Two-coin problem
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53 members have voted
Quote: MarilynSay that a box contains two coins. One coin has heads on both sides. The other coin has heads on one side and tails on the other. A coin is selected at random, and the face of one side is observed. If the face is heads, what is the probability that other side is heads?
Quote: WizardA 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?
I'm not here to gripe about Marilyn taking my wording without attribution but to bring up the problem for discussion. The question for the poll is what is the answer?
A) I believe this was already discussed here. (And I had to take time to wrap my head around it then too!)
B) Don't be angry with Marilyn. It's not exactly such a unique ponderable that you had to be the source she (or whoever posed the question to her) used. There's very few ways to word this same problem.
C) I'm sensing something of a theme for polls that do not use the maximum of ten choices.
Quote: WizardToday's Ask Marilyn features the two-coin problem. The wording is almost the same as on my mathproblems.info site. In fact here is a comparison:
I'm not here to gripe about Marilyn taking my wording without attribution but to bring up the problem for discussion. The question for the poll is what is the answer?
I picked 1/2 as there are 3 sides remaining, 1 tails and 1 head. But that doesn't matter. There is a 1/2 probability that you have the 2 headed coin. So there is a 1/2 chance the other side is a head.
Of course I am guessing I am somehow incorrect......
You bump into your colleague in a train on the weekend. He travels with a boy who is one of his two children.
What is the probability the other children is a girl ?
Quote: AZDuffmanQuote: WizardToday's Ask Marilyn features the two-coin problem. The wording is almost the same as on my mathproblems.info site. In fact here is a comparison:
I'm not here to gripe about Marilyn taking my wording without attribution but to bring up the problem for discussion. The question for the poll is what is the answer?
I picked 1/2 as there are 3 sides remaining, 1 tails and 1 head. But that doesn't matter. There is a 1/2 probability that you have the 2 headed coin. So there is a 1/2 chance the other side is a head.
Of course I am guessing I am somehow incorrect......
If you are seeing one head, then there is a 2 out of 3 chance that you are looking at the two headed coin, because you could be looking at either side of it. That's why the answer is 2/3.
Quote: MangoJThis may (or may not) be an equivalent problem:
You bump into your colleague in a train on the weekend. He travels with a boy who is one of his two children.
What is the probability the other children is a girl ?
If I recall this problem correctly, the answer is different if you are told the boy is his eldest child.
Quote: MangoJThis may (or may not) be an equivalent problem:
You bump into your colleague in a train on the weekend. He travels with a boy who is one of his two children.
What is the probability the other children is a girl ?
i am not savvy at how to link, but I previously asked that very question here on WOV site. Someone with more acumen can find the link.
Quote: SOOPOOi am not savvy at how to link, but I previously asked that very question here on WOV site. Someone with more acumen can find the link.
SOOPOO's question
Done!
There are two options for what's face up in the bag: HH and HT (the first one is always heads and the second one is 50/50).
If we evaluate what can happen with each possibility, this is what we get:
50% bag containing HT * 50% chance of drawing a heads * 100% chance it is the double sided coin given that a head was drawn = 25%.
50% bag containing HH * 100% chance of drawing a heads * 50% chance it is the double sided coin given that a head was drawn = 25%.
Therefore, the answer is 50%.
Quote: AsswhoopermcdaddySince there are only 2 coins, we're dealing with a probability of 1 out of 2.
Instead of thinking of two coins, it's more instructive to think of four sides. You could be looking at EITHER side of the two headed coin, or one side of the one headed. So, in two of the three instances, the other side will be heads.
Quote: MoscaInstead of thinking of two coins, it's more instructive to think of four sides. You could be looking at EITHER side of the two headed coin, or one side of the one headed. So, in two of the three instances, the other side will be heads.
I like your explanation the best. Quite simple to understand.
Event 1: What is the probability of first picking the double-headed coin and then seeing that it's heads? It's (1/2)(1) = 1/2.
Event 2: What's the probability of first seeing a randomly-selected coin as heads and then finding out that the coin is double-headed? It's (3/4)(x).
