Quote:All numbers on a per individual hand basis.
If I'm interpretting that properly, does this mean I can do calculations as if I'm playing a single-line game with that variance? I'm interested in finding the variance for determining risk / bankroll requirements / full-kelly.
Here's an example, perhaps it'll make more sense with this.
5% advantage, 9/6 JOB, $5 single-line or $0.05 100-line. Single line variance is 19.51 and 100-line variance is 214.18. Question is what is minimum BR requirement to play.
Kelly:
(Edge / Variance) * BR = Wager
:.
Wager * Variance / Edge = BR
Initially, I was doing this:
$5 Single line:
($25 * 19.51) / 0.05 = BR = $9,755
This would mean my BR requirement is $9,755 to play the $5 single line 9/6 JOB with a 5% edge.
$0.05 100-line:
($25 * 214.18) / 0.05 = BR = $107,090
This would mean my BR requirement is $107,090 to play $0.05 100-line 9/6 JOB with 5% edge. However, this doesn't quite make sense. So I looked around, sent some PMs, and still couldn't get it figured out. Surely, you wouldn't need that large of a BR to play that game. And most certainly, you wouldn't need a larger bankroll to play a game with less swings.
And I finally re-read the "per individual hand basis" and it clicked -- at least I think so.
So now I'm thinking this is the proper way to do the math:
$5 single line -- same thing.
$0.05 100-line:
($0.25 * 214.18) / 0.05 = BR = $1,070.90
Each hand is $0.05 x 5 = $0.25 per hand (times 100 is $25/spin total).
Meaning my BR requirement is $1070. This makes a whole hell of a lot more sense to me.
So, am I doing the math correctly in the second spoiler?
Granted, I understand using that formula isn't 100% accurate but is a "good estimator":
Quote:. However, for bets with more than one outcome, that can be hard to determine. Most gamblers use advantage/variance as an approximation, which is a very good estimator.
I also understand the whole "that is the MINIMUM bankroll required, meaning if you're below that BR, then you shouldn't play, so you really need a BR above that to actually play" thing as well.
https://wizardofodds.com/gambling/kelly-criterion/
https://wizardofodds.com/games/video-poker/appendix/3/
BR = (Edge / Variance)/wager
?
assuming your original formula is correct
Quote: odiousgambitwhy isn't it:
BR = (Edge / Variance)/wager
?
assuming your original formula is correct
If you divide both sides by Wager, you have to divide both sides by BR -
1/BR = (Edge / Variance) / Wager
1/BR = (Edge) / (Variance * Wager)
BR = (Variance * Wager) / Edge
Quote: odiousgambitwhy isn't it:
BR = (Edge / Variance)/wager
?
assuming your original formula is correct
I don't know why. What's the point?
Quote: RSI don't know why. What's the point?
I'm not really all that boned up on variance but if I had the choice between single line $5 or nickel hundred play I would take the hundred play every time.
Quote: mickeycrimmI'm not really all that boned up on variance but if I had the choice between single line $5 or nickel hundred play I would take the hundred play every time.
And I agree, I'd rather play 100-play than 1-play for the same total amount wagered. Unfortunately, it isn't always that simple. Most of the time nothing will be consistent. Maybe you have a $5 8/5 BP single line or $2 9/6 JOB 5 lines or $0.10 100-line DDB.
So, if someone is able to and knows how to do the math......is the math I did in the OP (in the second spoiler), the correct way to do it?
odious:
Quote: odiousgambitwhy isn't it:
BR = (Edge / Variance)/wager
?
assuming your original formula is correct
You have:
B = (E/V) / W
:.
B = (E/V) / (W/1)
= (E/V) * (1/W)
= (E*1) / (V*W)
= E / (V*W)
The original is W = (E/V)*B or W = (E*B)/V
If your formula is correct, then, replacing my W (E*B/V) into your calculation, when we cross cancel, we should get 1 == 1
B = E / (V*[E*B/V])
B = E / (V*E*B/V)
B = E / (E*B)
B = 1/B
B^2 = 1
B^2 != 1, so your formula, B=(E/V)/W is not correct.
Not that any of this matters, though, as that's not what I'm asking at all.
I don't know how accurate this is but I was told this from a well respected long time VP pro. For ten play, 3 times single line.
Quote: HunterhillI can't answer your variance question but for bankroll requirements a rough guide for 50 play is that you need 7 times what you need for single line. So for a $1 50 play you need 7 times what you need for a single line $1 play.
I don't know how accurate this is but I was told this from a well respected long time VP pro. For ten play, 3 times single line.
Thanks.
Doing the math, according to the way I believe is correct, it says you need 8x for 50-line than you would on 1-line. And 10-line would need about 2x than 1-line. (for the same denom). So this makes sense.
Quote: HunterhillI can't answer your variance question but for bankroll requirements a rough guide for 50 play is that you need 7 times what you need for single line. So for a $1 50 play you need 7 times what you need for a single line $1 play.
I don't know how accurate this is but I was told this from a well respected long time VP pro. For ten play, 3 times single line.
Based on a proper interpretation of the tables, for job the distribution of outcomes is more than 17 times wider for 50 play. But this misses the point. The bankroll requirements depend heavily on your edge on the game. Imagine the extreme case when you get 10% cash back. Then you need a tiny bankroll for both of the two options.
Quote: HunterhillI can't remember what he said for 100 play. What would your numbers say?
Not at my computer anymore, but if I remember it correctly, it was only 10x or 11x.
Ie: single line requires $5,000. 100 line requires $50,000-55,000.