Topic: Just-Dice.com : now with added CLAMs : Play or Invest - page 172. (Read 454769 times)

I've derived the expectation for any two roll sequence, house edge included. I'm assuming in this two roll sequence that different bet amounts (a1 and a2) and different probability games (p1 and p2) are used for each roll.


Expected profit, 1 roll sequence
Roll 1, win:  0.99*a/p-a   with probability = p
Roll 1, lose:  -a   with probability = (1-p)

Expected profit = (0.99*a/p-a)*p - a*(1 - p)
                = - 0.01*a ..... (1)

Expected profit, 2 roll sequence.

Roll 1, win:  0.99*a1/p1-a1   with probability = p1
Roll 1, lose:
   roll 2, win: 0.99*a2/p2-a2 - a1  prob = p2*(1 - p1)
   roll 2, lose: -(a1 + a2)   prob = (1 - p2)*(1 - p1)

Expected profit = (0.99*a1/p1-a1 )*p1+ (0.99*a2/p2-a2-a1)*p2*(1 - p1) -(a1 + a2)*(1 - p2)*(1 - p1)
                = - 0.01*a1 - 0.01*a2*(1-p1) ........ (2)


Since a1, a2 are always positive, and p1 is less than one and positive, the expected profit for a two roll sequence in the presence of a house edge will always be negative, and does not depend on the probability of the second roll.

When will the expectation for the multi-roll strategy be more than for the single roll?



 - 0.01*a  <  - 0.01*a1 - 0.01*a2*(1-p1)
           a  >  a1 + a2*(1-p1)


So, compared to a single roll, if the amount bet on the first roll plus the amount bet on the second roll multiplied by the probability of a loss on the first roll is less than the amount bet on any game, then you can expect to lose less.

For a non-martingale two roll sequence I guess you're right, dooglus!

If the probability is greater than zero, I can always move forward in the series of bet outcomes until I find a win. In an infinite series of wagers, there are an infinite number of wins yet to be won.

Every time I figure out how to explain what I mean, DC jumps in and does it for me! Grr.

Anyway, he's right. the Martingale sequence must end in a win or it's not a Martingale sequence.

The bounded version is the same except when the number of losses is greater than some bound, after which the total loss is the sum of the bets until that point. The derivation you helped me with results in an expectation of zero for the bounded sequence.

I guess that's the question.  Isn't it possible to lose forever?

In the limit as N->infinity, the chance of losing N in a row tends to zero, but the amount you've risked so far tends to infinity.

So your expected profit tends to 0 * infinity?  What's that?  0 times x = 0, but x * infinity = infinity.

I think it must be equal to 1...  

Edit: what I mean is your expectation is (0.99999...)*1 - 0.000...*inf = 1 - (0)*(inf) = 1 - 1 = 0

A sequence of martingale losses followed by a final win (in a 50/50/double up scenario):

Lose 1+2+4, Win 8 = +1
Lose 1+2+4+8+16+32+64+128+256+512, Win 1024 = +1

A series of bets with an infinite possible number of bets will always win a single bet unit. However in the real world a limited maximum bet or limited gambler bankroll means that the last bet might not be able to be placed - if the last bet above could not be placed, the gambler would bust -1023. The gambler wins many single bet units but is as likely to go broke from an unplaceable final bet as to double their money from many +1s (house edge makes the former more likely).

I don't like it.  Bound it at any finite point and expected profit is zero.  But let it go that extra "little bit" to infinity and the expectation leaps to +1?

The expectation of an n-step martingale sequence is:

E = p(win_any_bet)*1 - p(lose_all_bets)*sum_of_bets
= (1 - 0.5^n)*1 - (0.5^n)*(2^n - 1)
= (1 - 0.5^n) - (1 - 0.5^n)
= 0

In the limit as n->infinity, E=0->0

Where's the mistake here?

This is intriguing, with a little discrete hypervariate statistics, one could find the maximum probabability amplification obtainable in the set of possible bet sizes and possible payout multipliers on just-dice. If an optimum can be found for a series of two bets, it can be found for a longer series of bets also.

