Trading Lessons/Insights From Coin Flipping

[u21c3f6]

Quote from nLepwa:

Here's how you get 100%:

You either choose two numbers greater than 50 or two numbers less than 50. And you give me the one closer to 50.

With the strategy you described I would loose everytime.


There is indeed a strategy with positive expectancy for me.
Here's a hint: the strategy's win probability has something to do with 1/e.

Ninna

Maybe I am misunderstanding the rules.

Here is how I would proceed based on my understanding of the rules. If I can set-up a 50-50 situation, then I should profit due to the unbalanced payments. I would choose two numbers either higher or lower than 50. Then I would randomly select which of those two numbers to give you which should have a long-term expectation of your selecting whether or not the other number is higher or lower a 50-50 proposition. If you don't agree, please explain why.

Joe.

 

[tradingjournals]

Quote from nLepwa:

Would you play this game with me?

You are allowed to choose any two numbers between 1 and 100.
You reveal to me one of the two numbers.

Now, I must say whether the second number is greater than the first one.

If I guess correctly, you owe me $99.
If my guess is wrong I owe you $101.

So, would you play this game with me?

Ninna

If you have access to a random generator, I see how you can have a positive expectancy. The key is that you see one of the numbers and that you have access to a random generator. Do I get it right? I did not want spell out the method to allow others the opportunity to first think about it.

Guys: the problem posed has a positive expectancy. The person is not wasting your time, in case you want to think about it.

 

[masterjaz_99]

Quote from tradingjournals:

If you have access to a random generator, I see how you can have a positive expectancy. The key is that you see one of the numbers and that you have access to a random generator. Do I get it right? I did not want spell out the method to allow others the opportunity to first think about it.

Guys: the problem posed has a positive expectancy. The person is not wasting your time, in case you want to think about it.

I don't see this game as a coin flip, maybe it is just me, only because of the two-human aspect. Say I am picking the numbers and pick 1, 2. I tell you, the guesser, '2'. Logically you will say higher as ~97% of the time you would be right. But because I am human, slightly sneaky, and I want your money, you'd be wrong because the second number is actually lower...

Maybe I missed something in the rules, but I don't think this is a coin flip. UNLESS, you use a coin to determine hi/lo...

My two cents...

 

[walterjennings]

Quote from u21c3f6:

Maybe I am misunderstanding the rules.

Here is how I would proceed based on my understanding of the rules. If I can set-up a 50-50 situation, then I should profit due to the unbalanced payments. I would choose two numbers either higher or lower than 50. Then I would randomly select which of those two numbers to give you which should have a long-term expectation of your selecting whether or not the other number is higher or lower a 50-50 proposition. If you don't agree, please explain why.

Joe.

Or simpler, choose N1=50 and show that number, flip a coin to decide if N2 is 51 or 49. It should be impossible for them have a chance other than 50% of guessing if N2>N1. Which will lead to positive expectation with the imbalanced payout.

 

[Picaso]

Quote from nLepwa:

Would you play this game with me?

You are allowed to choose any two numbers between 1 and 100.
You reveal to me one of the two numbers.

Now, I must say whether the second number is greater than the first one.

If I guess correctly, you owe me $99.
If my guess is wrong I owe you $101.

So, would you play this game with me?

Ninna

I can choose from 100 numbers. If I choose 1 or 100 as one of the numbers, obviously I'm not going to reveal it to you, which means I will reveal to you two of 98 numbers. However, once I've revealed any number between 2 and 99 the chances of the other number being higher or lower are pretty much even (yes, if I reveal 2, there is only one number lower and 98 higher, but since <i>I choose</i> whether I pick a higher or a lower number, it doesn't matter.

So yes, I would play.

 

[masterjaz_99]

Quote from abattia:

I'm on for it ... where do we start?

My current system is purely noise, I've decided. It celebrated its first full year of live implementation and came out 126 wins, 125 losses with avg win of $288 and avg loss of $290. About as zero-sum as can be done.

The system uses front month NQ contract and the 5dma of 4pm close. If above the 5dma, short the next day, if below, go long. Simple...

So, how do I improve it?

Feel free to backtest, QQQQ may be easiest for historical OHLC data. There is slippage due to time of day entry (cannot peg the open and close prices), but should even out (but I can say it is negative by a small amount).

masterjaz

 

[masterjaz_99]

Quote from walterjennings:

Or simpler, choose N1=50 and show that number, flip a coin to decide if N2 is 51 or 49. It should be impossible for them have a chance other than 50% of guessing if N2>N1. Which will lead to positive expectation with the imbalanced payout.

WJ,
I thought of N1 = 50 as well, but then it does become a coin-toss game, which is what we are after, but why even pick two numbers then. You can just pick a single # and ask if it is higher or lower. The second number, to me, makes the game incomperable to a coin flip game. I am also excluding the unbalanced payout, so...yes, your way does work as well. I think we are on the same side of the coin!

M

 

[masterjaz_99]

Quote from nLepwa:

Here's how you get 100%:

You either choose two numbers greater than 50 or two numbers less than 50. And you give me the one closer to 50.

With the strategy you described I would loose everytime.

If, and only if, you pick the logical answer...if you were able to reverse engineer my picking strategy, then you could win 100%...hence, not a real coin flipping example, but my guess is that it was not supposed to be...


There is indeed a strategy with positive expectancy for me.
Here's a hint: the strategy's win probability has something to do with 1/e.

Ninna

 

[Mike805]

Quote from masterjaz_99:

My current system is purely noise, I've decided. It celebrated its first full year of live implementation and came out 126 wins, 125 losses with avg win of $288 and avg loss of $290. About as zero-sum as can be done.

The system uses front month NQ contract and the 5dma of 4pm close. If above the 5dma, short the next day, if below, go long. Simple...

So, how do I improve it?

Feel free to backtest, QQQQ may be easiest for historical OHLC data. There is slippage due to time of day entry (cannot peg the open and close prices), but should even out (but I can say it is negative by a small amount).

masterjaz

Let's try a few simple tests.

Categorize the trades by entry in one hour increments.

So the stats would look like

630am-730am : PnL, Avg. P, Avg. L, #Trades
730am-830am : PnL, Avg. P, Avg. L, #Trades

Etc...

If its daily then find the VIX closing on entry and the day:

10 >= VIX < 15 : PnL, Avg P/L Etc
15 >= VIX < 20 : PnL, Avg ...

Day of week entry:
Monday : PnL, Avg P/L etc
Tuesday: PnL

etc etc

Mike

 

[tradingjournals]

Quote from walterjennings:

My mistake. I didn't see that I "choose" two numbers. I assumed I randomly generated two numbers between 1-100 and choose a single one to show.

Walter: I think it is what he meant. You choose two numbers that you put in two envelops for instance, s/he opens one of the two envelops, and s/he decides whether that number is the largest of the two, or the number in the other envelop is the largest. S/he can have a positive expectancy, because s/he will use the information from the opened envelop to decide (more intelligently) on the number in the unopened envelop. Think of extremes as an example. If number revealed is 1, then he has to move to number 2. If number revealed is 100, then he take the one revealed.