Talking about probability on the Celtic manager thread got me thinking of old maths/probability questions. Here's a few I can remember in case any of you are as sad as me (each harder than the last): You're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, dog turds. You pick a door, and the host, who knows what's behind the doors, opens another door, and shows you a dog turd. He then offers you the choice of sticking with your door or switching to the other unopened dooe. Do you switch? Suppose that you are worried that you might have a rare disease. You decide to get tested, and suppose that the testing methods for this disease are correct 99 percent of the time. Suppose this disease is actually quite rare, occurring randomly in the general population in only one of every 10,000 people. If your test results come back positive, what are your chances that you actually have the disease? Someone flips a coin, if it comes up heads they put a white marble into a box, if it comes up tails they put a black marble in. They do this 10 times. You know this has happened but do not see it take place. You then reach into the box, take a marble out, look at the colour and replace it. You do this 10 times and pull out a white marble each time. What is the probability that all 10 marbles in the box are white?
With that first one, statistically you should switch - when you initially chose you had a 33.333 % recurring chance of the car. Now you have a 50% chance.
First one, you should always switch. If you stick you have a 1 in 3 chance of winning the car. If you switch you have a 2 in 3 chance of winning it. T'others - me brain hurts too much.
And the second one I think is 99% - the prevalence of the disease doesn't make any difference when you know how accurate the test is?
Actually it's not 50%, it's 66.67% as Jay says. http://www.bbc.co.uk/news/magazine-24045598 If that doesn't make your brain hurt nothing will.
My head hurts... if one of the doors is removed from the choice then you would have a one in two chance of getting the car irrespective of which door you chose ? unless I've misunderstood.
You choose a door. That is 1/3 to have the car behind it. The chance of it being behind one of the other 2 doors is 2/3. The host will always be able to open one of the other 2 doors to show a dog turd regardless of where the car is. Therefore the chance of it not being behind your door is the same, meaning it is 2/3 to be behind the other door. Think of it another way, there are 100 doors. You pick a door and the host shows you 98 dog turds and leaves one door closed. Would you be more tempted to switch now? Same principle.
[TABLE="class: wikitable"] <tbody>[TR] [TH="bgcolor: #F2F2F2"]behind door 1[/TH] [TH="bgcolor: #F2F2F2"]behind door 2[/TH] [TH="bgcolor: #F2F2F2"]behind door 3[/TH] [TH="bgcolor: #F2F2F2"]result if staying at door #1[/TH] [TH="bgcolor: #F2F2F2"]result if switching to the door offered[/TH] [/TR] [TR] [TD]Car[/TD] [TD]Goat[/TD] [TD]Goat[/TD] [TD]Car[/TD] [TD]Goat[/TD] [/TR] [TR] [TD]Goat[/TD] [TD]Car[/TD] [TD]Goat[/TD] [TD]Goat[/TD] [TD]Car[/TD] [/TR] [TR] [TD]Goat[/TD] [TD]Goat[/TD] [TD]Car[/TD] [TD]Goat[/TD] [TD]Car[/TD] [/TR] </tbody>[/TABLE]
Reight, I'll have a go at second one. It's about 1% It's because in such a rare disease false positives greatly outnumber genuine cases. In a sample of a million folk you'd expect 100 to have the disease (10,000 x 100). 99 of those will be correctly diagnosed. But if you tested the other 999,900, 1% of that lot will be incorrectly diagnosed. 1% of 999,900 is 9999. That's how many false positives you'd get. So you're far more likely to have a false positive than a genuine diagnosis. Genuine diagnosis from your million sample = 99 False positive = 9999 (99/9999)*100= 0.99 or about 1%
For the second one isn't it that the test will be wrong 1% of the time, so for 10,000 tests it is wrong 100 times and 1 test will be the correct positive so the answer is 1 in 101
If I have been shown that there are 98 doors that are the wrong answer then from that point on they are eliminated from the scenario. There are only two doors now to choose from and therefore either has an equal chance of having the car behind it. Any other answer is a set of lah-di-dah smug academics trying to convince you that they are clever...
Can't do third one as there are too many parameters and I've never done maths, so I don't know how probability formula work. Well, I did it in school if you can call trying to cop off with the lass sat behind you as maths. I thought making her laugh was the way forward, which I did, but it never worked. Thinking back, remembering what she was like, I would have impressed her a hell of a lot more if I'd just done the ******* maths. Unqualified and no snogging behind the bike sheds. Way to go Jay. The ignorance of youth. And adulthood...
isn't the third one 0.98% (50%*50%*50%...... and so on for the 10 tosses ) each coin toss had 50% probability of being heads and the product of all those outcomes is 0.98% ?
This was used as a plot device in Ian McEwan's last novel Sweet Tooth. And I think I got it. But there is ome disagreement in the maths community about apparently. But I'm dense at maths. http://en.wikipedia.org/wiki/Monty_Hall_Problem
It would be in isolation, but the fact that you have done the pully marble thing gives you extra information. The probability of there being 10 white marbles in there given you pulled a white one out 10 times is different to just the probability of the coin coming up heads 10 times.