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 07-18-2010, 05:15 AM #1 Extreme Ways adept   Join Date: Jan 2008 Posts: 883 I got an interesting probability question Let's say I have a 100 000 \$ bankroll I decide instead of making a **** ton of money playing poker I want to be stupid and play 0 - 00 roulette I have to either go broke, or get to 200 000 \$ bankroll playing it. I have to bet on red or black and 200\$ every spin. How often am I gonna go broke, and how often am I gonna hit 200 000 \$ and retire?
 07-18-2010, 05:44 AM #2 plaaynde Poker Historian     Join Date: Jan 2010 Location: Local Group Posts: 15,453 Re: I got an interesting probability question You almost certainly go broke. That is because you only put one 500th of the sum you want to double on the table at a time, which gives the house plenty of time to pick you off.
07-18-2010, 06:28 AM   #3
AlbertoKnox
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Re: I got an interesting probability question

Quote:
 Originally Posted by plaaynde You almost certainly go broke. That is because you only put one 500th of the sum you want to double on the table at a time, which gives the house plenty of time to pick you off.
Yup, someone will probably do the math for you, but your odds are close to zero this way.

If you do it all in one go your odds are just under 50%.

07-18-2010, 10:13 AM   #4
Actually Shows Proof

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Re: I got an interesting probability question

Quote:
 Originally Posted by Extreme Ways Let's say I have a 100 000 \$ bankroll I decide instead of making a **** ton of money playing poker I want to be stupid and play 0 - 00 roulette I have to either go broke, or get to 200 000 \$ bankroll playing it. I have to bet on red or black and 200\$ every spin. How often am I gonna go broke, and how often am I gonna hit 200 000 \$ and retire?
Your highest chance to win is to put all your money on a single spin. Then you have a 47.4% chance to get to \$200,000. Any other series of bets will be less, and the more times you bet the lower the chance gets. On average you will lose 5.26% of every bet you place.

The chance of you getting there with \$200 bets that never increase, is essentially zero. The actual chance is 1 over a number that has over 100 digits in it. You'd have a much better chance with some kind of progressive betting (but still almost surely go broke).

Last edited by spadebidder; 07-18-2010 at 10:20 AM.

 07-18-2010, 08:21 PM #5 Geniius journeyman     Join Date: Feb 2010 Location: Dallas Posts: 367 Re: I got an interesting probability question If you want to get to 200k, you will almost certainly go broke, yet the expected values of betting \$200, 500 times, or \$100,000 once, are equal. The higher variance you can tolerate, the more likely it is that you can acheive 200k. Sklansky actually talks about this situation in DUCY, about a casino giving him a \$1000 chip that he must gamble. In this case, given equal expected values, you should choose the most volatile option to maximize long term profits.
07-18-2010, 08:35 PM   #6
statmanhal
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Re: I got an interesting probability question

Quote:
 Originally Posted by Geniius If you want to get to 200k, you will almost certainly go broke, yet the expected values of betting \$200, 500 times, or \$100,000 once, are equal. The higher variance you can tolerate, the more likely it is that you can acheive 200k. Sklansky actually talks about this situation in DUCY, about a casino giving him a \$1000 chip that he must gamble. In this case, given equal expected values, you should choose the most volatile option to maximize long term profits.
Interesting. I posted a thread on the Business forum stating the following:

Suppose you had to invest a fixed amount periodically in a stock fund or money fund such as for a 401 K plan where your company makes a fixed contribution for you every payday. Then if you have two or more alternatives that have the same expected value at any future period, select that which has the highest volatility or variance. This is so simply because of reciprocal arithmetic. When the fund price is low you buy a lot of shares and when it is high you buy fewer. But you get more shares at a low price than the number of shares you lose at a higher price for equal deviations from the mean.

The key of course is that at any time in the future when you want to cash out some or all of your holdings, the expected price for each alternative is the same. In that case, you will have the most shares for the fund with the highest variance.

