Randomly Varying Bets at +TC

[blackjack avenger]

I Am With the Rhino

Again, I can't prove it with "math", but I think this causes problems.

brass tacks: betting directly proportional to your advantage is the gold standard for betting. Any deviation from that is going to have negative consequences, on win rate, or on variance, or both.

Let's say at any given TC, let's say TC +3, you're not betting the exactly proportional amount, but instead a "dispersion cloud" of possible bets, which center on that exact number. That means that sometimes you're going to be underbetting (decreasing win rate), and sometimes you're going to be over betting (increasing variance), and sometimes you're going to have it nailed on the head.

It's sounds like the effects of other forms of cover (like only increasing after a win / decreasing after a loss, etc).

Yep

 

[Unshake]

Again, I can't prove it with "math", but I think this causes problems.

brass tacks: betting directly proportional to your advantage is the gold standard for betting. Any deviation from that is going to have negative consequences, on win rate, or on variance, or both.

Let's say at any given TC, let's say TC +3, you're not betting the exactly proportional amount, but instead a "dispersion cloud" of possible bets, which center on that exact number. That means that sometimes you're going to be underbetting (decreasing win rate), and sometimes you're going to be over betting (increasing variance), and sometimes you're going to have it nailed on the head.

It's sounds like the effects of other forms of cover (like only increasing after a win / decreasing after a loss, etc).

I didn't say there was an advantage, if you read Zen's post that I quoted he said from a true count of -1 to +1. At these counts you are (more than likely ) playing either at a disadvantage or a very close to dead even game. <-- Obviously depending on the rules but we'll say the rules give a .5% house edge.

Varying bets here, while maintaining the same average bet, should not have (at least in the long run) a significant change (or any at all) in RoR.

 

[Kasi]

Again, I can't prove it with "math", but I think this causes problems.
brass tacks: betting directly proportional to your advantage is the gold standard for betting. Any deviation from that is going to have negative consequences, on win rate, or on variance, or both.

If Wise Frog hadn't just made me cry tonite, I'd give you my version of a math proof. Which, of course, as alot of my math proofs, doesn't mean it's worth crappola lol.

But, basically, if you encounter the same +TC 3 times, and, in the first case, bet 1 unit, 2 units and 3 units and, in the second case, bet 2 units the same 3 times, I think your EV is the same and your total standard dev in dollars would be the same over the same 3 hands.

Doesn't matter whether your optimally betting or not.

Also I hope you don't mean by "betting directly proportional to your advantage" you mean you would bet $100 with a $10K roll with a 1% advantage. That is not an optimal Kelly bet.



That's what I still think tonite anyway lol.

 

[sagefr0g]

If Wise Frog hadn't just made me cry tonite, I'd give you my version of a math proof. ...

<snicker>

 

[QFIT]

But, basically, if you encounter the same +TC 3 times, and, in the first case, bet 1 unit, 2 units and 3 units and, in the second case, bet 2 units the same 3 times, I think your EV is the same and your total standard dev in dollars would be the same over the same 3 hands.

If squares weren't involved in the calculation of variance, that would be true. But squares are involved. Therefore in the end, risk increases. As an example, suppose at TC +1 you bet 100 hands at 2 units. Or, you bet 99 hands at 1 unit and 1 hand at 101 units. The average bet is the same. But one method is far more risky.

 

[Kasi]

If squares weren't involved in the calculation of variance, that would be true. But squares are involved. Therefore in the end, risk increases. As an example, suppose at TC +1 you bet 100 hands at 2 units. Or, you bet 99 hands at 1 unit and 1 hand at 101 units. The average bet is the same. But one method is far more risky.

Well what if one always rotated betting 1 unit, 2 units and 3 units at the exact same TC compared to betting 2 units 3 times at the exact same count?

Anyway that's what I was trying to say implying the frequency of the lower bets would be the same as the frequency of the higher bets and middle bet.

Would then betting 1,2,3 be the same as betting 2,2,2?

Anyway - thanks for your help - I'm here to learn too.

 

[EasyRhino]

Let's look at a simplified example. Only TWO bets.

Incidentally, this relies on the gigantic assumption that I actually know how to compute variance and standard deviation correctly. But if I understand this googled page (Archive copy) correctly, I think I got it. (basically, you takel of the individual deviations from the average, then square them, then add them up, then divide by the number of samples).

You can choose 2,2, or 1,3. Same average bet, and the EV is 0, right?

Here are the possible outcomes for 2,2:
Win 1st, Win 2nd = +4 -> squared = 16
Win 1st, Lose 2nd = 0 -> squared = 0
Lose 1st, Win 2nd = 0 -> squared = 0
Lose 1st, Lose 2nd = -4 -> squared = 16

Sum of squares = 32
Divide by 4 samples, variance is 8
Standard deviation is 2.82

Here are the possible outcomes for 1,3
Win 1st, Win 2nd = +4 -> squared = 16
Win 1st, Lose 2nd = -2 -> squared = 4
Lost 1st, Win 2nd = +2 -> squared = 4
Lose 1st, Lose 2nd = -4 -> squared = 16

Sum of squares = 40
Divide by 4, variance is 10
Standard deviation is 3.16

Therefore, you just increased variance with no corresponding increase in EV. Congratulations, you just increased your risk of ruin.