If you think about it for a while, you'll realize that the probability of event 1 is the same as the probability of event 2. So, 1/2 = (3/4)(x). Solve for x.
Quote: CrystalMathI just cast my vote for 1/2.
There are two options for what's face up in the bag: HH and HT (the first one is always heads and the second one is 50/50).
If we evaluate what can happen with each possibility, this is what we get:
50% bag containing HT * 50% chance of drawing a heads * 100% chance it is the double sided coin given that a head was drawn = 25%.
50% bag containing HH * 100% chance of drawing a heads * 50% chance it is the double sided coin given that a head was drawn = 25%.
Therefore, the answer is 50%.
Correction: Must divide by the likelyhood that we drew a heads in the first place (75%), so the answer is really 50%/75% = 2/3. Too bad I can't rescind my vote.
The wording is key: "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?"
We've selected one coin and we observe that the up-side is Heads. On THIS coin, what is the probability that the other side is heads? This coin can either be the two headed coin or the fair coin. since we already know that the up-side on THIS coin is heads, then the down side of THIS coin can either be heads if it's the two headed coin or tails if it's the fair coin.
Quote: s2dbakerThe answer is 1/2
The wording is key: "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?"
We've selected one coin and we observe that the up-side is Heads. On THIS coin, what is the probability that the other side is heads? This coin can either be the two headed coin or the fair coin. since we already know that the up-side on THIS coin is heads, then the down side of THIS coin can either be heads if it's the two headed coin or tails if it's the fair coin.
Nope- since ONCE you have determined that the coin has AT LEAST one head, while only looking at one side, there is a 2/3 chance it is the two headed coin. Read Mosca's answer. It is so simple.
Then record how often you see the A on the other side as well. A hundred tries should suffice.
This is how you're thinking:Quote: SOOPOONope- since ONCE you have determined that the coin has AT LEAST one head, while only looking at one side, there is a 2/3 chance it is the two headed coin. Read Mosca's answer. It is so simple.
| Up Side | Down Side |
|---|---|
| Heads | Heads |
| Heads | Heads |
| Heads | Tails |
| Tails | Heads |
| Up Side | Down Side |
|---|---|
| Heads | Heads |
| Heads | Heads |
| Heads | Tails |
| Heads | Tails |
Quote: thecesspitIf I recall this problem correctly, the answer is different if you are told the boy is his eldest child.
Right. In the original, there's a 2/3 chance his other child is a girl. In the amended version, it's 1/2.
Quote: s2dbakerThis is how you're thinking:
But that's not the way that I think the problem is presented. You have to pick one of the two coins from the box. You have a 50% shot at picking the two headed coin and a 50% shot at picking the fair coin. The presenter never explains what happens in the event that you see a tail on the upside. You're assumption is that they throw out 25% of the results and start over. My assumption is that the presenter flips the fair coin back over to the head's up side which gives us this set:
Up Side Down Side Heads Heads Heads Heads Heads Tails Tails Heads
Up Side Down Side Heads Heads Heads Heads Heads Tails Heads Tails
That's a heck of an assumption!
In a case like this, an assumption doesn't have to be right, just plausible. Obviously the problem is not with our analysis but with the wording of the original problem.Quote: cclub79That's a heck of an assumption!
Quote: s2dbakerIn a case like this, an assumption doesn't have to be right, just plausible. Obviously the problem is not with our analysis but with the wording of the original problem.
No, the problem is worded ok. Where people get misled is in the way they perceive the world. You have chosen one COIN, but you only observe one FACE. You throw out all instances of "tails" through the definition of the problem, leaving 3 possibilities: you are looking at the top head of coin 1, you are looking at the bottom head of coin 1, or you are looking at the head of coin 2. In two of those three possibilities, the other side is heads.
People get all mcguffined by, "Two coins, so 1/2." But the problem isn't about coins, it is about faces of coins. The coins are just the vehicle.
