## Stubborn bettor

For the discussion of math. Duh.

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Dopefish
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### Re: Stubborn bettor

If you're just playing to have fun and since the expected return is always going to be 0, why bother increasing your bet when you win? Without thinking too hard about this, you'd basically just be simulating a 1D random walk, which is rather well understood in terms of what influence twerking the various parameters has.

Tangential question: I've never been to a casino, but do they actually have games with an expected return of 0 (e.g. 50:50 odds with the pay out described as in this thread)? I was rather under the impression that the whole reason casinos are profitable was due to the expected return of everything being in the casinos favour. I suppose some might focus on pushing alcohol and similar though so they profit even if the games themselves have a net 0 ER?

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### Re: Stubborn bettor

In casinos when you bet simple the expected value is always negative.
Depending on the rules (european or american) this value itself changes.
You have just to google it to understand how it works.
As I said before I designed the method for another kind of gambling games.
The model presented here is very simple.
You could adapt it to many things in a real life not specially gambling.

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### Re: Stubborn bettor

Someone who has cancer at its final stage was asked :
If you take this medication you will have 98% chances to live 2 or 3 months more unless you will die tomorrow. So there is only 2% chances that the medication fail.
Do you accept to take it?

What did you think he did?

PeteP
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### Re: Stubborn bettor

A better analogy would be:
A healthy person was asked: "You probably have a few decades left to live, take this medicine and in 98% of cases your live span will be a few months longer. But in 2% of cases it just kills you now. Will you take it?"

Your cancer analogy doesn't preserve that the possible reward is much smaller than the investment because for someone dying a few extra months can be much.

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### Re: Stubborn bettor

I will try to improve the method after trying it (virtually first and in real if it holds).
Thank you for all.

Wildcard
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### Re: Stubborn bettor

PeteP wrote:A better analogy would be:
A healthy person was asked: "You probably have a few decades left to live, take this medicine and in 98% of cases your live span will be a few months longer. But in 2% of cases it just kills you now. Will you take it?"

Your cancer analogy doesn't preserve that the possible reward is much smaller than the investment because for someone dying a few extra months can be much.

To me the interesting aspect of this discussion is how unexciting this option sounds even to me. I think it would sound even more unexciting to people who don't get the math behind expected return.

Don't get me wrong, expected return value is a very important mathematical concept, but I think if you surveyed people who don't understand it (e.g. anyone who buys lottery tickets), I suspect they would not take the medicine. I think there is an emotional concept of winnings/losses which is disconnected from Actual Math.

Probably the way the math works in their head is:

Code: Select all

`if (possibleWinnings > possibleLoss) {  takeTheBet();} else if (possibleLoss < possibleWinnings) {  dontTakeIt();} else {  //whatever, who cares}`
And maybe by the formula:

Code: Select all

`expected_return_value = winnings - cost`
But you know the old saying. Lottery tickets are a tax on people who are bad at math.

<.<
>.>

That's all I had to say.
There's no such thing as a funny sig.

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### Re: Stubborn bettor

Assuming that your results are right the probability to loose all the 2090 dollars (your capital) is NOT equal to 2%. You can not reach your goal 10 dollars in 2% of the cases but you can reach 9,8,7.....1,0,-1,-2.......

>-)
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### Re: Stubborn bettor

Do you mean 490? I'm pretty sure the chance of losing \$490 is 2%. You keep betting until you have \$10 or go broke so there's no chance of reaching \$9, 8, 7, or so on. The only outcomes are -\$490 or +\$10.

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### Re: Stubborn bettor

>-) wrote:Do you mean 490? I'm pretty sure the chance of losing \$490 is 2%. You keep betting until you have \$10 or go broke so there's no chance of reaching \$9, 8, 7, or so on. The only outcomes are -\$490 or +\$10.

Please can you explain me why?
I have a trouble to understand your claim.
490 dollars was changed to 2090 dollars based on what Pete found.
There is still misunderstanding about the capital needed.

Thank you.