BTW, I'm not saying this entire idea is a bad one. It just has a cost (like most any cover play), and that cost is variance.

 

[Kasi]

Incidentally, this relies on the gigantic assumption that I actually know how to compute variance and standard deviation correctly.

I struggle all the time with it. Obviously

I just thought if you bet the 123 thing compared to a 2 2 2 thing that each hand would have been played the same number of times and the total variance over the 3 hands would be the same and the total SD would be the same. But I suppose I'm full of it lol. Not the first time. Either way, glad the thread continued to Blue's original question.

So probably, whether you're doing it right or not, I guess your instincts are good! And that ain't bad!

Tomorrow I'll give you a good laugh at my version lol.

And, hey, if you can take a stab at this, and I think it's great you tried, whether right or wrong - that's how I started too - maybe take a stab at a session of BJ results lol!

 

[QFIT]

Let's look at a simplified example....

Close. But you need to treat the two hands separately squaring the results separately. So it's +2 and +2 for 4 and 4 summing to 8 compared with +1 and +3 for 1 and 9 summing to 10. That's for a coin flip problem which is enough to illustrate the problem. Of course for BJ, you need to include splits, doubles, BJs, Ins and Surr and the correct proportions of each possible result. This increases the difference.

 

[vonQuux]

Therefore, you just increased variance with no corresponding increase in EV. Congratulations, you just increased your risk of ruin.

Right, but I acknowledged this as the cost of providing cover in the first post:

Of course this will have the undesired effect of amplifying standard deviation like a b*tch.

vQ

 

[EasyRhino]

Close. But you need to treat the two hands separately squaring the results separately.

Oh, so I lucked into the right answer, with the wrong methodology. My algebra teacher would kill me. Thanks.

standard deviation like a b*tch

Missed the metaphor, my bad

 

[Unshake]

Let's look at a simplified example. Only TWO bets.

You can choose 2,2, or 1,3. Same average bet, and the EV is 0, right?

Here are the possible outcomes for 2,2:
Win 1st, Win 2nd = +4 -> squared = 16
Win 1st, Lose 2nd = 0 -> squared = 0
Lose 1st, Win 2nd = 0 -> squared = 0
Lose 1st, Lose 2nd = -4 -> squared = 16

Sum of squares = 32
Divide by 4 samples, variance is 8
Standard deviation is 2.82

Here are the possible outcomes for 1,3
Win 1st, Win 2nd = +4 -> squared = 16
Win 1st, Lose 2nd = -2 -> squared = 4
Lost 1st, Win 2nd = +2 -> squared = 4
Lose 1st, Lose 2nd = -4 -> squared = 16

Sum of squares = 40
Divide by 4, variance is 10
Standard deviation is 3.16

Therefore, you just increased variance with no corresponding increase in EV. Congratulations, you just increased your risk of ruin.

I guess this is how I originally saw it. I saw two scenarios:


Scenario one ==> Player 1 bets 15 units

Two Possible Outcomes:

Player 1 wins => +15 ---- Squared = 225
Player 2 loses => -15 ---- Squared = 225

Sum of squares = 550
Divide by number of outcomes= 225 (Variance)
Square root = 15 (Standard Deviation)


Scenario Two ==> Player 2 bets 10 or 20 units


While betting 10 units (50% of time) two possible outcomes:

Player 2 wins => +10 units ---- Squared = 100
Player 2 loses => -10 units ---- Squared = 100


Sum of squares = 200
Divide by number of outcomes= 100 (Variance)
Square root = 10 (Standard Deviation)


However when betting 20 units (50% of time) Player 2 experiences:

Player 2 wins => + 20 units ---- Squared = 400
Player 2 loses => -20 units ---- Squared = 400

Sum of squares = 800
Divide by number of outcomes= 400 (Variance)
Square root = 20 (Standard Deviation)

Since Player 2 experiences a standard deviation of 10 50% of the time and 20 50% of the time, wouldn't his average standard deviation be 15 (the same as player 1)? I've never really taken courses on statistics, so this is more than likely wrong...Does grouping both the 10 unit bet and the 20 unit (or in your case 1 and 3) bet make sense since they aren't related or placed at the same time?

In this case your risk of ruin does not change because your standard deviation (or more commonly per 100 hands) would remain the same. Obviously this is assuming a coin flip situation where the odds of winning are just as likely as losing.

 

[vonQuux]

Since Player 2 experiences a standard deviation of 10 50% of the time and 20 50% of the time, wouldn't his average standard deviation be 15 (the same as player 1)? I've never really taken courses on statistics, so this is more than likely wrong...Does grouping both the 10 unit bet and the 20 unit (or in your case 1 and 3) bet make sense since they aren't related or placed at the same time?

