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[Financisto]

Dear Friends,

Watching that so far discussion about martingale stuff reminded me about an interesting text I once read.

The topic is well known (progressive systems do not work the way we all want), but the math is strong in order to explain that.

It's named: "Mathematical Proof that Progressions cannot overcome Expectation."

I leave here the direct link to it:

http://www.bjmath.com/bjmath/progress/unfair.htm

And there is still more about progressive systems here:

http://www.bjmath.com/bjmath/progress/progress.htm

Enjoy it!

[willphase]

Dear Friends,

Watching that so far discussion about martingale stuff reminded me about an interesting text I once read.

It is possible to reduce the casino edge by just betting less, as per the 2-composite bet example Dooglus explained above.  I asked a friend to formalise the proof - here is his paper [link].

I hope this puts the subject of whether progressive betting is better than one single bet, to rest

Will

[dooglus]

Dear Friends,

Watching that so far discussion about martingale stuff reminded me about an interesting text I once read.

The topic is well known (progressive systems do not work the way we all want), but the math is strong in order to explain that.

It's named: "Mathematical Proof that Progressions cannot overcome Expectation."

I leave here the direct link to it:

http://www.bjmath.com/bjmath/progress/unfair.htm

And there is still more about progressive systems here:

http://www.bjmath.com/bjmath/progress/progress.htm

Enjoy it!

I've read them before.

Note that I'm not saying martingale overcomes expectation.  The expectation is that you lose 1% of everything you bet.  What I am saying is that martingale allows you to achieve your goal of turning X into Y without having to bet the whole amount X every time you try.  Since you expect to bet less than X, you expect to lose less than 1% of X, whereas if you bet the whole X in one go, you expect to lose 1% of X.

Having read the those links, do you agree with me, or everyone else?  It feels like I'm on my own here (but also right).  I wonder if this is how Dabs feels?  

[dooglus]

It is possible to reduce the casino edge by just betting less, as per the 2-composite bet example Dooglus explained above.  I asked a friend to formalise the proof - here is his paper [link].

If only you hadn't said it reduces the edge...  The edge is 1% and not subject to change.  It's possible to bet less, and so reduce expected losses.

[willphase]

It is possible to reduce the casino edge by just betting less, as per the 2-composite bet example Dooglus explained above.  I asked a friend to formalise the proof - here is his paper [link].

If only you hadn't said it reduces the edge...  The edge is 1% and not subject to change.  It's possible to bet less, and so reduce expected losses.

and now you quoted me so I can't fix it   Argh I just quoted you quoting me!!

Argh Q U O T E C E P T I O N

Will

[Zaih]

I tried explaining this to him for a very long time..

It won't work. He's a persistent bugger, he's having fun though, so fair enough.

Do you still think Martingale is worse than the equivalent single bet?

I thought we had been through that before and that you finally agreed with me.

We were going to bet on it, until you realised I was right, no?

I agree, but the way Dabs is doing it, he would be better off just doing a one off bet?

[mechs]

I tried explaining this to him for a very long time..

It won't work. He's a persistent bugger, he's having fun though, so fair enough.

Do you still think Martingale is worse than the equivalent single bet?

I thought we had been through that before and that you finally agreed with me.

We were going to bet on it, until you realised I was right, no?

I agree, but the way Dabs is doing it, he would be better off just doing a one off bet?

But where would the fun be in that?

[Financisto]

(...)

Having read the those links, do you agree with me, or everyone else?  It feels like I'm on my own here (but also right).  I wonder if this is how Dabs feels?  

Yes, I do!

(...) It is possible to reduce the casino edge by just betting less...

No, it's not.

Only by cheating... The misconception that progressive betting systems may overcome (or even reduce) the house's edge is just one more gamblers fallacy.

Money management ("betting less") may keep you longer in the game, the same for the 1% (lower) house edge.

I hope this puts the subject of whether progressive betting is better than one single bet, to rest

Sure it is.

But it does not turn the odds in your favor (and against the house's edge). It just keeps you alive a little longer.

You can get some "luck" (profit) in the short term if you surf the "waves" of an advantageous variance. But that's not going to happen every time with every gambler. Variance is tricky!

BTW, take a look at this article/simulation: http://vegasclick.com/gambling/million-dollar-bets.html

[superresistant]

No matter what complicated strategy you use, you're better off making a single straight-up bet for the amount you aim to win.

