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Comparing gambling systems


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49 replies to this topic

#11 bonanova

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Posted 20 November 2013 - 07:58 PM

Consider a gambling machine A. When you put in $X and pull the handle, it will spit out (equally likely) either $0.7*X, $0.8*X, $0.9*X, $1.1*X, $1.2*X, or $1.5*X.
 
Now consider the following two ways of playing this machine:
 
Put in $1, pull the handle, and keep whatever you get. Repeat.
Initially, put in $1. Pull the handle, then put in whatever you get. Repeat.
Can you win money with this machine? Which is the better way to play? How can this be?



The first two answers have been given (post 2) and verified (post 5) by a 100,000-pull simulation.

The discussion question elicited several thoughts (posts 5 7 8) that made the answers intuitive.

Finally, the answers can be changed (post 7) by changing the payoffs.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#12 Prime

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Posted 20 November 2013 - 08:26 PM

Spoiler for Not convinced

The six equally likely payoffs are .7 .8 .9 1.1 1.2 1.5.
Their arithmetic mean AM is 1.033333 ...
Their geometric mean GM is 0.9996530325.
 
In Method 1 you pay $1.00 for each pull and remove your winnings.
In Method 2 you pay $1.00 once and keep your stake at risk. Thus,
 
AM (>1) gives you the expected return per dollar per pull for Method 1. (Winnings are added.)
GM (<1) gives you the expected return per dollar per pull for Method 2. (Winnings are multiplied.)
 
Post 3 paid $1.00, pulled the handle twice and then removed the winnings.
This was done 36 times with the results averaged. This is not Method 2.

That's not what I said in Post 3.
 
It's actually Method 1 for a new set of payoffs: the 36 pairwise products of the original six payoffs.

No, it is not Method 1. On the second turn the entire bankroll is staked -- not the original betting amount.
This is a winning game, with an average return on a dollar bet of AM2.
AM2 is in fact the arithmetic mean (AM) of the new set of payoffs.
So this is one more example of a winning Method 1 result.
 
To test Method 2 for 72 pulls of the handle, you bet $1.00 once and keep the result at risk for all 72 pulls.
That final result is the just product of the 36 pairwise products, which is 0.9753235719 = GM72.
And that is the expected result for Method 2: each pull multiplies your stake by GM.
 
Again, why does Method 1 win and Method 2 lose?
Because AM applies to added winnings and AM>1; GM applies to multiplied winnings and GM<1.
 
Simulation: Pull the handle 100,000 times with random results:


  • Method 1:
    Average return on each $1.00 bet is 1.032955 <--> AM = 1.0333 ...
  • Method 2:
    Stake goes from $1.00 to $1.414863x10-31 = GM100000.
    The 100,000th root of 1.414863x10-31 is 0.9992899. <--> GM = 0.9996530325.

 

 
There is an interesting thought provoking point there, presenting a kind of pseudo-paradox in this problem.
I stand firmly by my solution in the post 3, though I am beginning to have some doubts about its straightforwardness property. The illustration I gave in support of the solution was not a proof. And since bonanova has misinterpreted my illustration, it must have been unintelligible. (It happens to me from time to time. In my head a statement I make seems precise and clear, but other people can't make out the ends of it.)
So in the interest of befuddlement, perplexity, and creative thought...

Spoiler for mystification... and proof

For anyone who is still not convinced, I am ready to play the game with a fair die, staking my entire bankroll on each roll of the die. (To speed up the process, we could roll 12 dice at a time.)


Edited by Prime, 20 November 2013 - 08:34 PM.

  • 0

Past prime, actually.


#13 bonanova

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Posted 20 November 2013 - 08:39 PM

Spoiler for what is expected


Do you need an infinite stake to play Method 1, as you would in a Martingale scenario?
Can't you feed in dollar bills until your stake exceeds a certain amount, then play with house money?

