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.
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.
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.
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.
Before I ask anymore questions, you did it from a draw of 5000? And the first card was 1 second card was 10?
You are thinking the numbers are going from 1 to 5000. If you knew that, then yes, you would turn the next card. But you don't.
No,I tested the original statement that was made: 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. 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.
%% 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]
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?