Stubborn bettor

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Re: Stubborn bettor

Postby Wildcard » Fri Jun 05, 2015 6:43 am UTC

You could try plopping the output results into a big averaging function. That should make the analysis pretty easily. (Alternatively you could do some fancy statistical analysis for standard deviation and what have you, but given that it's only simulated data that seems like it would be pretty silly.)

Now you could change the comparative constant to .49 and see if the average of the results is higher. Probability says it will be.

Probability also says you'll already have thought of these things by the time you read them here, but on the off-chance you didn't... :P

(Interesting...what language is that? I'm decent at reading simple code regardless of language, but I'm very spotty on knowing the specific language just from syntax. It's obviously from the C family, but it's not Java...is it JavaScript?)

Also, I want to say it's very refreshing to have a conversation wherein the other person actually takes a suggestion like "you could do a simulation" and runs with it, and observes the results, rather than just blindly asserting that they are right and they know what they are talking about. :) Kudos to you, Xanthir!
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Re: Stubborn bettor

Postby quantropy » Fri Jun 05, 2015 9:30 am UTC

If the casino could make money on an even game, then I, who get to choose when to bet, could do the same. Most people quit when they've lost their $10 or when they've made a large amount of money. I go in with a large amount of money and quit when I'm $10 ahead or when I've lost all my money. This seems to be what goathead52 is suggesting. The consensus is that it won't make any money.

Even if people go on betting after each win, the casino only has finite resources (or a house limit). After a sufficiently long run of wins the gambler won't be able to bet any more.

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Re: Stubborn bettor

Postby PeteP » Fri Jun 05, 2015 10:21 am UTC

The error is similiar to the one with the strategy where you always double your bet until you win. That strategy as well as playing until you have no money left requires infinite games to be sure, as well as infinite capital in the first case on your part in this case on the casinos.

Or to look at it another way, it doesn't really matters from the casinos perspective who is playing (as long as they can pay) whether it's a player who just lost all their money or one who already won a good amount or an entirely new player. They might as well be betting with a blackbox. With a bet of the same size the result changes their balance in the same way. And so it comes down again to summing up the expected values of the individual games.

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Re: Stubborn bettor

Postby Goahead52 » Fri Jun 05, 2015 12:20 pm UTC

@wildcard

The expected value is based on the hypothesis of infinite runs.
Why?
Because it depends on THE PROBABILITY.
What is the probability?
Some say that a coin is "fair".
How it is supposed to be fair.
If you toss a coin it has one chance out ot 2 to give either hail either head.
We toss a coin a finite number of times = for example 1000000 times.
We then analyze the outcome by using a battery of tests (from Diehard tests to others).
We then conclude that this coin is fair.
That means that if you toss ad infinitum you will 50% to have head or tail.
We start from finite number of tosses and we extend our conclusion (or results) to the infinite.
What represent 1000000 comparing to the infinite? nothing!!!
If you toss a coin 10 times you have at most 10 outcomes happening (you know it for sure).
How many outcomes are possible (2^10=1024)?
You can not predict the 10 but for sure it exists a subset of the 2^10 elements which is more probable than the others.
The sum of the 1`s vary from 0 to 10.
It is highly unlikely that the sum will be 0.
It could be zero because 000000000 and 0101010101 have the same probability to appear.
It could not be because zero but more likely 5 because there is only one outcome 0000000000 and lot of sequences summing up to 5.
All those contradictions make me "suspicious".
Anyway I have to go because I`m busy all this morning starting 9h am.
Have a good day.

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Re: Stubborn bettor

Postby gmalivuk » Fri Jun 05, 2015 12:41 pm UTC

Xanthir wrote:I did, though with a 0% chance of leaving as long as you had money. That's what they depend on, obviously - that most people will keep playing when they're up, until they get back to even or lower. As long as the amount of cash entering the system from people losing their money and new people rotating in is larger than the cash leaving the system from people quitting while they're ahead, the casino wins.

