gambling question - changing house edge by altering bet size?

[Fjordbit]

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.

[DannyHamilton]

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.

[SRoulette]

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.  

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.

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 .

[dooglus]

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.

[AvL42]

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*.

[Isokivi]

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

[dooglus]

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.

[dooglus]

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.

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

[AvL42]

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

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]

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.

[darkmule]

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]

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

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...

[AvL42]

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]

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]

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.