Does the answer to this quiz provide any insight into profitable trading?

Someone writes a bunch of random numbers down on paper and hands you three of them face down. You flip them over one at a time and have to stop when you think you have identified the largest of the three.

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Flip over the first two.

If the second one is larger, stop.

If the second number is smaller, flip over the third one.

Performing this way gives you a .66 chance of winning.

Martingale? And when you have more than 3 cards? Or the number of cards is unknown? The trading looks like the situation when the number of cards is unknown.
 
I think the illustration was showing "to take what you can get".

ES

Martingale? And when you have more than 3 cards? Or the number of cards is unknown? The trading looks like the situation when the number of cards is unknown.
 
Let's see if this holds true. Someone writes down 5000 numbers and hands you 3. The first one you flip over is 1. The second you flip over is 10. Do you have a 66% chance 10 is the highest number of the 3?
Yes, for situations where the 5000 numbers have no limit. You don't know what the highest and lowest numbers are. You are assuming the numbers start at 1 and are all positive integers.
 
Let's see if this holds true. Someone writes down 5000 numbers and hands you 3. The first one you flip over is 1. The second you flip over is 10. Do you have a 66% chance 10 is the highest number of the 3?

I did a simulation in Excel out of curiosity and I must admit that the 66% are very close to my results after a few thousand of random numbers. You can do the test yourself.
 
Yes, for situations where the 5000 numbers have no limit. You don't know what the highest and lowest numbers are. You are assuming the numbers start at 1 and are all positive integers.

Dude, you are flawed here.

5000 numbers. With no starting point, no baseline, you cannot even establish a test? You are encroaching on infinity. Think about it.
 
I did a simulation in Excel out of curiosity and I must admit that the 66% are very close to my results after a few thousand of random numbers. You can do the test yourself.
Before I ask anymore questions, you did it from a draw of 5000? And the first card was 1 second card was 10?
 
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Before I ask anymore questions, you did it from a draw of 5000? And the first card was 1 second card was 10?

No,I tested the original statement that was made:
  1. Flip over the first two.
  2. If the second one is larger, stop.
  3. If the second number is smaller, flip over the third one.
  4. Performing this way gives you a .66 chance of winning.
I was always around the 66% when I did these steps. If the second one is not bigger then the first one, chances are 66% that the third one will be bigger then the first one.
What you tell is completelly diffferent. I didn't see at start that you changed the original statement completelly.
 
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Let's see if this holds true. Someone writes down 5000 numbers and hands you 3. The first one you flip over is 1. The second you flip over is 10. Do you have a 66% chance 10 is the highest number of the 3?
%%
That a tough question to answer because it depends; if those numbers are stock prices, [SPY, stocks,QQQ S&P500....... ]seasons/seasonals most liklely will matter for a profit.As far as a math answer ; that a different post.[To answer another part of your/1a/b2/c3 trader question; 99-100 % chance i'm not trading/investing a $1 stock]:cool::cool:
 
I did a simulation in Excel out of curiosity and I must admit that the 66% are very close to my results after a few thousand of random numbers. You can do the test yourself.

How this 5K numbers relate to three cards? The field of outcomes shrink to 3 after cards are chosen. Take 5K or 5000K doesn't matter. Or am I getting something wrong?
 
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