I came to the same conclusion though I never got an explanation why it was incorrect. After reading the article that the game was supposedly taken from, my original assumption that the game was incorrectly (unclearly) stated was turned out to be correct. Specifically the line "You are allowed to choose any two numbers between 1 and 100.;You reveal to me one of the two numbers.", implicitly implies that you are allowed to 'choose' which card to show. Whereas the example she cited, the 2nd person randomly picks which card to show, which turns it into a completely different game. And prevents you from doing the trick that you and I described (this could have been explained 3 pages ago....) So in the end, the problem is miscommunication. Which is what you get when someone posing a question does not address why possible answers are correct or incorrect, ie, expose their understanding/assumptions of the game. *shakes fist*
Walter: I believe that if you do a search on what is known as the Secretary Problem, you would be able to find the information related to the point of the problem posed. Regards
The problem is you couldn't find the solution. I don't deal with ego issues, sorry, I'm only good at trading. This is why I will ignore your replies from now on. As for the game, the rules are clear and there are no misunderstanding. "Choose" doesn't mean "draw randomly". 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%. The "trick" that you described doesn't change this fact. Ninna
Right again. It is an application of the optimal stopping problem. Here's a link that explains the problem. http://www.americanscientist.org/issues/feature/knowing-when-to-stop/1 Ninna
Don't know if this has been mentioned in your thread, but trading is not about price movement as much as it is about market sentiment. When you look a price you are looking at an end effect. Consider study in corollaries to price. The coin in the real market is not evenly weighted
Ninna: Both your posts and the posts by Walter are excellent. It would be a waste of talent if both of you enter in some sort of arguments. I suggest that you post the solution at some point. I did not want to post it because you deserve posting it as you introduced the problem, and it is a very insightful problem/solution. Regards
I read your post before my above answer. You deserve credit for introducing the problem! Also thanks to Walter and others who discussed it. All you guys are an outstanding group of people. Keep it coming! I have some ideas and follow up questions related to this, but I want to let others moving it forward first. If they are not discussed by others, I would discuss them in future posts. So stick around.
Well that was needlessly condescending. Anyways, I've given up hope on the original poster being able to give a useful answer to my question. Hopefully some other helpful reader can explain to me how a game which can be reduced to a series of coin flip bets (based on the original given rules) and be made profitable for someone betting on 101:99 risk reward by using stopping strategy, which as far as I know does not apply to simple coin flip bets. If I am incorrect in my understanding, I would love to know if anyone would be so generous as to explain it to me.
N = 2 Surprise Now, suppose you must decide when to stop and choose between only two slips of paper or two cards. You turn one over, observe a number there and then must judge whether it is larger than the hidden number on the second. The surprising claim, originating with David Blackwell of the University of California, Berkeley, is that you can win at this game more than half the time. Obviously you can win exactly half the time by always stopping with the first number, or always stopping with the second, without even peeking. But to win more than half the time, you must find a way to use information from the first number to decide whether or not to stop. (Readers take comfort: When mathematicians first heard this claim, many of us found it implausible.) Here is one stopping rule that guarantees winning more than half the time. First, generate a random number R according to a standard Gaussian (bell-shaped) curve by using a computer or other device. Then turn over one of the slips of paper and observe its number. If R is larger than the observed number, continue and turn over the second card. If R is smaller, quit with the number observed on the first card. How can such a simple-minded strategy guarantee a win more than half the time? If R is smaller than each of the two written numbers, then you win exactly half the time ( p / 2 of the unknown probability p in Figure 4); if it is larger than both, you again win half that time ( q / 2 of q, also in Figure 4). But if R falls between the two written numbers, which it must do with strictly positive probability (since the two numbers are different and the Gaussian distribution assigns positive probability to every interval) then you win all the time. This gives you the edge you need, since p / 2 + q / 2 + 1âpâq is greater than ½, because 1 - p - q is greater than zero. For example, if the two hidden numbers are 1 and ð, this Gaussian method yields a value for p about .8413 and q about .0008, so the probability that it will select the larger number is more than 57 percent. Of course if the number writer knows this Gaussian strategy, he can make your winnings as close to ½ as he wants by writing numbers that are very close. If the number writer is not completely free to pick any number, but instead is required to choose an integer in the range {1,2,â¦,100}, say, then he cannot make your probability of winning arbitrarily close to ½. In this case it also seems obvious that the number-writer would never write a 1, since if you turn over a 1, you will always win by not stopping. But if he never writes a 1, he then would never write a 2 either since he never wrote a 1, and so on ad absurdum . Interested readers are invited to discover for themselves the optimal strategy in this case, and the amount more than ½ one can guarantee to win on the average.
Sorry, but I am still failing to see how this can be used to turn a series of coin flip bets into positive expectation. Would mind you explaining it in your own words addressing the specific coin flip example I gave?