gambling question - changing house edge by altering bet size?

AvL42

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Quote from: dooglus on December 06, 2015, 10:03:23 PM

Quote from: AvL42 on December 06, 2015, 06:36:53 PM

Not sure, why this showed up as new here. Did you edit it?

No, it's a new post.
I saw you saying I was fantasizing in a fantasy world, and wanted to prove that I wasn't, and that this +EV casino was real!

Maybe you just missed the smiley after my "fantasize" statement from back then?

Of course I knew about that site - we had been talking about that bug in the site's forum.

I really thought I'd have read that very reply long ago. Maybe I mixed it up with some reply in the respective site's own forum.

dooglus

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Quote from: AvL42 on December 06, 2015, 06:36:53 PM

Not sure, why this showed up as new here. Did you edit it?

No, it's a new post.

I saw you saying I was fantasizing in a fantasy world, and wanted to prove that I wasn't, and that this +EV casino was real!

Sorry it took me 3 years to reply to your post. I can see how that might be confusing.

AvL42

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Quote from: dooglus on December 05, 2015, 05:49:44 PM

I don't think the game was provably fair, and suspect it was probably not actually fair at all - which is why they could afford to pay out on zero...

Not sure, why this showed up as new here. Did you edit it?

Anyway, it seems like the mentioned casino no longer exists any more.

dooglus

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Quote from: AvL42 on November 27, 2012, 07:17:02 PM

Quote from: dooglus on November 27, 2012, 12:47:10 PM

Now just suppose you had found a roulette game, for example, where they accidentally
paid out 3x on a 13-in-37 shot (paying out as if the probability of winning was 0.33333
when it's really 0.35135). The odds are slightly in my favour, but I only have 100 chips.

Your mind must be back in some fantasy-world, when you fantasize about such a roulette
game Wink

Actually I found exactly such a roulette game. luckybitcoincasino.com used to pay out 2-to-1 on any multiple of 3. It's meant to pay out on 3, 6, 9, ..., 36, but it also paid out on 0. I told them about the bug and they said they didn't have time to fix it.

Here's some screenshots showing it pay out 2-to-1 on zero:

I don't think the game was provably fair, and suspect it was probably not actually fair at all - which is why they could afford to pay out on zero...

darkmule

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Quote

Now just suppose you had found a roulette game, for example, where they accidentally paid out 3x on a 13-in-37 shot (paying out as if the probability of winning was 0.33333 when it's really 0.35135). The odds are slightly in my favour, but I only have 100 chips. What's my optimum betting strategy to minimise my risk of ruin, and maximise my expected return?

Some gamblers use Kelly criterion betting. Many consider pure Kelly criterion betting too aggressive and as taking too high of a risk of ruin, so bet something like half a Kelly bet.

While the math is a little gnarly to do at the table especially if just playing for fun, I think the general principle is sound and can be done more by feel than by pure math. I.e., if I had a slight edge, I might bet 5% of my roll on each trial, adjusting for current bankroll size.

Since a casino has less tolerance for losing its entire bankroll (a nonprofessional gambler is usually just wagering disposable income), the casino should set its maximum bet at a value considerably below what Kelly would suggest.

Risk of ruin is more of a theoretical than an actual concern where most casinos are concerned, because the casino is likely to take action well before ruin is actually reached, rather than simply continuing to spread the same bets while going bankrupt.

(Needless to say, Kelly criterion does you no good if you don't have an edge. You're going broke regardless of how you bet.)

dooglus

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Quote from: flooraccount on November 26, 2012, 11:42:21 PM

Well, not necessarily true. Odds are Odds. But when other people are playing this could increase or reduce your result.
I would compare this to satoshidice. Odds, Luck, and amount of players can manipulate the result if that makes sense.

I don't think satoshidice is a good example. Whether I win or not is completely determined by the txid of my bet and the day's secret. It doesn't matter when during the day I send my bet, how lucky I am feeling at the time, or how many other people are playing.

Whether my transaction is a winning one or not is already determined by the laws of mathematics.

At any fair casino it shouldn't matter how many other people are playing at the time either. I should have a 12 in 37 chance of winning 3x my money whenever I play a column in (single zero) roulette whether I'm playing alone or with 5 other people. That's kind of what "fair" means.

AvL42

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Quote from: dooglus on November 27, 2012, 12:47:10 PM

Now just suppose you had found a roulette game, for example, where they accidentally
paid out 3x on a 13-in-37 shot (paying out as if the probability of winning was 0.33333
when it's really 0.35135). The odds are slightly in my favour, but I only have 100 chips.