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### Re: Stubborn bettor

>-) wrote:Do you mean 490? I'm pretty sure the chance of losing \$490 is 2%. You keep betting until you have \$10 or go broke so there's no chance of reaching \$9, 8, 7, or so on. The only outcomes are -\$490 or +\$10.

Juste a precision : "until you have 10 dollars or MORE"
Do not forget more.

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### Re: Stubborn bettor

Do not get me wrong : I know what the expected value means so please do not try to divert the subject to another one.
It is better to be not good in maths and millionaire than good in maths and poor.
You do not know where I`m heading so please stay focus on the problem I presented.

Thank you.

PeteP
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### Re: Stubborn bettor

The 98% number results from playing until you either have reached the goal or lost all your capital. If you don't pursue the strategy until the end you don't get the 98% chance, which is why the two percent are for losing the whole initial capital. If you want to limit it to a maximum number of coin tosses per game you probably won't reach your 98%. If I limit it to 100 tosses per game you have a chance of about 74% to reach your goal and more capital won't raise your chances (you can't use more than 100 initial capital during 100 tosses with your method.)

@>-):you get 490 if you hit the goal exactly by limiting the last bet, if you don't do that and there is a chance you will overshoot the goal you need more initial capital.

Anyway bored of this now, so I will step out.

@wildcard
Spoiler:
Wildcard wrote:
PeteP wrote:A better analogy would be:
A healthy person was asked: "You probably have a few decades left to live, take this medicine and in 98% of cases your live span will be a few months longer. But in 2% of cases it just kills you now. Will you take it?"

Your cancer analogy doesn't preserve that the possible reward is much smaller than the investment because for someone dying a few extra months can be much.

To me the interesting aspect of this discussion is how unexciting this option sounds even to me. I think it would sound even more unexciting to people who don't get the math behind expected return.

Don't get me wrong, expected return value is a very important mathematical concept, but I think if you surveyed people who don't understand it (e.g. anyone who buys lottery tickets), I suspect they would not take the medicine. I think there is an emotional concept of winnings/losses which is disconnected from Actual Math.

Probably the way the math works in their head is:

Code: Select all

`if (possibleWinnings > possibleLoss) {  takeTheBet();} else if (possibleLoss < possibleWinnings) {  dontTakeIt();} else {  //whatever, who cares}`
And maybe by the formula:

Code: Select all

`expected_return_value = winnings - cost`
But you know the old saying. Lottery tickets are a tax on people who are bad at math.

<.<
>.>

That's all I had to say.

Well expected value isn't the whole picture. For one it doesn't take into account how much it's worth to you. The more you have the less extra value it gives you. If you lose 90% of your money that is big and life changing, if you gain 90% more money that is still big but it probably won't influence your life as much as losing that amount. Also expected value works best if you can replenish the resource or have enough of it that you can repeat it frequently enough so that the average result approaches the expected value. If like in the life example one lose leaves you without the resource or a good way to replenish it even a positive expected value with a high victory chance might be considered to risky.

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### Re: Stubborn bettor

Even an expected value > 0 does not guaranty you that you will NOT loose all of your capital no matter how big is.
There is always a tiny chance that you will loose it.
This problem of stubborn strategy was very useful to me.
It lead me to something I have never imagine to think about it.
Big thanks to PeteP and Measure.
I have now a new project!

Xanthir
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### Re: Stubborn bettor

>-) wrote:Do you mean 490? I'm pretty sure the chance of losing \$490 is 2%. You keep betting until you have \$10 or go broke so there's no chance of reaching \$9, 8, 7, or so on. The only outcomes are -\$490 or +\$10.

Please can you explain me why?
I have a trouble to understand your claim.
490 dollars was changed to 2090 dollars based on what Pete found.
There is still misunderstanding about the capital needed.

Thank you.

If you do the absolutely naive strategy first described - at the start and every time you lose, bet \$1, every time you win, rebet what you just won - then you need \$2090 to have a 98% chance of getting a \$10 profit.

