 #### Posers and Puzzles wolfgang59 Posers and Puzzles 29 Jun '09 09:20
1. 29 Jun '09 09:20
I toss two coins, lets say a £1 coin and a £2 coin they land unseen by me.

A friend tells me that I have tossed "at least one Head"

What are the chances that the coins are different? (ie H & T)

Easy!
My friend's information tells me that the result is one of these three.

The chances that the coins are different is therefore 2/3

Of course the same argument applies if he tells me that at least one is a tail.

Therefore providing he tells me I have at least one of something (which obviously will be all the time) my chances of tossing a head and a tail are 2/3.

But we all know its 1/2.
🙄
2. 29 Jun '09 15:04
After he tells you that you have at least one of something (and he specifies which thing) that rules out the fourth possibility which would be two of the other thing. Thus increasing p from 1/2 to 2/3.
3. 29 Jun '09 15:27
This depends on the conditional probability of him telling you that it's head given any of the outcomes. I got this question in stat class year and thought it sucked, just because this conditional probability is not given. How does your friend decide what side he is going to tell you is at least one of. A logical way would be to just pick one of the two at random and then tell what side it is. However, this would make it twice as likely that you have thrown two heads compared to either combination of head tales, if he tells you that at least one is head.
However, the way you are answering this question is that he is just giving a true or false answer to the question: "Is there a head among the coins?". The whole paradox is created because people logically expect the 'value' and thus the information of the sentence to be the first, however due to unclear formulation the latter is also an option and since statisticians like to work with Events that are either occuring or not, they expect you to follow the second way of interpreting the question. I'd say this problem is flawed.
4. 29 Jun '09 15:37
This is the problem, you may not like it but it is not flawed.

I toss two coins, lets say a £1 coin and a £2 coin they land unseen by me.

A friend tells me that I have tossed "at least one Head"

What are the chances that the coins are different? (ie H & T)

Obviously my logic in solving it is flawed ... but why?
5. 29 Jun '09 16:301 edit
Originally posted by wolfgang59
This is the problem, you may not like it but it is not flawed.

I toss two coins, lets say a £1 coin and a £2 coin they land unseen by me.

A friend tells me that I have tossed "at least one Head"

What are the chances that the coins are different? (ie H & T)

Obviously my logic in solving it is flawed ... but why?
Because there is a different in saying "If there is at least one head, then there is a 2/3 chance that they're different"
and
"If I tell you what one of the coins is, then there's a 2/3 chance that they're different."

In the first case, 1/4 of the tosses will result in that toss being rejected, whereas in the second statement, there will never be any coin tosses eliminated, therefore the probability is 1/2.

Identifying what one of the coins is does not affect the probability of the other coin being heads or tails. However, eliminated all tosses where both coins are tails DOES.
6. 29 Jun '09 17:483 edits
It's again an issue of defining the information contained in the statement.

The probability of both heads given at least one is heads is:
P(HT|#H>0) = P(HT and #H>0)/P(#H>0) = P(HT)/P(#H>0).

P(HT) = 0.5
P(#H>0) = 0.75
=> P(HT|#H>0) = 2/3

The probability of both tails given at least one is tails is:
P(HT|#T>0) = P(HT and #T>0)/P(#T>0) = P(HT)/P(#T>0) = 0.5/0.75 = 2/3

So what are you saying when you have at least one of something? Basically, you're simply saying that you're in either the first or the last case. But you have to weigh each case according to it's probability! Unless you assume the friend prefers to declare heads (or tails) when they are different (so he has a choice), then he is equally likely to declare one or the other.*

So we have:
1/2*P(HT|#H>0)*P(#H>0)+1/2*P(HT|#T>0)*P(#T>0) = 1/2*2/3*0.75 + 1/2*2/3*0.75 = 0.5

And so we don't have to throw away probability theory yet!

*This is a subtle point. However, even if he prefers to declare one or the other, note that the sum does not change: pi*2/3*0.75 + (1-pi)*2/3*0.75 = 0.5. I'm being a bit imprecise here, as I also didn't prove that the case where there are only tails or only heads will not affect this, but I can be more precise if need be.
7. 30 Jun '09 17:09
Palynka I agree with you entirely ... so why do we get the old chestnut of:-

A man has two children.
What is the probability he has one son and one daughter?
We all agree 0.5

You hear him on the phone chatting to one of his children called MARY.
now what is the probabilty he has a son and a daughter?
0.5 (I hope Palynka agrees) but so many puzzles, quizzes, books and (dare I say it teachers) say 0.67 !!!!!!!!!!
8. 01 Jul '09 09:501 edit
Originally posted by wolfgang59
Palynka I agree with you entirely ... so why do we get the old chestnut of:-

A man has two children.
What is the probability he has one son and one daughter?
We all agree 0.5

You hear him on the phone chatting to one of his children called MARY.
now what is the probabilty he has a son and a daughter?
0.5 (I hope Palynka agrees) but so many puzzles, quizzes, books and (dare I say it teachers) say 0.67 !!!!!!!!!!
Interesting. I also bought and happily ate that chestnut until now...

