Two-coin problem
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53 members have voted
Quote: NareedDoes he flip coins with the Universe?
He did until he tried to Martingale the flips and lost an infinite amount of money, and got booted out of house by an angry Cosmos.
Quote: thecesspitHe did until he tried to Martingale the flips and lost an infinite, and got booted out of house by an angry Cosmos.
Hm. If God lost an infinite amount of money, how much money would he have left? :P
Quote: MathExtremistLet's do this whole thing a different way. We know we have two coins, one is normal and the other has two heads. That means if you randomly pick one coin, look at one side, and then flip it over, you have a 75% chance of seeing heads on the other side. We all agree on that. But then we can break that 75% chance-to-see-heads-on-the-flip-side into its component parts, one when you start with tails and one when you start with heads.
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.
I know I said I was done with this but I just cant give up (I know I like the punishment)
Like I said before I understand the math on how you get 2/3, But somehow I still think the answer is 1/2 (I know, I know)
Everyone is doing the math although the coin faces are independent in someway. Giving the example that if you observe heads there are 3 faces left to be on the other side and 2 are heads therefore the answer is 2/3. Buy why I disagree is....If I pull a coin and observed heads there is only 2 faces available at that time that can be on the other side of the coin 1 heads and 1 tails. The other heads that is on the "fair" coin is no longer a possibility. Because for that heads to appear on the other side I would have to observe tails.
Well, there are only two kinds of faces that you can observe, heads and tails, but there are three faces available.Quote: vert1276Buy why I disagree is....If I pull a coin and observed heads there is only 2 faces available at that time that can be on the other side of the coin 1 heads and 1 tails.
Quote: vert1276The other heads that is on the "fair" coin is no longer a possibility. Because for that heads to appear on the other side I would have to observe tails.
Well, there you are right -- the opposite side of the coin you are looking at is definitely not the heads side of the fair coin. The other three faces are possible: (1) tails of the fair coin, (2) head #1 of the two-headed coin, and (3) head #2 of the two-headed coin. Two out of three of those are heads, so the probability of finding heads on the other side is 2/3.
Quote: vert1276I know I said I was done with this but I just can give up (I know I like the punishment)
Like I said before I understand the math on how you get 2/3, But somehow I still think the answer is 1/2 (I know, I know)
Everyone is doing the math although the coin faces are independent in someway. Giving the example that if you observe heads there are 3 faces left to be on the other side and 2 are heads therefore the answer is 2/3. Buy why I disagree is....If I pull a coin and observed heads there is only 2 faces available at that time that can be on the other side of the coin 1 heads and 1 tails. The other heads that is on the "fair" coin is no longer a possibility. Because for that heads to appear on the other side I would have to observe tails.
Okay, let's make the sides independent. There are 4 coins in a bag. Three are heads/heads, one is tails/tails. You remove one and see that it is heads. What are the odds that the other coins are 2heads/2heads/2tails? 100%. The fact that the faces are connected is irrelevant. All these coins are in a bag. Same problem. The fact that the sides are on a single coin is what leads everyone astray.
Quote: vert1276I know I said I was done with this but I just cant give up (I know I like the punishment)
Like I said before I understand the math on how you get 2/3, But somehow I still think the answer is 1/2 (I know, I know)
Everyone is doing the math although the coin faces are independent in someway. Giving the example that if you observe heads there are 3 faces left to be on the other side and 2 are heads therefore the answer is 2/3. Buy why I disagree is....If I pull a coin and observed heads there is only 2 faces available at that time that can be on the other side of the coin 1 heads and 1 tails. The other heads that is on the "fair" coin is no longer a possibility. Because for that heads to appear on the other side I would have to observe tails.
Quite the opposite, the math is entirely based on the fact that the faces are *dependent*. Want an even more straightforward analogy? Here's one:
Take a six-sided die. Erase the 1 and 6, and put Heads on 2,3,4 and Tails on 5. Viewed from one of the blank faces, the die has:
H
H H
T
So now the question is this: if you roll the die and observe heads face-up, what is the probability that the face-down side is also heads?
