Reality based coin-tosser method that beats 95% of traders in the world.

Whisky · Oct 20, 2009

Quote from AyeYo:

The coin toss is break-even over time, it comes out 50/50, we all get that.

Really?. Can you GUARANTEE that if you go into a casino that has a roulette without zeros and the payoff for red and black are the same as a real casino, and you bet on red all night, you will definitely 100% sure GUARANTEED that you will come out breakeven?

What if you bet more or less with each roll?. Any guarantees?.

LMAZO

Whisky · Oct 20, 2009

Quote from AyeYo:

Then it's NOT break-even?

No it's not breakeven. You have to substract costs. That's a GUARANTEE.

LMAZO

Badoit · Oct 20, 2009

Quote from Whisky:

EV of first toss: 0.5 X - 0.5 X -costs
EV of second toss: 0.5 Y - 0.5 Y-costs
...
EV of Nth toss: 0.5 N - 0.5 N-costs

Sum of EV: 0 - sum of costs

X, Y, ..., N are what you call the size of wins and losses.

That's the best I can do. Good night.

p.s.: EV is "Expected value".

http://en.wikipedia.org/wiki/Log-normal_distribution

Whisky · Oct 20, 2009

Quote from Badoit:

http://en.wikipedia.org/wiki/Log-normal_distribution

Are you suggesting my coin is lognormal or the bum is lognormal?.

That would be abnormal. In any case:

"You will have to prove to yourself and others that your suggestions bring in at least 1 tic per day on average extra profit over the original method rules."

AyeYo · Oct 20, 2009

Quote from Whisky:

No it's not breakeven. You have to substract costs. That's a GUARANTEE.

LMAZO

Yea, I get that. You're saying break-even MINUS costs, I get that. I left off the minus costs to simplify and save myself typing.

You're saying it's break-even before costs? Prove it.

AyeYo · Oct 20, 2009

Quote from Whisky:

Really?. Can you GUARANTEE that if you go into a casino that has a roulette without zeros and the payoff for red and black are the same as a real casino, and you bet on red all night, you will definitely 100% sure GUARANTEED that you will come out breakeven?

What if you bet more or less with each roll?. Any guarantees?.

LMAZO

Strawman. You're avoiding the issue.

TheAngryHermit · Oct 20, 2009

Quote from Whisky:

Exactly!. It beats 95% of traders. That's the point.

Can you prove that it beats 95% of traders?

AyeYo · Oct 20, 2009

Quote from Whisky:

EV of first toss: 0.5 X - 0.5 X -costs
EV of second toss: 0.5 Y - 0.5 Y-costs
...
EV of Nth toss: 0.5 N - 0.5 N-costs

Sum of EV: 0 - sum of costs

X, Y, ..., N are what you call the size of wins and losses.

That's the best I can do. Good night.

p.s.: EV is "Expected value".

That doesn't address the question.

The underlying win/loss amount is variable by (for all practical purposes) an infinite amount. Show me how you can claim the system is break-even before costs.

All you're showing is that there's a 50/50 chance of losing or gaining EACH SPECIFIC AMOUNT. Show me how, when added all together, the sum of ALL the amounts comes out to 0. That is what's required for the system to be break-even before costs.

Listen, I get what you're trying to do and the thread concept is great. But if you're going to be a dick to people trying to offer suggestions, I figured someone should at least point out that your basic premise is flawed.

Badoit · Oct 20, 2009

Quote from Whisky:

Are you suggesting my coin is lognormal or the bum is lognormal?.

which is more likely:

i) Tomorrow, the SPX will close at 2 times its present value

ii) Tomorrow, the SPX will close at 0

cfd_trader · Oct 20, 2009

Hey All!

my first post here, after countless hours of reading, enjoying, learning, and laughing at times.

In answer to Mike's post below:

at first look at the numbers an obvious "YES" comes to mind.

Wayyy to obvious.

On second thoughts, and that's probably what you were aiming at, the answer is actually NO.

Why?

The answer lies in this simple question:

how many runs of this strategy over the same set of historical data did it take to get this neat Net Profit?

Busted!

Quote from Mike805:

Whisky,

It seems you've touched a nerve with some posters here. I'll go ahead and pinch that nerve.

"Fooled by Randomness" is not only a good book but the term is perhaps nowhere as readily apparent as it is in this forum. It is somewhat satisfying to know how many completely clueless people think that they understand a market dynamic, let alone the foundation of a true edge. Let them trade, the more the merrier I say . That said, I know exactly who is on the other side of my trades and why.

My earlier system result posting was meant to demonstrate how easy it is to create a percieved edge. In fact, if you open the attachement below you'll find another manifestation of this random process at work. All I've done is introduced a couple serial dependencies and trivial ones at that. The "results" are neat, eh?

Some poster from 2003 who shall remain nameless mentioned that this exercise proves nothing. Frankly, I'm suprised he's still around trading... understanding randomness is way more important than attaining an edge. In fact, if one doesn't understand randomness to the point of being able to understand the attached results, then, one will never attain a real edge.

Mike

To everyone else here: would you trade this system if you saw only the results report? Be honest.