What is the best system for a +ev occurance contest.

Hypothetical contest :

You are given +105 odds on a coin flip. There will be 1,000 coin flips you can bet on individually. You are given a starting bankroll of $1,000 of which you can bet any amount no matter how big (max. being current bankroll) or how small.

You must devise a system that will determine from roll 1 through roll 1000 on how much you will bet each flip. This system can be based on a % of roll, a set amount, a sliding scale, it can even pass on some flips, it can even quit after some amount has been reached.

So here is the contest - your system will be implemented for the 1,000 coin flips, and will be simulated 5,000 times through a credible random simulator which derives its results through noise or however it's done most realistically. Whoever has the most aggregate/average profit after the 5,000 simulations will win. However, if you go broke during any of the 5,000 simulations, you are completely disqualified.

So what system would be the best entry in this contest. I know that even with 5,000 simulations, some small amount of luck is involved, and yes, a great system might have 1 out of 5,000 simulations run so far outside of deviation that he get's disqualified, but that's the parameters of the game, and I'm curious what would be the best strategy to return maximum profit on a repeat +ev 50% proposition with almost zero risk of going broke.

the Wizard has answered this a long time ago. in any +'ve EV game, the optimal betting strategy is the Kelly's

http://en.wikipedia.org/wiki/Kelly_criterion

But there is one problem: "if you go broke during any of the 5,000 simulations, you are completely disqualified." this may call for some strategy change.

It occurs to me that if "no matter how small" is meant literally, it's possible, trivial even, to adapt any system at all, even flat betting or Martingale, to one with zero risk of going broke, although it'd be useless in the real world. Or actually, that's not even necessary - even if the units are discrete, it's enough that you can quit when some amount has been reached.

In the long run, the more you bet, the more you'll make, so what you want is to maximize the size of your total bet over 1,000 flips. My naïve strategy would be to simply bet half my current bankroll on every flip, with, again, the advantage that it's literally impossible to go broke, although it's possible I might end up with a bankroll of just over .000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 009¢. I imagine the best strategy would be something along these lines, although there might be a better fraction than 1/2.

Quote: 24Bingo

It occurs to me that if "no matter how small" is meant literally, it's possible, trivial even, to adapt any system at all, even flat betting or Martingale, to one with zero risk of going broke, although it'd be useless in the real world. Or actually, that's not even necessary - even if the units are discrete, it's enough that you can quit when some amount has been reached.

In the long run, the more you bet, the more you'll make, so what you want is to maximize the size of your total bet over 1,000 flips. My naïve strategy would be to simply bet half my current bankroll on every flip, with, again, the advantage that it's literally impossible to go broke, although it's possible I might end up with a bankroll of just over .000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 009¢. I imagine the best strategy would be something along these lines, although there might be a better fraction than 1/2.

I'm assuming there is a minimum bet, if so:

I think that strategy is less than exceptional. Let's say that you are putting your money on Heads and Heads/Tails alternate, we'll go ten flips.

$10,000 ($5000)

$5,000 + $2500

$7500 ($3750)

$3,750 + $1875

$5,625 ($2812.50)

$2,812.50 + $1,406.25

$4,218.75 ($2109.38)

$2,109.37 + $1054.69

$3,164.06 ($1,582.03)

$1,582.03 + $791.01 = $2,373.04

Alternatively, if you do the same thing, but start with a win:

$10,000 + $5,000

$15000 ($7500)

$7,500 +3750

$11,250 ($5,625)

$5,625 + $2812.5

$8437.50 ($4218.75)

$4,218.75 + $2109.38

$6328.15 ($3164.06)

$3164.05 + $1,582.02

$4,746.07 - $2373.03 = $2373.04

The, "Half bankroll," strategy will simply destroy you. In fact, $2,373.04 * 1.5 * 1.5 * 1.5 * 1.5 = $12013.52, which means that it would take four consecutive wins at this point to put you back at your starting point. If you lost four consecutive times at this point: $2,373.04/2/2/2/2 = $148.32, it would then take you eleven consecutive wins to put you back over the 10k mark.

