Two scenarios:

Scenario 1:
Say you're doing
4x $100 hands as opposed to 1x $100 per deal

Scenario 2:
4x $25 hands as opposed to 1x $100 per deal

Would scenario 1 constitute a $400 bet or something lower? I'm just curious to plug it into this calculator: http://www.beatingbonuses.com/simulator.htm

thanks

A rough guide is sqrt(N) where N is the number of hands.

So 2 hands of $100 carries roughly the same variance as 1 hand of $150, 4 hands of $100 carries the same variance as 1 hand of $200, and 4 hands of $25 carries the same variance as 1 hand of $50.

Nope. Same upcard.

Scenario 2 *reduces* variance. It gets you to the long run expectation quicker for the same amount of money wagered. Which is not really a good thing in a negative EV game, where variance is your friend. You should only use this method in advantage play (counting).

Thanks. I only provided scenario 2 as a point of reference. I'm more interested in what the 'true' bet size of scenario 1 is (cali suggested $200)

Would you mind elaborating? Are you suggesting scenario one isn't the equivalent variance as 1 x $250 bet?

When each bet isn't completely independent (which in this case they are not since they are tied together to the dealer's outcome of that 1 hand) there is another variable that comes into play called covariance. That's why callipygian said sqrt(N) is a *rough guide* and not the exact answer, yet David felt it necessary to jump on that anyway.

The actual variance will be somewhere between sqrt(N) and N... covariance is not easy to calculate.

That's the factor by which variance changes with N, not the variance itself.

Correct, I meant that :-)

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Could you dumb this down please. I don't speak math. Are you the saying the EV of the 4x$100 is within a close range of $200 or could the approximation range be completely wrong?

We're not talking about EV at all.

EV is the slope of your graph in the long run (for blackjack, this can be tens or hundreds of thousands of hands).

Variance is how often your short term results deviate from your long term results.

So let's say you flip a fair coin 10 times. You expect 5 heads and 5 tails. But it's possible you get 6 heads and 4 tails. It's also possible, but less likely, you get 7 heads and 3 tails. It's also possible, but very unlikely, you get 10 heads and 0 tails. 5 heads is your EV. Your mathematical variance is calculated mathematically, and I believe it ends up being 0.5. Typically, a measure called standard deviation (SD) is used, which is sqrt(variance). The rarity of an event is expressed as SDs from the norm. So 1 SD = yawn (middle 68%), 2 SDs = heh (middle 95%), 3 SDs = wow (middle 98%), 4 SDs = omg (middle 99.8%), etc.

It's a little weird because "variance" can mean the general concept of deviation from expectation, or it can be a precise mathematical definition.

Mathematical variance and EV increase linearly with independent trials. So if you flip a coin 1,000 times, your EV is 5*100 = 500 heads and your SD is 0.7*sqrt(100) = 7. So while flipping 7 heads out of 10 is a rare event (3 standard deviations above expected), flipping 700 heads out of 1000 is ultrarare (30 SDs above expected).

Co-variance is a concept which takes into account not all trials are independent. In blackjack, the dealer's upcard is the same across many trials. The results from two hands at the samr table will more likely match than the results from two hands at different tables.

Continuing our example, this would be like flipping a coin but always starting with the same face up. If you flip it in the same spot with the same strength, it's more likely to land the same way every time.

As someone noted, co-variance in the SD is difficult to quantify. It's between N independent trials (SD*sqrt(N)) and an N-sized bet (SD*N), but there's no easy formula.

In practicality, it's closer - much closer IIRC - to the former than the latter.

tl;dr: "rough guide" sqrt(N)

I expect 6 of one and 4 of the other; that's the most likely outcome (41% - next best is 25%). This is a pretty sick prop bet opportunity that I have exploited in the past.

Obviously this is only true when the 6 can be either heads or tails, which a smart gambler would realize. The chance of exactly 6 heads is less than the chance of exactly 5 heads, just as intuition tells us.

That said, callipygian's statement was mathematically correct, in that the "expectation" is for 5 heads. You used the word in a nonstandard way to mean the most common outcome (the mode), instead of the average outcome (the mean).

Expectation means the latter in probability, if not in prop bets.

I try and avoid prop-betting smart gamblers.

Well, one could talk about the set of 11 discrete outcomes, not a continuum of outcomes. There is no mean. There's no ambiguity, just a natural bit of cognitive dissonance on the part of the mark by using "x of one, y of the other" instead of "x heads, y tails" which is deliberately misleading.

Of course I'm telling you stuff you already know, but it is important to distinguish between the cases where language is ambiguous "one of the sisters is called blah" vs this type of case where the mark leaps to the wrong conclusion.

Seems like two hands works best with one other player at the table playing one hand. No idea why, its just something Ive noticed.

Of course you have.

That's only on Tuesdays with a full moon.