With a slightly more conservative strategy (if you've just won, and rebetting your winnings would take you to *more* than \$10 profit if it wins again, instead only bet enough to exactly hit \$10 profit), you only need \$490 starting capital to have a 98% chance of getting a \$10 profit. This is because the naive strategy always loses *all* of its winnings on a loss; that last toss that would take you over the goal still only has a 50% chance of succeeding. By dialing it down, you "bank" the excess, so that if you lose all the excess will fuel your continued betting.

As an intuitive way to think about this betting strategy, each time you start after a loss, you spend \$1 to try and start a winning streak. Your goal is to hit a sufficiently large winning streak to make a certain profit. At the start, you only need a streak of 4 wins to hit the goal - \$1, doubled four times, is \$16 winnings. This, of course, only happens 1/16th of the time. The other 15/16ths of the time, you need to spend another dollar to restart the chain from zero. Of course, after you've lost six times (losing 6 dollars from your initial bank), spending that seventh dollar means you now need to get a streak of *five* wins to recover the losses and still have a \$10 profit (with only a 1/32 chance of winning). After losing 15 more times, you now need a streak of *six* wins (with only a 1/64 chance of winning). This continues to decline; as you keep betting, it makes the chances of recovering less and less likely. You're most likely to "win" at the very beginning, after that you're chasing a continually-dropping win chance with a climbing likelihood of bankrupting entirely.

But for serious, there is no way to "win" at casinos. If you find betting enjoyable, great, it's no worse than any other expensive hobby. But there's no such thing as a "strategy" that will make you more likely to win. Your expected result is always negative. It might very be "better to be not good in maths and millionaire than good in maths and poor", but if your strategy is to win big by being bad at math in a casino, you're not going to end up a millionaire. Casino owners are millionaires *because* they are good at math and their patrons aren't.
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Xanthir
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### Re: Stubborn bettor

Dopefish wrote:Tangential question: I've never been to a casino, but do they actually have games with an expected return of 0 (e.g. 50:50 odds with the pay out described as in this thread)? I was rather under the impression that the whole reason casinos are profitable was due to the expected return of everything being in the casinos favour. I suppose some might focus on pushing alcohol and similar though so they profit even if the games themselves have a net 0 ER?

I don't know of any casinos that offer games with 0 expected value per-game, but casinos can still easily make money on such a game. The important thing is just that the casino has a larger bank than any individual bettor, so that the size of a streak that the casino can stand is larger than what the bettor can stand. This unbalances the random walk - winnings will wander over the positive and negative range, and people who are winning *tend to* keep betting (and the casino has enough bank to keep letting them), so they'll typically return to the expected value (zero), but people who lose all of their purse have to stop, leaving the balance in the casino's favor. Casinos could even offer games with a *slightly positive* expected value and still profit, due to this.

Using a game with negative expected value just makes it *more* profitable, as do things like betting limits (which, because they limit the amount bettors can win per game, encourage them to play more games instead to hit their profit goal, and thus be more likely to regress to the mean or have a negative streak that wipes them out).
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### Re: Stubborn bettor

@Xanthir

When I was talking about not good at maths but millionaire I was not talking about the roulette.
Re-read me I talked about another kind of gambling game.
There are many unorganized "gambling" games where there are no "bank" no "dealer".
When you buy something and you sell it it becomes gambling as soon as the value of what you bought is "uncertain" (art for example).

Thank you for your comment anyway.

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### Re: Stubborn bettor

If you're playing skill games, and are legitimately better than your opponents, then yes, you can make money. But for most of them you don't get good by being bad at math, either - good poker playing is fairly mathy, for example.
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quantropy
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### Re: Stubborn bettor

Dopefish wrote: I suppose some might focus on pushing alcohol and similar though so they profit even if the games themselves have a net 0 ER?