I think QFIT is right in that "outliers" (numbers further away from the central body of distribution) are amplified because they're squared.

The result being that not only would standard deviation (and thus RoR) be amplified, they'd be amplified quite heavily. To the point of it not being worth the cover unless you have a massive bankroll which, unfortunately, I do not. =)

vQ

 

[QFIT]

Since Player 2 experiences a standard deviation of 10 50% of the time and 20 50% of the time, wouldn't his average standard deviation be 15 (the same as player 1)?

You can’t average standard deviations. It would be sort of like averaging risk. Say sitting at home has a near zero risk of death and jumping off the Empire State Building has a near 100% chance of death. If you do both, do you average 100 and 0 and get a 50% chance of death? Nope, you die.

 

[EasyRhino]

Incidentally, I've dimly read that statistical risk under-measuring is part of the recent problems in the stock and credit markets. you know how you read news bites of people saying that credit activity are "once in 1,000 year" events? Well, those assumptions may be assuming that the distribution of chance in the stock market is a normal distribution, like you'd have in, say, a card game. However, it seems that the markets distributions tend to be less-than-normal, with more extreme events on either side. Which leads to hedge funds imploding, banks collapsing, etc.

Bear Stearns probably though they could randomly vary bets without any consequence as well.

 

[vonQuux]

You can’t average standard deviations. It would be sort of like averaging risk. Say sitting at home has a near zero risk of death and jumping off the Empire State Building has a near 100% chance of death. If you do both, do you average 100 and 0 and get a 50% chance of death? Nope, you die.

That's a hell of a good way to illustrate the point.

vQ

 

[Unshake]

You can’t average standard deviations. It would be sort of like averaging risk. Say sitting at home has a near zero risk of death and jumping off the Empire State Building has a near 100% chance of death. If you do both, do you average 100 and 0 and get a 50% chance of death? Nope, you die.

To me that example sounds like you are averaging the probabilities of two events isn't that different? For example if you have a 4/52 of chance of drawing an ace and a 13/52 chance of drawing a spade. The probability of drawing of an ace OR spade would be:

(chance of drawing ace + chance of drawing spade) - (chance of drawing ace of spades or overlap) ... or (4/52 + 13/52) - (1/52) = 16/52


For your example, the death one would be:
(chance of dying sitting + chance of dying jumping off building) - (chance of dying from both sitting and jumping or overlap)...or (100% + 0%) - (0% <-- is 0 because you could never die from sitting) = 100% so your chance of dying is 100%.

Averaging the odds, like you pointed out, is obviously flawed. On a side note I probably don't know nearly enough about statistics to be trying to argue anything :grin: Anyways, that being said maybe I should just take your word for it.

 

[Sonny]

Since Player 2 experiences a standard deviation of 10 50% of the time and 20 50% of the time, wouldn't his average standard deviation be 15 (the same as player 1)?

As Qfit said, SDs are not additive so you can’t really average them together. However, variance is additive so we can work with that. In you post you showed that the variance was 100 for betting 10 units and 400 for betting 20 units. That makes sense since in an even money game we would expect the SD to be 1 unit ($10 and $20 respectively). Your example used dollars instead of units but that makes no difference. The answers are still correct.

So to find the SD for Player 2 we just take the square root of the average variance:

Sqrt((100+400)/2)
= sqrt(250)
= 15.81

That’s almost $1 higher than Player 1. The difference is fairly small because the two players are using very similar bets, but the difference is there. If the bet sizes were much different, say $100 vs. $25/$175 then the variance would increase much more. In either case the EV has not changed but the risk has increased by betting more erratically.

-Sonny-

 

[Kasi]

As Qfit said, SDs are not additive so you can’t really average them together. -Sonny-

Thanks Sonny. And Qfit. That's exactly what I did wrong,
Examples always help me lol.

 

[bluewhale]

You can’t average standard deviations. It would be sort of like averaging risk. Say sitting at home has a near zero risk of death and jumping off the Empire State Building has a near 100% chance of death. If you do both, do you average 100 and 0 and get a 50% chance of death? Nope, you die.

thanks for explaining to people that variance goes up. i'm very good at math and as soon as i thought of the random bet thing i realized the implications (increased variance and in turn RoR).

So now back to my original question. Is this increase in variance worth it? I'll answer it for what seems like the majority of posters here: NO! all you red chippers who are betting less than 100 max its a clear no... stick to optimal betsizes as best you can. I wouldn't even consider this strategy until you're well over 20k.

However i mentioned that we are already rolling in a pretty huge hourly and if we adopt this strategy our overall ramp would decrease in response to the increased variance. It wouldn't be a massive decrease. Something like an hourly of 100 going down to 90 is a tradeoff i'm willing to accept for the lower risk of ban (ROB?).

what we've all but decided to do is vary our bets a lot at TC 1 (upto$100 more)... and for all other ramp bets each counter can vary + or - $50 of the actual bet he is supposed to make.