Anyway, I was blasted by most posters – dollar cost averaging is dumb, etc. etc. Here, we have no choice but to do the averaging – call it Required Periodic Investment-- and if so, then I think I am right.

Here's the link for anyone nterested

https://forumserver.twoplustwo.com/30...rategy-762428/

 07-18-2010, 08:44 PM #7 spadebidder Actually Shows Proof     Join Date: Aug 2008 Location: This looks interesting. Posts: 7,903 Re: I got an interesting probability question We should keep in mind that while the most volatile strategy (which is a single bet in roulette) has the highest chance to double your money, it also has the highest chance to go broke. In poker some of the biggest winners do so playing a high variance game style, but that group also includes the biggest losers. Volatility works both ways, in stocks and in gambling. Standard deviation measures risk.
 07-18-2010, 08:58 PM #8 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 3,931 Re: I got an interesting probability question I agree that risk is directly related to standard deviation. However, in my investment model, I put somewhat of a control on this by having the expected future prices all be equal. So, yes at some time in the future, when you want to cash out you may have to risk having to sell your volatile stock when it is at the low end of its distribution. But, normally, you don't have to sell all at once. And, also, you can use a form of reverse dollar cost averaging and that is to cash in a fixed dollar amount, not fixed number of shares. Anyway, didn't mean to change the thread's subject but the comment about volatility in Sklansky's book was interesting to me.
 07-18-2010, 09:17 PM #9 Geniius journeyman     Join Date: Feb 2010 Location: Dallas Posts: 367 Re: I got an interesting probability question I liken a lot of high-variance gambling to derivatives. When you buy a call option, you have a limited loss at whatever that call option costs. I'm guessing that those who prefer to trade options instead of stocks would also prefer to play roulette on one number than a set of 12 or 18. It's easier to gamble when you are aware that you are a heavy favorite to lose.
07-19-2010, 12:32 AM   #10
Geniius
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Location: Dallas
Posts: 367
Re: I got an interesting probability question

Quote:
 Originally Posted by brotupelo fyp
If this was true, MTTs in poker, parlays in sports betting, and slots jackpots wouldn't be nearly as popular as they are. Variable interval schedules for a possible high payout is what the true addict looks for. "I'm not losing, I'm just due to hit the jackpot!"

07-19-2010, 10:18 AM   #11
BruceZ
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Join Date: Sep 2002
Posts: 11,877
Re: I got an interesting probability question

Quote:
 Originally Posted by Extreme Ways Let's say I have a 100 000 \$ bankroll I decide instead of making a **** ton of money playing poker I want to be stupid and play 0 - 00 roulette I have to either go broke, or get to 200 000 \$ bankroll playing it. I have to bet on red or black and 200\$ every spin. How often am I gonna go broke, and how often am I gonna hit 200 000 \$ and retire?
The probability that you succeed is about 1 in 7.6 x 1022. That is 1 in 76 BILLION TRILLION.

From the classical risk of ruin formula, the probability that we go broke is

(1 - 0.9500) / (1 - 0.91000)

where 0.9 comes from (18/38)/(20/38), the ratio of the probability that we win a spin to the probability that we lose a spin. We seek to win 500 bets, and 1000 is our bankroll plus the number of bets that we seek to win. See this post for my derivation of this classic formula. Take 1 minus the risk of ruin formula derived there, and evaluate from the house's point of view to get the formula above. This is about 1 - (1.32 x 10-23), so our probability of success is about 1.32 x 10-23 or about 1 in 7.6 x 1022.

Last edited by BruceZ; 07-19-2010 at 03:26 PM.

07-19-2010, 11:08 AM   #12
Actually Shows Proof

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Re: I got an interesting probability question

Quote:
 Originally Posted by BruceZ The probability that you succeed is about 1 in 7.6 x 10^22. That is 1 in 76 BILLION TRILLION.
That's a greater chance than I thought with a back-of-a-napkin rough estimate. Thanks for the formulas. In the real world it's still essentially zero, however.

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