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Comparing gambling systems


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#31 plasmid

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Posted 23 November 2013 - 08:32 AM

I will now prove that calculating a geometric mean does not necessarily lead to an accurate conclusion, by counterexample.

I offer you a game where you pay $1 to have a 99/100 chance of winning $1000 and a 1/100 chance of losing your wager. The geometric mean of the possible outcomes is zero, so it would be foolish to play in such a game if you repeatedly bet your bankroll(?)

You can argue that playing an infinite number of times would guarantee that you lose your wager, but for any sensible number of plays it'd be a no-brainer. I'd even argue that after a Large number of plays, the infinitesimal chance you'll win those games times the payoff if you do win (which is very large as far as large numbers go) would make the game worthwhile.

Edited by plasmid, 23 November 2013 - 08:36 AM.

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#32 bonanova

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Posted 23 November 2013 - 10:47 PM

Each bet of $1 gives me a .99 chance of winning $1000 for an expectation of $990.
990 is the AM of one 0 and ninety nine 1000s.
So I will play the game (method 1) as long as I can stay awake.

There really is no practical way to show a failure for Method 2 other than to say eventually you will hit a zero payoff and lose your $1 bet, although the numbers get very large. Your stake grows by a factor 1000 each time the payoff is not zero. And you have to apply the payoffs 69 times for a 50% chance of getting 0. By then your stake is $10207. Applying the payoffs 916 times reduces the survival chances of your stake (and by then it's $102748) to .01. It's tempting to say you'd play that game multiple times as well. But that's still method 1.

The reality is your stake remains finite and thus vulnerable to being wiped out by a 0 payoff. But the other reality is that so long as n remains finite, getting a 0 payoff is not a certainty. That is, (99/100)n is never zero. The strongest statement I can think of now is that as n increases you have a vanishingly small chance of winning an astronomical amount of money. But when your winnings will equal the number of electrons in the universe, the probability of taking them home will still not be zero. Even though applying the payoffs once per second might take you the age of the universe to get your stake that high.

I didn't buy the (31/30)n argument in previous posts, because I sensed method 1 lurking in the reasoning. It's too easy to automatically think about playing a game multiple times and ending up ahead, vs playing one game forever. Forever is such an impractically long time. So I don't completely buy the current 990n growth rate either. But I look for an expression that goes to zero in the limit, and can't find it.

In my post 30, there also was a small but non-zero chance of being ahead after 2 million pulls.

Bravo. I fall on my sword.


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#33 Prime

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Posted 24 November 2013 - 12:10 AM

When playing indefinitely, you cannot cash in your winnings and spend the money on this side of Infinity. And on that other side, who will care how much you've won, or what fraction of $1 you have remaining?

 

Strange things happen at the infinity:

1. You are expected to end up with nothing, if you ride your winnings infinite number of times.

2. That is because the most likely outcome is the one where each of the six possible roll values will have appeared the same number of times. The probability of that most likely outcome is zero. (That requires a proof though. See the spoiler.)

3. Your average payoff after infinite number of rolls should be infinite, if you bet just $1 on every roll.

4. Your average payoff if you ride your winnings is going to be even more infinite. In fact, it is going to be infinitely greater than that from the previous point (3). That is because when riding your winnings, the infinite variable is an exponent whereas when betting a fixed amount on each roll the infinite variable is a multiplier.

5. To see your average payoff, you must play infinite number of sets of infinite number of rolls each. You need not live forever to go on that gambling binge. Just roll an infinite number of dice every time and roll them infinitely fast.

 

Spoiler for The likelihood of the median outcome

 

Coming back from the Infinity, as I said here before, for all practical purposes, it is impossible to lose in this game. And you must ride your winnings, if you want to win big.

The Expected Value (average payoff) for a 1000 consecutive rolls is (1.0333...)1000 = 1.74*1014. You cannot run enough experiments to confirm it empirically. Nonetheless, the winnings are good.

In order to help its patrons to win large sums of money, the casino added a small modification to the game. Namely, the minimum bet requirement of $1. So while rolling the dice and riding your winnings, if your total falls below $1, you must add enough cents to make your next bet at least $1.

