Suppose you could beat the casinos, how many buyins would you need for 95% confidence . . .
August 8th, 2013 at 3:34:14 AM
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Okay, Wizard of Odds hive mind. Here's my question:
Suppose you can win 60% of your sessions in a given casino game. And suppose further that the result of each session will take one of only two forms: (a) you'll lose your entire buy-in or (b) you'll double your buy-in and quit. (So, as a hypothetical example, say you always buy in for $200 when playing no-limit hold 'em, and you either lose the entire $200 or walk with exactly $200 in profit.) Now, in this example, how many buy-ins would you need to have, as a "lifetime bankroll," in order to assure yourself a 95% probability of not losing your entire bankroll?
Thanks!
Suppose you can win 60% of your sessions in a given casino game. And suppose further that the result of each session will take one of only two forms: (a) you'll lose your entire buy-in or (b) you'll double your buy-in and quit. (So, as a hypothetical example, say you always buy in for $200 when playing no-limit hold 'em, and you either lose the entire $200 or walk with exactly $200 in profit.) Now, in this example, how many buy-ins would you need to have, as a "lifetime bankroll," in order to assure yourself a 95% probability of not losing your entire bankroll?
Thanks!
August 8th, 2013 at 4:21:37 AM
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Don't mean to sound like a smarta**, but what's the purpose of this question? Thanks.
Fighting BS one post at a time!
August 8th, 2013 at 7:02:59 AM
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Sounds simple to me.
With a 60% success rate, you have a 6.4% chance of three consecutive failures. You have a 2.56% chance of four consecutive failures. So I'd say the answer is $800.
FYI: You have a 1.024% chance of five consecutive failures.
With a 60% success rate, you have a 6.4% chance of three consecutive failures. You have a 2.56% chance of four consecutive failures. So I'd say the answer is $800.
FYI: You have a 1.024% chance of five consecutive failures.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
August 8th, 2013 at 7:09:53 AM
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Quote: DJTeddyBearSounds simple to me.
With a 60% success rate, you have a 6.4% chance of three consecutive failures. You have a 2.56% chance of four consecutive failures. So I'd say the answer is $800.
FYI: You have a 1.024% chance of five consecutive failures.
You need to consider the case of 1 win 5 losses, and all the other sequences which have a 4 more losses than wins.
For example 32 wins and 36 losses is about a 1% probability.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
August 8th, 2013 at 9:11:44 AM
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The formula for Risk of Ruin is e^(-2*WR*BR/(SD^2))
You want a RoR of <= 5%
Your per session WR is 40
Your per session SD is 195.95918
So 0.05 = e^(-2*40*BR/(195.95918^2))
BR ~ $1438
Call it an even $1500. You'll go bust with less than 5% probability.
You want a RoR of <= 5%
Your per session WR is 40
Your per session SD is 195.95918
So 0.05 = e^(-2*40*BR/(195.95918^2))
BR ~ $1438
Call it an even $1500. You'll go bust with less than 5% probability.
August 8th, 2013 at 9:36:23 AM
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I agree too.
This is old school stuff in Poker and Blackjack (forgot to add sports betting)
(only works for positive EVs)
look for a BruceZ archive for a proof (no link you can Google it, I have it in a spreadsheet)
using ev and sd per session (I used 1 for your session buy-in)
ev = 0.2
sd = 0.979795897
[sd = for even money = 2* SQRT(P*(1-P))]
ror = 0.05
Bankroll
B = [-(s.d.)^2 x ln r]/2ev
Excel: = (-(st_dev)^2*LN(ror))/(2*ev)
7.19 units (agrees with above post results)
RoR
r = e^[(-2ev x B)/(s.d.)^2]
Excel: =EXP((-2*ev*bankroll)/(st_dev^2))
use 7 unit Bankroll = 5.4114%
Good Luck
This is old school stuff in Poker and Blackjack (forgot to add sports betting)
(only works for positive EVs)
look for a BruceZ archive for a proof (no link you can Google it, I have it in a spreadsheet)
using ev and sd per session (I used 1 for your session buy-in)
ev = 0.2
sd = 0.979795897
[sd = for even money = 2* SQRT(P*(1-P))]
ror = 0.05
Bankroll
B = [-(s.d.)^2 x ln r]/2ev
Excel: = (-(st_dev)^2*LN(ror))/(2*ev)
7.19 units (agrees with above post results)
RoR
r = e^[(-2ev x B)/(s.d.)^2]
Excel: =EXP((-2*ev*bankroll)/(st_dev^2))
use 7 unit Bankroll = 5.4114%
B units prob ruin
8 0.035673993
9 0.023517746
10 0.015503854
11 0.01022077
12 0.006737947
Good Luck
winsome johnny (not Win some johnny)
August 8th, 2013 at 9:45:09 AM
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So, really you're looking at a infinite series of plays where you have less than a 5% probability of reaching that particular state over a period of time.
Once you've hit that state, you can no longer bet, and the series is over.
My guess is 9 units.
Once you've hit that state, you can no longer bet, and the series is over.
My guess is 9 units.
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You want the truth! You can't handle the truth!
August 8th, 2013 at 9:51:42 AM
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Why do you assume the Standard Deviation is that number?
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You want the truth! You can't handle the truth!
August 8th, 2013 at 10:05:33 AM
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Quote: boymimboWhy do you assume the Standard Deviation is that number?
He doesn't assume.
[(.6)(200)^2 + (.4)(-200)^2 - (40)^2)]^0.5 = 195.96
August 8th, 2013 at 11:38:39 AM
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Quote: boymimboSo, really you're looking at a infinite series of plays where you have less than a 5% probability of reaching that particular state over a period of time.
Once you've hit that state, you can no longer bet, and the series is over.
My guess is 9 units.
I concur. I believe that there is a 4.3025% chance of ruin with a 9 unit bankroll.
I heart Crystal Math.
August 8th, 2013 at 11:45:37 AM
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Doesn't the standard deviation change with the number of trials: ((np(1-p))^.5
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You want the truth! You can't handle the truth!
August 8th, 2013 at 11:54:51 AM
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Does not sound right.Quote: CrystalMathI concur. I believe that there is a 4.3025% chance of ruin with a 9 unit bankroll.
Just using the Gambler's Ruin Formula
9 unit bankroll
.6 = success
to
double a
$9 bankroll
97.46471899%
ruin = 2.53528101%
hit 50 units
97.3987705%
ruin: 2.6012295%
simulation to hit 1,000 units before ruin
97.46%
ruin: 2.54%
close to the 2.3517746 % value returned by the RoR formula
winsome johnny (not Win some johnny)
August 8th, 2013 at 12:25:20 PM
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Quote: 7crapsDoes not sound right.
Just using the Gambler's Ruin Formula
9 unit bankroll
.6 = success
to
double a
$9 bankroll
97.46471899%
ruin = 2.53528101%
hit 50 units
97.3987705%
ruin: 2.6012295%
simulation to hit 1,000 units before ruin
97.46%
ruin: 2.54%
close to the 2.3517746 % value returned by the RoR formula
Yep, I had a mistake. There is a 2.6012% chance of ruin with a 9 unit bankroll (I match exactly your number for hitting 50 units).
I heart Crystal Math.
August 8th, 2013 at 12:32:52 PM
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Quote: boymimboDoesn't the standard deviation change with the number of trials: ((np(1-p))^.5
Yes, but then your WinRate changes too. If you wanted to look at it over, say 4 trials, your SD increases by a factor of 2, but your WR increases by a factor of 4.
In the RoR formula, there is a WR/SD^2 component. So the same ratio is maintained.