Quote: s2dbakerThis is how you're thinking:
But that's not the way that I think the problem is presented. You have to pick one of the two coins from the box. You have a 50% shot at picking the two headed coin and a 50% shot at picking the fair coin. The presenter never explains what happens in the event that you see a tail on the upside.
Up Side Down Side Heads Heads Heads Heads Heads Tails Tails Heads
No, that's precisely the way the problem is presented. In proper terms, the question is "what is the conditional probability of the other side being heads given that you have observed a head on your coin?" You have thrown out that condition and, as a result, created an entirely different question (with a different answer, but not the one you think).
In any event, consider this restatement. Suppose you stood the coins on end on two tables, as follows, where positions A, B, C, and D are places you could be standing, and the arrows indicate what you observe:
A --> HT <-- B
----------------------
C --> HH <-- D
If you stand in positions A, C, or D, you see heads. If you stand in position B, you see tails. It should be clear that this scenario is equivalent to the original. But then, the question is reduced to this:
"You are standing in one of position A, C, or D chosen at random. What is the probability that the opposite position (respectively, B, D, or C) observes heads?"
Then it's trivial. B does not, C and D do. That's 2/3.
Quote: s2dbaker
You have to pick one of the two coins from the box. You have a 50% shot at picking the two headed coin and a 50% shot at picking the fair coin.
That is not true. You still pick each coin with 50% chance, but you then discard all coins showing tails, since you never see tail on the upside.
Statistically there is no difference between voluntarily discarding all upside-tail coins, or seeing a single coin with heads up for that specific coin.
I think the answer is 1/2 and yes I know the smart money is on 2/3 but........
If the question was what are the ODDS of you reaching into a box and picking a coin, looking at it, it showing heads, AND THEN, other side being heads? The answer would be 2/3 Because the times you observe tails you would start again.
But thats not how the question was worded. It say you pick a coin you are holding it, and you observe heads. what are the chances the other side is heads. It's 50/50 or 1/2
I know a lot of you are craps players on here and so am I....Think of it this way... Once you have the coin, and you observe heads, you have have ready made your come out roll and you have established a point. From this point there are only 2 outcomes. one would be heads one would be tails. You can not take the advantage you had prior to establishing the point, and apply it to the outcome once the point is established.
If this post made sense..... you understand the inner workings of my weird sometimes dumb mind LOL
Quote: vert1276Well im gonna ramble here so if you can make sense of it good for you LOL.
I think the answer is 1/2 and yes I know the smart money is on 2/3 but........
If the question was what are the ODDS of you reaching into a box and picking a coin, looking at it, it showing heads, AND THEN, other side being heads? The answer would be 2/3 Because the times you observe tails you would start again.
But thats not how the question was worded. It say you pick a coin you are holding it, and you observe heads. what are the chances the other side is heads. It's 50/50 or 1/2
I know a lot of you are craps players on here and so am I....Think of it this way... Once you have the coin, and you observe heads, you have have ready made your come out roll and you have established a point. From this point there are only 2 outcomes. one would be heads one would be tails. You can not take the advantage you had prior to establishing the point, and apply it to the outcome once the point is established.
If this post made sense..... you understand the inner workings of my weird sometimes dumb mind LOL
You guys are really over-thinking the "if it's tails you start again..." IT'S NOT TAILS in this problem...the problem specifically says "IF THE FACE IS HEADS...what is the probability..." It doesn't say, but sometimes it's tails and you start over...
In your analogy, it would be like me asking you what are the odds that IF THE POINT IS TEN, you'll make the point before 7ing out. You aren't going to say, "Well what about all the times when the point isn't ten?" That's not the question. The parameters are clearly spelled out.
Quote: cclub79You guys are really over-thinking the "if it's tails you start again..." IT'S NOT TAILS in this problem...the problem specifically says "IF THE FACE IS HEADS...what is the probability..." It doesn't say, but sometimes it's tails and you start over...
In your analogy, it would be like me asking you what are the odds that IF THE POINT IS TEN, you'll make the point before 7ing out. You aren't going to say, "Well what about all the times when the point isn't ten?" That's not the question. The parameters are clearly spelled out.