Quote: MathExtremistQuite the opposite, the math is entirely based on the fact that the faces are *dependent*. Want an even more straightforward analogy? Here's one:
Take a six-sided die. Erase the 1 and 6, and put Heads on 2,3,4 and Tails on 5. Viewed from one of the blank faces, the die has:
H
H H
T
So now the question is this: if you roll the die and observe heads face-up, what is the probability that the face-down side is also heads?
Hey! That's what I said last night!
Quote: JohnzimboThe wizard has to be reading this thread and cracking up
I am!
Quote: vert1276I know I said I was done with this but I just cant give up (I know I like the punishment)
Like I said before I understand the math on how you get 2/3, But somehow I still think the answer is 1/2 (I know, I know)
Everyone is doing the math although the coin faces are independent in someway. Giving the example that if you observe heads there are 3 faces left to be on the other side and 2 are heads therefore the answer is 2/3. Buy why I disagree is....If I pull a coin and observed heads there is only 2 faces available at that time that can be on the other side of the coin 1 heads and 1 tails. The other heads that is on the "fair" coin is no longer a possibility. Because for that heads to appear on the other side I would have to observe tails.
If you REALLY think the answer is 1/2, I'm wondering why you don't do a trial experiment, say 500x? You could have done 1000s of times by now! (You actually would only have to do it 3x if you "rigged" the experiment to show each possible outcome...)
Quote: MathExtremistQuite the opposite, the math is entirely based on the fact that the faces are *dependent*. Want an even more straightforward analogy? Here's one:
Take a six-sided die. Erase the 1 and 6, and put Heads on 2,3,4 and Tails on 5. Viewed from one of the blank faces, the die has:
H
H H
T
So now the question is this: if you roll the die and observe heads face-up, what is the probability that the face-down side is also heads?
Thanks for debating with me, im sure im wrong but just cant see it in my mind. Let me now set up a scenario ok?
Lets assume we have 2 coins. There are 4 faces to the 2 coins(thats a given LOL) three of the faces are stamped "H" and one face is stamped "T". Therefore one coin has to have "H" on both sides. So far this is exactly like how the question is set up.
Ok...SO I reach in the box and pull a coin.... and observe one of the faces its stamped "H". What are the chances the other side is also stamped "H"? (again this is the same as the original question)
Now, math would dictate there are 2 "H" stamps left to be viewed and 1 "T" stamp left to be viewed, therefore the chances the other side is stamped "H" is 2/3 BUT.........
I would say there are not 2 "H"' stamps left to be viewed. There is only 1 "H" left to be view, because at least 1(maybe 2) "H" stamps have to be left behind in the box that I pulled the coin from. No matter what coin I pull there will always be at least 1 "H" left in the box so that can be eliminated from being a possibility. And since I have "observed" a "H" stamp already. There are only 2 possibilities left 1 "H" stamp and 1 "T" stamp.
Quote: vert1276Quote: MathExtremistQuite the opposite, the math is entirely based on the fact that the faces are *dependent*. Want an even more straightforward analogy? Here's one:
Take a six-sided die. Erase the 1 and 6, and put Heads on 2,3,4 and Tails on 5. Viewed from one of the blank faces, the die has:
H
H H
T
So now the question is this: if you roll the die and observe heads face-up, what is the probability that the face-down side is also heads?
Thanks for debating with me, im sure im wrong but just cant see it in my mind. Let me now set up a scenario ok?
Lets assume we have 2 coins. There are 4 faces to the 2 coins(thats a given LOL) three of the faces are stamped "H" and one face is stamped "T". Therefore one coin has to have "H" on both sides. So far this is exactly like how the question is set up.
Ok...SO I reach in the box and pull a coin.... and observe one of the faces its stamped "H". What are the chances the other side is also stamped "H"? (again this is the same as the original question)
Now, math would dictate there are 2 "H" stamps left to be viewed and 1 "T" stamp left to be viewed, therefore the chances the other side is stamped "H" is 2/3 BUT.........
