If you mean by the original poster myself, could please you indicate where you asked me to give an answer. If you asked a question, I must have missed reading it? Please clarify. Thanks.
It does not. The two problems are different. The coin flipping has no terminal conditions where you can work backward to decide on a strategy.
What are the terminal conditions in this game? I am still seeing this game as completely analogous to a series of coin flip bets.
"Sorry, but I am still failing to see how this can be used to turn a series of coin flip bets into positive expectation" Bingo. The obligatory coin flip is an excellent primer for a junior high school statistics exercise but IMHO has no bearing on consistent profitability in trading markets. You can have streaks of heads or tails but in the long run it's a scratch. It is not correct to equate a successful, professional trader as a superior manager of random events. By strategic design, the successful trader has distilled the market into an event predicament he can exploit at an expectation greater than randomness. I am not aware of any person who successfully makes a living gambling red or black at the roulette wheel - which is analogous to the veritable coin flip. But we are all aware of the professional Black Jack and Poker gamblers who can manage to make a living at it. Superior strategy and a better market with more variables are the difference between the examples.
Here a way to think about it. Since we know the range of the numbers (1 to 100), if you were to know the first number, the probability to decide on the second card does not have to be 50-50. To see it, consider the case of picking as a first number in the first card, number 100. The chance of the second card beating the first number is ZERO. If someone were to always pick card 2, they implicitely give a nonzero chance to card 2 , when the knowledge of card tells us that the chance of card 2 to yield a higher number is zero (i.e. 100 is revealed in the first card). Therefore, when the inspection of the first card is allowed, it is already not a 50-50 game, because of the conditional probability is not the same as the original probability. The random generator just automates the process to cover all cases. For instance if you were to pick 100 in the first card, the random number generated cannot beat that, which is an implicit reflection of the second card cannot beat the first one if 100 is revealed in first card. Note that it is not about choosing whether the second card contains the highest number, but it is about whether the card contains a higher number.
Little confused by this statement. Its not about guessing if the second card has a highest number, its about guessing whether the card (which card?) has a higher number? I think my brain is just poorly wired to deal with this problem. I still think if I choose to show a card with the number X: (1 < X < 100) on it, then flip a coin to decide if the hidden card is either X-1 or X+1. Then, knowing the value of the card being X, does not give any extra information wrt guessing if the hidden card is X-1 or X+1, thus making guessing if X is greater or less than the hidden card come down to 50:50. Then I can do this for every game, causing me to win overall. This is still how I am seeing it... Could you tell me why this is wrong or is violating the rules of this game?
I think you are correct. There is no advantage to be gained by the guesser if your numbers are together. The advantage comes from two numbers that are separated so that using a random number generator, anytime the random number is between the two numbers, you will always select the higher card while the rest of the trials will be 50-50. Joe.
But by the description of 'the game', we can choose our numbers every game, so they are not random, not separated. Which would mean the game is beatable. In fact, you could do that strategy but use 2 < X < 99 and then choose the second card to be X+2 or X-2 so there is separation. So that shouldn't make a difference. And what do you mean 'always select the higher card'? That is a little confusing to me. From my understanding of the game, they are given a known card, and have to guess if that is higher or lower than the unknown card, 'selecting the higher card' doesn't seem to be a meaningful decision for them, when they are limited to guessing 'card2 is higher, card2 is lower'
IMO yes, based on the instructions as I understand them. I would like someone to explain if they are not in agreement with this. Joe.