After your stake reaches $2 you're basically starting over, with a free dollar to use.
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#14 bonanova

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Posted 20 November 2013 - 09:00 PM

 

Spoiler for Not convinced
The six equally likely payoffs are .7 .8 .9 1.1 1.2 1.5.Their arithmetic mean AM is 1.033333 ...Their geometric mean GM is 0.9996530325.
 In Method 1 you pay $1.00 for each pull and remove your winnings.In Method 2 you pay $1.00 once and keep your stake at risk. Thus,
 AM (>1) gives you the expected return per dollar per pull for Method 1. (Winnings are added.)GM (<1) gives you the expected return per dollar per pull for Method 2. (Winnings are multiplied.)
 Post 3 paid $1.00, pulled the handle twice and then removed the winnings.
This was done 36 times with the results averaged. This is not Method 2.

That's not what I said in Post 3.
 It's actually Method 1 for a new set of payoffs: the 36 pairwise products of the original six payoffs.
No, it is not Method 1. On the second turn the entire bankroll is staked -- not the original betting amount.This is a winning game, with an average return on a dollar bet of AM2.AM2 is in fact the arithmetic mean (AM) of the new set of payoffs.So this is one more example of a winning Method 1 result.
 
To test Method 2 for 72 pulls of the handle, you bet $1.00 once and keep the result at risk for all 72 pulls.That final result is the just product of the 36 pairwise products, which is 0.9753235719 = GM72.And that is the expected result for Method 2: each pull multiplies your stake by GM.
 Again, why does Method 1 win and Method 2 lose? Because AM applies to added winnings and AM>1; GM applies to multiplied winnings and GM<1.
 Simulation: Pull the handle 100,000 times with random results:

  •  
  • Method 1:
    Average return on each $1.00 bet is 1.032955 <--> AM = 1.0333 ...
  • Method 2:
    Stake goes from $1.00 to $1.414863x10-31 = GM100000.
    The 100,000th root of 1.414863x10-31 is 0.9992899. <--> GM = 0.9996530325.
 
There is an interesting thought provoking point there, presenting a kind of pseudo-paradox in this problem.I stand firmly by my solution in the post 3, though I am beginning to have some doubts about its straightforwardness property. The illustration I gave in support of the solution was not a proof. And since bonanova has misinterpreted my illustration, it must have been unintelligible. (It happens to me from time to time. In my head a statement I make seems precise and clear, but other people can't make out the ends of it.)
So in the interest of befuddlement, perplexity, and creative thought...
Spoiler for mystification... and proof

For anyone who is still not convinced, I am ready to play the game with a fair die, staking my entire bankroll on each roll of the die. (To speed up the process, we could roll 12 dice at a time.)

It seemed clear what I implied about the pairwise case, but maybe it wasn't.

On the second pull the entire stake is at risk. Correct.
You subject your $1 to two pulls rather than to a single pull.
What I did not say explicitly was that's equivalent to a single pull with a payoff equal to the product of the two payoffs.
Then after the second pull you withdraw your winnings. They are never again at risk.

What I said was that is equivalent to method 1, just using a different set of payoffs.

You had 72 pulls, but you didn't bet just $1. You bet $36.
To represent method 2, you can't put into play a fresh $1, and take out your winnings, every other pull
You must leave your stake at risk for all 72 pulls.
If you do, eventually the .7s will bring you down.


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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#15 Prime

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Posted 20 November 2013 - 09:07 PM

Spoiler for what is expected

Yes, there it is -- Expected Value vs. Expected Outcome.

I fancy myself as an avid gambler. And in practice I would choose the mix of the two methods, as Rainman has suggested here. (Playing infinite number of times is rather impractical.)

I have run an experiment playing 422 consecutive rolls 10 times with the following results: 

4 times I had less than $0.01 left of my initial $1 bankroll.

3 times I had between $0.02 and $0.35 left.

3 times I won with the largest win of $83.62.

That left me very much ahead (more than $80, while putting at risk just $10).

The new question here is, what is the optimal string of consecutive rolls?

3 times I


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Past prime, actually.


#16 Prime

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Posted 20 November 2013 - 09:17 PM

....