I think it's almost always the other way round :

From Surely you're joking, Mr. Feynman
It was just wonderful for a man who didn't gamble, because I was enjoying all the advantages - the rooms were inexpensive, the meals were next to nothing, the shows were good and I liked the girls

Tirian
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### Re: Stubborn bettor

Xanthir wrote:I don't know of any casinos that offer games with 0 expected value per-game, but casinos can still easily make money on such a game. The important thing is just that the casino has a larger bank than any individual bettor, so that the size of a streak that the casino can stand is larger than what the bettor can stand. This unbalances the random walk - winnings will wander over the positive and negative range, and people who are winning *tend to* keep betting (and the casino has enough bank to keep letting them), so they'll typically return to the expected value (zero), but people who lose all of their purse have to stop, leaving the balance in the casino's favor. Casinos could even offer games with a *slightly positive* expected value and still profit, due to this.

Using a game with negative expected value just makes it *more* profitable, as do things like betting limits (which, because they limit the amount bettors can win per game, encourage them to play more games instead to hit their profit goal, and thus be more likely to regress to the mean or have a negative streak that wipes them out).

Just to veer off-topic a little, I believe casinos do offer a game with positive expectation: blackjack. At least I think so. Back when I was a sprog, I wrote a simulation of the basic strategy from "Beat The Dealer" (i.e. no card counting to vary wage sizes or strategies) and became convinced that it was mildly in the player's favor even before adding in the nuances of card-counting which make the game so favorable to the player that you'll get kicked out as soon as they realize that you're doing it. I don't believe the casinos changed the rules of blackjack in response to that revelation (or rather, they did for a while but then realized that the average player doesn't have a winning strategy and you make more money from selling them the allure that clever people can beat the game).

But to the larger point, I think you misunderstand expectation and random walks. Yes, it is virtually certain that a casino will eventually break a player with a much smaller bank at a fair game, but that doesn't mean that they're making money at it. Imagine such a gambling game with ten seats. The casino will be winning five and losing five every round (on average over time). Every once in a while, a player will leave the table because his bank is gone, but that seat will be refilled and the casino is still winning five and losing fiver per round (i.e. not making money).

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### Re: Stubborn bettor

Here is in fact the core of the problem.
The expected value is always in favor of the organizers.
I`m talking about the known and organized gambling (horse racing sports casino~games roluette blacjack etc...).
Many players bet randomly without any strategy. Few players try to play "smartly".
Few are making money (lucky?) and the vast majority are loosing (either dumb or unlucky?).
The question is :
If we play smart I mean using all the tools (maths, statistics, etc...) can we be among the what we call "lucky bettors"?
If yes than there is for sure a method allowing us to make money.
If no than there is only one thing to pray for : is that the luck will smile us.
I do not believe in luck and I think that it exists a method to win.
My conjecture is based on the complexity of the strategies. An infinite numbers of scenarios that we can not master them.

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### Re: Stubborn bettor

Goahead52 wrote:Here is in fact the core of the problem.
The expected value is always in favor of the organizers.
I`m talking about the known and organized gambling (horse racing sports casino~games roluette blacjack etc...).
Many players bet randomly without any strategy. Few players try to play "smartly".
Few are making money (lucky?) and the vast majority are loosing (either dumb or unlucky?).
The question is :
If we play smart I mean using all the tools (maths, statistics, etc...) can we be among the what we call "lucky bettors"?
If yes than there is for sure a method allowing us to make money.
If no than there is only one thing to pray for : is that the luck will smile us.
I do not believe in luck and I think that it exists a method to win.
My conjecture is based on the complexity of the strategies. An infinite numbers of scenarios that we can not master them.
You're not playing against other players (in most games), you're playing only against the house, so the luck or strategy (or lack thereof) that other players use is totally irrelevant to what you're expected to win with your own money. And even though most casinos could still make a profit if they allowed some positive-expectation games for those players who use a smart strategy, they can always make more profit by making sure every game at every table is expected to make them money, regardless of player strategy.