Your mind must be back in some fantasy-world, when you fantasize about such a roulette
game Wink

Quote from: dooglus on November 27, 2012, 12:47:10 PM

What's my optimum betting strategy to minimise my risk of ruin, and maximise my expected return?
When I played yesterday I was thinking it was best to bet 2 when my balance was over 73, and 1 otherwise.

The strategy isn't really all that bad: it at least statistically prolongs broke-ness slightly, compared to
"fixed-bet-2". What would be more interesting would be comparing your mixed strategy with "fixed-bet-1"
strategy in simulations, and also compare the average time to broke-ness in that course.
Without having calculated it myself, I'd predict still a slighly earlier broke-ness with your strategy
(compared to fixed-bet-1). Otoh, your strategy might perform better (than fixed-bet-1), if you also
set another line (e.g. on 200) which on reaching would make you terminally switch to
fixed-bet-0 (that is: quit&withdraw).

Feel free to pick a real casino or your fantasy one for simulations.

dooglus

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Quote from: AvL42 on November 27, 2012, 11:38:03 AM

The reason, why my utterings do NOT contradict the "each bet has an expected return of 0" is,
that you are only caring for those points in time, when an equilibrium of losses and wins is reached.

Yes, that's right. The reason I was caring most about those points in time is I was thinking those were both the most common points, and also the "average" points. The most common difference between "heads" and "tails" is 0. And for every positive difference, there's an equally likely equal and opposite negative difference. So if I'm losing at this central common point, then the +ve and -ve on either side cancel each other out, and I have a net -ve expectation. That was my reasoning, but the bolded part is incorrect. Because the profit when I have N more wins than losses is approximately twice the size of my loss when I have N more losses than wins. So the +ve swamps the -ve and makes up for the loss I suffer at the central "common" point.

Quote from: AvL42 on November 27, 2012, 11:38:03 AM

In real life, however, you just don't get infinite credit, so at some time you will be just broke,
and no longer able to continue playing. And that's where you'll indeed *never ever* get another
lose/win-equilibrium.

Right. In real life you get 100 chips.

Now just suppose you had found a roulette game, for example, where they accidentally paid out 3x on a 13-in-37 shot (paying out as if the probability of winning was 0.33333 when it's really 0.35135). The odds are slightly in my favour, but I only have 100 chips. What's my optimum betting strategy to minimise my risk of ruin, and maximise my expected return? When I played yesterday I was thinking it was best to bet 2 when my balance was over 73, and 1 otherwise. But I kept crossing the 73/74 line, and after a couple of hundred spins I had won and lost the same number of spins, but my balance had gone from 100 down to 70 or so. That's why I started this thread - I couldn't get my head around having had luck that felt like it should be break-even, but I'd managed to bet my way into a loss with it.

There are 20 ways in 6 bets of having 3 wins and 3 losses:

Code:

WWWLLL  2  4  6  4  2  0
WWLWLL  2  4  2  4  2  0
WWLLWL  2  4  2  0  2  0
WWLLLW  2  4  2  0 -2 -1
WLWWLL  2  0  2  4  2  0
WLWLWL  2  0  2  0  2  0
WLWLLW  2  0  2  0 -2 -1
WLLWWL  2  0 -2 -1  0 -2
WLLWLW  2  0 -2 -1 -2 -1
WLLLWW  2  0 -2 -3 -2 -1
LWWWLL -2 -1  0  2  0 -2
LWWLWL -2 -1  0 -2 -1 -2
LWWLLW -2 -1  0 -2 -3 -2
LWLWWL -2 -1 -2 -1  0 -2
LWLWLW -2 -1 -2 -1 -2 -1
LWLLWW -2 -1 -2 -3 -2 -1
LLWWWL -2 -3 -2 -1  0 -2
LLWWLW -2 -3 -2 -1 -2 -1
LLWLWW -2 -3 -2 -3 -2 -1
LLLWWW -2 -3 -4 -3 -2 -1

5 break even
9 lose 1
6 lose 2
0 bring a profit

Here are the other combinations of 6 bets. First the ones where I win more times than I lose:

Code:

WWWWWW  2  4  6  8 10 12
WWWWWL  2  4  6  8 10  8
WWWWLW  2  4  6  8  6  8
WWWLWW  2  4  6  4  6  8
WWLWWW  2  4  2  4  6  8
WLWWWW  2  0  2  4  6  8
LWWWWW -2 -1  0  2  4  6
WWWWLL  2  4  6  8  6  4
WWWLWL  2  4  6  4  6  4
WWWLLW  2  4  6  4  2  4
WWLWWL  2  4  2  4  6  4
WWLWLW  2  4  2  4  2  4
WWLLWW  2  4  2  0  2  4
WLWWWL  2  0  2  4  6  4
WLWWLW  2  0  2  4  2  4
WLWLWW  2  0  2  0  2  4
WLLWWW  2  0 -2 -1  0  2
LWWWWL -2 -1  0  2  4  2
LWWWLW -2 -1  0  2  0  2
LWLWWW -2 -1 -2 -1  0  2
LLWWWW -2 -3 -2 -1  0  2
LWWLWW -2 -1  0 -2 -1  0

and then the ones where I lose more times than I win:

Code:

WWLLLL  2  4  2  0 -2 -3
WLWLLL  2  0  2  0 -2 -3
WLLWLL  2  0 -2 -1 -2 -3
WLLLWL  2  0 -2 -3 -2 -3
WLLLLW  2  0 -2 -3 -4 -3
LWLWLL -2 -1 -2 -1 -2 -3
LWLLWL -2 -1 -2 -3 -2 -3
LWLLLW -2 -1 -2 -3 -4 -3
LLWWLL -2 -3 -2 -1 -2 -3
LLWLWL -2 -3 -2 -3 -2 -3
LLWLLW -2 -3 -2 -3 -4 -3
LLLWWL -2 -3 -4 -3 -2 -3
LLLWLW -2 -3 -4 -3 -4 -3
LLLLWW -2 -3 -4 -5 -4 -3
WLLLLL  2  0 -2 -3 -4 -5
LWWLLL -2 -1  0 -2 -3 -4
LWLLLL -2 -1 -2 -3 -4 -5
LLWLLL -2 -3 -2 -3 -4 -5
LLLWLL -2 -3 -4 -3 -4 -5
LLLLWL -2 -3 -4 -5 -4 -5
LLLLLW -2 -3 -4 -5 -6 -5
LLLLLL -2 -3 -4 -5 -6 -7

Summary:

Code:

#W < #L: 22 combinations,  losing a total of  83 chips
#W = #L: 20 combinations,  losing a total of  21 chips
#W > #L: 22 combinations, winning a total of 104 chips

dooglus

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Quote from: Isokivi on November 27, 2012, 11:43:33 AM

Atleast in physical blackjack without an autoshuffler counting cards and alering bets accordingly is a valid strategy to attack the houses edge.

Of course, if the odds change as you play then you want to bet more when the odds are in your favour, and less when they're not. But what I'm talking about here is a fixed odds game.

Isokivi

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Items flashing here available at btctrinkets.com

Atleast in physical blackjack without an autoshuffler counting cards and alering bets accordingly is a valid strategy to attack the houses edge.

AvL42

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The reason, why my utterings do NOT contradict the "each bet has an expected return of 0" is,
that you are only caring for those points in time, when an equilibrium of losses and wins is reached.

Using systematically variable bet-amounts will let it seem that you're losing or winning
(depending on actual strategy) at each such win/lose-equilibrium point, but that doesn't
say anything about streaks that *never ever* meet the equilibrium point. *Never ever*?
Now, that is of course statistically impossible while you keep the assumption of infinite credit.

In real life, however, you just don't get infinite credit, so at some time you will be just broke,
and no longer able to continue playing. And that's where you'll indeed *never ever* get another
lose/win-equilibrium. You'll be stuck at the total loss of budget, and that will be most likely at
some larger total number of losses than that of wins. While your money at these equilibrium-
points may grow or decline, the probability of end of game always compensates that for a total
expected return of 0.

PS: if there really was unlimited credit (and also unlimited amount per bet), then the expected
return would become mathematically incalculable (like zero divided by zero, or like an infinite
sum of positive and negative values as in (∞ + -∞). These types of incalculable values behave
like being every value at the same time as in 0*x=0 and ∞+x=∞ is correct for every x.)
So, for unlimited credit&bets, the expected return becomes *any value*.

dooglus

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Quote from: payb.tc on November 26, 2012, 11:59:24 PM

Quote from: dooglus on November 26, 2012, 05:52:45 PM

Quote from: AvL42 on November 26, 2012, 05:04:54 PM

Just play it reverse: When you're below the 1000-line, play two coins, and when above you play only one coin per toss. -> winning strategy *lol* *JK!*

Yes, it sounds stupid, but probably is just as valid as my argument. With your scheme it takes 2 losses to fall to 999 from 1001, but only one win to recover. So it is a winning strategy.

Now that's got to be wrong... Just wish I could see why!

wouldn't it take 2 wins to go from 999 to 1001? you still have to pay for each bet, so each win of 2 only increases your balance by 1.

