If you are allowed to choose which number to show first, then this game is beatable. However, I think this is where the instructions are unclear. I think the intent of the game is to show one of the two numbers randomly which then gives the guesser the advantage even if the numbers are together. The advantage gets larger for the guesser the further apart the two numbers are. What I mean by always selecting the highest card when the random # is between the two numbers, assume the two numbers are 30 and 40. We will also assume that 30 and 40 are both shown first 100 times and that the random # gen will select each number 1-100 once. The results are as follows: When 40 is shown first and the random # is 1-40, you keep the 40 for 40 wins and when the random # is 41-100, you switch for 60 losses. When 30 is shown first and the random # is 1-30, you keep the 30 for 30 losses and when the random # is 31-100, you switch for 70 wins. The bottom line is you have 110 wins and only 90 losses. This happens because anytime the random # is 31-40 you win regardless of which number is shown first. So you wind up with 20 guaranteed winners and the balance of the 180 trials are split 50-50 for 90 losses and 90+20 winners. Hope that was clear. Joe.
That makes sense if the card that is shown is picked at random. But from the poster of the game's message. "Whether you pick which number to reveal randomly or not doesn't matter. In both cases there is a strategy that brings the other side to a win probability higher than 50%." And by "keep" I assume you mean, "assign as the high card guess", people seem to throw out random words thats seem to have nothing to do with the problem we are discussion, confusing me to no ends
We have two advantages over the roulette player. We have access to a wider range of betting instruments (i.e. options) and our random process (the market) has a skew. The coin-flip system becomes very important for estimating this skew. And options are a way to capture the profit left on the table. Ninna
The solution I had in mind when answering Ninna's post is actually different. I am wondering what would be a method that maximizes the edge? The method shown above gives an edge, but it does not discuss whether it is the maximum edge. My method (if it works) might give the maximum edge, or at least that is the aim. Could you guys think of other ways (or ask your friends or others in forums to find alternative solutions)?
Anything you don't understand may look to you like coin flipping, ... or even worse, like an example of Murphy's Law applied ... Markets are no flipping coins.
I completely disagree ... Options are no betting instruments ... Market is not random, and has no skew ... Options are by no means "a way to capture the profit left on the table" ...
The difference between the market skew and normal is basically your profit. You need to find a way to measure that difference. Ninna
For anyone who is still interested in this problem. I double checked with some other mathies. Turns out our initial intuition was correct, if you allow player A to choose which card to show and they employ the correct numbering strategy, you give them the opportunity to arbitrarily force the possibility of player B guessing correctly to exactly 50%, by literally turning it into a coin flip bet for player B, this making it a losing game for player B. Phew, thought I didn't understand maths for a second there. *Puts his degree back up on the wall*
Not sure why anyone would compare trading to coin-flipping. All you have to do is look at a chart to see that markets trend....