That's a highly unusual case, though. It doesn't seem like it to our human intuition, but HTHTHT is no more likely than HHHHHH, and since the bet goes up after every win and down after every loss, it's a very ugly string for such a strategy. The point is to bet as much as possible while avoiding as much as possible the risk of not having very much to bet.

Quote: 24Bingo

That's a highly unusual case, though. It doesn't seem like it to our human intuition, but HTHTHT is no more likely than HHHHHH, and since the bet goes up after every win and down after every loss, it's a very ugly string for such a strategy. The point is to bet as much as possible while avoiding as much as possible the risk of not having very much to bet.

HHHTTT

$10000-$15000-$22500-$33750-$16875-$8437.5-$4,218.75

TTTHHH

$10000-$5000-$2500-$1250-$1875-$2812.50-$4,218.75


It doesn't matter how you do it, you need a winning percentage of well over 50% for this to be in anyway sustainable.

You're looking at this over very short intervals, and assuming exactly 50%. Over a very long interval, this strategy, like any other, cannot help but show a profit, so the very fact that all your thought experiments show no profit shows they are atypical. (Not to mention that they neglect the fact that wins pay 21:20.)

Regardless, the Kelly criterion would recommend a similar strategy, but with 1/42 (...EYPHKA...) instead of 1/2.

Quote: 24Bingo

You're looking at this over very short intervals, and assuming exactly 50%. Over a very long interval, this strategy, like any other, cannot help but show a profit, so the very fact that all your thought experiments show no profit shows they are atypical. (Not to mention that they neglect the fact that wins pay 21:20.)

Regardless, the Kelly criterion would recommend a similar strategy, but with 1/42 (...EYPHKA...) instead of 1/2.

You multiply your bankroll by 1.525 (The .5 paying 21:20 = .5/20*21= .525) for every win and you divide it by 2 with every loss. With regard to the intervals, showing a huge loss over a short interval doesn't seem promising for the long haul. You lose in every scenario except Heads outright winning more than Tails, and you have to do this over 5,000 trials without ever busting.

If you can subdivide your bets indefinitely (which the initial rules essentially said you could), that's not a problem. And I don't care if all 20 possible combinations of 3 heads and 3 tails are losers, a positive expectation game will win in the long run however you bet, provided you don't go broke, the only question being how much. Although 1/42 would be ideal, apparently, which I suppose makes sense - it's a strategy of rapid upswings punctuated by long downswings, as you show somewhat, so the amount of the downswings needs to be slim enough to keep feeding the edge substantially, without too badly hurting the upswings.

And best of all, you'll never dip below .000003¢!

Quote: 24Bingo

If you can subdivide your bets indefinitely (which the initial rules essentially said you could), that's not a problem. And I don't care if all 20 possible combinations of 3 heads and 3 tails are losers, a positive expectation game will win in the long run however you bet, provided you don't go broke, the only question being how much. Although 1/42 would be ideal, apparently, which I suppose makes sense - it's a strategy of rapid upswings punctuated by long downswings, as you show somewhat, so the amount of the downswings needs to be slim enough to keep feeding the edge substantially, without too badly hurting the upswings.

And best of all, you'll never dip below .000003¢!

The problem is that your betting style detracts from it being a positive expectation game. The main purpose of a system such as Martingale or Labouchere is to make it so that you actually need win a lesser percentage of the time than is expected for the system to complete to success. In the case of your system, you actually need to win more than expected.

It's simple, on average, each loss will negate 2/1.525 = 1.31147540983 wins.

So it is possible for a system to negate a positive edge, over, say, a billion trials. Just so we're absolutely clear, that is what you're saying.

(I've already admitted the system of half your bankroll is non-optimal for the challenge, and under the alternative, by your logic, each loss would only "negate" 35280/36121 wins, so you're somewhere, but I don't think you're quite there.)