AvL42 was suggesting that I bet 2 coins when my balance is less than 1000. When I'm at 999 and so bet 2 coins and then win, I make a profit of 2 coins, taking my balance to 1001.

SRoulette

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Quote from: DeathAndTaxes on November 26, 2012, 05:06:52 PM

Quote from: dooglus on November 26, 2012, 04:51:33 PM

I've always thought that it's not possible to change the expected return of a game with fixed odds by altering your bet size. I've seen this stated all over the place.

It isn't.

Quote

I was betting the 3-36 line over and over, starting with 100 chips. I was betting 2 chips each spin, but whenever I fell below 70 chips I was reducing my bet to 1 chip per spin, and increasing it back to 2 chips per spin when I got back to 70 chips.

I counted my wins and losses. Each time my number of wins was exactly equal to half my number of losses, I noticed my balance was a little lower than the previous time it happened.

You are merely observing the effect of a small sample size. First your number of wins shouldn't be exactly half. Due to the house odds it should be slightly less than half (the 0 & 00 numbers). The fact that it was exactly half was just conicidental. You simply won less of the larger wagers and more of the smaller wagers. Had the reverse been true would you believe you found a method to always beat the house? If you played for quadrillions upon quadrillions of spins and had infinite amount of money to lose your expected loss = (house odds) x (total amount bet). The more you play the more it aproaches this value, in the short term there can be anomoloies. Note: a hundred spins isn't statistically valid. Try doing 100,000 spins or more.

Quote

Suppose you're tossing a fair coin, and getting fair odds, but decide to play 2 chips per flip when you have 1000 or more chips, and 1 chip per toss when you have less than 1000 chips.

Your expected return is 0; you'll neither win nor lose in the long run, since you're getting fair odds. We can suppose the house is willing to give you unlimited credit, so going bust isn't a concern.

If I have 1000 chips and lose, I now only have 998 chips and will start playing for 1 chip each flip. ...

Your penny scenario is equally flawed because you assume you will lose on the first trial. What if you win, or win 30 times in a row. By simply looking at only the scenario's which begin below break even obviously your expected outcome will always be below breakeven. Once you lost to 998 going forward your expected gain is 0 and thus 998 is the break even. The coin has no memory that in the past you lost a bet and thus are "owed" one extra win.

You can simulate your penny game rather easily in software using a random number generator. You have it flip trillions of times and try every possible betting combination you can think of. In the end total return would be 0 (+/- margin of error).

What casinos should be careful of is that a large bet range coupled with a low house edge can increase amounts of rounds (and bank) required before the houses average earn smooths out.
Some gambling systems we are simulating require more than 1 trillion rounds for the casino to get its average earn matching the house edge within 10% each round.
[/quote]

Here is our output from running against under 32000 on SD.
A round for our simulator is starting at the minimum bet (0.001 btc) and doubling until either it wins or hits the wall (250 btc).
This sim ran for 100'000'000 rounds or 204816982 bets.

Code:

204816982 bets totaling 2219210.39558984. Profit: -54538.5617772679, -2.46066769%. Before Loss Average: rounds 165015.501650165, profit 0.1777. Best Streak: rounds 892527, profit 931.860691995724.

We are collecting some interesting statistics here as well:

Profit: How much the simulated player won / lost
Profit Percent: This should end up matching the house edge but you need to simulate enough rounds for the data to finally smooth out, eg 1 trillion.
Rounds Before Loss Average: how many successful rounds were played before it failed.
Profit Before Loss Average: Average amount won before system fails.
Best Streak: count of the biggest successive win for the system.
Best Streak profit: the profit made from best streak.

Thankfully our simulator is now threaded so we can grunt through some heavy stats quickly but it still needs a lot of work

.

DannyHamilton

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Lets consider the possible outcomes of your coin game assuming that we start with 1000 and therefore bet 2 coins on the first toss. We'll take a look at 4 consecutive tosses just to see what the possible outcomes are:

WWWW = Win 8 chips
WWWL = Win 4 chips
WWLW = Win 4 chips
WWLL = Even
WLWW = Win 4 chips
WLWL = Even
WLLW = Lose 1 chip
WLLL = Lose 3 chips
LWWW = Win 2 chips
LWWL = Lose 2 chips
LWLW = Lose 1 chip
LWLL = Lose 3 chips
LLWW = Lose 1 chip
LLWL = Lose 3 chips
LLLW = Lose 3 chips
LLLL = Lose 5 chips

If we add up all the scenarios that win, we get a total of 22 chips
If we add up all the scenarios that lose, we get a total of 22 chips

Sounds like there is no advantage to this betting scheme to me.