I would say there are not 2 "H"' stamps left to be viewed. There is only 1 "H" left to be view, because at least 1(maybe 2) "H" stamps have to be left behind in the box that I pulled the coin from. No matter what coin I pull there will always be at least 1 "H" left in the box so that can be eliminated from being a possibility. And since I have "observed" a "H" stamp already. There are only 2 possibilities left 1 "H" stamp and 1 "T" stamp.
Nooooooooooo................
By that logic, if you had 100 marbles in a bag, 99 red and 1 blue, and you pulled a marble, there would be a 50/50 chance of it being blue because there have to be 98 red ones still in the bag....... . . ... .... . . . . . . . .
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 disagree because you are already given the first outcome which is heads. "a coin is selected at random. If the face shows heads, what's the probability the other side is heads" One event has occurred (the question itself offers the hypothetical if you already know one side is heads). They're asking for the probability of the other side. 1 out of 2.
Quote: vert1276No matter what coin I pull there will always be at least 1 "H" left in the box so that can be eliminated from being a possibility. And since I have "observed" a "H" stamp already. There are only 2 possibilities left 1 "H" stamp and 1 "T" stamp.
Ah... But WHICH "H" have you viewed! Since YOU DON'T KNOW, the answer is 2/3. There are TWICE AS MANY CHANCES THAT YOU HAVE THE H/H COIN. Because observing one H is part of the initial structure of the question. But you don't know which H it is, one of two on one coin, or the lone one on the other.
Quote: MoscaAh... But WHICH "H" have you viewed! Since YOU DON'T KNOW, the answer is 2/3. There are TWICE AS MANY CHANCES THAT YOU HAVE THE H/H COIN. Because observing one H is part of the initial structure of the question. But you don't know which H it is, one of two on one coin, or the lone one on the other.
Mosca, are you with me that this is some prank on the Wizard or us or something? Why are they resisting attempts to just do a test? Thanks for fighting the battle with me!
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.
Why don't you just close your eyes before drawing? Equally plausible no?
Quote: MoscaNo, 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.
The problem is about the faces of the coin. But consider this:
1.) I take the fair coin and the rigged coin and I place them on a table, heads up. I make you choose 1 of the 2 coins. What's the probability you choose the double headed coin? It's still 1 out of 2.
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.
What do you mean if it's tails you start again? That's a separate dis-joint probability. The question does not ask the probability of the combined event. You are not being asked to draw 4 times. It's already drawn heads for you, and THEN you are asked the probability.
Quote: MoscaAh... But WHICH "H" have you viewed! Since YOU DON'T KNOW, the answer is 2/3. There are TWICE AS MANY CHANCES THAT YOU HAVE THE H/H COIN. Because observing one H is part of the initial structure of the question. But you don't know which H it is, one of two on one coin, or the lone one on the other.
It doesnt matter which "H" i have observed. I have Observed one of the possible 3 "H" stamps and on "H" stamp has to be left in the box I pull the coin from. Therefore there are only 2 stamps left to be viewed.
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!
Makes more sense if we're only dealing with a 4 sided piece of paper and drawing from one of the sides. But the 2 separate coins means 2 separate outcomes. 1 of 2 as oppose to 3 of 4. At least that's my interpretation.
Quote: cclub79Quote: MoscaAh... But WHICH "H" have you viewed! Since YOU DON'T KNOW, the answer is 2/3. There are TWICE AS MANY CHANCES THAT YOU HAVE THE H/H COIN. Because observing one H is part of the initial structure of the question. But you don't know which H it is, one of two on one coin, or the lone one on the other.
Mosca, are you with me that this is some prank on the Wizard or us or something? Why are they resisting attempts to just do a test? Thanks for fighting the battle with me!
I just want someone to go, "Aha! Now I get it!" But it doesn't look like that is going to happen.
