am sitting at this table of 7 players. (Although Boot may have been referring to my individual box in his post.) And, nobody mentioned the dealer hand - the 8th hand.

Presumably, variance-math also applies to the dealer hand. Therefore, the dealer hand is just as prone to variance swings as us other 7 wannabees, all waiting in hope for our lucky streak to descend from heaven, now!

We all want our unfair share of 'only good' variance for us good guys, and we all wish 'only bad' variance for the dealer hand. We wish. But boyoboy, you won't hear a peep out of us when The Dreaded One (V) is going good for us; but don't we squeal like cut pigs when The Dread has swung in favor of the dealer! Now what could possibly cause these pendulum-swings of the Dread?

...just like a ploppy.

Wine and more whine,

Cobbson

You don't say, JN. Why is it less? By how much? And how come? Time for the geeks again!

I'm not sure what the poster meant when he said the dealer's variance is less than the players. However, if we assume seven players, then the variance would be less for the house than it would be for each of the individual players. This is because the money is spread out over seven individual results. If you are a single player, you will either lose, win or push the entire amount you have bet with each hand you play. The end result for the house will be less drastic on most rounds. More often than not, the house will win some of the hands and lose some of the hands. The highs and lows will be flatter in terms of the total amount of money won or lost for each round. In that sense, the flow of money for the house will resemble the Great Plains more than it will the Rocky Mountains.

Bootlegger, It's always good to read your usually clear expession, whatever you write, which can only add to our understanding. The eternal quest for more, clearer and more meaningful understanding of this wild and unruly game is well served by guys such as you, who freely give. Thanks pal.

To JN. What precisely did you mean by your (nt) post?

Boot, you say the variance will be less for the house because the money is spread out over 7 individual results. By how much will it be less? Is there math for this, or will it depend?

And, : "More often than not, the house will win some, lose some..." So that makes for the flatter Great Plains. Right.

However, when the house most often wins some, loses some; unfortunately (or fortunately) for me - the card counter mixed up somewhere among those 7 players - only some of us are going to be 'lucky' I guess, whatever the count is! Too bad for the 'unlucky' ones. Put it down to a very, very capricious variance.

Thanks again Bootlegger. PS I bet you are doin a lot better out of blackjack than you ever did out of moonshine whiskey!

"Boot, you say the variance will be less for the house because the money is spread out over 7 individual results. By how much will it be less? Is there math for this, or will it depend?"

First, I'm not sure exactly what Boot had in mind, but, clearly, if one is the house, playing against, say, a full table, the variance for the house will surely be a lot MORE than the variance for any one single hand on the table.

From p. 142 of Griffin, we see that the formula for the variance for n simultaneous hands of blackjack is, depending on rules:

vn + cov*n(n-1), where v = the variance of one hand of blackjack (typically, 1.26 squared units, according to Griffin, but more for games with doa and das), cov = covariance between two hands = .50, and n = number of hands.

For a full table, the variance would, thus, be:
1.26(7) + 0.50(7)(6) = 29.82 squared units, assuming the same one-unit bet on each of the 7 hands.

In terms of the more familiar standard deviation, we see that, for one hand, s.d. = sqrt(1.26) = 1.12, and for the seven hands, s.d. = sqrt(29.82) = 5.46. As 5.46 is, clearly, less than 7 x 1.12, one might correctly conclude that the s.d. for seven simultaneous hands is less than seven times the s.d. for one hand, but that statement is somewhat different from saying that the dealer experiences less variance than the individual player.

Frankly, I'm not sure exactly what Boot meant by that.

Don

Don, Thanks for your input. You wrote, "...clearly, if one is the house, playing against, say, a full table, the variance for the house will surely be a lot MORE than the variance for any single hand on the table."

Now that seems to be the opposite opinion to what Boot said... But, you wheeled out some Griffin-math to reinforce your view, which makes you hot favorite in this.

Assuming that what you say is correct, then the dealer-hand's performance-chart should look like a side-view of the Rocky Mountains; with FEW Great Plains at all, and lots of mountain peaks. Could this be a more accurate description than Boot's model?

Far be it from me to get into a math dispute with either you or Griffin, but you wondered what I was thinking, so here it is.