This still seems only a probability amplification for a particular goal; the EV doesn't change as all bets are individual. Any "martingale"-ing of such a thing would allow you to plan the best bet sequence for goal=win max bet or similar goal, but still it would be a futile exercise in creatively sending your money to just-dice investors.

I agree with the statement, but not that it's a modification of the original. Nitpicky bugger, aren't I? 

A martingale sequence, for example a 50% chance to win martingale with a starting bet of 1 btc, zero house edge and no bounds will always result in a profit of 1 btc at the end of the sequence - regardless of the length of the sequence. It won't double or lose any coins. It will have an infinitely better chance of success than an equivalent single bet.

A bounded martingale sequence with the same parameters as above has an expectation of zero profit. It will have exactly the same expected return as a single bet.

If we can agree on that, I'll get to work on deriving the expectation of a Martingale sequence when there is a house edge (which I think will be a mite trickier).

If you have an infinite bankroll, odds-on that the casino will too, along with an infinite max bet.

... then you run into the house's max bet.

I can't imagine the house would ever offer a max profit more than about - uh - 463 BTC.  That would be crazy.

Read this, you are going to like it.
http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel

Well yeah, definitely. However, there is different ways of approaching infinity in these manners.
For example- I have an infinite amount of marbles, I give you all of them, and now I have 0. infinity-infinity=0
I have an infinite amount of marbles, and I give you all of them except for 2. Infinity-infinity=7
I have an infinite amount of marbles and give you every other one. infinity-infinity=infinity.

"There is no single answer to this, as infinity is not a number in the strictest sense. 
Infinity is a "limit".  It is something you can get closer and closer to, but you never
quite reach it.  What matters is how fast you approach the first infinity (call it
infinity01) versus how fast you approach the second infinity (call it infinity02).  If
you approach both at exactly the same speed, always the same distance from each, then
infinity01-infinity02=0.  If you are always a little closer to infinity01, then
infinity01-infinity02 is a positive number.  If you move toward infinity01 twice as
fast as how you approach infinity02, then infinity01-infinity02=infinity.  The behavior
of limits is not quite the same as the behavior of numbers.

Dr. Ken Mellendorf"

So although there is no concrete 'proof' within it, given the nature of infinity...it seems 'possible' that in a never-ending martingale with infinite bankroll, you may be able to bust and end up with a 0 bankroll.

Infinite - Infinite is undefinable.

Google for several proofs of this (or proofs that it is either 0, combined with a proof that it is Infinity, proving that it's neither or both at once, blablabla, undefinable)

I was contemplating something earlier. Touches more on the theoretical than anything, but may shed some light.

If you had an infinite bankroll, and let a martingale run, is it possible to end up at 0? Or maybe inevitable? Or would you always come out +EV?

If you had an infinite bankroll, then your martingale, in theory, can go on infinitely. This brings out a few possibilities. If truly infinite, you would encounter an infinite cycle of losses, and an infinite cycle of wins.

If you actually ended up with an infinite amount of losses would you end up with a 0 bankroll? Since we are talking about a set, theoretical amount of infinity, would infinity-infinity=0 in this situation? Or would, in this situation infinity-infinity=infinity, since the bankroll x, which = infinity constantly fluctuates with wins and losses? If it's the latter than the martingale would always yield +EV, but if the former, it may be inevitable to end up with 0.

I need to modify my claim to exclude zero house edge games.  When the edge is zero, every strategy which ends up either doubling or losing your coins has a 50% chance of doubling and a 50% chance of losing everything.

Given that modification do we still disagree?

I doubt it.  Imagine the house edge was 90% instead of 1%. Would you thus be able
to turn an arbitrarily unfair game into a fair one?
I guess the limit depends on the initial house edge and is strictly positive.

If someone wants to work  it out, better first find out the optimal proportion
to break your capital into two bets. Then extend this to n bets; then take the limit n to infinity.
It may be a nice exercise in calculus.

by the way, while doing one martingale sequence is better than doing a single bet, doing
many martingale sequences is strictly worse than doing one. And this - "grinding the martingale" -
is what our players seem to most enjoy doing.
Of course, saying "it's more rational to bet just once" is just as useless as saying "it's more rational
not to bet at all," since it doesn't take into account the "fun" part.

A sad fact: more rational = less fun.

I think you're right - but for the wrong reasons.

