Zigarreten-I’m sorry If I was snappish. At the moment I’m rather upset with myself. Not only have I been wrong in public, there’s a record of it! The really annoying thing is that I’m actually pretty good at math.
It is 2/3
There are 3 blue sides. Call them A, B, and C. A, and B are on the same coin and are thus opposite eachother. C is opposite a red side. The visible blue side is either A, B, or C. If it is A, the other side is blue. If it is B, the other side is blue. If it is C, the other side is red. Out of 3 possibilities, 2 are blue. That's 2/3 or 66.66666%
For the purposes of the OP, it is irrelavent that sides A and B are technically different sides. That together they make up the all blue coin (and therefore will show a blue side if that coin is used, is what counts.
I feel like the main issue of the debate here is what the person is looking for.
Well…that seems to me to be asking about a specific side. There are three blue sides. There are two coins having blue. But we’re not looking for coin, we’re looking for side. If you wanted coin, it’s 1/2, if you wanted side, it’s 2/3. That’s my figuring though.
What is the probability that the bottom is blue also?
If the OP is asking for the probability of the bottom also being blue, he is essentially asking what is the probability that this is the all blue coin as the sides (or faces) cannot act independently from each other.
The 2/3 answer assumes that while in the jar the faces have the ability to mix and match. We are dealing with coins. That the OP asks the probability of the color of a face, does not alter that fact.
So, I’ve almost got myself to 2/3 (although I will say some of the argument presented was not relevant - caveat coming).
The arguments I didn’t take to above were those that took into consideration all three coins (or six sides, if you will). Once we’ve eliminated Mr. Red All Over, we’re at the case I stated above:
blues/# possibles * # remaining blues/ # remaining possibles = probability blue
3/4 * 2/3 = .5 probability blue?
So, ultrafilter, you seem a lot handier with probabilities than I. Where’d I screw that up? I’m thinking that we should not consider the conditional case (caveat here), as above, and take it from step two alone.
It’s 2/3. The halfers are forgetting that the very fact that you’re looking at a blue face to start with has changed your perception of whether you’ve drawn an all-blue coin or not. (In math lingo, seeing a blue side up has changed your Bayesian priors.)
Allow me to illustrate with a little thought experiment. Suppose you have two million-sided die. (Hey, I said it was a thought experiment.) One of the dice is a “normal” die with all the numbers from one to a million on it. The other die has the number “1” on every one of its faces.
The two dice are hidden in a bag. You grab in the bag, take a die at random, and drop it on the table. You look at the number on top. (No, you can’t look at the numbers on the sides.;)) The die shows a “1”.
Given you see a “1” showing, what is the chance that you’ve grabbed the normal die from the bag? 50%? That doesn’t seem right. After all, if you grabbed the normal die, it’s literally a million-to-one shot that you’d roll a “1” with that die. It’s much, much more likely that you’ve picked up the die with “1” on all of it’s sides.
A similar thing is going on in the coin example, just with different probabilities. Seeing a blue side up is a big clue that you probably have not picked the half red/half blue coin.
For the dope on Bayesian probability, consult the intro at:
I suppose it’s just how you see things then. You and the 1/2 family think in terms of coins. I and others of the 2/3 family think in terms of sides. I was convinced over pretty quickly. Wumpus stated well I think, your perception becomes flawed by observation.
I’m not positive I’m reading you correctly, but let me try.
I take it that by “# blues/# possibles (= 3/4)” that you are referring to the initial probability of the blue face coming up in the first place. I’m also thinking that “# remaining blues/ # remaining possibles (= 2/3)” refers to how many blues are left (after the initial blue).
The problem here is the 3/4 shouldn’t be there–we’re assuming it’s blue face up to begin with; that’s our entire “universe” of possibilities, it’s given that blue has come up, we know that with probability one.
For the second part, that’s basically the idea–there are three different possible “down” sides to the coin, one red, two blue.
So the 3/4 shouldn’t be there, only the probability of 2/3.
Just to work this all out according to Bayes’ Rule…
Our question is, “what is the probability of the coin we’ve drawn being the all-blue coin, given we’ve got a blue face showing?”
Bayes’ Rule says: P(A|B) = P(B|A)*P(A)/P(B)
where “P(A|B)” means “the probability of event A happening given event B has already happened.”
So, P(all-blue coin|blue face showing) = P(blue face showing| all-blue coin)*P(all-blue coin)/P(blue face showing)
Now let’s figure out the terms on the right-hand side of the equation. The first one’s easy–the probability we have a blue face showing given we’ve chosen the all blue-coin is obviously 100%, or 1.0.
The unconditional probability of choosing the all blue coin–i.e the probability at the very beginning of the experiment, before any coins are drawn–is 1/3. Remember, there was an all-blue coin, an all-red coin, and the split-color coin.
The unconditional probability of having a blue face showing is 1/2. Of the three coins described above, half the faces are red, and half are blue.
So the P(all-blue coin|blue face showing) = 1.0*(1/3)/(1/2) = 2/3.
So, the conditional case of two coins, 1 r/b and 1 b/b, results in a second coin face viewing probability of a blue half the time when viewed from the initial condition, but 2/3 of the time once one blue is known on the first roll.
I’m a converted lurker. I was on the 50% side for a while, then started to write a program to prove it.
I set up a case statement for each of the possible six sides being face up.
1 = First side of red/red coin
2 = Second side of red/red coin
3 = Red side of red/blue coin
4 = Blue side of red/blue coin
5 = First side of blue/blue coin
6 = Second side of blue/blue coin
I realized I could eliminate the first three possibilities as they all had a red side facing upward. Since the remaining three possibilities would be equally distributed, each one gets a 33.3% chance of coming up. Realizing the the bottom two would account for two of those possibilities, thus totalling 66.6%, I stopped writing the program.
Yeah, sounds like you got it. If we have the two coins (b/r and b/b), the probability of the top face being blue AND the bottom face also being blue is 1/2, which seems to be what you were calculating earlier. If we assume that a blue face is already visible, then there’s a 2/3 chance the hidden face is also blue. Am I right in thinking that’s what you were saying?
Basically, the only relevant thing is the number of possible outcomes. This sort of situation shows up in a number of problems–Cecil did a column that deals with a problem very much like this.