I was thinking in terms of percentages of the total amount of money bet, as opposed to a single hand. Obviously, a high roller could come in and screw the whole thing up for the house, but on average, I'd say the house has less variance than an individual player does because it is spreading its money out over several hands.

If we assume a group of players who are all betting the same amount, let's say $10 to make it easy, then the dealer (house) is risking $70 on the initial bet for each round. That amount would go up slightly for blackjacks, doubles and splits, which would add to the variance for the house. But, on the other hand, more often than not, the house will win some and lose some on each round. The final total result is often going to be some percentage of the total money bet by all of the players. So, when you are adding up the possible outcomes to compute the variance, you're going to have an outcome that the individual players won't have. They will have a total loss, a push, a win and the 3:2 result for a natural. The dealer will have those results, keeping in mind that he is paying the extra for the natural, but he will also have something in-between on most rounds of play. I think that something in-between will reduce his variance when the total amount bet on the seven hands in considered. In effect, the average distance from the expected value for each round is what matters to the house, not for each hand. Those in-between results will shorten that distance.

That's what I had in mind. I could be wrong. I have been before. :-)

I should have made this clear in the post above, but you mentioned the formula in Griffin for multiple hands. This may be a little different because the dealer is playing one hand against multiple hands. Griffin's formula deals with playing multiple hands against the dealer, which is different from what the dealer is doing and estimates a variance and a covariance for playing multiple hands from the player's perspective. I'm not sure the dealer has a covariance since he is playing one hand and since there are different factors for the dealer's hand, I'm not sure the variance would be the same for the dealer as it is for the player even if they were playing head's up.

In short, I'm not sure that Griffin's formula for playing multiple hands applies in this case.

One formula for variance is: average (result squared) - (average expectation) squared. The signs change for the dealer, but the result is the same. the square of a negative is the same as the square of the positive.

ETF

"I'm not sure that Griffin's formula for playing multiple hands applies in this case."

My point is that the dealer is not playing multiple hands against a single hand, as the players would do. He is playing one hand against multiple hands. Would there be a co-variance, which is what the formula Don cited includes? For the dealer, there are not multiple results, there is one result at the end of the round after all of the player's bets are paid or collected.

Boot, you're just not right about this.

If you consider overall e.v., and deviation from the mean, that it a separate consideration, which we could discuss, if you like. But, I showed you how the s.d. for the seven hands is much greater than the s.d. for one hand, while less than seven times the s.d. for one hand. If you'd like to discuss a different concept, that's fine, and we can introduce e.v., percentages of money bet, etc. You'll have to carefully define what it is you'd like to quantify.

All I answered was the concept of variance. If you play seven hands, your variance is much, much greater than if you play one hand. And, since the dealer's results are simply the opposite of those for the player(s) (he is, after all, booking all the action), the dealer's variance against a full table will clearly be much greater than the variance for any one of the players at the table.

Surely, you see that, right?

Don

Here's a simple way to look at what you're probably trying to express.

Just assume that the e.v. is zero, to make the math easy. If each of seven players bets one unit, the e.v. for the dealer on that round is 0, and his s.d. for that round, as calculated in my previous post, is 5.46 units. The s.d., expressed as a percentage of all the seven units bet, is, obviously, 5.46/7 = 78%.

Now, for a single player, the e.v. is still zero, but the s.d. is 1.12 units. So, as a percentage of his single bet, the s.d is a (larger) 112%.

But, in absolute terms, if the dealer experiences a one-s.d. result on the round, he will win or lose 5.46 units, while, if the player experiences a one-s.d. event, he will win or lose 1.12 units.

So, for any given round, the dealer's variance, in absolute terms, is much greater than that of any one player, as I have been pointing out. But, as, perhaps, you meant to express, the dealer's s.d., as a percentage of the total wagers, will be less than the percentage s.d. of a single wager.

Clear?

Don

It's Thanksgiving afternoon and I've got a full belly and sleepy eyes, so this will be short. Maybe we could have a longer discussion some other day. I understand what you are saying, but I think I am taking a different look at this. To me, the dealer's unit is 7 times larger than the unit of each individual player. For example, if each player is betting $10, the dealer's unit is $70. I also don't think the Griffin formula given on page 142 of his book applies to the dealer. That formula deals with one player playing multiple hands against the dealer's single hand. The dealer is playing one hand against the multiple hands of the players.