I understand what you are say but I dont think you understand what im saying...........
Thats say I reach into the box pick out a coin. Observe that face and it is heads. FROM THAT POINT FORWARD what are the chances the other side is heads? the answer would be 1/2. Because I can only be holding 1 of 2 coins. not 1 of 4 faces.
Quote: vert1276I understand what you are say but I dont think you understand what im saying...........
Thats say I reach into the box pick out a coin. Observe that face and it is heads. FROM THAT POINT FORWARD what are the chances the other side is heads? the answer would be 1/2. Because I can only be holding 1 of 2 coins. not 1 of 4 faces.
You are assuming that you are looking at the "Front" of one of the two coins, so the "back" has to be either tails or heads. But you could also be looking at the "back" of the double heads one. 2 of the 3 times, you will flip over the coin and see another heads. You should really do an experiment if you are unconvinced.
I was hoping to use Monty Hall with "more doors" to help explain, but it won't work in this situation. How about this. You have a die between your thumb and forefinger, and you are spinning it, looking at the 4 sides that are labeled H, H, H, and T. You spin it and spin it with your eyes closed, and then you stop. You are now looking at an H. What are the odds that there's another H on the opposite side of the die? This is exactly the same as two coins.
Ok, I don't think this explanation made any sense. Sorry, lol.Quote: cclub79You are assuming that you are looking at the "Front" of one of the two coins, so the "back" has to be either tails or heads. But you could also be looking at the "back" of the double heads one. 2 of the 3 times, you will flip over the coin and see another heads. You should really do an experiment if you are unconvinced.
Quote: HotBlondeOk, I don't think this explanation made any sense. Sorry, lol.
Leave the classroom for a couple of years and I get rusty I guess. The four sides was usually good enough to convince most of my students, but those that weren't needed the experiment.
I don't know how much simpler it can be than the fact that there are 3 sides that have Heads, and 2 of them are on the same coin. So if you are looking at a Head, you have a 2/3 chance of holding the coin that has two heads, therefore there's a 2/3 chance that there's a head on the other side!
Quote: vert1276I understand what you are say but I dont think you understand what im saying...........
Thats say I reach into the box pick out a coin. Observe that face and it is heads. FROM THAT POINT FORWARD what are the chances the other side is heads? the answer would be 1/2. Because I can only be holding 1 of 2 coins. not 1 of 4 faces.
You are viewing one of four faces, and HOLDING two of four. What is on the other side? If you are holding the two headed coin, then there are TWO POSSIBILITIES: one chance you are viewing side A, and turn it over to see side B; and one chance that you will be viewing side B, and turn it over to see side A. And there is also one chance that you are holding the heads/tails coin. Hence, 2/3.
How can it be a 2/3 chance that there's a head on the other side when you've already eliminated the first option on either coin? That only leaves the coin with the head on the other side or the coin with the tail on the other side.Quote: cclub79Leave the classroom for a couple of years and I get rusty I guess. The four sides was usually good enough to convince most of my students, but those that weren't needed the experiment.
I don't know how much simpler it can be than the fact that there are 3 sides that have Heads, and 2 of them are on the same coin. So if you are looking at a Head, you have a 2/3 chance of holding the coin that has two heads, therefore there's a 2/3 chance that there's a head on the other side!
But I guess that is the whole purpose of this thread, it sure makes for an interesting discussion!
Quote: HotBlondeHow can it be a 2/3 chance that there's a head on the other side when you've already eliminated the first option on either coin? That only leaves the coin with the head on the other side or the coin with the tail on the other side.
But I guess that is the whole purpose of this thread, it sure makes for an interesting discussion!
There's not a first option on either coin. There's only a first option on ONE coin. You've only eliminated 1/4 of the options. Again, I recommend reading about the Monty Hall problem. You will think the answer to that is 1/2 also, but you'll find the correct answer and then this problem will be clearer...