On the second pull the entire stake is at risk. Correct.
You subject your $1 to two pulls rather than to a single pull.
.....

 

I was not implying any actual "pulls" in my illustration. Nor did I say anything about withdrawing or adding any amounts to the stake.

I did not mean it as an actual trial in the game. I meant it as the enumeration of all possible variations with their respective probabilities.


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Past prime, actually.


#17 Rainman

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Posted 20 November 2013 - 10:26 PM

 

Spoiler for what is expected

Yes, there it is -- Expected Value vs. Expected Outcome.

I fancy myself as an avid gambler. And in practice I would choose the mix of the two methods, as Rainman has suggested here. (Playing infinite number of times is rather impractical.)

I have run an experiment playing 422 consecutive rolls 10 times with the following results: 

4 times I had less than $0.01 left of my initial $1 bankroll.

3 times I had between $0.02 and $0.35 left.

3 times I won with the largest win of $83.62.

That left me very much ahead (more than $80, while putting at risk just $10).

The new question here is, what is the optimal string of consecutive rolls?

3 times I

 

It seems your post was cut off unfinished.

 

Expected outcome is the true value for any gambler who seeks to eliminate the "gambling" part. The probability is 1 that your own outcome approaches the expected outcome as the number of games approaches infinity. So in the long run, you are practically guaranteed to win if your expected outcome is positive. The same is not true for expected value.

 

Expected value is too influenced by the extremely high payoffs in the extremely unlikely variations. In this case, if you play for example 100 times, the extremely lucky variation where you hit 1.5 every time would yield a net payoff of roughly +$406,561,177,535,215,236. On the other hand, the extremely unlucky variation where you hit 0.7 every time would yield a net payoff of roughly -$1. The average net payoff, or expected value, would be (31/30)100-1, or roughly $27. So the variance (actual payoff - average payoff) is 406,561,177,535,215,209 for the luckiest variation and only -28 for the unluckiest variation. As follows, the expected value is extremely tilted by the high variance from impossibly lucky scenarios. You would need to be insanely lucky just to get anywhere close to the EV.

 

Your own experiments illustrate this perfectly. The average net payoff for running 422 consecutive games 10 times is 10*(31/30)422-1, or roughly $10,220,338. Your actual net payoff was just more than $80, falling way short of the EV. Had you kept your entire bankroll running for all 4220 games, you would have lost almost everything. This was not a case of bad luck, but rather quite expected.

 

Had you instead bet half your bankroll every time, with a starting bankroll of the same $10, for 4220 games, your expected outcome would have been ~10*1.049244220/6, or +$4,800,000,000,000,000. Feel free to simulate it, you will end up somewhere around that number. Your EV would of course be even higher, 10*(61/60)4220-1 or roughly +$2,000,000,000,000,000,000,000,000,000,000.


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#18 bonanova

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Posted 20 November 2013 - 11:51 PM

 

....


On the second pull the entire stake is at risk. Correct.
You subject your $1 to two pulls rather than to a single pull.
.....

 

I was not implying any actual "pulls" in my illustration. Nor did I say anything about withdrawing or adding any amounts to the stake.

I did not mean it as an actual trial in the game. I meant it as the enumeration of all possible variations with their respective probabilities.

 

 

I misunderstood.

I inferred to to be an assertion that Method 2 was a winning strategy.


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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#19 Rainman

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Posted 21 November 2013 - 12:29 AM

 

Spoiler for what is expected


Do you need an infinite stake to play Method 1, as you would in a Martingale scenario?
Can't you feed in dollar bills until your stake exceeds a certain amount, then play with house money?

After your stake reaches $2 you're basically starting over, with a free dollar to use.

 

What I meant is, if your bankroll is finite, there is a risk that you will run out of money and be unable to continue with method 1. You could hit a freak streak of consecutive losses. The probability is non-zero so it has to be accounted for, when calculating expected outcome in the long run. With method 1, the expected outcome for N games equals N*31/30 only because your bankroll is assumed to be infinite. With a finite bankroll B, the expected outcome for N games would equal N*31/30 for small values of N, but once B-0.3N<1 the expected outcome would drop below N*31/30. Trying to calculate it exactly with respect to both B and N would be way too much for my brain, and I don't think the formula would be pretty.