Spoiler for EMPIRICAL RESULTS from Casino Games

 

To sum up:

In this game having a bankroll of $20, and in reasonable amount of gambling time (few days)

1) If you ride your winnings, you most likely end up with the winnings of millions of $$.

2). When betting $1 at a time, in the same amount of time, you'll most certainly walk away with few hundred of $$. (Frankly, not fun and not worth your time.)


Edited by Prime, 24 November 2013 - 12:19 AM.

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Past prime, actually.


#34 plasmid

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Posted 24 November 2013 - 04:25 AM

Now's my turn to disagree with Prime.
If you ride your winnings, the most likely outcome is not that you will win millions. The most likely outcome is that you will lose money. It's just that your payoff if you do win is much larger than your odds of losing, like a favorable lottery.

The mean outcome of always letting everything ride is greater than your initial $1 entry wager.
The median outcome of always letting everything ride is less than your initial $1.
If there's doubt about this, it could be tested by simulating a bunch of runs and calculating both the mean and median.

Which of those two, the mean or median, is "most important" is a philosophical argument.
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#35 Prime

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Posted 24 November 2013 - 09:08 AM

Now's my turn to disagree with Prime.
If you ride your winnings, the most likely outcome is not that you will win millions. The most likely outcome is that you will lose money. It's just that your payoff if you do win is much larger than your odds of losing, like a favorable lottery.

The mean outcome of always letting everything ride is greater than your initial $1 entry wager.
The median outcome of always letting everything ride is less than your initial $1.
If there's doubt about this, it could be tested by simulating a bunch of runs and calculating both the mean and median.

Which of those two, the mean or median, is "most important" is a philosophical argument.

But I have run a bunch of simulations. And some numeric analysis as well. Enough to convince myself. See my posts 12, 20, 22, and 33 inside the spoilers.

Riding your winnings for 1000 rolls, seems like a good choice. That won millions in my simulations in very short order of time.

 

At this point I feel that for further discussion, we need to solve analytically what is probability of winning X $$ or more while riding for N consecutive rolls. E.g., $1,000,000 or more after 1000 rolls. For that one must find exact number of 1000-roll variations yielding more than 1,000,000 and divide that number by 61000.

I don't feel up to the task at the moment. But I can provide any such data for up to 12 rolls.


Edited by Prime, 24 November 2013 - 09:09 AM.

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Past prime, actually.


#36 BMAD

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Posted 24 November 2013 - 03:33 PM

It always amazes me when people can take what I thought to be a simple fun question and derive much complicated analysis out of it. Surprisingly this question came from a discussion with my 7 year old child.

Edited by BMAD, 24 November 2013 - 03:37 PM.

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#37 plasmid

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Posted 24 November 2013 - 04:20 PM

Oh sorry, I didn't see that you already had the spoiler in post 22 showing that there's a <50% chance of ending up with $1 or more after repeatedly letting everything ride. So that shows that the median and most likely outcome is in fact a loss.
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#38 Rainman

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Posted 24 November 2013 - 04:22 PM

Prime, if you have a bankroll of $20 and can run method 2 twenty times, then of course you're going to win. No one here is arguing against that. But method 2 in OP does not say anything about you being able to reset your stake at $1 whenever you feel like it. You get one shot. One. Shot. For all purposes your bankroll is $1. If you lose you lose. The most likely result is you will lose. Your simulations all show that. If you run 10 simulations of 1000 rolls each and 6 of them are losing, it means 6/10 people will lose. The other four people will win. As N grows larger the probability of winning approaches 0, as I have proven mathematically. Which means, if all 7,000,000,000 people in the world run method 2 for long enough, the probability approaches 1 that they will all lose.
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#39 Prime

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Posted 25 November 2013 - 01:06 AM

It always amazes me when people can take what I thought to be a simple fun question and derive much complicated analysis out of it. Surprisingly this question came from a discussion with my 7 year old child.

 

What makes these topics so interesting to me, is my lack of education. Math education in particular.

 

Oh sorry, I didn't see that you already had the spoiler in post 22 showing that there's a <50% chance of ending up with $1 or more after repeatedly letting everything ride. So that shows that the median and most likely outcome is in fact a loss.

 

Perusing your argument from the post 24, imagine a casino offered you a choice between two games, 1000 dice roll each:

1) Stake $1 with a 5% chance of winning $1,000,000 or more; OR

2) Deposit $300 with a 50% chance of winning something between $16 and $50, while in the extremely unlikely cases losing your entire $300 or winning $500.