Quote: MoscaYou are viewing one of four faces, and HOLDING two of four. What is on the other side? If you are holding the two headed coin, then there are TWO POSSIBILITIES: one chance you are viewing side A, and turn it over to see side B; and one chance that you will be viewing side B, and turn it over to see side A. And there is also one chance that you are holding the heads/tails coin. Hence, 2/3.
Look I understand the math of it an how you get to the 2/3. The sad part is I have degree in econ and am pretty good at math LOL and am a working economist and it pretty sad I cant wrap my head around this if im wrong(which it looks like I am LOL)
Im gonna ramble here then im done, read if you want ok?
I understand if there were 4 one sided coins in a box 3 had heads one had tails. and I picked out one that had a head on it. what would be my chances that The next one I picked would also be a heads? The answer would be 2/3. Three coins left and 2 of 3 have heads. and this is how you are figuring the question in this thread. because I can either pick the "A" or "B" side of the coin with 2 heads on it.
But the coins have 2 sides and you are picking up 2 sides at once regardless of what side you are looking at. for example there are 2 ways to roll a 3......... a 1+2 or 2+1. but there is only one way to roll a 2 a 1+1. You cant say there is 2 ways to roll a 2 because you can switch the 1's around.(if that makes any sense LOL) The same logic holds for the coin with 2 heads on it. there is only on way to pick it up(you choose that coin) therefore there is a 50% probability you can pick the 2 headed coin. Therefore once you have picked a coin and observed heads there is still only a 50% probability you are holding the coin with 2 heads.
Quote: Wizard... One coin is selected from the box at random and the face of one side is observed. If the face is heads, ....
Quote: vert1276Therefore once you have picked a coin and observed heads there is still only a 50% probability you are holding the coin with 2 heads.
The 50% probability of having the two-headed coin applies before you look at the face recognize what the face of the coin looks like; i.e., at the end of the first sentence above in the Wizard's statement of the problem. Once the "If the face is heads, ..." portion of the statement is added, you are dealing with a conditional probability, which is different from the initial, unconditional (also called the "prior") probability. In this case the unconditional/prior probability is 1/2, while the conditional probability is 2/3. The condition eliminates 25% of the initial possibilities; i.e., that you looked at the face and it was tails.
It's perhaps appropriate to make passing reference to Bayes's Theorem.
Note: edited with strike-through and italics to say what I really meant the first time.
If the coin was shown to be tails, we would know the other coin with 100% certainty. The fact that it's heads tells us nothing. Heads does not affect the measure of probability, whereas we gain perfect knowledge with a tails.
Quote: KelmoIt's 1/2.
If the coin was shown to be tails, we would know the other coin with 100% certainty. The fact that it's heads tells us nothing. Heads does not affect the measure of probability, whereas we gain perfect knowledge with a tails.
If heads tells you nothing, by the same argument tails tells you nothing.
However you know the other side when shown tails - and hence you know something more when shown heads.
Let's reverse it for a second. You've chosen a coin from the two, and look at a side and see that it is tails. What are the chances the other side is heads? Is it still 1/2? The logic you used gives that answer.
You are answering the problem as if it were asked, "There are two coins, one heads/heads and the other heads/tails. What are the odds of picking one or the other at random?" But it wasn't asked that way. You've seen one head. There are three heads you could have seen, not two; two on one coin, and one on the other.
Quote: vert1276Look I understand the math of it an how you get to the 2/3. The sad part is I have degree in econ and am pretty good at math LOL and am a working economist and it pretty sad I cant wrap my head around this if im wrong(which it looks like I am LOL)
Im gonna ramble here then im done, read if you want ok?
I understand if there were 4 one sided coins in a box 3 had heads one had tails. and I picked out one that had a head on it. what would be my chances that The next one I picked would also be a heads? The answer would be 2/3. Three coins left and 2 of 3 have heads. and this is how you are figuring the question in this thread. because I can either pick the "A" or "B" side of the coin with 2 heads on it.