Fjordbit

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firstbits.com/1kznfw

You can illustrate that you are incorrect through induction. On the first flip you start with 100 coins and have 50% chance of losing or 50% chance of winning. If you lose, then your expected outcome is 998, if you win then you expected outcome is 1002. Because .5*998 + .5*1002 = 1000, you see that the expected outcome is the same as your starting coins.

Now assume that after k rounds, your expected value is 1000.

Prove for k+1. This is pretty simple, you know that E[k+1] = E[k] + .5 * -bet + .5 * bet, thus E[k+1] = E[k]. This holds true for any bet so long as they are equal size because the positive balances the negative.

Therefor by induction, the expected value after any arbitrary number of rounds is 1000.

payb.tc

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Quote from: dooglus on November 26, 2012, 05:52:45 PM

Quote from: AvL42 on November 26, 2012, 05:04:54 PM

Just play it reverse: When you're below the 1000-line, play two coins, and when above you play only one coin per toss. -> winning strategy *lol* *JK!*

Yes, it sounds stupid, but probably is just as valid as my argument. With your scheme it takes 2 losses to fall to 999 from 1001, but only one win to recover. So it is a winning strategy.

Now that's got to be wrong... Just wish I could see why!

wouldn't it take 2 wins to go from 999 to 1001? you still have to pay for each bet, so each win of 2 only increases your balance by 1.

flooraccount

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Quote from: dooglus on November 26, 2012, 11:30:50 PM

Quote from: flooraccount on November 26, 2012, 11:14:42 PM

Just a thought: Are you taking into account that other people are playing at the same time as you?

I'm assuming each turn is independent of all other turns, so I don't care if anyone else is playing or not.

Well, not necessarily true. Odds are Odds. But when other people are playing this could increase or reduce your result.
I would compare this to satoshidice. Odds, Luck, and amount of players can manipulate the result if that makes sense.

dooglus

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Quote from: flooraccount on November 26, 2012, 11:14:42 PM

Just a thought: Are you taking into account that other people are playing at the same time as you?

I'm assuming each turn is independent of all other turns, so I don't care if anyone else is playing or not.

flooraccount

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Quote from: dooglus on November 26, 2012, 10:59:04 PM

Quote from: crowe on November 26, 2012, 10:55:54 PM

Your expected value on any/every flip is 0. No matter whether you are betting 1 or 2, the EV of each flip is exactly the same. Thus, your long term EV is also the same. Everything else is just noise and variance.

I realised that. But I also had found an argument that convinced me that I had a negative expectation, which contradicted what I already knew. So I wanted to clear things up and understand what was wrong with my argument.

Turns out my argument was just saying that the most likely result was that I would lose, which is true, and doesn't contradict that the EV is 0.

Just a thought: Are you taking into account that other people are playing at the same time as you?

dooglus

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Quote from: crowe on November 26, 2012, 10:55:54 PM

Your expected value on any/every flip is 0. No matter whether you are betting 1 or 2, the EV of each flip is exactly the same. Thus, your long term EV is also the same. Everything else is just noise and variance.

I realised that. But I also had found an argument that convinced me that I had a negative expectation, which contradicted what I already knew. So I wanted to clear things up and understand what was wrong with my argument.

Turns out my argument was just saying that the most likely result was that I would lose, which is true, and doesn't contradict that the EV is 0.

crowe

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Your expected value on any/every flip is 0. No matter whether you are betting 1 or 2, the EV of each flip is exactly the same. Thus, your long term EV is also the same. Everything else is just noise and variance.

As long as each event is 50/50 and the amount you win is equal to the amount you can lose, the EV is 0. You will win half your bets of 2, and lose half your bets of 2. Same when wagering 1. You will lose half, and win half.

dooglus

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(07:39:32 PM) deb: You people really get off on this shit don't You?
(07:39:57 PM) Chris: it's not a sexual thing
(07:40:04 PM) deb: i mean no disrespect, i just can't fathom where making charts to work out gambling strategies that don't really matter is a good thing to do
(07:40:04 PM) Chris: it's more of an obsessive thing
(07:40:12 PM) Chris: I couldn't relax before I understood wtf was going on

dooglus

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Quote from: MPOE-PR on November 26, 2012, 07:40:00 PM

I think this is incorrectly modeled as a "single game". Your overall expectation is to make it out even on all games where you bet 1 coin, and on all games where you bet 2 coins. Overall, they should both even out and so their sum should also even out.

However, if you do something like 50 games A lost + 50 games B lost + 40 games A won + 60 games B won, the situation isn't correctly described as "100 won, 100 lost".

