a trading problem for mathematicians
- Thread starter trend2009
- Start date
I would have thought by now that he would have figured it out by hand. If he knew how often it happened and his rules are so simple, I really can't believe this has gone on for days. He could have come to the conclusion using a spreadsheet if he didn't have any other way to figure it out.
50%
50%
I know you say this is wrong. Actually it is you who are wrong. Initially your probability was 1 in 3 because you didn't know what was behind any of the three doors. After you learned what was behind one door, your probability turned into 1 in 2. Those who maintain that new knowledge doesn't change probabilities have a very strange concept of probability IMO. What if you suddenly gained x-ray vision? Then you would have perfect knowledge of what was behind the two closed doors and you would have 100% chance of picking the door with the car. New knowledge changes probabilities.
And before you dismiss the x-ray vision thing as impossible, what if you saw one door slightly move because the goat brushed against it, or what if you heard the faint rumble of a stomach growl because the goat was hungry? New knowledge, almost as good as having x-ray vision. Expert poker players use their opponents' tells to change their chances of winning all the time.
Yeah, I read it and don't buy it.
Here's their best argument:
As Keith Devlin says (Devlin 2003), "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
Here's the counter argument: The probability that the car is behind your door and any other door is 2/3. A door with a goat is opened. Now the probability that the car is behind your door and the unchosen door is 1. Have you learned anything at all new about your door or the unchosen door? No. So either is equally likely to have the car.
Acting like the new knowledge of opening one door affects only the unchosen door is capricious and arbitrary. New knowledge affects EVERYTHING.
Here's the flaw in the 2/3 argument:
"Your choice of door A <s>has</s> had a chance of 1 in 3 of being the winner. <s>I have not changed that</s> That absolutely changed when I opened a goat door."
I LOVE this thread. euclid is the ONLY guy speaking some sense (, and there are a couple of quite successful traders in here, too, LOL!) 
For some reason, I am reminded of the survey which found that people who rank the poorest in mathematical skills ranked themselves as being super-confident that they were correct. Dunning-Kruger effect : http://en.wikipedia.org/wiki/Dunning–Kruger_effect
> 36/(36 + 16)
[1] 0.6923077
Good night.
For some reason, I am reminded of the survey which found that people who rank the poorest in mathematical skills ranked themselves as being super-confident that they were correct. Dunning-Kruger effect : http://en.wikipedia.org/wiki/Dunning–Kruger_effect
> 36/(36 + 16)
[1] 0.6923077
Good night.
I don't know, I think this model might not be correct, System B cannot be assume to be applied to the balls selected by System A if they're uncorrelated. This point has probably been made, I don't know. I think the model is more like:
1) suppose there are a lot of price bars
2) system A tags some of the price bars. 60% of those tagged are winners, 40% are losers
3) system B independently tags some of the price bars. 60% are winners, 40% are losers
My probability book would probably say: how many of the price bars were both tagged by A and B? Then of those, how many agreed (this would be total number of trades)? Then of those, how many won? That would be the win rate: winners / number of bars with agreed signals.
In one extreme, suppose there were 50,000 price bars, System A only gave a signal on 100, System B on another 100. Very likely there is no overlap between the 2 systems. In the other extreme, they give a signal for every bar, then every bar has overlapping signals. Most likely it's somewhere in between, I think you need to find the overlap number. Then you could estimate the agreement rate, then the win rate.
I don't know the theory or formulas well enough to estimate that, I always get the exercises wrong, it seems everyone has a different opinion too; instead I'd make a computer simulation and get some numbers from that. (In addition, my book says that's another way to get a probability estimate.)
you do not need to buy it, though you should put the work into it. If you ever put the minimum effort into something that really puzzled you then you would have set up a simple monte carlo simulation and would have discovered that the probabilities of hitting the price indeed switch from 1/3 -> 2/3 , they double.
Your posts on the OP's problem did not make sense nor do your comments on Monty Hall make sense. As said before it delights me to know that the number of ignorant and lazy people does not decrease but is safely floored.
Your posts on the OP's problem did not make sense nor do your comments on Monty Hall make sense. As said before it delights me to know that the number of ignorant and lazy people does not decrease but is safely floored.