But the coins have 2 sides and you are picking up 2 sides at once regardless of what side you are looking at. for example there are 2 ways to roll a 3......... a 1+2 or 2+1. but there is only one way to roll a 2 a 1+1. You cant say there is 2 ways to roll a 2 because you can switch the 1's around.(if that makes any sense LOL) The same logic holds for the coin with 2 heads on it. there is only on way to pick it up(you choose that coin) therefore there is a 50% probability you can pick the 2 headed coin. Therefore once you have picked a coin and observed heads there is still only a 50% probability you are holding the coin with 2 heads.
Agreed and well put.Quote: vert1276I understand what you are say but I dont think you understand what im saying...........
Thats say I reach into the box pick out a coin. Observe that face and it is heads. FROM THAT POINT FORWARD what are the chances the other side is heads? the answer would be 1/2. Because I can only be holding 1 of 2 coins. not 1 of 4 faces.
Quote: s2dbakerAgreed and well put.Quote: vert1276I understand what you are say but I dont think you understand what im saying...........
Thats say I reach into the box pick out a coin. Observe that face and it is heads. FROM THAT POINT FORWARD what are the chances the other side is heads? the answer would be 1/2. Because I can only be holding 1 of 2 coins. not 1 of 4 faces.
I think I disagree, but I want to be certain we are not describing different questions. Usually, this situation would be described in this manner:
Given that one coin has been selected at random and that the one face that has been observed is heads, what is the probability that the opposite face of that coin is heads? (Note: this wording implies that you have reached the "point" that a coin has been selected and a face has been identified as heads.)
Do you view that wording as equivalent to vert1276's wording of "FROM THAT POINT FORWARD what are the chances the other side is heads?"
Given the wording I presented, I contend this is definitely a conditional probability (as I described on the previous page), and the correct answer would be 2/3, while the prior probability (probability of holding a particular one of the coins without specifying the nature of the observed face) is 1/2. However, you may be asking a different question, so I need to know whether you consider these two wordings equivalent. That should establish whether we are looking at different questions or disagreeing about the answer to the same question.
To further expound, let me express why I do not like the logic that "...the answer would be 1/2. Because I can only be holding 1 of 2 coins." Taking an absurd distortion to prove the point: suppose that one coin weighed the equivalent of a quarter and the other weighed a ton. I look in the box, deliberately select one coin, and (without telling you which coin it is) I hold it high over my head and ask you to guess which one it is. There are only two coins that I could have chosen, so there is a 50% probability that I am holding a one-ton coin over my head. Is that right?
The point is that the fact that there are only two coins does not at all mean that, after a condition has been imposed, there is still a 50% probability for each of them. The condition could be that a face has been observed as heads, or it could be that I am able to hold the coin over my head. In either case, that condition impacts the probability of which coin has been selected -- it is a conditional probability, different from 50% in each of these cases. In the original problem, seeing a head on the face makes the conditional probability of a head on the other face equal to 2/3; in the absurdly-distorted version, my ability to hold the coin over my head makes the conditional probability of the coin being the one-ton coin equal to zero.
Quote: DocTaking an absurd distortion to prove the point: suppose that one coin weighed the equivalent of a quarter and the other weighed a ton. I look in the box, deliberately select one coin, and (without telling you which coin it is) I hold it high over my head and ask you to guess which one it is.
Simpler: If you see tails, is the probability still 1/2? Same two coins, right? But seeing one side changes the probability, doesn't it?
75% heads-on-flip-side overall =
1/4 chance to get tails * 100% chance heads is on the flip side (because we know there's always a head on the other side if you have the fair coin, and there aren't any headless coins here)
+
3/4 chance to get heads * N% chance heads is on the flip side.
Solve for N.
God does not play dice with the Universe!Quote: NareedSuppose we mix this problem with Schroedinger's Cat. If you first see heads, what are the odds the cat is dead? :P
Quote: s2dbakerGod does not play dice with the Universe!
Not only does God play dice, but he sometimes throws them where they cannot be seen.
(Edited to get the quote right.)
The trivial case of a posterior probability is: What is the probability of picking the head/head dice, under the condition that both sides of the coin show head ?
Of course this is 1, although the a priori probability of picking that dice is 1/2.
Now the transition to the original question is not that difficult.