The reason isn't because martingales are better value than single bets in any absolute sense, but because of the way in which the house edge is applied on J-D (it's 1% of amount staked not a percentage of the winning chance).  So on a 50/50 bet the player wins 1.98 times their stake - i.e. 2% less than they would in a game with no edge.  On a 98% win bet the player only wins just over 1.01 their stake - i.e. nearly 50% less than they would in a game with no edge.

It's the impact of that which produces the effect you describe - as a multi-bet martingale to simulate a single bet has to be at higher odds which represent better value on J-D.  

You can see this point if you consider a bet that isn't allowed on J-D - and apply J-D's method of applying edge.  Consider a bet with a 99% win chance.  Such a bet would pay exactly what you wagered any time you won - and would give precisely the standard 1% house edge.  Obviously that bet isn't allowed as it's plainly unfair (you could never make a profit from it) - but the reason WHY it's unfair applies to other 'odds-on' bets as well, just not to the same extent  (and applies all the way down the spectrum with the bets with a tiny chance to win being the best value).  And that is why J-D is martingale-friendly - because martingales use better value bets to achieve the same target profits from same starting capital.  

My instinct tells me that the best value series will be the longest martingale you can afford to BR that will reach your target profits in 1 successful cycle - once it take more cycles your expectation reduces as you end up placing more in total wagers than you would in a single bet.  As house edge is a percentage of amount wagered you need to minimise the amount wagered whilst ending either at your target or at zero.  Your example does that in 2 bets - on average wagering significantly under 1000 (71% of time you wager 1000, other 29% of time you only make the first bet).  If you then applied same principle to split the other two bets into 2-step martingales amount wagered would decrease and percentage of the time you won increase.  In practice this is constrained by your target profits having to be lower than max win and number of steps being limited by the minimum bet size (which is reached pretty quickly as at each division the odds for previous step have to be higher for the bet to be optimal).

Martingales are NOT a 'winning' strategy - there's no such thing in general.  But there are specific circumstances where betting on a house-edge game CAN make sense (where you have X coins and if you had Y more you do something with more extra value to you than the cost of the house edge on making that extra).  And in those circumstances optimised martingales CAN be the best option - with the goal always being to reach one of bust or your target with the smallest average amount wagered.

Yes. If it were just about my posts or the content of them, I wouldn't be bothered. This one seems personal. What I really want to know is, what is actually bothering him? It's probably not the discussion.

The reason why martingale strategies are alive today, is that they actually work. The limitations are often used as an excuse why they won't work (such as you need unlimited money.) You don't need unlimited money, you just need enough to survive as long a losing streak as you have reasonably (or unreasonably) predicted. Or, if I may be allowed to correct myself, they work until they fail. But that's a chance you've got to take.

Magic seeds are just my way of saying that while I don't completely understand why it should work the way I predict it will happen, I am willing to bet on it with my own coins. The fact that I'm asking other people to join tells me I'm not alone (or they couldn't be bothered to think about it too much and feel that "it's only a bitcoin".)

The Gambler's Fallacy exists because it always infects people, like a disease. I'm just repeating what it says "hey, I just lost 6 in a row, maybe, just maybe, this 7th bet will win."

It's annoying, because I can be proven wrong, except that I might win (because I still have an 87.7779% chance to win at minimum) anyway, and people will think, ... ... and start debating this over and over.

For the rest, if you're going to do a simulation, at least use the same lucky number generation method of the site you are trying to simulate.

I know I'll live to regret this, but I find myself disagreeing with you on this point. You pointed out that the expected value of any bounded Martingale sequence is 0, and I assume you're talking about a bounded sequence.

Can you derive the expectation for me? I'm   happy to be    ok with being    will grudgingly accept being wrong, but at the moment I can't figure out exactly what you mean - any derivation of expectation for a martingale sequence and some simulations I just tried all show me the expected value of any bounded martingale sequence (in absence of a house edge) is zero -  the same as for a single bet.

For what it's worth, although your posts are often infuriating and hard to understand I've never even considered asking you not to post or wanted to ignore you.  I don't have much faith in most of your assertions about probability, but it seems it's us two against the world when it comes to defending the poor old martingale strategy.