Quote: 24Bingo

So it is possible for a system to negate a positive edge, over, say, a billion trials. Just so we're absolutely clear, that is what you're saying.

(I've already admitted the system of half your bankroll is non-optimal for the challenge, and under the alternative, by your logic, each loss will only "negate" 35280/36121 wins, so you're somewhere, but I don't think you're quite there.)

It's not really a system, in the sense that I would use the word. It's not any kind of Progression, it's simply betting half of your bankroll each time. Your average losing bet is going to cost you more than your average winning bet gets you. If you lose half of your bankroll on Tails, it's going to take you multiple consecutive wins to reach the same point or better.

You're very simply betting in a fashion that causes you to have to win more often than you lose. I cannot determine why this is not self-apparent when a loss divides your total bankroll by 2 and a win multiplies it by 1.525.

I know another system that could negate your positive edge. You could run an infinite Reverse Martingale on it. You just let every win ride until you eventually lose.

Furthermore, I don't see where you get that you can never fall below $0.03, off of $10,000, twenty consecutive losses puts you under $0.01. $0.00953674316, to be specific.

The odds of that happening in twenty flips are .00000095367 or 1,048,581:1

You're talking about 5,000 sets of, "First Twenty," flips, but the fact always holds. In other words, at any point at which you are below $10,000, twenty (or less) losing flips will wipe you out. There are going to be 1,000 flips per trial, but the last nineteen won't count for these purposes as starting a set because you can't have a series of twenty losing flips if the trial ends before that. We're talking about 981 flips that could potentially lead to twenty losses in a row.

981 * 5,000 (trials) = 4,905,000 possible sets of twenty losing flips.

If you take 4,905,000/1,048,581 = 4.67775 expected sets of twenty losing flips consecutively and within one trial for the entire test.

The question now becomes whether or not you'll be below $10,000 at such time that one of these sets comes to fruition. There may also be longer sets. For example, twenty-one losses in a row would break you if you had $15,000. That's less likely than twenty losses in a row, but must be considered.

Hey guys, I'm Glen, a math student from Australia (1st post)

With no minimum bet: The optimal betting strategy is the Kelly criterion as you can never go broke and it will produce the greatest gains. This has been well and truly proven so I feel I shouldn't have to say any more on that.

With a minimum bet: You could simply adjust the Kelly criterion so that you will never fall below the minimum bet over any given 1000 coin flips, pretty easy to work out and it will give you close to optimal results. However this insures that you would survive even if every flip landed tails which has a probability of 1 in 2^1000 which is not practical when you only need to survive 5000 rounds. I would recommend being more aggressive and taking on "some" chance of going broke. I would recommend setting the probability to going broke equal to 1 in (the number of rounds)^2 or in this case 1 in 5000^2. Doing this is just an extension of the kelly criterion

Quote: Mission146

I know another system that could negate your positive edge. You could run an infinite Reverse Martingale on it. You just let every win ride until you eventually lose.

No betting system/strategy can negate a (positive or negative) edge. This is the gambler fallacy. The fact that in this day and age an administrator of a forum of this kind could possible still believe in the gamblers fallacy is a little shocking.

Furthermore, running an infinite reverse martingale on it still gives a positive expectation. Understanding orders of infinity is essential in understanding why this is so.

If you ran an infinite number of infinite reverse martingales on this, at least one of these would never actually land on tails and you would gain an infinite amount of money which can be shown to be infinity times more than the infinite amount of money you would lose from the rest of them. This means that you would net an infinite amount of money... hence you can't negate the positive edge.... think about it.... or ask someone with mathematical knowledge to explain it to you further.

Hi Glen (and others)

Kelly is certainly not the answer here.

For example, here is a superior strategy to Kelly:

For each trial, follow Kelly for the first 999 flips. Then, on the 1000th flip, bet (remaining bankroll - epsilon).

Clearly, this raises your aggregate EV significantly, with no risk of going broke in any trial.

Kelly is the correct answer long-term. However, this is not long-term. It has an ending.