If your goal is to gain 20% per month with a high-probability Martingale-style strategy, your chances are improved by just making a single bet at 1.2x profit each month with the entire capital. The higher the probability you set in your strategy, the less likely it is to hit a losing streak long enough to break your bank, but the more runs you need to do to achieve the same profit, offsetting the reduced chance to lose.

You can run the numbers and see this quite clearly, but there's also an intuitive way to think about it. JD takes, on average, 1% of the amount you bet. If you make a single bet, you have an expected 1% loss on the amount wagered. If you use a Martingale-style strategy, you bet the same money more than once. To achieve, say, 20% profit, your total amount wagered will be more with a Martingale-style strategy than with a single 1.2x profit bet. And since in the long run, you expect your losses to reach 1% of the total amount wagered, the Martingale-style strategy loses. (the 20% profit is an example, it holds true for other values as well)

tldr: Dabs, if you set a monthly profit goal, I can outperform your strategy by placing a single bet each month (if I had magic beans, special seeds, dragons and all the yada yada of course).

That is what the Dubins and Savage law (1956) say but it apply on infinite number of bets and profit. The point of martingale is to reduce the chance of loosing when aiming for smaller profit.

Dabs is now aiming for about 10% profit : Martingale is the best way to reach the goal and I think his maths are right if we just consider that small profit area.

[organofcorti]

I think people are confusing a "true" martingale sequence with a realistic one. A true martingale betting sequence has no limits on either profit or bankroll. Or maybe "expected value" is confusing.

Ignoring the house edge, the expected profit value of a "true" martingale sequence is always a*(1 - p)/p where p = probability of winning and a = the starting amount. It's what defines a martingale sequence.

Edit: I originally had an error in the next section, but dooglus showed me the error of my way. What follows is updated, corrected, and AFAICT correct.


When a bound is set the result is different. For example, if a bettor plays a 50% chance to win game and starts with a = 1 and stops betting after the n = 9th loss in a row, the total expected profit is the expected profit from up to 8 losses (and then a win) + the expected profit of greater than 8 losses in a row:

Code:

E(profit) = Pr(losses in a row <= 8 )*(1 - p)/p - Pr(losses in a row  > 8)* sum(2^0+2^1+...+2^8)
          = (1-(1 - 0.5)^9) - (1 - 0.5)^9 * 511
          = (1-1/512 -  1/512 * 511)
          = 0

So in this particular case, the expected profit is -1, (even before the house edge is taken into account).

This result can be generalised as you would expect. The geometric CDF for losses in a row = 0, 1, 2, ... n is:

Code:

Lower tail = Pr(losses in a row <= n): (1 - (1 - p)^(n+1))
Upper tail = Pr(losses in a row > n): (1 - p)^(n+1)

Using the above and the definition of a martingale sequence, the expected profit of a "realistic" martingale sequence (still ignoring the house edge), where a bettor stops betting after n losses in a row is:

Code:

E(profit for losses in a row <= n): (1 - (1 - p)^(n+1)) * a * (1 - p) / p
E(profit for losses in a row > n): -(1 - p)^(n + 1) * a*(((1 - p)^(-n) - 1)/p + 1)
E(profit) = (1 - (1 - p)^(n+1)) * a * (1 - p) / p -(1 - p)^(n + 1) * a*(((1 - p)^(-n) - 1)/p + 1)
          = 0

The interesting thing here is that no matter the largest number of losses in a row you can manage, the expected profit for a bounded martingale sequence will always be 0. This means that unless both you and the casino have an infinite bankroll and bets are unlimited, your expected profit will always be 0, even at a casino with no house edge.

tl;dr:
1. The expected profit of a martingale sequence is a*(1 - p)/p
2. The expected profit of a bounded martingale sequence is 0.

[Dabs]

Edit: and am also a little concerned that the peopleperson most aligned with my views according to recent posts is Dabs!  

Having read the those links, do you agree with me, or everyone else?  It feels like I'm on my own here (but also right).  I wonder if this is how Dabs feels?  

I agree.

Thanks for the support! You can email me at ... ... ...

Ok, seriously, I don't really know. I just read all the answers and all the links and all the math.

