Hi all! I have a small bet going, and I was wondering if anyone could settle it for me.
A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads?
Everyone but me (usually meaning I’m wrong) says that you have two possibilities: H/T coin or H/H coin. Odds therefore are 1/2.
I’m still stuck on another train of thought. My coin sides are:
1H, 1T, 2H, 2H (where 1 and 2 are coins and T/H are Tails/Heads). So I pull a coin and see a Head. This means I’m looking at either (1H, 2H, or 2H [other side]). So odds are 2/3 that I’m looking at coin #2 and therefore 2/3 that the other side is a head as well.
[ul][li]You draw the two-headed coin and see face 1.[/li][li]You draw the two-headed coin and see face 2.[/li][li]You draw the normal coin and see heads.[/li][li]You draw the normal coin and see tails.[/ul][/li]
In three of the four events, you get heads, so the probability is 3/4.
Condtion that the face observed is already heads. Therefore, 3 possible outcomes, 2 of which will be heads (the double head coin). So the chance is 2/3.
ultrafilter: The OP wasn’t “what is the probability of drawing heads?”, but “given that we’ve drawn heads, what is the probability that the other side of the coin is heads?”
CanTak3: I’m pretty sure you’re right. There are four possibilities, all with equal likelyhood: You view the tails side of the mixed coin (other side heads); you view the heads side of the mixed coin (other side tails); you view heads side #1 of the two-headed coin (other side heads); you view heads side #2 of the two-headed coin (other side heads.) The first possibility isn’t being considered, so there are three possibilities, all with equal likelyhood. And the other side of the coins is heads in two of them.
The proper way to take advantage of this, of course, is to set this game up, say with two index cards, and offer to play this game with your friends. Say that you’ll give them $6 every time a head comes up and the other side is tails, and they’ll give you $5 every time a head comes up and the other side is heads. Then laugh all the way to the bank.
The answer is 1/2. The reason is that you have already established an initial condition of the side facing up of the selected coin is heads. That takes away one of the heads on the double coin, and the head on the normal coin, leaving you with this: Select a coin…what are the chances the side you can’t see is heads? Since you’ve established that you’ve pulled the coin out with heads showing, you have only that one of the coins has heads, the other has tails, giving a 1/2 probability.
Now, if the question was: “What is the probability of picking out a coin, that shows heads when you pull it out, then flipping it and showing heads on the other side?” it changes a bit…
Then there are 4 sides that can show coming up initially: H, H, H, T. So, you have a 3/4 probability of meeting the first condition. Then, provided the first condition is met, you have a 1/2 proability of meeting the second, giving you a total probability of 3/8.
Your initial question makes the first condition 1/1 probability, leaving only the 1/2 probability after the coin shows heads to begin with.
Yeah, wrong problem. 2/3 is the right answer–the only outcome that’s ruled out is the one where you draw the normal coin and see tails, and the others are equiprobable.
Incorrect…since you’ve already established that the face observed is heads, you have TWO possible outcomes, H or T. You must eliminate the ‘showing heads’ for both coins, not just the one…
You must not confuse this with the “Monty Hall” problem, where you learn one of the ‘doors.’ The revealing of a door in that situation only eliminates that door, whereas the revealing of a heads in this situation is tied to the probability of another heads being on the reverse. This is a situation where knowing one of the sides alters the probability of the remaining sides showing up, while in a ‘door choice’ problem, it doesn’t.
Jman: I get what you’re trying to say, but I don’t think you’re explaining it well.
If the two sides of a coin are distinguishable–say one’s red and one’s blue, and you can flip to the blue side before you look at whether it’s heads or tails–and you always look at the same side, then the probability is 1/2.
But if you draw a coin, flip it, and look at whichever side comes up, the probability is 2/3. So the OP is not entirely specified.