To be a little more realistic, I just did some trials with a chance of them leaving. Once they go above twice their starting bank, they have a 50% chance to leave; this increases until they're guaranteed to leave at thrice their starting bank. The results are really swingy and hard to analyze, but hm, initial results look pretty balanced around 0 casino profit. Interesting.
The casino's bank doesn't care who's betting. If you assume each player that leaves (because they're broke or because they' ve tripled their money or because they've been at this for a 70-year winning streak and have now died) is replaced with another player, then it may as well just be one player with arbitrarily high starting capital, and you just get a random walk.
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Re: Stubborn bettor

Postby Tirian » Fri Jun 05, 2015 5:13 pm UTC

Xanthir wrote:The results are really swingy and hard to analyze, but hm, initial results look pretty balanced around 0 casino profit. Interesting.


Okay, it's a relief to think that the simulation supports my model of what's going on.

This seems like one of those Monty Hall arguments where two smart people have different arguments that seem equally reasonable but only one of them is right. Rather than just say "I'm right because a simulation says so", let me see if I can go back and find the flaw in your argument.

Xanthir wrote:Say everybody enters the casino with $10, and plays a coin-toss for $1 a throw. When they run out of money, instead of leaving and being replaced, their friend hands them another $10 bill to keep going with. (This should be identical to replacing them with a new player with a fresh bank.) The limit behavior is that everyone hovers around $10, and getting an occasional cash infusion to "reset" your bank to $10 doesn't affect this - you still hover at $10, because gambling doesn't have memory. The end result is as if everyone came in with varying bank sizes, but the larger your bank was the worse you did, so that everyone leaves with $10 left. The rest of their money had to go somewhere, and the casino has it.


I believe that your perspective is skewed by focusing on the moment when the player runs out of money (at which point the casino has self-evidently made that player's money). That seems natural, since we're ideally assuming that the player plays until the bank runs out and that the casino could not possibly run out, so it's the only event from the player's perspective that's (almost) certain. But the casino is still playing a fair game, so their expected profit over its entire lifetime will still be 0. Gambler's ruin says that an unbiased random walk on the number line will eventually reach every integer at some point -- the fact that the casino's profit will eventually hit +10 will not keep it from returning to 0 and even -1,000,000 at some point in time.

I think the difference between our views is a little like supertask. Let's imagine a casino with an infinite capacity playing a fair game with one entrance and one exit, and an infinite number of players lining up to enter the casino with $10, which they will play until they lose and then leave (and not leaving until then). Every time the casino plays a round with everyone in the casino, they admit a new player. You're standing outside the exit and observing that everyone eventually leaves with no money, so the casino must be making money. I'm standing inside the casino seeing that the house wins half their games and loses half their games and that the wealth of all of the losing players has largely been transferred to the winning players, not the casino. But even though the wealthiest player at any given moment is (almost) certain to lose it all at some point in the future from your perspective, the casino's profit is still a random walk starting at 0 and at any moment they are equally likely to have a net profit as a net loss.

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Re: Stubborn bettor

Postby Adam H » Fri Jun 05, 2015 5:56 pm UTC

A similar logic problem is this classic:

There's a culture where parents must continue to procreate until they have one boy. What is the boy/girl ratio?

The correct answer is 1:1. "Stopping" at a predetermined point won't skew the probabilities, just like in this gambling problem.
-Adam

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Re: Stubborn bettor

Postby gmalivuk » Fri Jun 05, 2015 7:32 pm UTC

Adam H wrote:A similar logic problem is this classic:

There's a culture where parents must continue to procreate until they have one boy. What is the boy/girl ratio?

The correct answer is 1:1. "Stopping" at a predetermined point won't skew the probabilities, just like in this gambling problem.

gmalivuk wrote:Yeah, this is something like the counterintuitive probability question of what the sex ratio would look like if all parents continued having babies until they had exactly one boy and then stopped. Most people naively think that it would upset the balance, because there are lots of families with multiple girls but no families with multiple boys, but really it remains 50/50 (or whatever it was to start).
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Re: Stubborn bettor

Postby Wildcard » Fri Jun 05, 2015 7:47 pm UTC

Goahead52 wrote:It is highly unlikely that the sum will be 0.
Right, but it's exactly as likely to be less than 0 as it is to be more than 0. That's the point. Winning is exactly as likely as losing. On any given flip you are exactly as likely to win the amount of your bet as you are to lose the amount of your bet. You can't change that by changing the amounts of your bets. Numbers aside, you are still exactly as likely to win money as to lose money.

