Hi all!
Not having taken statistics in school, I’ve got no idea what the answer to the following question is. The first correct answer gets a virtual cookie (it’s a cookie jar you see.)
You have 3 coins in a jar:
one is red on both sides,
one is red on one side, blue on the other,
and the last one is blue on both sides.
You pull one coin out of the jar at random and slap it on a table. You see that the top is blue. What is the probability that the bottom is blue also?
I’ve asked this question of another group and got different answers: 50%, 66% and I figured out a way to calculate a 75%. What’s the real answer.
My trial of 38 throws has yielded 21 blue-up with 14 of these being blue-down and 7 being red-down. 66%. I do, however, think that 38 throws is a rather limited test.
Your total outcomes number 6- 2 outcomes have red up, red down, 1 outcome has red up, blue down, 1 outcome has blue up, red down, and two outcomes have blue up, blue down.
Of the blue ups, two have blue down, one has red down.
So seeing a blue means there’s a 2/3 probability the bottom
is blue.
~66.666%.
Another way to look at this is that when you pull out the
all blue coin, you’re set. When you pull out the red/blue,
you still have another ‘hurdle’ to overcome to see the blue
face - placing it so blue shows.
So the blue/blue shows blue twice as often as the red/blue.
True there are 3 coins in the jar, but that doesn’t enter in to the problem.
If the visible side is blue, then you can not be looking at the red/red coin. The coin is either blue/blue or blue/red. Thus, the other side is either blue or red. That’s 1 possible outcome out of 2. That’s 50%.
For this problem, math is more useful than empirical testing. It is possible to toss a penny a million times and get heads every time. But the odds of getting heads are always 50/50.
When you take the coin out there six possible results; note that each side of each coin must be dealt with separately, even when they are the same color:
Top Bot
R-1 R-2
R-2 R-1
R B
B R
B-1 B-2
B-2 B-1
Since the top side is blue, the first three possible results are eliminated. Of the remaining three possibilities two have a blue bottom. 66%
The question is worked like this: IF you have a blue side facing you, what is the other side? This means you have already pulled a coin out of the jar and put it down, so the odds involved in pulling a blue side out of the jar are irrelevant.
So, given that there is a coin on the table with a blue side facing you, there are equal odds that the coin is the blue/blue one or the blue/red one, there being one of each.
That is the question at hand. The odds of drawing a coin with a blue side or having a coin land blue side up are not relevant. The coin IS on the table. A blue side IS up. Those are conditions given in the OP.
Under those conditions, the coin must be either r/b or b/b/. If I turn the coin over, either I get a blue side or I get a red side. Again, given the conditions stated in the OP those are the only possible outcomes. One outcome of a possible 2 is blue. That is 50/50.
P denotes the probability of a given outcome, and [sup]c[/sup] denotes the opposite of a given outcome. P(A|B) is the probability that A happens given that B has already happened. In this case, A is the event that the bottom is blue, and B is the event that the top is blue.
So P(A) is 1/2, and P(A[sup]c[/sup]) is 1/2. P(B|A) is 2/3, and P(B|A[sup]c[/sup]) is 1/2. See zigaretten’s post to see where I got these from, using the formulas P(A|B) = P(A and B)/P(B) and P(A[sup]c[/sup]) = 1 - P(A).
Plugging this in, I get 2/3, which is consistent with zigaretten’s answer (and his/her method is great for conditional problems). It also meshes with the OP’s trials–for an exercise, try to figure out the probability of getting 66% as an estimate when the true value is 50%.
For those who want more info, see Sheldon Ross’s Introduction to Probability Models, 7th ed.
No, you have two coins with a total of three blue sides. Imagine that you take a magic marker and actually write 1,2 and 3 on the three sides. Forget about the times that the red side comes up. When a blue side is on top: 1/3 of the time you will get a 1 on top; 1/3 of the time you will get a 2 on top and 1/3 of the time you will get a 3 on top.
Of these three possibilities; two of them have blue bottoms (so to speak) and one has a red bottom.
This is a trap people often fall into; it’s very similar to the old Monty Hall problem.
What can happen in drawing a coin and placing it down on the table?
Draw R/R, red face up–1/3
Draw R/B, red face up–1/6
Draw R/B, blue face up–1/6
Draw B/B, blue face up–1/3
Obviously, we’re only concerned with 3 and 4. Let’s scale the probabilities of 3 and 4 so they add up to 1 (because we know for 100% that either 3 or 4 happens). Scale them by multiplying by 2:
Draw R/B, blue face up–1/3
Draw B/B, blue face up–2/3.
So if you end up having a blue face up, 1/3 of the time you drew the R/B coin, 2/3 of the time you drew the B/B coin.
bup - I’m gonna leave this to you since I seem to be just endlessly repeating you. Good luck !!
(You may need it. I once argued for an hour with a guy who insisted that the odds of rolling a six on a die were 50/50. It either will be a six, or it won’t; 50/50!!)
<<If the visible side is blue, then you can not be looking at the red/red coin. The coin is either blue/blue or blue/red. Thus, the other side is either blue or red. That’s 1 possible outcome out of 2. That’s 50%.>>
The fallacy here is in thinking that one side is blue so the other side is blue or red. Sounds good, but mis-states the problem.
Instead of blue/red, number all the sides:
R1 and R2 are the all red coin
R3 and B3 are the red/blue coin
B1 and B2 are the all blue coin
Now, when you already have a blue side up, you’re saying that the other side must be red R3 or blue B1. But the other side could be blue B2, if you see what I mean. There are three possible blue sides that could be visible, not just two. This leads to the two-thirds answer, and points the fallacy of the 1/2 answer.
I ran a simulation, and wound up with 8 Red and 5 Blue, conclusively proving … um… ah, never mind.
Yes, you had the same answer, but you didn’t spell out the sample space, and that’s what people like me need. I had to go through and fix the probabilities three times before I got it right. Anyways, my apologies for overlooking your contribution.
(Though the comment was not particularly aimed at yourself; it was rather meant to be a general comment on how difficult it is to get people to change their minds once they get the wrong answer into their heads. Look at how long people have been arguing over the Monty Hall problem.)
For those who still think it may be 50/50, take it to a logical extreme:
Say we don’t have three coins, say instead we have three dice with a million sides each. One die is all red, another is all blue, the third die has all but one face red, the remaining face blue.
A die is pulled out and rolled. All you’re given is it came up blue. What’s the probability that it was the all blue die?
Remember, if you pull out the 999,999 red, 1 blue die, it’s extremely unlikely it will come up blue. The chances are overwhelming that the all blue die was the one drawn.
We have the exact same principle working here–if the coin is blue face up, it’s more likely that the all blue coin was drawn, because it’s more likely to come up blue than the B/R coin.