Tirian wrote: I think you misunderstand expectation and random walks. Yes, it is virtually certain that a casino will eventually break a player with a much smaller bank at a fair game, but that doesn't mean that they're making money at it. Imagine such a gambling game with ten seats. The casino will be winning five and losing five every round (on average over time). Every once in a while, a player will leave the table because his bank is gone, but that seat will be refilled and the casino is still winning five and losing fiver per round (i.e. not making money).
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

In horse racing or sports bet you are playing against other players.
The case of the roulette is particular.
On each table not all the people run out of money.
Some (few) are winners.
Let us simulate the game.
n players around the table each one with different capital bet.
Each player use either some "method" (starting from the martingales to the astrology) either bet randomly.
Is there any "method" to play smartly such as you could make money?
My answer is yes.
Using the stubborn method will guaranty you in 98% of the cases to reach 10 dollars with a capital of 490 dollars. You loose only in 2% of the cases.
If one player use this method against others assuming that the others bet randomly then for sure he will loose less.
The casino will always win for sure.
So even in the casino games (roulette black jack etc...) there is ALWAYS a room for playing more smartly than others.
It is hard to find a method win-win because there are a lot of strategies to study.
The stubborn strategy presented above is simple.
Rebetting is assumed to be 100% but if you change smartly the percentage to bet it will lead to different outcome.
If 2 players bet (one of them black and the other red) sharing their losses and their goal it will lead to different outcome.
If a robot with some AI bet than it will be another plus (playing online allows it). When I say robot I mean some program assisting the player to bet. It could handle on real time what is going on.

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### Re: Stubborn bettor

Goahead52 wrote:Using the stubborn method will guaranty you in 98% of the cases to reach 10 dollars with a capital of 490 dollars. You loose only in 2% of the cases.

But you lose a lot more when you lose. 98% of the time you win \$10, 2% of the time you lose \$490. The expected value of this game is 0.

Actually the expected value is less than that, because the probability of winning on a roulette wheel is slightly less than 50% due to the 0 (and sometimes 00) spaces on the wheel which are always wins for the house. But that is probably a separate discussion.

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### Re: Stubborn bettor

If we consider 2 players A and B :
- player A bet on head or tail randomly one dollar per toss.
- player B using the stubborn strategy

The 2 players start with the same capital 490 dollars.
Does the player have 98% chances to win 10 dollars or more and 2% chances to loose all his capital (490 dollars).

If the 2 players have the same probability then the stubborn strategy worth nothing!
If not there are 2 outcomes : either the player A has more chances to survive (the strategy stubborn will be then negative) either the stubborn is better
Last edited by Goahead52 on Thu Jun 04, 2015 8:36 pm UTC, edited 1 time in total.

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### Re: Stubborn bettor

Gwydion wrote:
Goahead52 wrote:Using the stubborn method will guaranty you in 98% of the cases to reach 10 dollars with a capital of 490 dollars. You loose only in 2% of the cases.

But you lose a lot more when you lose. 98% of the time you win \$10, 2% of the time you lose \$490. The expected value of this game is 0.

Actually the expected value is less than that, because the probability of winning on a roulette wheel is slightly less than 50% due to the 0 (and sometimes 00) spaces on the wheel which are always wins for the house. But that is probably a separate discussion.

Even if the expected value is > 0 nothing will guaranty you that you will not loose all of your money.
A sequence of TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT .....T which has the probability > 0 to happen will make you loose all.

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### Re: Stubborn bettor

Goahead52 wrote:If one player use this method against others assuming that the others bet randomly then for sure he will loose less.
It's not "for sure" in any sense of the word. You have a 2% chance of losing \$490 and a 98% chance of winning \$10, over the course of many games. I might just bet \$490 all at once on the first go, on something with 50/50 odds, and have a 50% chance of losing all of it, which is admittedly a much higher chance than your 2% chance over time, but I also have a 50% chance of winning \$490, which is a lot more money than your \$10, even if you are a lot more likely to win that much.

In the end, we each have an expected win of \$0. The only thing different strategies do is change the shape of the outcomes around \$0, but \$0 is always the average.
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### Re: Stubborn bettor

We diverge on the real meaning of what an average is.
Expected value is an average based on the hypothesis on the infinity of the tosses.
The number of tosses is finite. Hence there is a high probability that 0 will not happen as frequently than minus or plus something.
2 hypothesis at the origin of all calculations are always wrong :
- the infinity of tosses
- the machine (built by human being and checked by human) is supposed to reproduce truly randomness

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### Re: Stubborn bettor

So... what? You're relying on the fact that no one plays eternally or can guarantee true randomness to make your strategy better?

