Some fun Counter-Intuitive puzzles.

MrIndigo

Consider this. I have a box with different coloured balls in it. There are 5 red balls, and 7 blue balls in my box. You reach into the box and take out a ball at random, and put it in your pocket without looking at it. What is the probability the ball in your pocket is red?

Without putting the first ball back, I take another ball out of my box at random and show it to you. It is red. What is the probability the ball in your pocket is red?


Another scenario. A boy comes home to his girlfriend, and she tells him that she took a routine pregnancy test when she was at the doctor, and it came up positive. The boy can't believe it, because both of them use contraceptives; he uses condoms, and she is on the pill. He immediately goes to find the boxes of each contraceptive, and finds that they both have a success rate of 99%. How unlucky they must have been! The girl looks at the print on the back of the pregnancy test pamphlet and finds that the specified false positive rate (i.e., the chance that you get a positive result if you are not pregnant) is 5%, and the false negative rate (the chance you get a negative result if you are pregnant) is 2%. The girl has not been sleeping with anyone else. What is the probability the girl is pregnant?
 
Consider this. I have a box with different coloured balls in it. There are 5 red balls, and 7 blue balls in my box. You reach into the box and take out a ball at random, and put it in your pocket without looking at it. What is the probability the ball in your pocket is red?

Without putting the first ball back, I take another ball out of my box at random and show it to you. It is red. What is the probability the ball in your pocket is red?
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you don't mention if red and blue are the only colored balls in your box, without this information, this is impossible to find a probability. if it really is as simple as only 12 balls, then i guess, yes we can find a probability.
 
kd24 said:
you don't mention if red and blue are the only colored balls in your box, without this information, this is impossible to find a probability. if it really is as simple as only 12 balls, then i guess, yes we can find a probability.
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Yes, the list of balls I provided is complete. Only 12 balls.
 
MrIndigo said:
Both correct, although I'm interested in how you worked out the second one.
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I was too lazy to do the third one, but eyeballing second it was just (5 Red - 1 Red = 4 Red Remaining + 7 Balls = 11 Total Balls For 4 (Red left) / 11 (Total Balls)
 
MrIndigo said:
Both correct, although I'm interested in how you worked out the second one.
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In the second case: Since I have seen a red ball, there are eleven balls I have not seen, of which 4 are red. One of those 11 is in my pocket. EDIT: Goddamn ninja'd

I got 0.20%, rounding up, but I'm assuming you calculated correctly. Again, how did you calculate it?
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I drew a tree diagram, which is a bit difficult to do on the forum. EDIT: Since it's only two steps, I could have done a grid, to get the same result.

Chance of being pregnant = chance of both contraception methods failing = .0001

p(pregnant) = .0001
---p(test+) = .98
---p(test-) = .02
p(not pregnant) = .9999
---p(test+) = .05
---p(test-) = .95

We know the test is positive, so we're only interested in those cases

p(pregnant, test+) = 9.8 x 10^-5
p(not pregnant, test+) = .049995
p(test+) = sum of all ways to get a positive test result = p(pregnant, test+) + p(not pregnant, test+) = .050093

Thus, p(pregnant) = p(pregnant, test+) / p(test+) = 0.20% (OK, I did misround, oops)

(Sidenote: Failure rates of contraception are usually quoted on the basis of a couple using the method for a year. Thus, using the failure rates directly may not be appropriate. But I ignored that issue.)
 
MrIndigo said:
Both correct, although I'm interested in how you worked out the second one.
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Are you sure? The probability that the ball was red when you picked it was 5/12, it should remain 5/12 no matter what new information that you get. Isn't that the case with the Monty Hall problem? The entire point is that probability remains constant from the time of choosing.
 
no it does not, as you gain more information you can refine the probability, consider if after secretly putting the ball in the pocket he withdrew all 5 red balls from the hat, now what is the probability that the ball in the pocket is red?

by your logic the answer would be 5/12 but the answer is clearly 0 since all red balls can be seen out of the pocket.
 
It's also understood that when he showed you the red ball, he didn't put it back in.
 
Even if he did put it back in(although I guess that's the trendy stats class way of saying it so nm!!), you picked the ball before he showed you one from the box. If he showed you a Red Ball you know you didn't pick it, so as was posted, the odds are refined. You picked one of twelve balls, each with the possibility of being red or blue. You take one, and he shows you one, both still come from the same set of twelve balls. You had a 5/12 then, but you now know that you can't have that Red ball that was shown, so it is removed from the set of 12 to make a set of 11 with 4 red/7 blue that you could potentially have in your pocket
 
That gives me an idea for an extension to the first problem.

Consider this. I have a box with different coloured balls in it. There are 5 red balls, and 7 blue balls in my box. You reach into the box and take out a ball at random, and put it in your pocket without looking at it. What is the probability the ball in your pocket is red?

Without putting the first ball back, I take another ball out of my box at random and show it to you. It is red. What is the probability the ball in your pocket is red?
Click to expand...
Now, I put the red ball I have back in the box, shake them around, and draw yet another ball out of the box at random. It is also red.Now what is the probability the ball in your pocket is red?

(Note the answer may or may not still be 4/11. I haven't checked myself yet.)

EDIT: Just solved it. But I'm gonna keep tight lipped about the answer ;-)
 
cantab said:
That gives me an idea for an extension to the first problem.

Now, I put the red ball I have back in the box, shake them around, and draw yet another ball out of the box at random. It is also red.Now what is the probability the ball in your pocket is red?

(Note the answer may or may not still be 4/11. I haven't checked myself yet.)

EDIT: Just solved it. But I'm gonna keep tight lipped about the answer ;-)
Click to expand...

I'm getting 31%.
 
Lhefriel_Medies said:
Are you sure? The probability that the ball was red when you picked it was 5/12, it should remain 5/12 no matter what new information that you get. Isn't that the case with the Monty Hall problem? The entire point is that probability remains constant from the time of choosing.
Click to expand...

Actually, it's the reverse. The Monty Hall problem changes the probability of your door based on the new information.

Think about it this way:

Let's say I have two balls in my box, one red and one blue. If you draw a ball out and put it in your pocket, there is a 50% chance it is red. If I draw the second ball out, and it is red, then obviously the probability of the ball in your pocket being red has to change (to 0%).
 
Generalization, given A red balls and B non red balls, one is secretly pulled and put into a pocket, then n red balls in a row are subsequently randomly revealed from the pile and then replaced, what is the probability that the ball in the pocket is red?
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I realize that it is not clear from the picture but the n-1 is an exponent.
 
Here's one for you guys. There are 6 balls in a bag, 3 red 3 blue. I take out one without looking. Someone takes two balls out of the bag, and they are both red. He puts them back in, shakes it, takes two more balls out and one is red and one is blue. What is the probability that my ball is red?

Correct Latios
 
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