We could replace the coin flip game with another game, wherein you bet $19 and roll a 20-sided die. If it comes up 20, you lose your $19. Otherwise, you win $1. This accomplishes your purpose of having a "high likelihood of making a small profit", but you see that it also gives you a small possibility of losing a lot of money.

Let's say you have a 50 sided die. You bet $490. If the number 50 comes up, you lose all $490. Otherwise, you win $10.

All your strategy (as modified to cap winnings at $10) does is convert the coin flipping game into being equivalent to the 50 sided die game. The only difference in practice is that the strategy can take many hundreds of flips before it reaches the equivalent result to a single die roll.

Perhaps someone else can work out a single die roll game that is exactly equivalent to your unmodified strategy. It would have to have probability 98% of winning $10 or more, and would consequently have 2% chance of losing $490 or more. But I believe all these odds could be exactly enumerated with a sufficiently large die? Anyone?
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Re: Stubborn bettor

Postby PeteP » Fri Jun 05, 2015 11:03 pm UTC

His unmodified strategy needs about 2090 as start capital so the highest possible is to win 12 times in a row after you were down to 43, 11 wins will land you at 2089 and the twelfth at 4137. Now you can't reach all values from 2090 to 4137 but there are around 1100 possible end results I think. And you would need to determine the probability for each of them before creating your giant die. Doable of course but a pain in the ass and kinda pointless.

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Re: Stubborn bettor

Postby Goahead52 » Sat Jun 06, 2015 3:19 pm UTC

The stubborn strategy will surely work if we add this constraint : let us say that we bet on some gambling game where we are SURE (100%) that the number of heads is always > to the number of tails for any range of 100 consecutive tosses.
What will be then the capital needed to reach 10 dollars or more with the probability of 99,99% success?
In fact it exists a gambling game (not with coins) satisfying these conditions. Better than the roulette there is no LIMIT for betting.

For @Wildcard
Your example of expected value need to be clarified. When you roll a die there are 2 things unclear : probability of success ou failure (are we talking about betting on 2, 3 numbers or more out of 6? are we talking about return 2,3,5..???).
I can not compute your expected value through the give-me give-you data.
Thank you for your comment.

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Re: Stubborn bettor

Postby PeteP » Sat Jun 06, 2015 3:25 pm UTC

You mean this?
Wildcard wrote:
You say you understand expected value. Okay. Prove it. Let's say I roll a single 6-sided die. If the result is:
1: You give me $6.
2: I give you $24.
3: You give me $15.
4: You give me $75.
5: I give you $3.
6: I give you $9.

What is the expected value of your winnings (or loss) for a single roll?

With an unbiased die the chance for each side is 1/6. Everything you need is given . If you can't calculate the expected value for that you really have no idea what it is.

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Re: Stubborn bettor

Postby Goahead52 » Sat Jun 06, 2015 3:36 pm UTC

PeteP wrote:You mean this?
Wildcard wrote:
You say you understand expected value. Okay. Prove it. Let's say I roll a single 6-sided die. If the result is:
1: You give me $6.
2: I give you $24.
3: You give me $15.
4: You give me $75.
5: I give you $3.
6: I give you $9.

What is the expected value of your winnings (or loss) for a single roll?

With an unbiased die the chance for each side is 1/6. Everything you need is given . If you can't calculate the expected value for that you really have no idea what it is.

The chance of betting one number is 1/6.
What about betting on 2 numbers? and winning if one of them appears?
What about the amount to bet and the return?
Sorry.
It is unclear for me.
Otherwise I do not know what is the expected value.
If I do not know what is the expected value then I will stop posting and I will hire a math teacher.
You will never see me again until I become an expert on the probability field.

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Re: Stubborn bettor

Postby measure » Sat Jun 06, 2015 5:01 pm UTC

Spoiler:
Goahead52 wrote:
PeteP wrote:You mean this?
Wildcard wrote:
You say you understand expected value. Okay. Prove it. Let's say I roll a single 6-sided die. If the result is:
1: You give me $6.
2: I give you $24.
3: You give me $15.
4: You give me $75.
5: I give you $3.
6: I give you $9.

What is the expected value of your winnings (or loss) for a single roll?

