Two coins each with a head one one side and a tail on the other are tossed in the air but you do not see them land.
You are told that one of the coins landed showing heads. What are the chances that the other coin also landed showing heads?
25%
Isn't it still 50-50? What side the one coin landed on has nothing to do with the other coin. I'm not real confident about this, as you can see.
Quote from: En_Route on May 14, 2012, 12:58:19 AM
Two coins each with a head one one side and a tail on the other are tossed in the air but you do not see them land.
You are told that one of the coins landed showing heads. What are the chances that the other coin also landed showing heads?
You again!?! >:(
Quote from: Ecurb Noselrub on May 14, 2012, 02:24:30 AM
Isn't it still 50-50? What side the one coin landed on has nothing to do with the other coin. I'm not real confident about this, as you can see.
Don't be lazy people, you assume too much, do the permutations, there are only 4 of them.
1/2*1/2= 1/4
P1 - H, T
P2 - T, H
P3 - T, T
P4 - H, H
P1,2,4 all have one coin with H
With the other coin being H once or T twice
So 1/3 = 33.333%
seems the trick can be with the wording, perhaps I'm not interpreting it the same way.....
the probably that it will "also land on heads", implies "and," so the probability that it is heads is just a simple probability calculation, which is 50% heads each flip, which is just 25% for HH
HH, HT, TH, TT all have an equal chance of occurring.
Although, perhaps because the first heads is known already, the probability of the second coin landing on heads is independent of the first toss, and therefore, 50%
I can't fathom anything other than 50% or 25%, the 33.33%; just does not compute without a further explanation.
Quote from: Jimmy on May 14, 2012, 05:14:28 AM
seems the trick can be with the wording, perhaps I'm not interpreting it the same way.....
the probably that it will "also land on heads", implies "and," so the probability that it is heads is just a simple probability calculation, which is 50% heads each flip, which is just 25% for HH
HH, HT, TH, TT all have an equal chance of occurring.
Although, perhaps because the first heads is known already, the probability of the second coin landing on heads is independent of the first toss, and therefore, 50%
I can't fathom anything other than 50% or 25%, the 33.33%; just does not compute without a further explanation.
The original question states
"also landed showing heads?"
This is past tense.
So both coins have already landed.
A person has told you that one of them was heads.
There are 4 possible permutations but only 3 that contain at least one head.
So we only have 3 possible permutations.
Of those only one contains another head. The HH permutation.
The other two perms are HT and TH.
So chances of a second heads, given that one of the coins is known as heads is 1/3 = 33.333%
I think the question is stated very clearly.
Quote from: Jimmy on May 14, 2012, 05:14:28 AM
seems the trick can be with the wording, perhaps I'm not interpreting it the same way.....
the probably that it will "also land on heads", implies "and," so the probability that it is heads is just a simple probability calculation, which is 50% heads each flip, which is just 25% for HH
HH, HT, TH, TT all have an equal chance of occurring.
Although, perhaps because the first heads is known already, the probability of the second coin landing on heads is independent of the first toss, and therefore, 50%
I can't fathom anything other than 50% or 25%, the 33.33%; just does not compute without a further explanation.
Scratch that, I got it......wording....TT rule out only leaves three options, 1/3H to 2/3 T on second toss
I
50-50. Neither coin is in any way affected by the fall of the other.
Quote from: OldGit on May 14, 2012, 10:33:40 AM
50-50. Neither coin is in any way affected by the fall of the other.
If he'd told you one was heads before the other had been tossed it would be 50-50
But he hasn't, you've been given enough information to conlude the fall is head/head, head/tail or tail/head, there's three opions. It's the devil's 66.6% thing again, well 33.3% this time but same devilish principle.