But I do agree that 2 bets are better than 1, and 3 bets are better than 2. So 7 bets are better than 6, as long as you follow a similar progression. The output or result in the end will be the same (or very close) but the odds of surviving the longer sequence is much better than just one big giant bet.

The next bet (this so called 7th bet) can be considered a one off bet. The next 100k rolls is a martingale strategy that is programmed to survive if it wins the 6th. As per the Gambler's Fallacy, since I just lost one sequence of 6, the next sequence of 6 will probably not occur until at least 300k to 500k rolls. Half of that is 150k rolls. 100k rolls is a third of that. So the probability of encountering a 6 loss sequence ...

Also, don't confuse "true" martingale sequences and "real" martingale sequences with "modified" martingale strategies that involve sequences, but stop or reset at certain points, and adjust depending on other variables, as well as giant beanstalks and magic seeds, dragons, dung beetles and voodoo math.

Ok. Maybe we should just ask a third party to hold coins to bet against my JD bet? Are you willing to bet that Dabs will lose either this next roll, or within the next 100k rolls?

*edit* also, don't forget that we're allowed to bet zero, but has a time penalty (currently 3 seconds) for doing so. So I can lose a sequence betting zero or a few satoshis (so the wait is reduced to 1 second.)

[deepceleron]

You will find correct statistics on few gambling strategy sites. People who are competent in math don't gamble (unless it's gamble or death - let's say the mob will kill you if you don't pay back $100,000; if you gamble the $10,000 you do have, there's a 10% chance you won't be murdered). They also don't waste the time making strategy guides.

To answer a question whether it is better to bet one large or many small amounts, that depends on your goal. If you gamble for entertainment, the goal would be to extend your entertainment, not bet your bankroll on one roulette number - a 36 in 37 chance your fun is over in 30 seconds. Play bingo, it wastes all your time for little money. If your goal is to win money, then you should invest in this site's bankroll or open your own casino.

Just like interest compounded daily earns you more than interest compounded monthly, more bets of the same player bankroll earns the casino a compounding return.

Strictly speaking, one large bet has a better probability of a given return vs bust, if the goal is the single bet win payout.

What do I mean by "the goal" - this takes some explanation. Let's take the roulette table, with 2/37 house edge (5.26%). If my goal is to make 35x my bankroll, betting 100% of my bankroll on a number gives me a 2.6% chance of meeting my goal. Betting 1% of my bankroll on a series of number bets until I meet my goal or go bust is a near 100% chance of bust.

If your goal is simply to come out ahead after a set number of bets (where you have enough money to not go bust until you've placed all the bets):

A. the chance you will meet this goal betting a single number with one bet is only 2.6% - although it has a bigger payoff potential, there is a 97% you leave down (actually busted). However, if you bet 100 times on the single number, the chance you will leave the table ahead is 49.1%.

B. the chance you will meet this goal betting once on a color is 47.4%, but after betting 100 times in a row on a color you only have a 26.5% chance of being up at the end.

(If you place the 100 bets in sequence and bet more than 1/100th your bankroll on each bet, there is a chance you go bust before the bet sequence completes, and this further lowers your chance of ending ahead.)

Similarly, if I state my goal is either to double my money or go bust on just-dice, the chance that I double my money with one bet is 49.5%. If I bet 1/10th of that on the same bet until I either double my money or go bust, the chance is higher I go bust (not going to waste my time in R to tell you the odds...)


I'll add a conclusion to my post: gamblers like variance. People buy lottery tickets not because the house edge is 50%, but because there is miniscule chance of millionaire status. The more you bet, the more you reduce variance. If I have 10,000BTC and bet it once on double my money, 49.5% chance I double my money. If I bet 1000000 0.01 BTC bets, the variance is gone and payout approaches the house edge, I might as well be sending 10000 BTC to a site that pays 9900 back.

[organofcorti]

...

Similarly, if I state my goal is either to double my money or go bust on just-dice, the chance that I double my money with one bet is 49.5%. If I bet 1/10th of that on the same bet until I either double my money or go bust, the chance is higher I go bust (not going to waste my time in R to tell you the odds...)
...

This is what I plan on doing.  Figuring the odds for a given gain using martingale.  For the same risk I'm sure it is less than making a single bet.  The math is not easy.  I've done simulations and the probability distribution function is complicated.  I'm going to use the simulation to figure the odds.