Yes. Once I've started playing the 1 game (A?), I only play the 2 game (B?) if I've won more than I've lost at A. So when my wins and losses eventually match, A will have more wins than losses, and B will have more losses than wins, and so I'll be making a loss overall.

It's OK that the most common result (which is #wins == #losses) is a net loss, because the whole curve is skewed to the left. The most likely final balance is a little less than the starting 1000, but there's a good sized chunk of significant winnings to the right. Overall, the average is 1000:

Edit: Notice how the likely range of balances when I lose (600-1000) is half as wide as the likely range of balances when I win (1000-1800).

The red (losing) area is twice the size of the green (winning) area, so we're twice as likely to lose as to win. But it's much narrower, so if we lose, we don't lose much, but if we win, we're likely to win a decent amount.

MPOE-PR

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I think this is incorrectly modeled as a "single game". Your overall expectation is to make it out even on all games where you bet 1 coin, and on all games where you bet 2 coins. Overall, they should both even out and so their sum should also even out.

However, if you do something like 50 games A lost + 50 games B lost + 40 games A won + 60 games B won, the situation isn't correctly described as "100 won, 100 lost".

dooglus

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Quote from: DeathAndTaxes on November 26, 2012, 06:19:23 PM

You do realize that +16,000 is within the margin of error right. 16,000 deviation on 2 billion flips is 0.0008% of expected.

Yes, I realise. That's what I was getting at with:

Quote from: dooglus on November 26, 2012, 06:00:56 PM

I suspect that the 7484 crossings is insignificant compared to the amounts won and lost, and that's why the expected return is still 0. Could that be the case?

But I was wondering how it was possible that the expected return wasn't at least slightly lower than zero, given that crossing from 1000 to 998 effectively "wastes" a loss.

I think I have a reasonable explanation of that now:

The losses incurred each time we cross from 1000 to 998 are offset by the fact that winnings are slightly skewed upwards. For example, if we win our first 5 bets, we win 10 coins, but if we lose our first five bets, we lose only 6 coins. So the expected return can be thought of in a hand-waving way as:

(p(lots more wins than losses) * big_win) +
(p(lots more losses than wins) * smaller_loss) +
(p(similar wins and losses) * loss_due_to_crossings_from_1000_to_998)

where the first term is cancelled out by the other two.

It's very rough and ready, but I think that's enough to let me stop thinking about it!

AvL42

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Reversing the original plan (i.e. betting 2 when below and only 1 when above)
can even be extended further (and often is):

- When I lose once, I repeat the bet with double value.
- When I lose again, I double again
- ...
- When I win, I've compensated the previous losses.

Assuming the broke-free environment, then at some point every bad luck
streak would break, and you'd end up debt-free with just one lucky throw.

How does it work? Mathematics trolled? Nope.

It's the assumed broke-free environment that's crucial to the reverse "strategies".
(Also, in reality there's typically a max-limit on bets - primarily to cap such cascades.)

By increasing the bets for lower budget, you improve your payout-performance
but also boost your crash&burn broke-likeliness.

By lowering your bets for lower budget, you can avert broke-ness for somewhat
longer, but the average luck will give you less money.

What you found on the net about variable betting is most likely for random
variable betting-amounts. It's not true for such systematically chosen betting
amounts.

DeathAndTaxes

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Gerald Davis

You do realize that +16,000 is within the margin of error right. 16,000 deviation on 2 billion flips is 0.0008% of expected. Trying running it 10 billion, or a trillion, or a quadrillion you will notice the deviation will continue to shrink as a % of the expected. In other words as there are more and more trials the actual outcome approaches the expected outcome.

Expected Outcome is called that because it is EXPECTED but you rarely will ever hit it EXACTLY (i.e. actual outcome = expected outcome) as an ironic twist of math as the same size gets larger the actual outcome will approach the expected outcome HOWEVER the odds that it will be exactly equal to the expected outcome decrease.

This is easy to observe:
2 flips. expected outcome = 1 heads. Odds of actual outcome matching expected outcome = 2 out of 4 combinations (50%)
4 flips. expected outcome = 2 heads. Odds of actual outcome matching expected outcome = 4 out of 16 combinations (25%)

dooglus

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Quote from: DeathAndTaxes on November 26, 2012, 05:06:52 PM

You can simulate your penny game rather easily in software using a random number generator. You have it flip trillions of times and try every possible betting combination you can think of. In the end total return would be 0 (+/- margin of error).

I started with 1000 chips, and tossed the coin 1 billion times.