Quote from: Stevil on May 14, 2012, 05:25:57 AM
Quote from: Jimmy on May 14, 2012, 05:14:28 AM
seems the trick can be with the wording, perhaps I'm not interpreting it the same way.....
the probably that it will "also land on heads", implies "and," so the probability that it is heads is just a simple probability calculation, which is 50% heads each flip, which is just 25% for HH
HH, HT, TH, TT all have an equal chance of occurring.
Although, perhaps because the first heads is known already, the probability of the second coin landing on heads is independent of the first toss, and therefore, 50%
I can't fathom anything other than 50% or 25%, the 33.33%; just does not compute without a further explanation.
The original question states
"also landed showing heads?"
This is past tense.
So both coins have already landed.
A person has told you that one of them was heads.
There are 4 possible permutations but only 3 that contain at least one head.
So we only have 3 possible permutations.
Of those only one contains another head. The HH permutation.
The other two perms are HT and TH.
So chances of a second heads, given that one of the coins is known as heads is 1/3 = 33.333%
I think the question is stated very clearly.
The solution (as per your lucid exposition) is relatively evident to those with a stats background but counter-intuitive to those not initiated into those dark arts. The research suggests that the human brain is not wired in such away as to have an instinctive grasp of probability- rather the reverse.
Quote from: En_Route on May 14, 2012, 07:18:49 PM
Quote from: Stevil on May 14, 2012, 05:25:57 AM
Quote from: Jimmy on May 14, 2012, 05:14:28 AM
seems the trick can be with the wording, perhaps I'm not interpreting it the same way.....
the probably that it will "also land on heads", implies "and," so the probability that it is heads is just a simple probability calculation, which is 50% heads each flip, which is just 25% for HH
HH, HT, TH, TT all have an equal chance of occurring.
Although, perhaps because the first heads is known already, the probability of the second coin landing on heads is independent of the first toss, and therefore, 50%
I can't fathom anything other than 50% or 25%, the 33.33%; just does not compute without a further explanation.
The original question states
"also landed showing heads?"
This is past tense.
So both coins have already landed.
A person has told you that one of them was heads.
There are 4 possible permutations but only 3 that contain at least one head.
So we only have 3 possible permutations.
Of those only one contains another head. The HH permutation.
The other two perms are HT and TH.
So chances of a second heads, given that one of the coins is known as heads is 1/3 = 33.333%
I think the question is stated very clearly.
The solution (as per your lucid exposition) is relatively evident to those with a stats background but counter-intuitive to those not initiated into those dark arts. The research suggests that the human brain is not wired in such away as to have an instinctive grasp of probability- rather the reverse.
For certain :-X
Quote from: The Magic Pudding on May 14, 2012, 12:09:46 PM
Quote from: OldGit on May 14, 2012, 10:33:40 AM
50-50. Neither coin is in any way affected by the fall of the other.
If he'd told you one was heads before the other had been tossed it would be 50-50
But he hasn't, you've been given enough information to conlude the fall is head/head, head/tail or tail/head, there's three opions. It's the devil's 66.6% thing again, well 33.3% this time but same devilish principle.
But he may be lying. You didn't see the coin actually land. That's part of the fact scenario.
Quote from: En_Route on May 14, 2012, 12:58:19 AM
Two coins each with a head one one side and a tail on the other are tossed in the air but you do not see them land.
You are told that one of the coins landed showing heads. What are the chances that the other coin also landed showing heads?
Part of it kinda makes sense to me if the correct answer is 33.33%. Maybe. I dunno. Let's see... :D
According to this, I didn't see the coins land, but I was told that one of them landed showing 'heads'.
The only way the coins could have landed together would have been
1. Heads heads
2. Heads tails
3. Tails tails
So far, I'm following, because there are only those options and each of them has a 33.33% chance of being right before the coins land.
But AFTER one coin lands, how is there still 33.33% of a chance it'll be heads? There are now only 2 choices left: it'll either be heads, or tails. If I know that one coin has landed heads up, the combination CAN'T be tails tails, so it would have to be either heads heads (a chance the second coin has its head up) or heads tails (a chance the second coin has its tails up).
