Aruuba: Effect on ROR & SD with Multi-hands

[Aruuba]

Assume a negative expectation game where you are being dealt off the top of the shoe and the shoe is re-shuffled after every round.
You are always alone against the dealer and can play 1-5 spots.
Is there any difference at all in Risk of Ruin or Standard Deviation in playing one hand of $15.00 versus 3 hands of $5.00? Is there any reason I'd rather play 900 hands (or rounds) of the former vs 300 hands of the latter, or are they exactly the same? Or 1 hand of $10.00 vs 5 hands of $2.00, etc.
Or does co-variance somehow make it better (i.e. less SD or ROR) to do one vs the other?

I think I saw in an earlier post about playing 2 spots that there is less SD associated with multiple spots. Is that true?

[Don Schlesinger]

> Assume a negative expectation game where you
> are being dealt off the top of the shoe and
> the shoe is re-shuffled after every round.
> You are always alone against the dealer and
> can play 1-5 spots.
> Is there any difference at all in Risk of
> Ruin or Standard Deviation in playing one
> hand of $15.00 versus 3 hands of $5.00?

Yes. EV is the same, per round, but SD is smaller for the three hands, so variance is smaller, as well. ROR is a function of EV/var, but ROR is proportional to -EV/var, so when the EV itself is negative, I think we're saying that the ROR is increased by spreading to the three hands.

Someone correct me if I'm wrong. I may be getting tangled up with the signs here.

Don

[Double21]

> The real question is why anyone in their right mind would play this game!

[Aruuba]

Of course in the "real" world we live in, you never would want to play this game. Eventually you would lose all your money either way.
More specifically, and partially because there is no "Internet BJ" section here, I am thinking of when you, say, buy $200 and get a free $200 bonus as you do on the Internet. Before you can cash-out you have to wager, let's say $4,000.00. So I am really wondering, when given the choice, if I should make 5 $2 bets at a multi-play table or 1 $10 bet, until I have met the wagering requirements. So basically I'm asking, given a finite & fixed amount of wagering, if there is any difference.
From Don's answer, it sounds to me either way I would end up with the same EV ($380 if the HA is -.05%) but it would be a "smoother" ride (less SD) if I played more spots . If that is the case, then naturally, I'd rather avoid the large swings, if possible.
I'm still not clear why the Risk of Ruin is higher with multi-spots since it seems, intuitively, certainty, either way. But if the Risk of Ruin is higher playing multi-spots, does that mean I should endure a "bumpier" ride playing one spot because there is less chance I will bust out before meeting Wagering Requirements?
Maybe it's a function of how many decisions are going to be made?
I've always played one spot and I'm not going to play multi-hands without a good reason. Somehow I assume the casinos offer it because it works in their favor, though I don't understand how if EV is the same. Why else would an Internet casino offer multi-spots when they know card-counting is useless?
It sounds to me Don's reply says, if you play more spots, the Road to H*ll (Ruin) is paved more smoothly so you get there quicker.
Did I understand Don's reply correctly?

(Note - I couldn't figure how to do this as most bankroll calculators won't let you put in a negative EV!)

This is my first day here (first impression - alot better than bj21) and if Internet Gambling questions are not welcome, my apologies. All I know is I can't lose on the internet and I can't win in "real" life!

[ET Fan]

> Someone correct me if I'm wrong. I may be
> getting tangled up with the signs here.

By spreading to three hands of $5 instead of one hand of $15, you reduce variance, which in turn reduces ROR -- whether EV is positive or negative. EV/Var figures into a ROR formula I found in some book once, for infinite play:

ROR = e^(-2xBxEV/Var)

Note that if EV is negative, the expression returns values greater than 1, which is meaningless for a probability. This simply reflects the fact that ROR is 100% in a negative EV game. So long term ROR is indeed a function of EV/Var (for a given BR), but I don't think they're really "proportional" to one another.

Now presumably, since we're talking about a negative EV game, we're interested in ROR for some limited time frame. Here, a trip ROR formula (also in previously mentioned book) is appropriate. If you do a few examples, you will find that for a given BR, a given time frame, and any given EV -- be it positive or negative -- a larger variance brings on a larger ROR.

So nobody has any reason to love variance. ROR is vanishingly small for a decently managed casino. And it could be argued variance is the reason they exist. Yet, if they could find a way to reduce variance, while maintaining their take, they'd be all over it.

ETF

[Don Schlesinger]

I enjoyed your post. The discussion has a familiar ring to it :-) Comments follow.

> By spreading to three hands of $5 instead of
> one hand of $15, you reduce variance, which
> in turn reduces ROR -- whether EV is
> positive or negative.

For long-term ROR, with negative EV, you seem to demonstrate later that this may not be true. Values exceed 100%, which are meaningless, but, do they not get larger as variance gets smaller?

> EV/Var figures into a
> ROR formula I found in some book once, for
> infinite play:

> ROR = e^(-2xBxEV/Var)

Wise author! :-)

> Note that if EV is negative, the expression
> returns values greater than 1, which is
> meaningless for a probability. This simply
> reflects the fact that ROR is 100% in a
> negative EV game. So long term ROR is indeed
> a function of EV/Var (for a given BR), but I
> don't think they're really
> "proportional" to one another.

When EV is positive, and bankroll is fixed, ROR is a function of EV/var. If you hold those two variables in the same proportion, while changing them, ROR remains the same.

> Now presumably, since we're talking about a
> negative EV game, we're interested in ROR
> for some limited time frame. Here, a trip
> ROR formula (also in previously mentioned
> book) is appropriate. If you do a few
> examples, you will find that for a given BR,
> a given time frame, and any given EV -- be
> it positive or negative -- a larger
> variance brings on a larger ROR.