Even if we can't guarantee true randomness, it is not hard to come up with a pseudorandom generator that averages out to exactly 50/50 odds.

(Also, if we disagree about what an average is, it's because you're using it in some nonstandard way that is incorrect according to the rest of mathematics.)
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Wildcard
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### Re: Stubborn bettor

Goahead52 wrote:If we consider 2 players A and B :
- player A bet on head or tail randomly one dollar per toss.
- player B using the stubborn strategy

The 2 players start with the same capital 490 dollars.
Does the player have 98% chances to win 10 dollars or more and 2% chances to loose all his capital (490 dollars).

If the 2 players have the same probability then the stubborn strategy worth nothing!
If not there are 2 outcomes : either the player A has more chances to survive (the strategy stubborn will be then negative) either the stubborn is better

The problem here is that you don't actually understand what "expected value" means.

You think you do, but you don't.

Use a dictionary. Here's a definition from the New Oxford American:
a predicted value of a variable, calculated as the sum of all possible values each multiplied by the probability of its occurrence.

2% chance of losing \$490 --> 2/100 * (-\$490) = -\$9.80
98% chance of gaining \$10 --> 98/100 * (+\$10) = \$9.80
Expected value = -\$9.80 + \$9.80 = \$0.00

The expected value for the game you describe will not change no matter what strategy you follow. You can adopt a strategy that will give you a 99.999% chance of winning \$1, but that means that you will have a .001% chance of losing \$99,999—and it also means you will need to have \$99,999 to start with. And also, the length of time it will take to actually win the \$1 may be EXTREMELY long, and the possibility of a total loss is never ruled out. At every point in the game, the expected value for the money you will win if you keep playing is still \$0. If you don't understand this, you need to make up some examples for yourself using the definition of "expected value" given above. Otherwise you will just waste everyone's efforts to convince you.

Related reading: the Gambler's fallacy.

I hope the mods don't mind my very tiny use of red text to make a point clearer to someone who just isn't getting it.
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### Re: Stubborn bettor

I`m not trying to convince anyone.
I will use my stubborn strategy because I`m stubborn too (not in the roulette to be precise).
I will not show you my banking account.
I will keep it for me.
Thank you all for all of your comments.

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### Re: Stubborn bettor

Well on the positive side if you actually play it with 1 dollar as base bet losing significant amounts would take quite a while. So I guess as wishful thinking in regards to gambling strategies goes it should be one of the less costly ones.

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### Re: Stubborn bettor

Wildcard wrote:
Goahead52 wrote:If we consider 2 players A and B :
- player A bet on head or tail randomly one dollar per toss.
- player B using the stubborn strategy

The 2 players start with the same capital 490 dollars.
Does the player have 98% chances to win 10 dollars or more and 2% chances to loose all his capital (490 dollars).

If the 2 players have the same probability then the stubborn strategy worth nothing!
If not there are 2 outcomes : either the player A has more chances to survive (the strategy stubborn will be then negative) either the stubborn is better

The problem here is that you don't actually understand what "expected value" means.

You think you do, but you don't.

Use a dictionary. Here's a definition from the New Oxford American:
a predicted value of a variable, calculated as the sum of all possible values each multiplied by the probability of its occurrence.

2% chance of losing \$490 --> 2/100 * (-\$490) = -\$9.80
98% chance of gaining \$10 --> 98/100 * (+\$10) = \$9.80
Expected value = -\$9.80 + \$9.80 = \$0.00

The expected value for the game you describe will not change no matter what strategy you follow. You can adopt a strategy that will give you a 99.999% chance of winning \$1, but that means that you will have a .001% chance of losing \$99,999—and it also means you will need to have \$99,999 to start with. And also, the length of time it will take to actually win the \$1 may be EXTREMELY long, and the possibility of a total loss is never ruled out. At every point in the game, the expected value for the money you will win if you keep playing is still \$0. If you don't understand this, you need to make up some examples for yourself using the definition of "expected value" given above. Otherwise you will just waste everyone's efforts to convince you.