Fish
Dog
Bird
Cat
You draw a coin and you see the Fish. What are the odds that the Cat is on the other side? You'd say 1 in 3, right? Well, there's no difference between that problem and the Heads/Tails one:
Fish (representing Heads1)
Dog (representing Heads2)
Bird (representing Heads3)
Cat ((representing Tails)
If putting the heads/tails makes you change your answer to 50%, then you are getting hung up on the visual, not the math.
Quote: cclub79Quote: vert1276Quote: MathExtremistQuite the opposite, the math is entirely based on the fact that the faces are *dependent*. Want an even more straightforward analogy? Here's one:
Take a six-sided die. Erase the 1 and 6, and put Heads on 2,3,4 and Tails on 5. Viewed from one of the blank faces, the die has:
H
H H
T
So now the question is this: if you roll the die and observe heads face-up, what is the probability that the face-down side is also heads?
Thanks for debating with me, im sure im wrong but just cant see it in my mind. Let me now set up a scenario ok?
Lets assume we have 2 coins. There are 4 faces to the 2 coins(thats a given LOL) three of the faces are stamped "H" and one face is stamped "T". Therefore one coin has to have "H" on both sides. So far this is exactly like how the question is set up.
Ok...SO I reach in the box and pull a coin.... and observe one of the faces its stamped "H". What are the chances the other side is also stamped "H"? (again this is the same as the original question)
Now, math would dictate there are 2 "H" stamps left to be viewed and 1 "T" stamp left to be viewed, therefore the chances the other side is stamped "H" is 2/3 BUT.........
I would say there are not 2 "H"' stamps left to be viewed. There is only 1 "H" left to be view, because at least 1(maybe 2) "H" stamps have to be left behind in the box that I pulled the coin from. No matter what coin I pull there will always be at least 1 "H" left in the box so that can be eliminated from being a possibility. And since I have "observed" a "H" stamp already. There are only 2 possibilities left 1 "H" stamp and 1 "T" stamp.
Nooooooooooo................
By that logic, if you had 100 marbles in a bag, 99 red and 1 blue, and you pulled a marble, there would be a 50/50 chance of it being blue because there have to be 98 red ones still in the bag....... . . ... .... . . . . . . . .
thats not the same logic not even close.
Quote: AsswhoopermcdaddyThe problem is about the faces of the coin. But consider this:
1.) I take the fair coin and the rigged coin and I place them on a table, heads up. I make you choose 1 of the 2 coins. What's the probability you choose the double headed coin? It's still 1 out of 2.
But that's not the question. The way you have posed it, I get to see two coins, both heads. In the original problem, one coin is obscured. Let me ask you; if you saw a tail, would the odds still be 1/2? Of course not. But if you use the same logic you are using, then that's the answer you get.
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.
You're making it sound like separate probabilities...."one chance you are viewing side A, and another side B". How's this
Coin 1: Sides Heads and Tails
Coin 2: Sides Heads and Heads
I chose a coin at random and display the side with Heads. I have no idea if its Coin 1 or 2. What is the probability the other side is Heads? It's 1 out of 2
Let's try a different example:
Coin 1: Sides Heads and Tails
Coin 2: Sides Heads and Heads
Coin 3: Sides Heads and Heads
I chose a coin at random and I display the side with Heads. I have no idea if I have chosen Coin 1, 2, or 3. What is the probability the other side is Heads? 2 out of 3. In my opinion.
Let's try a different example:
Coin 1: Sides Heads and Tails
Coin 2: Sides Heads and Heads
Coin 3: Sides Heads and Heads
What is the probability of choosing a coin with the side of heads? 5 heads / 6 outcomes
Quote: vert1276Quote: cclub79Quote: vert1276Quote: MathExtremistQuite the opposite, the math is entirely based on the fact that the faces are *dependent*. Want an even more straightforward analogy? Here's one:
Take a six-sided die. Erase the 1 and 6, and put Heads on 2,3,4 and Tails on 5. Viewed from one of the blank faces, the die has:
H
H H
T
So now the question is this: if you roll the die and observe heads face-up, what is the probability that the face-down side is also heads?
Thanks for debating with me, im sure im wrong but just cant see it in my mind. Let me now set up a scenario ok?