With an unbiased die the chance for each side is 1/6. Everything you need is given . If you can't calculate the expected value for that you really have no idea what it is.

The chance of betting one number is 1/6.
What about betting on 2 numbers? and winning if one of them appears?
What about the amount to bet and the return?
Sorry.
It is unclear for me.
Otherwise I do not know what is the expected value.
If I do not know what is the expected value then I will stop posting and I will hire a math teacher.
You will never see me again until I become an expert on the probability field.
The question does not involve betting, just a gain or loss that depends on the result (i.e. every time a three is rolled, you have a net loss of $15). If each of these six possible results is equally likely, what is the expected value of your net gain for a single roll?

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Re: Stubborn bettor

Postby Goahead52 » Sat Jun 06, 2015 5:36 pm UTC

If a gain or less depends on the results of rolling a die is NOT betting then what it is?
Maybe it is time for me to hire an English teacher too.

Good luck to you.

As I`m an ignorant I resign.

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Re: Stubborn bettor

Postby PeteP » Sat Jun 06, 2015 5:46 pm UTC

It is a bet just not one were you bet on a number, there is a defined outcome for each number.

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Re: Stubborn bettor

Postby Sizik » Sat Jun 06, 2015 5:59 pm UTC

Goahead52 wrote:If a gain or less depends on the results of rolling a die is NOT betting then what it is?
Maybe it is time for me to hire an English teacher too.

Good luck to you.

As I`m an ignorant I resign.


It's betting when you choose which results give you a positive payout. When the payouts are predetermined (e.g. slot machines), it's not usually though of as "betting" (although the money you wager is often called a "bet").
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Re: Stubborn bettor

Postby Goahead52 » Sat Jun 06, 2015 9:19 pm UTC

Sizik wrote:
Goahead52 wrote:If a gain or less depends on the results of rolling a die is NOT betting then what it is?
Maybe it is time for me to hire an English teacher too.

Good luck to you.

As I`m an ignorant I resign.


It's betting when you choose which results give you a positive payout. When the payouts are predetermined (e.g. slot machines), it's not usually though of as "betting" (although the money you wager is often called a "bet").

English is not my mother tongue even if I 3 mother tongues.
Thank you for your explanation.
I still do not understand how you give me money and I give you money depending on the outcome of the die????!!!! And it is not betting.
Are the payouts predetermined?
It is like this
you roll die I give 1 dollar
you roll die I give you 1 dollar
....
each time you roll a die I give I dollar
(let us 30 times)
What is the expected value? (It is job and it has nothing to do with the outcome because the die is very heavy).
Are you talking about mean (statistical parameter?)?
La moyenne ponderee peut etre ponderee par des facteurs externes qui n`ont rien a voir avec le lancer de des.

Whatever is your answer the question is NOT clear and could be interpreted in 100000 of ways.

Cela dit, merci.

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Re: Stubborn bettor

Postby gmalivuk » Sat Jun 06, 2015 9:35 pm UTC

Goahead52 wrote:Whatever is your answer the question is NOT clear and could be interpreted in 100000 of ways.
The question is perfectly clear and can be interpreted in exactly ONE way by someone who understands what "expected value" means.

I roll a fair 6-sided die once.

If the result on the die is 1, you give me $6. (-6 for you)
If the result on the die is 2, I give you $24. (+24 for you)
If the result on the die is 3, you give me $15. (-15 for you)
If the result on the die is 4, you give me $75. (-75 for you)
If the result on the die is 5, I give you $3. (+3 for you)
If the result on the die is 6, I give you $9. (+9 for you)

Each of these values has a probability of 1/6, so the expected value for you is:
1/6*(-6) + 1/6*(24) + 1/6*(-15) + 1/6*(-75) + 1/6*(3) + 1/6*(9) = -10

This means that, on average, you can expect to lose $10 every time we play this game. While it's true that you might gain $24 or lose $75 on any given roll, if we play 100 times you'll be pretty close to down $1000.
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Re: Stubborn bettor

Postby Goahead52 » Sat Jun 06, 2015 11:10 pm UTC

You are right.
I thought that 1,2,3,....6 were the first roll the second and so on.
If you read my answers you will soon understand that I misread.
Sorry for all this.
Anyway I intended to leave anyway.


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