A simple question of probability

[Moschops]

It also implies that it can't be HT or TH, depending on which coin was looked at.

And if we do not have that information? If we do not know which coin was looked at, which is the intention (i.e. that's unknown).

[helios]

P(HH' | HT) is "what's the probability of NOT heads-heads, given that it actually is heads-tails".

This is not your original question.

I toss two fair coins. I look at them and then I tell you truthfully that the coins are definitely NOT both heads, or NOT both tails, depending on what I saw.

So, given this, what is the probability that they are both the same?

[Disch]

EDIT: crap actually I was misreading the problem. Hahahaha

It's still totally 50% though.

[helios]

P(HH') = 1 - P(TT').
Again I disagree. I think P(HH') is 3/4. I think P(TT') is 3/4.

Once again, HH' is not the same set as ¬HH. HH' and TT' must necessarily have a null intersection, otherwise it means that there's some cases where you're stating two things at the same time, which can't be true.

[Moschops]

This is not your original question.

Could well be. As I did say at the start, the answer was 1/3 and I have changed the wording, searching for a less ambiguous wording. Sounds like I didn't find it. The new wording is :"I toss two coins and tell you one or both of them is heads. What's the chances of heads heads?"

As stated previously, it's an ( obviously bad) attempt to turn this original question:

The question above is an attempt to make the following question more intuitive: "I've got two children and I promise you they're not both girls; what are the chances that they're both boys?" The answer is 1/3, but people seem to either insist on 1/2, or insist on altering the question until the answer is 1/2.

into something easier to intuit.

That said:

P(HH' | HT) is "what's the probability of NOT heads-heads, given that it actually is heads-tails".

Is that not true? And is there something wrong with my Bayesian calculation, given that I come to a different answer to you?

[Moschops]

Once again, HH' is not the same set as ¬HH.

I think HH' means "anything other than heads-heads". Is that true? If so, I think P(HH') is 3/4. If it's not true, what does HH' mean?

[Disch]

Actually no my program still works for this problem! Hooray.


Given what you said here Moschops:

Moschops wrote:

I'll look at one of them at random and tell you the coins are definitely not both <other side of the coin I looked at> (because that way, I also do not have any information that you don't - I don't know what they are any more than you do).

I challenge you to find the logical fallacy in my post here:
http://cplusplus.com/forum/lounge/88256/#msg473370

Or in my program proving this results in 50% here:
http://cplusplus.com/forum/lounge/88256/#msg473380

[helios]

As I did say at the start, the answer was 1/3 and I have changed the wording, searching for a less ambiguous wording.

The answer to the problem in the OP is certainly not 1/3. It's 1/2.
The answer to this problem:

What's the probability of HH given that ¬TT?

is 1/3. They sound very much alike, but they're entirely different. ¬TT gives a lot of information about the system. In the original problem, there was some uncertainty about the exact state the system was in because when you looked at the coins and said "not double tails" or "not double heads" there was some overlap at HT and TH. Without that overlap, the uncertainty is gone, and the probability does become 1/3:

P(HH | ¬TT) = P(¬TT | HH)P(HH) / (P(¬TT | HH)P(HH) + P(¬TT | HT)P(HT) + P(¬TT | TH)P(TH) + P(¬TT | TT)P(TT)) =
= P(¬TT | HH) / (P(¬TT | HH) + P(¬TT | HT) + P(¬TT | TH) + P(¬TT | TT)) =
= 1 / (1 + 1 + 1 + 0) =
= 1/3

Which is obvious. If it's not one of the four things, it has to be one of the other three, which still have uniform distribution.

I think HH' means "anything other than heads-heads". Is that true? If so, I think P(HH') is 3/4. If it's not true, what does HH' mean?

Like I said, this was the first thing I defined:

Let HH, TH, HT, and TT be the coin configurations, and TT' be you saying "it's not double tails".

My definition implies that even if the coins land on HT or TH, you may or may not say "it's not double tails".

[Moschops]

Disch, nothing doing I'm afraid. It was badly worded (which is what I'm working on) so I'm chucking the whole lot out now :) Hopefully the new wording works better (along the lines of "I've tossed two coins and I'm telling you it's not tails-tails, and I'm not telling you anything else at all. What are the odds it's heads-heads?" which then becomes "I've got two children and I'm telling you they're not girl-girl. What are the odds they're boy-boy?")

Like I said, this was the first thing I defined:
Let HH, TH, HT, and TT be the coin configurations, and TT' be you saying "it's not double tails".

Aha. P(TT') isn't the probability that it's not tails-tails. it's the probability that I SAY it's not tail-tails. Missed that.

[Disch]

Tell you what Moschops:

If I ever meet in you real life, we'll have to get together and do some gambling. You can flip the coins, tell me they're not HH or not TT, and I'll bet with 2:1 odds that the flips came out the same. I'll take that bet any day of the week.

[Volatile Pulse]

I'm sorry guys, I still respect you, but wtf?

@Moschops
Effectively, regardless of which coin you look at, only one of the two possibilities of the coins being different is possible. It's also only possible for them to be the same. A coin doesn't magically get three sides so the odds are 1:2

To put this simply. You have two possible same scenarios. HH or TT. We can rule one of them out, but you're also effectively ruling out one of the HT or TH possibilities too, without knowing you are.

No matter how you word any problem that involves two sets of two, there is only a 50% chance of the final outcome when one is revealed. Eliminating one of the two, leaves one item left with two possibilities, which is exactly where the 50% chance comes from. You can spin it, twist it, slice it up and dice it, but something that has only two possible outcomes can't magically grow a third.

@Disch
I have some gambling games if you're interested, but be warned, the odds aren't forever in your favor.