Related reading: the Gambler's fallacy.

I hope the mods don't mind my very tiny use of red text to make a point clearer to someone who just isn't getting it.

What is wrong in you computation of the expected is that :
You stop betting when you 10 dollars or more NOT EXACTLY 10 dollars.
Show me because I`m really dumb how are going to reach exactly +10 WITHOUT picking that money from your capital.
The rule was slightly modified (Measure signaled it). He assumed that you can adjust you rebet from what you just win.
If you have exactly +10 profit then your computation is right.
If not than the expected value is not zero.
No need even to make simulation to find the capital 490 dollars to win 10 dollars (98%). It means 49 dollars to win 1 dollar = 98%.
I know what the expected value means.
The way we compute is wrong. If you exclude the hypothesis that expected value computation is ASSUMING the infinity of tosses then you will reach other results.
What is the probability that in 100 tosses the expected value =0 (I mean exactly zero)?
Here is my conjecture : randomness will never give you some sort of PERMANENT symmetry.
I`m bettor and I always bet on asymmetry.
That is why 2 players (bet on the 2 outcomes one for black other for red) could be both winners.
How to prove it?
I do not know because I do not have the mathematical tools to prove it.

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### Re: Stubborn bettor

Goahead52 wrote:You stop betting when you 10 dollars or more NOT EXACTLY 10 dollars.
This has been addressed earlier. I was citing the modified strategy wherein you reduce your bet amounts to cap the possible winnings at \$10. For instance, if you start (with however much money) and win \$1, then rebet and win \$2, rebet and win \$4, rebet and win \$8, you would (with the modified strategy I was calculating with) only bet \$2 for the next bet, so that if you win you will have a total profit of exactly \$10.

If you don't reduce your bets in these circumstances, all you do is increase the possibility and probability of a greater loss of money. The expected value is still zero.

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?

Really. Answer this. Even if you "don't have the tools". Just apply the definition of expected value which you say you understand, and preferably show your work in your post.

Incidentally:
Spoiler:
What you say here:
Goahead52 wrote:I know what the expected value means.
The way we compute is wrong. If you exclude the hypothesis that expected value computation is ASSUMING the infinity of tosses then you will reach other results.
What is the probability that in 100 tosses the expected value =0 (I mean exactly zero)?
...shows a misunderstanding of "expected value".

Expected value does not mean "The value that you expect." It means
The New Oxford American Dictionary wrote:...the sum of all possible values each multiplied by the probability of its occurrence.
The expected value for the number of dots that will be facing upward after you roll a single die is 3.5. Even though there is no way possible for 3.5 dots to show up. You don't "expect" 3.5 dots to show up. That is the average of the possibilities. Likewise, the expected value for winnings after 100 coin tosses is exactly 0. The expected value for ONE toss is also exactly 0.
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### Re: Stubborn bettor

Goahead, it is possible that your theoretical two bettors can both be winners. This can only happen if one of them "wins" and quits, then the other goes on a hot streak afterward. An example of such a sequence would be BBBB (first player quits with 16, second has -4) RRRR (second player gets out with 12). A similar sequence can be drawn for most of the game states, though if both players get too deep in the hole, one player might run out of money before the other "wins" and quits.

Another likely scenario is that both players lose. Here is such a sequence: BRBRBR… every second toss sees both players down 1 from where they had been before. Another such sequence is BBRRBBRR… in which every player goes up 3, down 5, repeat. If there are no sufficiently long strings of consecutive B or R, then eventually they will both see their money dwindle to nothing.