I stated before that I thought that martingale had a smaller expected value than a single bet.  That thinking was not only wrong but the utility of EV for what I'm trying to show isn't even correct.

Here is what is correct.  The expected profit = -(house edge)*sum(every bet).  This is true no mater what bet sequence you use.

Betting on the house side is the way to have a positive profit.

What exactly do you mean by "expected profit"? How exactly are you calculating it? Is it the expected profit per sequence or per roll? The expected value is a weighted average - what are your weightings and their associated probabilities?

Not hassling, just interested - you haven't provided enough information for me to know what you mean.

[DomenicoRomano]

...
Here: suppose you have 3 BTC, bet 1 BTC at 2x, and if you lose, bet the other 2 BTC at 2x.  2x bets have a 0.495 probability of winning.  You win 1 BTC total unless both bets lose.  Both bets lose with p=0.505*0.505 = 0.255025, and so you win with p=0.744975

The equivalent single bet would pay out 99/74.4975 = 1.32890365448505x and would return 3.98671096345515 BTC.

That's less than the 4 BTC returned by the martingale sequence with the same probability.

OK. I just worked this little example out and you are right:

Strategy 1: Bet 3 BTC, p = 0.7425         - get 4 BTC with p = 0.7425
Strategy 2: Bet 1 BTC, p = 0.495,
                if lose bet 2 BTC, p = 0.495. - get 4 BTC with p = 0.744975

Strategy 2 has better odds, by p = 0.002475 or 0.3%.  Strategy 2 is martingale starting bet of 1, exit of 4 or bust.

This is explained by the fact that Strategy 2 doesn't bet 3 BTC unless the first bet loses whereas Strategy 1 bets 3 BTC.  Your losses are proportional to the total amount bet.  The average amount bet with Strategy 2 is 2 BTC.  This is an example of the principal that: The less you bet the less you lose.

The problem with my simulation is that I haven't put the exits in.  It's quitting after winning your 1 BTC that gives Strategy 2 an advantage.

On a personal note: I really didn't want to believe this, but now I do.

[organofcorti]

...

Similarly, if I state my goal is either to double my money or go bust on just-dice, the chance that I double my money with one bet is 49.5%. If I bet 1/10th of that on the same bet until I either double my money or go bust, the chance is higher I go bust (not going to waste my time in R to tell you the odds...)
...

This is what I plan on doing.  Figuring the odds for a given gain using martingale.  For the same risk I'm sure it is less than making a single bet.  The math is not easy.  I've done simulations and the probability distribution function is complicated.  I'm going to use the simulation to figure the odds.

I stated before that I thought that martingale had a smaller expected value than a single bet.  That thinking was not only wrong but the utility of EV for what I'm trying to show isn't even correct.

Here is what is correct.  The expected profit = -(house edge)*sum(every bet).  This is true no mater what bet sequence you use.

Betting on the house side is the way to have a positive profit.

What exactly do you mean by "expected profit"? How exactly are you calculating it? Is it the expected profit per sequence or per roll? The expected value is a weighted average - what are your weightings and their associated probabilities?

Not hassling, just interested - you haven't provided enough information for me to know what you mean.

Expected profit is how much money you make on average, Total winnings - Total wagers. 

I'm sorry, I'm still not getting it.

When you say "Expected profit is how much money you make on average", over what is it the average? Per roll or per sequence? Also "Total winnings - Total wagers" is not an average, it's a difference. So you've stumped me - I can't see how you're getting an average there.

Every bet will, on average, make -1% of the amount wagered.  This is true no matter what the sequence.  To get the total average profit just add up all the bets and multiply by -1%.

Not quite true - if you look a few posts above I derived the expected profit for both an unbounded and a bounded martingale sequence without house edge and it's not the sum of all bets, which is what I think you're saying it would be if there was no house edge.

[dooglus]

For example, if a bettor plays a 50% chance to win game and starts with a = 1 and stops betting after the n = 9th loss in a row, the total expected profit is the expected profit from up to 8 losses (and then a win) + the expected profit of greater than 8 losses in a row:

Code:

E(profit) = Pr(losses in a row <= 8 )*(1 - p)/p - Pr(losses in a row  > 8)* sum(2^0+2^1+...+2^9)
          = (1-(1 - 0.5)^9) - (1 - 0.5)^9 * 1023
          = (1-1/512 -  1/512 * 1023)
          = - 1

So in this particular case, the expected profit is -1, (even before the house edge is taken into account).