Sigh. I'm confused again. It looks to me like there should be a 50/50 chance on the second coin. I don't see how the first one being dropped affects whether or not heads comes up on the second coin. Can anyone explain, please?
There are two coins, each coin lands as either heads or tails this gives us 2x2 permutations
Coin 1, Coin 2
H , H
H , T
T , H
T , T
Both coins are tossed, both land. The result is known to a person but not you.
This person tells you that one of the coins is a H.
This means, of the four possible options, the T, T didn't happen because we now know at least one of the coins had to be H.
So there are three options that are possible (given that we know one coin turned up H)
Coin 1, Coin 2
H , H
H , T
T , H
We know one coin is H (we don't know if this was Coin 1 or Coin 2) but we know it was one of them. What was the other? Of those three options, two of them has the other coin being T and one of those options has the other coin being a H.
So for the other to be H there is one out of three options. 1/3 = 33.33%
Does this make it easier to think it through?
I always find it hard to explain my own thinking to someone else.
The important thing to realise is that we don't know which coin was heads.
If we knew Coin 1 was heads then there is a 50/50 chance that Coin 2 is heads because Coin 2 can either be H or T.
But the problem statement just says that one of the coins is heads. So this allows us to still include as a valid option the permutation where Coin 1 is Tails and Coin 2 is Heads which would have been ruled out if we had stated in the problem statement that Coin 1 is heads.
I understand your logic but I still can't accept it. Please can we have two volunteers from the audience to perform this trick, say, 10,000 times and log the results?
Stevil, thank you! Yes, that makes sense now! :) You explained it very well.
For me, it was a reading comprehension issue. Or a late night issue. Or both. :D
I was reading the problem and assuming somehow that the first coin was the one that landed heads up, but you're right, we don't know which coin landed that way. Funny how we read things into the text that aren't actually there. (well, funny how I do, anyhow; you didn't!)
That's where I was getting the 50/50 chance from -- I for some reason assumed I knew the first coin that landed was heads up.
Quote from: OldGit on May 15, 2012, 11:05:00 AM
I understand your logic but I still can't accept it.
Makes you sound like a theist.
Quote from: En_RouteMakes you sound like a theist.
Aaagh, not a theist! ;D No, I mean that I have a different argument which seems to me to be right - that neither coin can in any way be affected by the fall of the other. If you throw one coin, it's 50-50 and throwing another by its side can't affect that.
Also, I admitted it needs a trial.
Quote from: OldGit on May 15, 2012, 02:35:48 PM
Quote from: En_RouteMakes you sound like a theist.
Aaagh, not a theist! ;D No, I mean that I have a different argument which seems to me to be right - that neither coin can in any way be affected by the fall of the other. If you throw one coin, it's 50-50 and throwing another by its side can't affect that.
Also, I admitted it needs a trial.
The point is that what you are being asked in effect is the probability of having thrown two heads if you know that you didn't throw two tails.
The explanation becomes less tricky if you don't just use two coins. Okay we know a single coin flip has a 50% chance of being tails, and 50% of being heads. Now if you throw 10 coins in the air and 5 of them land under the sofa, so you can only see 5, and of those visible 5, 4 of them are heads. Knowing there is a 50% chance of heads or tails, that means the best probability is that the remaining 5 coins are 1 heads, 4 tails, bringing our total to 5 heads and 5 tails, or 50%. In other words, of the remaining 5 coins, you should find more tails than heads, even though each coin would individually have a 50% chance.
So if you have two coins and throw 1 head, the chances of throwing a second head reduce below 50%.