 

Which game would you play? (The percentages here are illustration -- not an actual calculation.)

 

Prime, if you have a bankroll of $20 and can run method 2 twenty times, then of course you're going to win. No one here is arguing against that. But method 2 in OP does not say anything about you being able to reset your stake at $1 whenever you feel like it. You get one shot. One. Shot. For all purposes your bankroll is $1. If you lose you lose. The most likely result is you will lose. Your simulations all show that. If you run 10 simulations of 1000 rolls each and 6 of them are losing, it means 6/10 people will lose. The other four people will win. As N grows larger the probability of winning approaches 0, as I have proven mathematically. Which means, if all 7,000,000,000 people in the world run method 2 for long enough, the probability approaches 1 that they will all lose.

 

I think, what we are arguing here is called Normal Distribution, or Gaussian Distribution, or Bell Curve, or some other such name.

Spoiler for NORMAL DISTRIBUTION

 

Spoiler for MORE NUMERIC ANALYSIS AND EMPIRICAL DATA

In view of the above, I think, the wining likelihood of a 1000-roll ride is being underestimated.


Edited by Prime, 25 November 2013 - 01:10 AM.

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Past prime, actually.


#40 Rainman

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Posted 25 November 2013 - 03:14 AM


In other words, even if you show that the probability of the ratio of variants within any given interval around the Median tends to 1, you'd still have to show that the resulting Payoff within that interval is zero.

 

Done that. Twice. If the ratios are all within the interval 0.1666<x<0.1667 then the resulting payoff is always less than 1, and it tends to zero as N tends to infinity. An upper bound for the payoff is: Payoff < (0.70.1666*0.80.1666*0.90.1666*1.10.1667*1.20.1667*1.50.1667)N < 0.9998N < 1. In plain words: over time, one hundred percent of the variations will be close enough to the median variation, that their payoff is zero.

 

To understand why this matters, do not concern yourself with the nominal values of the stakes and payoffs, but rather with their effect on your bankroll.

If your bankroll is $1, and you bet $1 and hit 0.7, your bankroll is reduced from 1 to 0.7, a change by the factor 0.7. Here the nominal change equals the bankroll change.

However, if your bankroll had been $2 instead, and you bet $1 and hit 0.7, your bankroll would be reduced from 2 to 1.7, a change by the factor 0.85. Because you only bet half your bankroll, the change is not as dramatic. If you bet half your bankroll instead of your entire bankroll, the payoffs would result in a change of 0.85, 0.9, 0.95, 1.05, 1.1, or 1.25. The geometrical mean of these changes is (0.85*0.9*0.95*1.05*1.1*1.25)1/6 ~ 1.008, which means you now have a positive expected effect on your bankroll. If you bet half your bankroll every time, you can expect it to grow exponentially (1.008N).

 

Now we might ask, when is the inflection point between a good game and a bad game? How much of your bankroll can you bet before you expect to lose money?

Well, if you bet the fraction x of your bankroll, the payoffs would be 0.7x, 0.8x, 0.9x, 1.1x, 1.2x, and 1.5x. Adding the 1-x you have remaining on the side, the change in bankroll would be 1-0.3x, 1-0.2x, 1-0.1x, 1+0.1x, 1+0.2x, or 1+0.5x. The geometrical mean would be GM = ((1-0.3x)(1-0.2x)(1-0.1x)(1+0.1x)(1+0.2x)(1+0.5x))1/6, which can be solved for x, at GM=1. The solution in the interval 0<x<1 is a bit over 0.989, which means the game is good for you as long as you bet between 0% and 98,9% of your bankroll.

 

But why am I not concerned with the nominal values of the winnings, but rather their effect on my bankroll? Consider this: if you had a $1,000,000 total fortune (this includes your house and valuables), would you risk it all for a 50% chance to triple it? Would you risk it all for an 80% chance to double it? Would you risk $999,000 of it for a 60% chance to double it? Very few people would be silly enough to take those risks, even though the EV is very good. The true difference between having one million and being dead broke is a lot more dramatic than the difference between having two millions and having one million, but the nominal difference is the same.


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