Lets assume we have 2 coins. There are 4 faces to the 2 coins(thats a given LOL) three of the faces are stamped "H" and one face is stamped "T". Therefore one coin has to have "H" on both sides. So far this is exactly like how the question is set up.
Ok...SO I reach in the box and pull a coin.... and observe one of the faces its stamped "H". What are the chances the other side is also stamped "H"? (again this is the same as the original question)
Now, math would dictate there are 2 "H" stamps left to be viewed and 1 "T" stamp left to be viewed, therefore the chances the other side is stamped "H" is 2/3 BUT.........
I would say there are not 2 "H"' stamps left to be viewed. There is only 1 "H" left to be view, because at least 1(maybe 2) "H" stamps have to be left behind in the box that I pulled the coin from. No matter what coin I pull there will always be at least 1 "H" left in the box so that can be eliminated from being a possibility. And since I have "observed" a "H" stamp already. There are only 2 possibilities left 1 "H" stamp and 1 "T" stamp.
Nooooooooooo................
By that logic, if you had 100 marbles in a bag, 99 red and 1 blue, and you pulled a marble, there would be a 50/50 chance of it being blue because there have to be 98 red ones still in the bag....... . . ... .... . . . . . . . .
thats not the same logic not even close.
I disagree. You want to change the probability because you say "No matter what, there's still at least 1 heads in the bag" So what? "Well it can't be the 1 still in the bag, so there's only 1 left for it to be" That's exactly the same as saying "It can't be the 98 still in the bag." I take back my desire for you to try the Monty Hall Problem, because clearly the 100 doors solution is not going to help!
We kind of could use some backup, Wizard....
Quote: cclub79My friend just suggested this example, that he did for his 4th grade class (!) You have two coins with four animals on them:
Fish
Dog
Bird
Cat
You draw a coin and you see the Fish. What are the odds that the Cat is on the other side? You'd say 1 in 3, right? Well, there's no difference between that problem and the Heads/Tails one:
Fish (representing Heads1)
Dog (representing Heads2)
Bird (representing Heads3)
Cat ((representing Tails)
If putting the heads/tails makes you change your answer to 50%, then you are getting hung up on the visual, not the math.
Thats an excellent example. And I like I said I have always understood the math of how you get to 2/3. I was saying I cant see it in my MIND.
Quote: MathExtremistNot only does God play dice, but he sometimes throws them where they cannot be seen.
(Edited to get the quote right.)
Just as you are about to make a 6pt fire, the dice remain suspended in mid-air. The voice of the pit (Satan) "shooter out".
Quote: vert1276Thats an excellent example. And I like I said I have always understood the math of how you get to 2/3. I was saying I cant see it in my MIND.
But now you can? That's all we ever wanted! :)
Quote: vert1276Quote: MathExtremistQuite the opposite, the math is entirely based on the fact that the faces are *dependent*. Want an even more straightforward analogy? Here's one:
Take a six-sided die. Erase the 1 and 6, and put Heads on 2,3,4 and Tails on 5. Viewed from one of the blank faces, the die has:
H
H H
T
So now the question is this: if you roll the die and observe heads face-up, what is the probability that the face-down side is also heads?
Assuming the order of the sides don't matter wouldn't it be 2 Heads Left divided by 2 Heads Left + 1 Tail + 1 blank side = 2/5
You've got our heads rolling on this one chasing our tails in the blankness of space.
Quote: MoscaBut that's not the question. The way you have posed it, I get to see two coins, both heads. In the original problem, one coin is obscured. Let me ask you; if you saw a tail, would the odds still be 1/2? Of course not. But if you use the same logic you are using, then that's the answer you get.
If I saw Tails first, what would the question be? That in and of itself is a different question.....or are we just arguing over the symantics that impact the probability?
Quote: vert1276Thats an excellent example. And I like I said I have always understood the math of how you get to 2/3. I was saying I cant see it in my MIND.
If you agree with 2/3, you minus well ask the question with marbles instead. 3 blue marbles and 1 red marble in a bag.