 

Your point is valid though, we might still have a positive expected outcome for method 1, with a large enough bankroll. It could be tested with computer simulations, but I don't have the programming skills to do those. Besides, the expected outcome is at most N*31/30, which is dwarfed by the ~1.049N/6 you would get by betting half your bankroll each time.


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#20 Prime

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Posted 21 November 2013 - 02:32 AM

 

 

Spoiler for what is expected

Yes, there it is -- Expected Value vs. Expected Outcome.

I fancy myself as an avid gambler. And in practice I would choose the mix of the two methods, as Rainman has suggested here. (Playing infinite number of times is rather impractical.)

I have run an experiment playing 422 consecutive rolls 10 times with the following results: 

4 times I had less than $0.01 left of my initial $1 bankroll.

3 times I had between $0.02 and $0.35 left.

3 times I won with the largest win of $83.62.

That left me very much ahead (more than $80, while putting at risk just $10).

The new question here is, what is the optimal string of consecutive rolls?

3 times I

 

It seems your post was cut off unfinished.

 

Expected outcome is the true value for any gambler who seeks to eliminate the "gambling" part. The probability is 1 that your own outcome approaches the expected outcome as the number of games approaches infinity. So in the long run, you are practically guaranteed to win if your expected outcome is positive. The same is not true for expected value.

 

Expected value is too influenced by the extremely high payoffs in the extremely unlikely variations. In this case, if you play for example 100 times, the extremely lucky variation where you hit 1.5 every time would yield a net payoff of roughly +$406,561,177,535,215,236. On the other hand, the extremely unlucky variation where you hit 0.7 every time would yield a net payoff of roughly -$1. The average net payoff, or expected value, would be (31/30)100-1, or roughly $27. So the variance (actual payoff - average payoff) is 406,561,177,535,215,209 for the luckiest variation and only -28 for the unluckiest variation. As follows, the expected value is extremely tilted by the high variance from impossibly lucky scenarios. You would need to be insanely lucky just to get anywhere close to the EV.

 

Your own experiments illustrate this perfectly. The average net payoff for running 422 consecutive games 10 times is 10*(31/30)422-1, or roughly $10,220,338. Your actual net payoff was just more than $80, falling way short of the EV. Had you kept your entire bankroll running for all 4220 games, you would have lost almost everything. This was not a case of bad luck, but rather quite expected.

 

Had you instead bet half your bankroll every time, with a starting bankroll of the same $10, for 4220 games, your expected outcome would have been ~10*1.049244220/6, or +$4,800,000,000,000,000. Feel free to simulate it, you will end up somewhere around that number. Your EV would of course be even higher, 10*(61/60)4220-1 or roughly +$2,000,000,000,000,000,000,000,000,000,000.

 

 

I am not disagreeing with the Expected Outcome concept. (And I have described the same in the post #12 under “PERSPECTIVE 1”, before I read your post #8. Actually, I should have read your post first. Now I invite everyone to read the spoiler in the post #12, where I derive the formula for the EV with a proof.)

However, in view of the question in the OP “Can you win in this game?” the Average/Expected Value seems very relevant for practical gambling.

I also like your system where you bet half of your entire bankroll on each turn. Although, I can't say I follow the 1.049N/6 formula for this method. I'd have to think about it.

Finding “optimal” system in terms of available time, initial bankroll, Expected Outcome, and Expected Value seems like more serious mathematical research.

Another interesting and, perhaps, simpler thing to solve would be: What are the chances of ending up ahead after N rolls when staking your entire bankroll?

(I could tell it's 17/36, or better than 47% for 2 consecutive rolls and 105/236, or better than 44% for 3 consecutive rolls.)

 

I played few more games against my computer and it owes me a lot of money.

Spoiler for COMPUTER SIMULATION

 


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Past prime, actually.





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