Here are my thoughts:
Let:
GB: if girl/boy
BB: if both boys
GG: if both girls
Db: Declared/revealed boy
Dg: Declared/revealed girl

There are 4 possible cases:
GB & Db: P(GB & Db) = P(Db|GB)*P(GB) = 0.5*0.5 = 1/4
GB & Dg: P(GB & Dg) = 1/4 (as above)
GG & Dg: P(GG & Dg) = P(GG) = 1/4
BB & Db: P(BB & Db) = P(BB) = 1/4
Note that they sum up to 1 as necessary.

Then it's clear to see that P(GB|Dg) = P(Dg|GB)*P(GB)/P(Dg) = 0.5*0.5/0.5 = 0.5!

Am I doing something wrong here? 😕 I don't think so. Note that this is fundamentally different from conditioning on Db on #B>1. Why? Because when the true state is GB you have a 50% chance of Db, but a 100% chance of #B>1! That is:

P(#B>1|GB)=1
P(Db|GB) = 0.5

So P(GB|#B>1) = 2/3 and P(GB|Db)=1/2.
9. 01 Jul '09 21:022 edits
Originally posted by wolfgang59
I toss two coins, lets say a £1 coin and a £2 coin they land unseen by me.

A friend tells me that I have tossed "at least one Head"

What are the chances that the coins are different? (ie H & T)

Easy!
My friend's information tells me that the result is one of these three.

The c ...[text shortened]... l the time) my chances of tossing a head and a tail are 2/3.

But we all know its 1/2.
🙄
let's say you specifically asked your friend to watch the coins and say whether or not there was at least one head.

then these are the scenarios:
HH - "yes, at least one is a head"
HT - "yes, at least one is a head"
TH - "yes, at least one is a head"
TT - "no, neither one is a head"

in this instance, you have clearly eliminated the TT possibility, and the chance of mixed coins is 2/3.

You could have also asked your friend to tell you whether or not there was a tail - but in both cases, the friend could only give info about the coin side you asked him about

now - we look at the scenario that actually occurred - we will assume that your friend would have given some sort of a response - and if the coins are mixed, that he would have been equally likely to talk about heads or to talk about tails

suppose we made 40 flips with each possibility happening an equal number of times

HH - "there is at least one head" (10)
HT - "there is at least one head" (5) and "there is at least one tail" (5)
TH - "there is at least one head" (5) and "there is at least one tail" (5)
TT - "there is at least one tail" (10)

you hear "there is at least one head" - this statement would occur 20 times during the 40 flips. Of those 20 times, 10 would have occurred after HH, while only 5 would have occurred after HT and 5 would have occurred after TH.

As a result, half the time you heard this statement, you'd have two heads, and half the time you'd have mixed coins

the key to the paradox, is that when the coins are the same, the friend is forced to mention that coin side, but when the coins are mixed, your friend has a choice of which coin side to mention
10. 01 Jul '09 21:171 edit
now if in the second scenario

let's say your friend adores the head side of the coin - and will always choose to make a statement about heads if at all possible. In each of the mixed coin cases, he would have a 100% chance of saying "there is at least one head" and a 0% chance of saying "there is at least one tail". In this case, he would say "there is at least one tail" ONLY in the case where both sides are tails

so if you hear "there is at least one head"

HH - 10 times
HT - 10 times
TH - 10 times
TT - 0

In this situation, mixed coins occur in 20 of the 30 cases where your friend says "there is at least one head" and as such the odds are 2/3. This would be an identical situation to the first scenario where you specifically asked your friend to say yes or no on the question of heads.
11. 02 Jul '09 12:18
But the chance of him adoring heads rather than tails is 0.5 .............
12. 02 Jul '09 13:22
Originally posted by wolfgang59
But the chance of him adoring heads rather than tails is 0.5 .............
I think that if you compute the likelihood, then you'd find 0.5 to be a gross overestimation.
13. 02 Jul '09 14:011 edit
Originally posted by wolfgang59
But the chance of him adoring heads rather than tails is 0.5 .............
I first discussed the case where your friend was forced to inform you only about heads - in that case, the odds of mixed coins was 2/3

I then discussed the case where the friend adored each side of the coin equally (0.5). In that situation, I showed that the odds of having mixed coins is 1/2.

I then considered the case where the friend adored heads 100% of the time -- and in that situation, the odds of mixed coins was 2/3 -- the same as if you had specifically instructed your friend to inform you only about heads. The same would be true for the opposite case where your friend adored tails 100% of the time and only mentioned heads if it was HH -- the odds of mixed coins would again be 2/3

Now -- suppose your friend adores heads at some level besides 0%, 50%, or 100% -- say he merely prefers heads or prefers tails. In these cases, the odds of mixed coins would fall somewhere between 2/3 and 1/2.
14. 02 Jul '09 14:22
Originally posted by Melanerpes
I first discussed the case where your friend was forced to inform you only about heads - in that case, the odds of mixed coins was 2/3

I then discussed the case where the friend adored each side of the coin equally (0.5). In that situation, I showed that the odds of having mixed coins is 1/2.

I then considered the case where the friend adored heads ...[text shortened]... prefers tails. In these cases, the odds of mixed coins would fall somewhere between 2/3 and 1/2.
I think it was a joke about platonic love or just being after tails...
15. 02 Jul '09 15:14
Originally posted by Palynka
I think it was a joke about platonic love or just being after tails...
LOL --- sounds like a lot more than platonic love is going on here 😀