I got 499985225 wins and 500014775 losses
29550 more losses than wins
and a final balance of -36034

Each of the 29550 excess losses should have lost me 1 chip, so I should be at -28550, but I'm 7484 chips short of that.

I got the simulation to print out each time the number of wins and losses were the same. The last time that happened before the billionth turn was after 612,215,220 turns, at which point there had been 306107610 wins and 306107610 losses, and the balance was -6484. ie. 7484 below where I started. This suggests to me that the balance crossed from 1000 to 998 7484 times, 'wasting' 1 win each time.

I suspect that the 7484 crossings is insignificant compared to the amounts won and lost, and that's why the expected return is still 0. Could that be the case?

Edit: I left the simulation running. After 2 billion tosses I was back in profit:

2000000000 turns; 1000008100 wins; 999991900 losses; 16200 more wins than losses; 6002 balance

the last time the numbers of wins and losses were the same:

1821704084 turns; 910852042 wins and 910852042 losses; -8477 balance

so now there have been 9477 crossings from 1000 to 998, and my balance is only positive because I have been lucky; even though I've ~~have~~ had 16k more wins than losses, I'm only 5k up.

Here's the Python script I used by the way: http://pastie.org/5439806

dooglus

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Quote from: AvL42 on November 26, 2012, 05:04:54 PM

Just play it reverse: When you're below the 1000-line, play two coins, and when above you play only one coin per toss. -> winning strategy *lol* *JK!*

Yes, it sounds stupid, but probably is just as valid as my argument. With your scheme it takes 2 losses to fall to 999 from 1001, but only one win to recover. So it is a winning strategy.

Now that's got to be wrong... Just wish I could see why!

DeathAndTaxes

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Gerald Davis

You can't start after an event and consider that part of the random set.

i.e.
Either starting at 1,000 he has equally chance of winning or losing. Expected outcome is 1,000 for an any number of flips going forward.

If he ALREADY (as in the past) lost 2 and only has 998 chips then 998 chips becomes the starting point and he has an equal chance of winning or losing and 998 is the expected outcome for any number of flips going forward.

if he is starting with 998 and wants to acheive 1,000 coins he is looking to gain two. then the odds of that ARE going to be less than zero. Lets take this to the logical extreme. Say he played his game until he had lost all but 2 chips. Now say from this point he plays until he either loses all chips or gets back to 1,000. The odds of that are <1%. Would you say the odds of having break even or better on a coin flip are <1%? Of course not. The odds are only so low because he is trying to turn 2 chips into 1,000 chips. What happened BEFORE he ended up at 2 chips is irrelivent. The coins don't "know" he originally started at 1,000 and lost it down to 2. The odds going forward are exactly the same as any other fair coin flip.

dooglus

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Quote from: DeathAndTaxes on November 26, 2012, 05:06:52 PM

Your penny scenario is flawed because you assume you will lose on the first trial. What if you win, or win 30 times in a row. By simply looking at only the scenario's which begin below break even obviously your expected outcome will always be below breakeven.

I'm thinking that since it's a fair coin, my path will be essentially a random walk through profit and loss. So if I do win the first trial, at some point in the future I'll be back to 1000 coins, and can effectively discount all the coin tosses in between, since the wins and the losses will be equal.

If I lose that first trial, or any trial when my balance is 1000, I'll lose 2 coins and 'waste' 2 wins getting back to even. Other than that everything seems equal, so it appears to me that this case, which will keep happening each time I cross from 1000 to 998, will result in me losing in the long run.

I agree that this is probably incorrect, but can't see the hole in my reasoning.

Quote

You can simulate your penny game rather easily in software using a random number generator. You have it flip trillions of times and try every possible betting combination you can think of. In the end total return would be 0 (+/- margin of error).

I will do. Here's a worked example of how I get 22 wins and only 20 losses, but break even:

Code:

balance results wins losses
1000
W,W,L,W,W,L,L,L 4 4
1000
L 0 1
998
W,L,L,W,W,L,W,W 5 3
1000
W,W,L,L 2 2
1000
L 0 1
998
W,W 2 0
1000
W,W,L,W,W,L,L,W,W,L,L,W,W,L,L,W,L,L 9 9
1000 -- --
22 20

AvL42

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Quote from: DeathAndTaxes on November 26, 2012, 05:06:52 PM

Quote from: dooglus on November 26, 2012, 04:51:33 PM

I've always thought that it's not possible to change the expected return of a game with fixed odds by altering your bet size. I've seen this stated all over the place.

It isn't.