Quote: glenandrew

No betting system/strategy can negate a (positive or negative) edge. This is the gambler fallacy. The fact that in this day and age an administrator of a forum of this kind could possible still believe in the gamblers fallacy is a little shocking.

I've already demonstrated a system that would negate the positive edge in the first page of this thread. If you know how, (I don't, but the Math is really simple) run a simulation of the bet half of your bankroll strategy within those parameters, the bankroll gets pounded.

Quote:

Furthermore, running an infinite reverse martingale on it still gives a positive expectation. Understanding orders of infinity is essential in understanding why this is so.

I am forced to question if you know what I mean by, "Infinite Reverse Martingale." That would simply mean running a Reverse Martingale until you lose. If you are doubling your bet until you lose, then you will eventually lose the base bet amount.

In this case, the odds were +105. We're going to say that you pull out the difference before doubling your bet:

100 Bet Wins 105 with a Difference of 5

200 Bet Wins 210 with a Difference of 10 and total difference of 15

400 Bet Wins 420 with a Difference of 20 and total difference of 35

800 Bet Wins 840 with a Difference of 40 and total difference of 75

1600 Bet Wins 1680 with a Difference of 80 and total difference of 155

As you can see, you have to win five in a row before pulling the difference compensates for a loss on the base bet. If you let the difference ride, then you simply can't win because you let the Reverse Martingale ride until a loss. I guess you could theoretically die before losing, then technically you've won...

By the way, the example above also ignores losses on the first attempt, which also must be compensated for.

Quote:

If you ran an infinite number of infinite reverse martingales on this, at least one of these would never actually land on tails and you would gain an infinite amount of money which can be shown to be infinity times more than the infinite amount of money you would lose from the rest of them. This means that you would net an infinite amount of money... hence you can't negate the positive edge.... think about it.... or ask someone with mathematical knowledge to explain it to you further.

I'll let you get started on that, then, it will probably take awhile.

Make sure that you don't go broke in 5000 simulations is the top priority. I don't have a calculator handy but I would try a reverse Martingale with whatever gets me enough chance to win X in a row with just under 50% probability in 1000 tries. The amount would be just enough so that I can expect not to go broke with 3 sigmas of confidence. If the bankroll gets to less th an or equal to one bet, I bet half aand half again if I lose, until I get my bankroll back to 2x bet. I'll come up with numbers later.

Quote: glenandrew

With no minimum bet: The optimal betting strategy is the Kelly criterion as you can never go broke and it will produce the greatest gains. This has been well and truly proven so I feel I shouldn't have to say any more on that.

With a minimum bet: You could simply adjust the Kelly criterion so that you will never fall below the minimum bet over any given 1000 coin flips, pretty easy to work out and it will give you close to optimal results.

You are mistaken, the essence of the Kelly criterion is not that you cannot bust your bankroll (bet ANY fraction of your bankroll, and you will never bust it).
Kelly bet is the bet which maximizes CEV (that is the estimated loarithmic bankroll). For a two-way bet (win or loss) the optimal bet size is given by the Kelly criterion.

Now if your bet has a minimum, you still simply maximize CEV. If the Kelly bet is a valid betsize (above minimum), then there is no argue about it will be the best bet. BBut if the Kelly bet is below the minimum - then just calculate CEV at the minimum bet and compare it to the CEV of "no bet". If CEV is positive, then the minimum bet is the best bet. (Assumed that you always have the option to decline the bet)

I'm somewhat disappointed that you come up with something like "over 1000 coin flips". Where does the "magic number" 1000 come from in the first place ?. As a math student this should immediatly raise some flag. The only natural numbers a math student usually needs is 0, 1, and 2.

Quote: MangoJ

You are mistaken, the essence of the Kelly criterion is not that you cannot bust your bankroll (bet ANY fraction of your bankroll, and you will never bust it).
Kelly bet is the bet which maximizes CEV (that is the estimated loarithmic bankroll). For a two-way bet (win or loss) the optimal bet size is given by the Kelly criterion.