I'm not sure how to answer your other question. The odds of getting 50 heads and 50 tails on 100 coin tosses is just a shade under 8%, but this is not the same as an expected value calculation. The expected value of betting on a coin toss with even odds is 0. This expected value is independent of the number of tosses, because they are independent events whose EV can be summed to find the EV for all tosses - in this case, 0+0+…+0 = 0. Wildcard addressed this differently just above.

It turns out that because of your strategy for changing wagers, the expected value might be slightly larger than 0 - as you said, you might end up with more than \$10 depending on how much you have when you start your eventual winning streak. My comment earlier was based on the modified betting strategy that decreases your bet on the toss before you would win, such that if you happen to lose you don't go all the way back to the start but can "bank" some of it to keep from going bankrupt quite as fast. Still, even with a positive expected value you could still lose a lot, and the value of such a system depends entirely on how much you value your \$490, and how much you value an extra \$10.

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### Re: Stubborn bettor

Gwydion wrote:
It turns out that because of your strategy for changing wagers, the expected value might be slightly larger than 0 - as you said, you might end up with more than \$10 depending on how much you have when you start your eventual winning streak. My comment earlier was based on the modified betting strategy that decreases your bet on the toss before you would win, such that if you happen to lose you don't go all the way back to the start but can "bank" some of it to keep from going bankrupt quite as fast. Still, even with a positive expected value you could still lose a lot, and the value of such a system depends entirely on how much you value your \$490, and how much you value an extra \$10.

Nah, it remains zero (as you just said yourself the bets are independent and the expected value of the strategy is the sum of their expected values.) Not lowering the bet also means that you drop much farther away from the goal when you lose that last bet, which results in a much higher needed capital compensating for the chance to win more than 10. (Also finishing a game with that strategy can take hundreds of tosses.)

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### Re: Stubborn bettor

gmalivuk wrote:
Tirian wrote: I think you misunderstand expectation and random walks. Yes, it is virtually certain that a casino will eventually break a player with a much smaller bank at a fair game, but that doesn't mean that they're making money at it. Imagine such a gambling game with ten seats. The casino will be winning five and losing five every round (on average over time). Every once in a while, a player will leave the table because his bank is gone, but that seat will be refilled and the casino is still winning five and losing fiver per round (i.e. not making money).
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).

No, I'm pretty sure this is wrong, and the casino does indeed make money on an even game. A quick naive simulation seems to support this, at least. And I think reasoning does, too.

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.
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

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### Re: Stubborn bettor

Xanthir wrote:...the casino does indeed make money on an even game.
...
Say everybody enters the casino with \$10...
...everyone leaves with \$10 left. The rest of their money had to go somewhere, and the casino has it.
Wait, what?
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### Re: Stubborn bettor

...did you miss the part where there's all the cash infusions from their friends?

Edit: In other words, they enter with \$10 and however much money their friends are carrying, they leave with \$10 (on average) and their friends are broke. All that money from their friends is now owned by the casino.
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

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### Re: Stubborn bettor

Okay. I don't see how that will make any difference. If it's an even game, the players will win probabilistically as often as they lose.

You could try a simulation, with a zero expected value game like a coin toss, and a class "Player" which is initialized with a certain bank value and a randomly initialized "quitting condition", i.e., if any player is broke they always quit, but for each player if they double their initial bank value they quit with randomly set probability (p), or whatever. And run it for 10000 visitors to the casino like 500 separate times and see what happens.

Then change the expected value of the game to 51% win condition for the casino and 49% win condition for the players and rerun the simulations.
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### Re: Stubborn bettor

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.

Code: Select all

`<!doctype html><script>var profit = 0;var trials = 1e3;var gamblers = 1e3;var startingBank = 2;for(var j = 0; j < trials; j++) {var casino = 1e9;for(var i = 0; i < gamblers; i++) {  var bank = startingBank;  while(bank > 0) {    if(Math.random() > .5) {      bank--; casino++;    } else {      bank++; casino--;    }    if(bank > startingBank*2 && Math.random() < (bank - startingBank)/(2*startingBank)) {      break;    }  }}profit += casino - 1e9;}console.log((profit/trials).toFixed(2));</script> `
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))