You've found a betting strategy for a 0% game which gives a non zero expectation?

When that happens, your math must be wrong.

In this case, the 9 losses in a row are 2^0, 2^1 .. 2^8, which sum to 2^9-1 or 511, not 1023.

E(profit) = (1-1/512 -  1/512 * 511) = 0

[dooglus]

Similarly, if I state my goal is either to double my money or go bust on just-dice, the chance that I double my money with one bet is 49.5%. If I bet 1/10th of that on the same bet until I either double my money or go bust, the chance is higher I go bust (not going to waste my time in R to tell you the odds...)

People keep saying that, but they are wrong.  If you want to double your money, you have a higher chance if you place multiple bets than if you place a single bet.  You have to pick the right multiple bets of course.

It's really quite easy to demonstrate:

If you place a single bet, then the chance of doubling your money is 49.5%.

Now consider this 2 bet sequence:

1. bet 0.41421356 BTC with payout 3.41421356x and chance 28.99642866% to win 1.41421356 BTC for a profit of 1 BTC
2. if you lose, bet 0.58578644 BTC at the same payout and chance to win 2 BTC for a net profit of 1 BTC.

Your overall chance of success is 49.58492857%.  That is higher than 49.5%.

(Note that those bets aren't exactly available on Just-Dice, since chance is only available to 4 significant figures, but that's just a nit-pick.  It's still possible to double your money with a higher than 49.5% chance using 2 bets, and not using 1 bet).

[dooglus]

Every bet will, on average, make -1% of the amount wagered.  This is true no matter what the sequence.  To get the total average profit just add up all the bets and multiply by -1%.

Not quite true - if you look a few posts above I derived the expected profit for both an unbounded and a bounded martingale sequence without house edge and it's not the sum of all bets, which is what I think you're saying it would be if there was no house edge.

Your derivation suffered from an off-by-one (fencepost) error.  The expected return from a game with zero house edge is zero however you play.

[organofcorti]

Hey doog, thanks for taking the time to address the post. I wasn't completely sure about the derivation so I'm glad you found time to check. Cheers!

For example, if a bettor plays a 50% chance to win game and starts with a = 1 and stops betting after the n = 9th loss in a row, the total expected profit is the expected profit from up to 8 losses (and then a win) + the expected profit of greater than 8 losses in a row:

Code:

E(profit) = Pr(losses in a row <= 8 )*(1 - p)/p - Pr(losses in a row  > 8)* sum(2^0+2^1+...+2^9)
          = (1-(1 - 0.5)^9) - (1 - 0.5)^9 * 1023
          = (1-1/512 -  1/512 * 1023)
          = - 1

So in this particular case, the expected profit is -1, (even before the house edge is taken into account).

You've found a betting strategy for a 0% game which gives a non zero expectation?

Nah, it was some French guy a couple of hundred years ago  If I discovered it I would have called it something way cooler than a "Martingale sequence".

When that happens, your math must be wrong.

Not for a martingale sequence. The expectation of a standard Martingale sequence (double up when betting on a coin toss) is greater than zero, isn't it? If you have unlimited funds and the casino does too? Every completed (standard) Martingale sequence has an expected profit of 1:



Sequence completed on:              Bet     Win      Total cost    Profit
                     First roll:     1       2         1           1
                     Second roll:    2       4         3           1
                     Third roll:     4       8         7           1


Maybe you're thinking I mean the expectation per roll rather than per sequence? The expectation per roll is, as you say, zero.

In this case, the 9 losses in a row are 2^0, 2^1 .. 2^8, which sum to 2^9-1 or 511, not 1023.

E(profit) = (1-1/512 -  1/512 * 511) = 0

The old counting from zero error, eh? Well spotted, that man! You are of course correct and when I get a moment I'll amend the post.

Still, I find it an interesting result.

The expected profit for a bounded Martingale sequence is the same as the expected profit per roll, and is less than the expected profit for the same unbounded sequence (which is 1). Or am I missing something? You usually have a better insight into these type of problems than I do.

[nimda]

Infinity does pose a bit of a problem here

How is the house supposed to break even after an unlucky streak if the streak never ends?