What's the probability of drawing blue? 3/4
What's the probability of drawing a blue first and then blue second? (3/4)*(2/3) = 6/12 = 1/2
What's the probability of drawing a blue marble after you remove one blue marble from the bag = 2/3.
Quote: cclub79
Hey! That's what I said last night!
Is it? That's what I get for skimming then.
Let me try to explain this way ok? I take out 2 quarters from my pocket and 4 postage stamps. 3 postage stamps have and American flag on them and one postage stamp has a liberty bell on it. I lick the postage stamps and stick one to each face of the quarters.
I throw the 2 quarters in a box. I reach into the box and pull out a quarter. I observe a American flags stamp on one of the faces of the quarter that I pulled out of the box. What do I know at this point RIGHT NOW?
1) I know there is at least one American flag stamp still in the box, correct?
2) I know the American flag stamp I am "observing" can not also be on the other side of the quarter, correct?
3) I still have 2 possible postage stamps that can be on the other side of the quarter, correct?
Now you all are saying there is a 66% chance the other side of the quarter has an American flag stamp on it. And Im saying there is only 2 possibility left now that I have observed one face and one of the 2 has to be the liberty bell stamp. please explain to me how there can be 3 possibilities left to be other the other side of the coin WITHOUT reaching back into the box?
I dont compare this to the Monty hall problem. And I fully understand the math of the money hall problem. But the Money hall problem is basses on assumptions, that Monty will never open one of the 2 remaining doors that has the car behind it, because it would ruin the show. Therefore it makes sense to switch doors. this question is not based on assumptions.
Quote: AsswhoopermcdaddyIf you agree with 2/3, you minus well ask the question with marbles instead. 3 blue marbles and 1 red marble in a bag.
What's the probability of drawing blue? 3/4
What's the probability of drawing a blue first and then blue second? (3/4)*(2/3) = 6/12 = 1/2
What's the probability of drawing a blue marble after you remove one blue marble from the bag = 2/3.
Yes I fully understand the math of that. But you are still drawing all the marbles from ONE bag in your problem. In the 2 coin question, I am not drawing the other side of the coin from a bag(or in our case, box). Once I observe one face being heads. I can then eliminate 2 of the 3 heads from being on the other side of the coin. Since there must be at least 1 heads still in the box that cant be on the other side of the coin. I am never reaching back into the box. Where you are reaching back into the "bag" bringing another blue marble back into play.
If you pull a coin from the box that has heads up, what is the chance that the other side will have heads? 2/3.
If you already have one of the two coins in your hand and it has heads up, what is the chance the other side is heads? 1/2.
The first scenario is a compound event (parlay). The second scenario is an independent trial.
Quote: kp
If you pull a coin from the box that has heads up, what is the chance that the other side will have heads? 2/3.
If you already have one of the two coins in your hand and it has heads up, what is the chance the other side is heads? 1/2.
I don't see where you've gained any additional information between these two examples. Either way, you're just looking at the heads side of one of two coins.
Quote: kpMy first answer was 2/3. Then I doodled on paper and came up with 2/3. Now a couple of days later I'm switching to 1/2.
If you pull a coin from the box that has heads up, what is the chance that the other side will have heads? 2/3.
If you already have one of the two coins in your hand and it has heads up, what is the chance the other side is heads? 1/2.
The first scenario is a compound event (parlay). The second scenario is an independent trial.
I feel like Darth Vader and im turning people to the dark side of 1/2 LOL!!!!!!!
Quote: vert1276LMAO I know you guys think im totally messin with you but im not.
And I was just foolin' when I accused you of being tuttigym. But seriously, I'm having flashbacks. =)
Your points are good enough that I'm going back to examine the original question, but then again, I'm one of the one's who think picking the other door in Monty is pointless =P
Edit: Nevermind. 10 seconds and I think 2/3 too. Still wouldn't pick the other door though.
Quote: vert12761) I know there is at least one American flag stamp still in the box, correct?
2) I know the American flag stamp I am "observing" can not also be on the other side of the quarter, correct?
