you are probably aware of this problem - you have 100 numbers written on 100 pieces of paper, your goal is to pick the highest without any prior info about the distribution. you see the first, then decide if you want to bet it is the highest or continue with the next one. once you see the next, you can't go back - you can either pick it to be the highest or continue. the optimal strategy is to open the first n/e, then pick the next highest.http://mathworld.wolfram.com/SultansDow ... m.htmlhere is a claim that i read somewhere - you have 2 randomly chosen real numbers written on 2 pieces of paper. you see the first, then you have to decide if it is the highest of the two. you can either pick it, or decide to see the next and pick the second. you have absolutely no information about the distribution from which the numbers are chosen, and it can be any two real numbers. can you improve your odds of guessing the highest number from 50-50 to something better?the article claims that you can - look at the first number, then draw a random number from some continuous probability distribution of your choice, say normal. if the number on the first sheet of paper is less than your random number, you bet on the second card, otherwise you bet on the first. the claim is that if your random number happens to be in the interval determined by the two numbers on the cards, then you guess correctly. if you random number is outside the interval, then it is 50-50 chance that you guess correctly. so by using some "test distribution" which gives a non-zero probability for any interval (such as the normal pdf), then you managed to improve your odds of guessing the highest number.can someone refute this or explain it better if they think it is a correct argument? it seems to me like seeing in the dark, almost like quantum interference

i can provide an amateurish view.. since the decision is conditional on knowing the rand no 'e' and the number on first card 'z0', for the argument to be true 'p(z>z0|e>z0)' should be meaningful, but i am unable to find any such meaning

P(Correct) = P(x > y)*P(z < x) + P(x < y)*P(z > x) = P(x >y)*P(z < x) + (1 - P(x >y))*(1 - P(z < x)) = 0.5*P(z < x) + 0.5*(1 - P(z < x)) = 0.5

isn't it P(x>y | z<x)P(z<x)?and if yes, can you say that it is equal to P(x>y)P(z<x), because these two don't seem independent to me since the order of x,y,z matters?

Suppose we define x,y,z as follows:x is the smaller of the two numbers written on paper chits, y is the greater of the two numbers written on paper chits. z = any randomly generated numberLet p1 = P(z<x), p2 = P(x<z<y), p3 = P(z>y); (p1+p2+p3 = 1)P(the proposed strategy gives a correct guess) = p1*(we pick the paper chit having 'y' on it) + p2*(we pick any paper chit) + p3*(we pick the paper chit having 'x' on it)=p1*0.5+p2+p3*0.5 = 0.5 + p2/2In beero1000's post, (correct me if I am wrong), it seems that in the two cases taken, x and y are getting interchanged, i.e. the x's in P(z<x) and P(z>x) are not the same: (in P(z<x), x is the larger number and in P(z>x), x is the smaller number, so they do not add upto one, i.e. P(z>x) != 1-P(z<x))

An afterthought:The formula P(correct)=0.5+p2/2 suggests that we can improve the probability of guessing correctly the higher number. But that is not exactly so.Note that p2 depends on the distribution from which we generate the number z, and as stilyo mentioned, that could be any distribution of our choice.But in order to be completely general in our approach, the only distribution that we can take is the one in which each real number is equally likely (i.e. the distribution from which x and y themselves are generated), and in this case, p2=0Suppose that we take any other distribution, say a gaussion whose 90% area is between 100 and 200, and suppose x and y were 1 million and 2 million resp. Here again, p2~0. In fact, p2 will be finite only if we are extremely lucky, or we already have some information about x and y (and in that case, they are not generated from the "equally likely" distribution).So, the probability that p2 is finite is zero.So the probability that we can improve our chances of guessing correctly is zero.