No, it’s not. The fact that the known side matches one of the other sides is irrelevent, since it is a starting condition, not a condition of choosing. The problem as phrased gives a starting condition of heads. Think of it this way: You have two coins on a table, both showing heads…one has a tail, the other has another heads: choose one: What’s the probability the coin you choose has a head on the other side. THIS IS the same problem, because the way the question is phrased, the chance of ‘pulling out the coin showing heads’ is 1. The way the question is phrased, you CAN NOT pull out a tail showing, which means that you are left with simply choosing one of two coins…one with a tail, one with a head, with the obverse of both being irrelevent, because that side will ALWAYS be showing.
Solve this using Bayes’ theorem: Let’s call the two coins C1 (two heads) and C2 (one head, one tail).
Let H = find heads on coin selected. We want P(C1|H). This is just P(H|C1)P(C1)/[P(H|C1)P(C1)+P(H|C2)P(C2)].
P(C1)=P(C2)=.5 since the selection of the coin was random.
P(H|C1)=1, P(H|C2)=.5. Plug everything into the formula to get:
P(C1|H)=1.5/(1.5 + .5.5) = .5/.75 = 2/3.
This is sorta like the Monty Hall problem, but simpler. But like the MH problem, the result is kind of counterintuitive.
Ok, god knows I’m no expert in probabilities, but I don’t think this is quite the same problem.
The question specifies that ONE of the two coins is showing heads, not that they both are.
As I read the original question, it says that if you pull out a coin and the side showing is tails you ignore that pull, not that pulling a tails is impossible.
seem to clearly imply the visible side can be tails.
Under those rules, seems like there are three possibilities for the unseen side of a coin showing heads: two heads, one tails.
Seems to me the odds should 2/3 for the unseen side being heads.
It is not the same as having two coins sitting on the table. Given that you see a head on the coin you draw, it is more likely that you have drawn the two-headed coin since two of the possibilities that result in you seeing a head are from one coin, and only one of the possibilities that result in you seeing a head are from the other coin. Since it is more likely given the statement of the problem that you drew the two headed coin, it is also more likely that when you turn the coin over it is a head (I know - that’s two different ways of saying the same thing). When you have the coins laying on the table and you point at one of them you are only basing the probability on one side of each coin (the underside) rather than two sides of one coin and one side of the other - hence the difference in probabilities.
Jman, the trick here is that there are two ways that you could pull a coin with a head on one side and a head on the other - head 1 of coin 1 and head 2 of coin 1. You don’t know which one you might be looking at.
The only possibilities that result in you looking at a head on the first side of the coin you draw are:
Head 1 on coin 1.
Head 2 on coin 1.
Head on coin 2.
Those three possibilities are all equally likely, and two of them result in the same coin having a second head on the reverse side. Thus 2/3.
Imagine two coins. One has faces numbered one and two, and the other has faces numbered three and four. A coin is drawn with a number less than 4 on the face you observe, what is the probability that the opposing face has a number less than 4 on it.
The initial condition of the number being less than 4 leaves 3 equally likely possibilities for the observed number: 1, 2, or 3.
Case 1: Observed 1. Other side 2 < 4. Success.
Case 2: Observed 2. Other side 1 < 4. Success.
Case 3: Observed 3. Other side 4 = 4. Failure.
Two succeses out of three possibilities = 2/3 probability of success.
When you draw a coin and look at one side, there are actually four possibilities (you are looking at 1 of 4 available faces):
1: Coin 1, Side 1 -> You see heads. The other side is also heads.
2: Coin 1, Side 2 -> You see heads. The other side is also heads.
3: Coin 2, Side 1 -> You see heads. The other side is tails.
4: Coin 2, Side 2 -> You see tails. The other side is heads.
All 4 cases are equally likely. If you first see heads, case 4 is eliminated. But the other 3 option are still equally likely. So the chance the other side is heads is now 2/3.
In a repeated test, each coin has a 1/2 chance of being selected, but if the tails coin comes up tails, then that trial is rejected from the sample. The fact that the H/T coin is rejected half of the times it is picked is what biases the probability toward the 2-headed coin.
(I concur…it’s 2/3, for reasons already clearly elicidated. You can demonstrated it by a truth table, an application of Bayes’ Theorem, or by running a sufficient number of trials, but the result is the same.)