3) I still have 2 possible postage stamps that can be on the other side of the quarter, correct?
Now you all are saying there is a 66% chance the other side of the quarter has an American flag stamp on it. And Im saying there is only 2 possibility left now that I have observed one face and one of the 2 has to be the liberty bell stamp. please explain to me how there can be 3 possibilities left to be other the other side of the coin WITHOUT reaching back into the box?
Don't confuse possibility with probability. There are only two possibilities because there are only two images -- flag or bell. But the flag is twice as likely as the bell.
This has been suggested before (bird/fish/etc. analogy) but when you distinguish the sides it becomes easy, I hope. If you number the heads 1, 2, and 3, but you apply them randomly to the coins, you can either have
H1/H2, H3/T
or
H1/H3, H2/T
or
H1/T, H2/H3
Let's suppose I do that and put the coins into a bag, and I chose the arrangement randomly with equal probability from the above possibilities. You reach into the bag and pull out a coin with H1 showing. What are the chances that the flip side is either H2 *or* H3?
Quote: MathExtremistDon't confuse possibility with probability. There are only two possibilities because there are only two images -- flag or bell. But the flag is twice as likely as the bell.
This has been suggested before (bird/fish/etc. analogy) but when you distinguish the sides it becomes easy, I hope. If you number the heads 1, 2, and 3, but you apply them randomly to the coins, you can either have
H1/H2, H3/T
or
H1/H3, H2/T
or
H1/T, H2/H3
Let's suppose I do that and put the coins into a bag, and I chose the arrangement randomly with equal probability from the above possibilities. You reach into the bag and pull out a coin with H1 showing. What are the chances that the flip side is either H2 *or* H3?
Said that Monday, too. We're just going to keep using the same arguments, so I'm with Mosca. The 4th graders that weren't convinced did the experiment and ended up with "Heads" around 65-70% of the time. If you don't want to do that, we're just going to be arguing in circles.
I haven't read ANY of the responses since I was the first responder on page 1.
As I see it, past results have no influence on future actions.
You either pulled the two-headed coin out or not.
So it's a 50% chance the flip side is heads.
I started reading the thread. About half-way thru page 4, the fog cleared.
I then went back to page 2, to re-read this little comment in ChesterDog's Post:
Yeah, phrased that way, it's obviously 2/3.Quote: ChesterDogWhat's the probability of first seeing a randomly-selected coin as heads and then finding out that the coin is double-headed?
Quote: DJTeddyBearMaybe I'm oversimplifying things but....
I haven't read ANY of the responses since I was the first responder on page 1.
As I see it, past results have no influence on future actions.
You either pulled the two-headed coin out or not.
So it's a 50% chance the flip side is heads.
If you pull a coin a million times, 25% of the time you will pull the normal coin on the tail side, 25% of the time you would pick it on the head side and 50% of the time you would get the other coin on head. These are your 4 possibilities:
Double headed coin, heads.
Double headed coin, heads.
Regular coin, heads.
Regular coin, tails.
Now let's take out the last possibility.
Double headed coin, heads.
Double headed coin, heads.
Regular coin, heads.
Regular coin, tails.
There are now 3 possibilities, two of which are that you picked the double headed coin.
Yeah, I know. I already did a follow-up.
And I've also come up with a simpler response:
There are three faces that are heads. You could have picked any one of them. Two of them have heads on the other face, therefore, 2/3 of the time it's the double headed coin.
Quote: cclub79Said that Monday, too. We're just going to keep using the same arguments, so I'm with Mosca. The 4th graders that weren't convinced did the experiment and ended up with "Heads" around 65-70% of the time. If you don't want to do that, we're just going to be arguing in circles.
And since I don't like arguing in circles, I'll simply offer this:
I'll gladly offer anyone a wager of $100 to my $130 that *tails* will show when we play the game. If the odds of heads or tails are truly 1/2, that yields a healthy +15% EV. I'll play for as many trials as you want. Any takers?
(Of course, if I'm right, I have a +18% EV...)
