Which case has higher variance?

Consider the following two cases:

(1) 2-deck, bet spread 1-5;
(2) 8-deck, bet spread 1-10.

Let's assume that the expectation of the two cases are the same. (Well, you may argue that their expectation may not be exactly the same. To remedy this, we change the bet spread of 8-deck to whatever is needed to make the two cases the same expectation.)

My question is: which one has a higher variance? Please present your reasoning. Thanks.

the reason is because there is a greater impact on the remainder of the deck with each single card removed. A +4 can quickly become a 0 count. Also, since the highs and lows tend to be shorter due to more frequent shuffling, you can get hit pretty hard when your max bets are out and then the high count is over because they have to shuffle.

I frequently find my bankroll going through 2 or more standard deviations with 2 deck. This doesn't seem to happen to me as often on shoe games.

Your answer seems anti-intuitive. My original guess was that 8-deck had more fluctuations.

Have you considered the fact that the max bet is higher for 8-deck?

Assuming both players are properly bankrolled, the percentage impact on their respective bankrolls should be the same for each max bet lost. For 8 decks, you need a greater bankroll since your max bet is higher. If both have the same ROR relative to their max bet, then the volatility of the 2-deck game can be the deciding factor.

At 8 decks, the effect of removal is less significant to the remainder of the deck. So, for example, 4 small cards come out in each game. This would mean a TC of +2 in a 2-deck but only a TC of +0.5 in the 8 deck. So the 2 deck player would put more money out at this point because he has an edge, whereas the 8 deck player doesn't and should still be at minimum bet. The frequency of occurrence for this situation means that the 2 deck player will bet more money more often and so that will make the average wager similar to the 8 deck that moves less frequently but more largely. However, short-term fluctuations from frequent shuffling, greater impacts, etc., will cause the 2 deck player to run through more standard deviations than the 8 deck.

Let me fine tune the problem to reduce ambiguity.

First, let's assume that 8-deck provides more liberal
rules (DAS and late surrender) so that the off-the-top
house edge is the same for 2-deck and 8-deck. In fact,
this is the case in real life.

Second, the minimum bet is the same for 2-deck and 8-deck,
i.e., 1 unit. Thus the 8-deck has higher maximum bet.

Third, the bankroll is the same for both cases. In fact,
to calculate variance, this condition is not needed.

(Be careful, please don't use ROR in your reasoning.
If you think that ROR is higher for 8-deck, you have
admitted that its variance is higher, because they
both have the same expectation.)

From my original post, we have agreed that the overall
expectation is the same for 2-deck and 8-deck.

One more thing is that the average bet is also approximately
the same for 2-deck and 8-deck, i.e., about 2 units.

Now, could you rethink the problem?

The SD/100 hands are as follows:

2D, S17,DAS, 62/104, 1-4: 20.85
2D, S17,DAS, 62/104, 1-6: 20.37
8D, H17,DAS,LS 6.5/8, 1-10 play-all: 25.33
8D, H17,DAS,LS 6.5/8, 1-10 back-counting: 23.33

Note that this assumes optimal-betting, which means the minimum bet will be different among the different scenarios.

Let me make some general comments

You asked to assume that thee expectations were the same. If we adjust the betting to give the same EV, then the better game will have lower variance.

However, we shouldn�t play that way. You should adjust the betting to give the same RoR. If you do that, then the higher EV game will have the higher Variance.

Usually the DD game will be better than the 8 deck game. If you bet optimally in both, then will experience greater swings in the DD game. Actually, the swings would turn out to be the same, you would just see them faster in the DD game. Your chance of loosing half the bank would be the same in both, but it would occur in the DD game.

If you try adjust the expectation, then you will either be over-betting in the 8 deck game, or under-betting in the 2-deck game. I think most players are more likely to over-bet the 8 deck game. It is possible to over-bets o much, that your Certainty Equivalent is negative. That is, your game could be so poor that you would be better off not playing.

then I think the answer lies in speed of play. The more hands you see, the more probable the chances of an unusual variation occurring.

The 2 deck game is much faster. Shuffling takes up less time, no mid-round entry in the 2 deck means there are fewer buy-ins to slow things down, whereas the 8-deck has people constantly buying in and playing. 8 deck is more prone to people coming in mid-round. The result is more hands per hour than the 8 deck.

So, if EV, avg. bet, are both equal to either game, then the hands per hour will be the deciding factor. You will approach the large number distribution more quickly in the 2 deck due to more hands per hour. Each hour, the needle swings greater in the 2 deck.

to a per-hour measure. I see the SD is lower for the 2 deck game, but I am certain that time is also a factor. A 2 deck player can get almost 3 hands for every 2 that the 8 deck player gets.

For example:

Let's assume the 100 hands represents 1 hour of 8 deck play. In that same time, the 2 deck player will get about 150 hands. As a result, the 2 deck player is subject to more variance per hour.

SD increases by the square root of the number of hands played. Therefore, the factor to convert from 100 hands to 150 hands is sqrt(150)/sqrt(100) = 1.22.

which is quite meaningless. Convert the units to dollars and see which has higher variance. You'll have a much bigger unit in a two deck game than an eight deck. If not, how do you justify betting 2.5X as much with the same edge in the 8D game? (the 10 unit bet compared to the 4 unit bet) MathProf's post below explains what this is about.