Actually it is. Crossing the borderline downwards means he lost 2 in one game, while
crossing it upwards means he won one in one win-game. If you then pair up the total
wins and losses, then those crossing the border will be (almost) equal (as you can't cross
the line twice in same direction without at least once the other direction inbetween),
but the upward crosses will have earned less than the downward crosses lost.

DeathAndTaxes

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Gerald Davis

Quote from: dooglus on November 26, 2012, 04:51:33 PM

I've always thought that it's not possible to change the expected return of a game with fixed odds by altering your bet size. I've seen this stated all over the place.

It isn't.

Quote

I was betting the 3-36 line over and over, starting with 100 chips. I was betting 2 chips each spin, but whenever I fell below 70 chips I was reducing my bet to 1 chip per spin, and increasing it back to 2 chips per spin when I got back to 70 chips.

I counted my wins and losses. Each time my number of wins was exactly equal to half my number of losses, I noticed my balance was a little lower than the previous time it happened.

You are merely observing the effect of a small sample size. First your number of wins shouldn't be exactly half. Due to the house odds it should be slightly less than half (the 0 & 00 numbers). The fact that it was exactly half was just conicidental. You simply won less of the larger wagers and more of the smaller wagers. Had the reverse been true would you believe you found a method to always beat the house? If you played for quadrillions upon quadrillions of spins and had infinite amount of money to lose your expected loss = (house odds) x (total amount bet). The more you play the more it aproaches this value, in the short term there can be anomoloies. Note: a hundred spins isn't statistically valid. Try doing 100,000 spins or more.

Quote

Suppose you're tossing a fair coin, and getting fair odds, but decide to play 2 chips per flip when you have 1000 or more chips, and 1 chip per toss when you have less than 1000 chips.

Your expected return is 0; you'll neither win nor lose in the long run, since you're getting fair odds. We can suppose the house is willing to give you unlimited credit, so going bust isn't a concern.

If I have 1000 chips and lose, I now only have 998 chips and will start playing for 1 chip each flip. ...

Your penny scenario is equally flawed because you assume you will lose on the first trial. What if you win, or win 30 times in a row. By simply looking at only the scenario's which begin below break even obviously your expected outcome will always be below breakeven. Once you lost to 998 going forward your expected gain is 0 and thus 998 is the break even. The coin has no memory that in the past you lost a bet and thus are "owed" one extra win.

You can simulate your penny game rather easily in software using a random number generator. You have it flip trillions of times and try every possible betting combination you can think of. In the end total return would be 0 (+/- margin of error).

AvL42

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Just play it reverse: When you're below the 1000-line, play two coins, and when above you play only one coin per toss. -> winning strategy *lol* *JK!*

dooglus

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I've always thought that it's not possible to change the expected return of a game with fixed odds by altering your bet size. I've seen this stated all over the place.

For the last hour or two I've been playing roulette at luckybitcoincasino, and seem to have found a way of reducing my expected return.

I was betting the 3-36 line over and over, starting with 100 chips. I was betting 2 chips each spin, but whenever I fell below 70 chips I was reducing my bet to 1 chip per spin, and increasing it back to 2 chips per spin when I got back to 70 chips.

I counted my wins and losses. Each time my number of wins was exactly equal to half my number of losses, I noticed my balance was a little lower than the previous time it happened.

I would expect that since a win pays out twice as much as a loss loses, having N wins and 2N ~~wins~~ losses should mean I was breaking even, but that wasn't the case.

I tried to understand why, and came up with this thought experiment:

Quote

Suppose you're tossing a fair coin, and getting fair odds, but decide to play 2 chips per flip when you have 1000 or more chips, and 1 chip per toss when you have less than 1000 chips.

Your expected return is 0; you'll neither win nor lose in the long run, since you're getting fair odds. We can suppose the house is willing to give you unlimited credit, so going bust isn't a concern.

If I have 1000 chips and lose, I now only have 998 chips and will start playing for 1 chip each flip.

I'm not going to get back to 1000 chips until my following sequence of plays has 2 more wins than losses (N losses, and N+2 wins, say), at which point I'm going to start playing for 2 chips per flip again.

Each time I lose at 1000 chips, it takes a total of N+1 losses (including the loss of 2 coins) and N+2 wins to break even.

Since in the long run I can expect my number of losses to be equal to my number of wins, doesn't that imply I can expect to not break even, long term, since each time I cross the 1000 threshold I need 1 extra unmatched win to make up my loss?

So what's wrong with the above argument? It looks to me like I've found a strategy that reduces my expected return, making it negative in a zero-expectation coin toss game. I must be missing something. Mustn't I?

Topic: gambling question - changing house edge by altering bet size? (Read 2616 times)