Now if your bet has a minimum, you still simply maximize CEV. If the Kelly bet is a valid betsize (above minimum), then there is no argue about it will be the best bet. BBut if the Kelly bet is below the minimum - then just calculate CEV at the minimum bet and compare it to the CEV of "no bet". If CEV is positive, then the minimum bet is the best bet. (Assumed that you always have the option to decline the bet)

ummm... yea I know... thats what I was implying

Quote: MangoJ

I'm somewhat disappointed that you come up with something like "over 1000 coin flips". Where does the "magic number" 1000 come from in the first place ?. As a math student this should immediatly raise some flag. The only natural numbers a math student usually needs is 0, 1, and 2.

ummm.... 1000 comes from the 1st post of this thread... I did not "come up" with anything

Quote: AxiomOfChoice

Hi Glen (and others)

Kelly is certainly not the answer here.

For example, here is a superior strategy to Kelly:

For each trial, follow Kelly for the first 999 flips. Then, on the 1000th flip, bet (remaining bankroll - epsilon).

Clearly, this raises your aggregate EV significantly, with no risk of going broke in any trial.

Kelly is the correct answer long-term. However, this is not long-term. It has an ending.

Yes, you seem to be onto the correct answer here.

Quote: Mission146

I am forced to question if you know what I mean by, "Infinite Reverse Martingale." That would simply mean running a Reverse Martingale until you lose. If you are doubling your bet until you lose, then you will eventually lose the base bet amount.

Yes I know what you mean... and I'm a afraid you are wrong... this still has a positive EV

You are implying that there is no chance that an infinite series can exist in which only heads appears. This implication is of course wrong and while the chance of it happening approaches nothing it still produces an infinite positive EV greater than all the loses combined.

Furthermore:
If you were to do your infinite reverse martingale an infinite number of times over 4 things would happen:
You would lose on an infinite number of them... totaling an infinite lose
At least one of them would only contain heads
This one would produce a gain greater than the infinite lose of the losers
Your total profit would be +infinity.

This is all really basic stuff, I still dont understand why you believe in the Martingale, I will feel really stupid if this whole thing is a practical joke.

I'll bet $ 0.002, and be paid $0.0021 for each correct guess.
I'll use a bet-ramp of 1-1-2 for three consecutive wins.
Win or lose the $0.004 bet, the next bet is $0.002, and the ramp repeats.
On each session I randomly pick H or T, and make that bet all 1000 times.

How well does this do 1000 per session x 5000 sessions?

EDIT: re wrote the answer after thinking this through better... all $10000 gets wagered.

Quote: glenandrew

Yes I know what you mean... and I'm a afraid you are wrong... this still has a positive EV

You are implying that there is no chance that an infinite series can exist in which only heads appears. This implication is of course wrong and while the chance of it happening approaches nothing it still produces an infinite positive EV greater than all the loses combined.

Furthermore:
If you were to do your infinite reverse martingale an infinite number of times over 4 things would happen:
You would lose on an infinite number of them... totaling an infinite lose
At least one of them would only contain heads
This one would produce a gain greater than the infinite lose of the losers
Your total profit would be +infinity.

This is all really basic stuff, I still dont understand why you believe in the Martingale, I will feel really stupid if this whole thing is a practical joke.

If you are committed to the Reverse Martingale, there is no gain until you stop. If you are determined to do it infinitely, then you will not stop. While your wins may continue to grow and grow and grow, they are continuously being risked on the next turn, (unless you are pulling the 5 of the +105) thus, you have not gained anything at any point. There are only two possibilites at any point, the first is losing the base bet and the second is having gained nothing because you win, however, never stop betting.

With all due respect, I think we are both right, just talking past one another, my apologies if that is the case.

Quote: Mission146

There are only two possibilites at any point, the first is losing the base bet and the second is having gained nothing because you win, however, never stop betting.

You are correct in saying this, totally agree with you. I only ever had a problem with "negating a positive EV" which is impossible.

Good point, I probably could have done a better job articulating what I was trying to express.