Quote from: fester30 on May 15, 2012, 10:51:11 PM
The explanation becomes less tricky if you don't just use two coins. Okay we know a single coin flip has a 50% chance of being tails, and 50% of being heads. Now if you throw 10 coins in the air and 5 of them land under the sofa, so you can only see 5, and of those visible 5, 4 of them are heads. Knowing there is a 50% chance of heads or tails, that means the best probability is that the remaining 5 coins are 1 heads, 4 tails, bringing our total to 5 heads and 5 tails, or 50%. In other words, of the remaining 5 coins, you should find more tails than heads, even though each coin would individually have a 50% chance.
So if you have two coins and throw 1 head, the chances of throwing a second head reduce below 50%.
That's not right, you are just trying to confuse us.
There is no relationship between the coins under the couch and those not under.
Odds aren't combined because an observer has never seen the results of the coins under the couch.
This reminds me of the Roulette wheel. Take out the green spaces, and you have the same number of black and reds. If you bet on black every time, doubling the bet every time you lose, and starting at the original bet every time you win, you'll net a gain of the original bet with every win. You might have several reds show up in a row, but eventually you'll get a black. Each individual spin is 50%, but as you accumulate reds, the overall chance of a spin being red decreases. This betting strategy doesn't work so well in the real world because of the two green spaces and because of maximum bets on the tables that allow you to double your bet usually only 5 times.
You're right, each coin isn't affected by each other coin, and in this case each spin isn't affected by each other spin. However, when talking statistics and probability, it's not spin by spin or coin by coin, but an accumulation of data.
The one thing that wasn't mentioned in the question, however... are these coins properly weighted to give exactly 50-50 chance? If not, that throws off the probabilities, just like a rigged Roulette wheel.
Quote from: fester30 on May 16, 2012, 07:58:42 AM
This reminds me of the Roulette wheel. Take out the green spaces, and you have the same number of black and reds. If you bet on black every time, doubling the bet every time you lose, and starting at the original bet every time you win, you'll net a gain of the original bet with every win. You might have several reds show up in a row, but eventually you'll get a black. Each individual spin is 50%, but as you accumulate reds, the overall chance of a spin being red decreases. This betting strategy doesn't work so well in the real world because of the two green spaces...
Hmmm, without the greens, and given enough spins of the wheel, you will break even.
With the greens you will eventually lose.
I can't for the life of me know why people play roulette at the casino.
But, some people have a system, that they believe in. I just do the math. With the green you lose, without the green you waste a lot of time and then break even.
Quote from: Stevil on May 16, 2012, 08:30:15 AM
Hmmm, without the greens, and given enough spins of the wheel, you will break even.
With the greens you will eventually lose.
I can't for the life of me know why people play roulette at the casino.
But, some people have a system, that they believe in. I just do the math. With the green you lose, without the green you waste a lot of time and then break even.
I was at a casino a few years back, and saw a woman at the roulette wheel who was praying out loud for Jesus to make her win. When I kinda raised my eyebrows and gave her a half-smile (the verbal equivalent is 'what on earth are you doing?') she said, very confidently, "In a dream last night God told me that I was to come here today and play, and I believe with all my heart that my Jesus will NOT let me down!"
So, that's why at least SOME people play roulette. :D
Me, I don't like the odds, either. If I want to plunk down a few dollars on the understanding that I'll probably lose, that's one thing. But it baffles me why anyone would think God would help them win, let alone tell them to go to a casino.
Quote from: Amicale on May 17, 2012, 12:18:57 AM
But it baffles me why anyone would think God would help them win, let alone tell them to go to a casino.
Maybe god wanted her to learn that a fool and their money are soon departed?
Cunning huh?
Quote from: Stevil on May 17, 2012, 12:39:26 AM
Quote from: Amicale on May 17, 2012, 12:18:57 AM
But it baffles me why anyone would think God would help them win, let alone tell them to go to a casino.
Maybe god wanted her to learn that a fool and their money are soon departed?
Cunning huh?
Trickster :D
I'm going to take the opportunity to mention that if you like Derren Brown, he did a whole show called "The System" (should be on YouTube) about some of this. He's an illusionist, so the show will end up not being what you expect, of course. ;D