> So nobody has any reason to love variance.

How about someone playing a negative-EV game, who wants to try to reach a certain goal, above the current bank level, with as high a probability as possible? Would more variance not be better for this individual, ascribing to the "maximum boldness" of Epstein, et al? If I'm trying to double 100 units at roulette, should I bet the 100 units one at a time, or all at once?

> ROR is vanishingly small for a decently
> managed casino. And it could be argued
> variance is the reason they exist. Yet, if
> they could find a way to reduce variance,
> while maintaining their take, they'd be all
> over it.

No, I doubt it. If variance could be drastically reduced in all the negative-EV games that the casinos offer, it would be the death knell of those games. If players lost with virtual certainty, with no chance at "lucky swings," I doubt that they would continue to play.

Just my take.

Don

[ET Fan]

Values exceed 100%, which are meaningless, but, do they not get larger as variance gets smaller?

With a negative EV, the values of e^(-2xBxEV/Var) get larger as variance gets smaller, as you say. But ROR does not get larger. It can't get larger than 1, so the formula is not meant to handle negative EVs.

When EV is positive, and bankroll is fixed, ROR is a function of EV/var. If you hold those two variables in the same proportion, while changing them, ROR remains the same.

Yup.

>How about someone playing a negative-EV game, who wants to try to reach a certain goal, above the current bank level, with as high a probability as possible? Would more variance not be better for this individual, ascribing to the "maximum boldness" of Epstein, et al? If I'm trying to double 100 units at roulette, should I bet the 100 units one at a time, or all at once?


You'd want to bet the 100 units all at once, not to increase variance, but to decrease N, the number of independant trials. Conversely, a positive-EV gambler wants to INcrease N by making his bets as small as possible. The law of large numbers works in favor of the positive EV gambler and against the negative EV gambler.

A neg-EV gambler may prefer (love might be too strong a word) a high variance gamble to a low variance gamble in some cases. But we all know what usually happens to neg-EV gamblers. When ruin hits, they'll hate variance with a special passion.

If variance could be drastically reduced in all the negative-EV games that the casinos offer, it would be the death knell of those games.

Yes, but that wouldn't be "maintaining their take," as per my hypo. No fair bringing the real world into the discussion. ;-)

ETF

[Don Schlesinger]

See how easy that exchange was? We should have tried it a few months ago, no?

There's something about this site that brings out the civility in people. :-)

Don

[Cyrus]

I happen to agree with everything you wrote in that post regarding Variance, negative-EV games, gamblers and casinos. (But you knew that!)

"How about someone playing a negative-EV game, who wants to try to reach a certain goal, above the current bank level, with as high a probability as possible? Would more variance not be better for this individual, ascribing to the "maximum boldness" of Epstein, et al?
If I'm trying to double 100 units at roulette, should I bet the 100 units one at a time, or all at once?"

You know it's "all at once" but ETF brings "the real world" into the discussion and spoils the mathematical neatness, all because the gambler who will bet his whole bankroll in one go and then calmly walk out the door is a rarer animal than a smart casino executive.

"If variance could be drastically reduced in all the negative-EV games that the casinos offer, it would be the death knell of those games. If players lost with virtual certainty, with no chance at "lucky swings," I doubt that they would continue to play."

Do not doubt that at all, please. You know that this is absolutely correct. It's no fun playing reverse ATM.

Casinos love Variance. (Rather, to appease ETF, make that, smart casino managers should love Variance. But of course casino managers actually hate Variance.)

--Cyrus

[Karel]

With a negative EV, the values of e^(-2xBxEV/Var) get larger as variance gets smaller, as you say. But ROR does not get larger. It can't get larger than 1, so the formula is not meant to handle negative EVs.

With a certain goal and negative expectation, you can get the proper formula similar to what I wrote on DD, Theory & Math. I will post a proof (derivation) of the formula on the same page later.

For maximing the goal with negative expectation game, the optimum is so-called "bold gambling". The good reference is Dubins & Savage, "How to Gamble If You Must", McGraw-Hill, NY, 1965.

Regards,

Karel

[ML]

Variance caused by bet ranging obeys Epstein's Theorem I and will cause a faster and more certain ruin as the equations show. However, internal variance (i.e. betting the number rather than red or black in roulette) does, in fact, slow the speed of ruin just as variance raises Nsubzero and slows the speed of win for the counter.

Betting sims demonstrate this well. Do you have my betting program?

[El Poet]

> Variance caused by bet ranging obeys
> Epstein's Theorem I and will cause a faster
> and more certain ruin as the equations show.
> However, internal variance (i.e. betting the
> number rather than red or black in roulette)
> does, in fact, slow the speed of ruin just
> as variance raises Nsubzero and slows the
> speed of win for the counter.

> Betting sims demonstrate this well. Do you
> have my betting program?
Hi Again ML,
I'M interested in your betting program.

[ML]

It is a GWBasic Program but can be modified.

On bjmath.com-Forums-Betting. Entitled "Program" 21 May 2001. You can search for ML.

To modify it for roulette, a number of changes have to be made. First, the real odds on roulette are bad enough results will cascade and all final results will be zero wins. To get usable results, you have to assume the zero hole is a quarter the size of a regular hole and modify the odds accordingly. In other words you have a wheel with 145 holes, four of which are each number and one of which is a zero. Betting a number with payout odds of 36! to 1 or 1 to 1 will show betting a number rather than a color loses more slowly which is the point I was trying to make.

I probably did not make clear this is for repeated plays. It would not overcome the maximum boldness strategy to double of making one bet on a color for all the bankroll.