Now I'm going to take a nap. Happy Thanksgiving. :-)

You're not understanding what I've written. That the dealer plays one hand is irrelevant. His RESULT is the opposite of the cumulative result of all the hands on the table. How do you propose to calculate the dealer's variance if not to suppose that it is the same as the variance of all the hands that he is playing against?

Don

I thought of what you're saying and you are probably correct. However, the dealer's result is not the same if you consider that the entire table's bets are equal to one bet for the dealer. The players will have a result that is, more often than not, either -1 unit, 0 units or +1 unit. If we calculate the total of the players bets as the dealer's single unit, the dealer's result will fall in a range from -1 unit to +1 unit, with most results falling somewhere in between (I know I'm simplifying this). The dealer's final result will be closer to his expecation than those of each individual player in most instances. That is why I am saying the variance for the dealer is smaller than it is for the players.

One comparison I might make would be to video poker. I can play a one-line $5 machine at $25 per hand. My variance will be high. I can play a $.05 100 line machine at $25 per hand for all 100 lines. My variance for my $25 unit will be lower on the 100 line nickel machine than it will be on the single line $5 machine. That's where I'm coming from.

I think we might be two ships passing in the night on this one.

What you say now: "The dealer's final result will be closer to his expecation than those of each individual player in most instances."

What you said before: "I'm not sure the variance would be the same for the dealer as it is for the player even if they were playing head's up."

It absolutely is the same. Forget about units. A unit is in the eye of the beholder. A unit can be anything. Think of variance in terms of dollars squared.

Or use units if you must. But then don't change the unit in the middle of the problem, in order to make Griffin's formula inapplicable. That would be cheating, and you don't believe in cheating, remember? ;-)

You also said: "In short, I'm not sure that Griffin's formula for playing multiple hands applies in this case."

It applies, if you use it correctly. The larger point is that the dealer's variance inexorably becomes much larger when more spots are played, just as Griffin's formula suggests.

Sorry, but your ship is off course, mathematically speaking.

ETF

I, for one, have read with interest concerning dealer variance; especially from three of the most experienced and informed guys around, who should know. A technical difference of opinion or two can only add to the spice of life... Who is right? Maybe each is right, to varying degrees. (Variance degrees?!)

Judging by the poor numbers (so far) of hits for posts on this thread, many are not interested in reading about it at all. Strange that, considering the dealer hand is King! We don't often see dealer-hand discussions such as this one on BJ sites, especially by three Big Guys.

I say the dealer hand is King because whatever you might do with your hand, (even bust) the DH is just patiently waiting it's opportunity to...kill you, or promote your bank account. What power the King has!

As for the Dealer-hand variance, or the player's-hand variance being able to be math-calculated, I seriously doubt if any card counter, mid-shoe is going to do that. Why would they, anyway; in the middle of a game? What possible benefit could that hold, for the next round for example?

I hope you guys continue the discussion. It is nothing if not entertaining.

Boot, in your post Nov 27 5:24 ..."since there are different factors for the dealer hand"... Now what different factors are these? (For my notebook on the DH.)

ETFan, "Variance helps us quantify luck." Great to see the Union of Science and Mysticism!

One more for you all: Say all 7 players are each flat-betting the same minimum. The dealer-bust rate will be close to 28.02%, after say, one million hands. What will be the bust-rates for each of the seven? Will they each turn out to be very close to one another, granting all 7 are playing perfect BS?

And we did not even get onto why 55% of double-downs are successful!

"One more for you all: Say all 7 players are each flat-betting the same minimum. The dealer-bust rate will be close to 28.02%,'

You meant 28.20%.

"after say, one million hands. What will be the bust-rates for each of the seven? Will they each turn out to be very close to one another, granting all 7 are playing perfect BS?"

I state in BJA3, p. 50, that I think the 15.9% bust rate that I pointed out for the BS player was the first such mention in print anywhere. I had never seen it before.

Don