Simple coin tossing problem?

[PrayingMantis]

This is a simple problem I thought of. Any help will be appriciated, because I'm a bit lost in it...

My friend has tossed a coin 4 times, and written down the results.

Now he tells me: "In this sequence, I got three heads, and one other result, which is either heads or tails. Guess what it was. I'll give you 100$ if you guess it right!"

Well, if I'm treating this in a non-conditional manner, the probability is of course 0.5 for each. So it makes no difference if I say heads or tails.

However, I know that for this 4 tosses sequence, the probability that there were only three heads, is 4:1 to the prob. there were four (i.e, there are 4 ways to get [3 heads, 1 tails], but only 1 to get [4 heads]). So it also makes some sense to assume that the answer is 0.8 for tails, 0.2 for heads. Therefore, I better say tails.

How should I solve this one?

[Ghazban]

Since each flip is totally indendent of all previous flips, the odds are still 50/50 for heads or tails even though 3 heads came on the previous 3 flips. The fact that its more likely to get 3 heads and 1 tails in 4 flips than 4 heads is totally irrelevant.

[TomCollins]

If he didn't tell you any information, and had you bet on 3 heads or 4 heads, then you are better off betting on 3 heads, since there are 4 times as many ways to pull this off.

But... he already told you the first results were heads. And I also bet on heads, because the previous results may indicate the coin is somehow deformed or shaped to land on heads more often.

Assuming a fair coin, its 50-50 for the 4th one.

[PrayingMantis]

[ QUOTE ]
But... he already told you the first results were heads

[/ QUOTE ]

The problem is that he didn't tell me that. He only told me "I got three heads, and one other result, which is either heads or tails". Nothing about first results and last result. Is it still 50-50?

And it's a fair coin.

[PrayingMantis]

[ QUOTE ]
even though 3 heads came on the previous 3 flips

[/ QUOTE ]

He didn't mention 3 heads came on the first three flips (i.e, on the previous 3 flips). He only said he got three "head" reasults, and one other result, that is either heads or tails. That could be in any order.

[kwyjibo00]

these guys are wrong. you should say tails. i'll let others elaborate.

[FloppedFlush]

It depends whether your friend was going to offer you this proposition regardless of what the results of the coin flipping was and what offer he was going to make in each situation. Say your friend went into the trial with the following rules:

if 4 heads - say 3 heads, what was the 4th?
if 3 heads - say 3 heads, what was the 4th?
if 2 heads - say 2 of whatever the first flip was, what was the 4th?
if 1 head - say 3 tails, what was the 4th?
if 0 heads - say 3 tails, what was the 4th?

then you've got the situation you describe in your post. Out of all the situations where your friend tells you he got 3 heads, the other flip being tails is much more likely than the other flip being heads.

On the other hand, say your friend had these rules:
if 4 heads - say 3 heads, what was the 4th?
if 3 heads - say 2 heads, what was the 4th?
if 2 heads - say 2 heads, what was the 4th?
if 1 head - say 2 tails, what was the 4th?
if 0 heads - say 3 tails, what was the 4th?

Now if he says he got 3 heads, the other flip is also a head 100% of the time.

Then there's the various stages in between to consider - when your friend's rules aren't quite so well defined (ie. say 3 heads 37% of the time and 2 heads 63% of the time, etc.).

So just like poker, it's at least as much about reading the player as it is about reading the cards.

[TomCollins]

Sorry for the misunderstanding. I took the 3 heads to mean 3 in a row.

The answer seems to depend on how your friend thinks.

So let's assume his method of telling you what he had was picking 3 of the 4 flips, and telling you those results. This will remove the element of trickery from your friend. It is 50-50 here (can you see why?).

If he decides to tell you what the majority was, you bet on tails, because there are 4x as many of these possibilities.

[Mano]

Definitely pick tails.

Possible outcomes (your probability space):

All 4 coins heads.
First coin tails, the rest heads.
Second coin tails, the rest heads.
Third coin tails, the rest heads.
Fourth coin tails, the rest heads.

All 5 of these outcomes is equally probable, so 80% (4/5) probability that there was a tail from information given.

[JTrout]

You've only got to figure out at what level your friend thinks.

Average guy thinks: " if I pose this question to my friend, he'll probably say tails."

guy thinking 1 higher level: "if I pose this question to my friend, he'll assume I'm trying to trick him- and will answer heads."

etc. etc.

[karlson]

Mano's got the right idea.

Take the simplest case. I flip a coin twice and tell you that one of them was heads. What's the chance that the other was heads?

It's 1/3, not 1/2.

[PrayingMantis]

[ QUOTE ]
Mano's got the right idea.

Take the simplest case. I flip a coin twice and tell you that one of them was heads. What's the chance that the other was heads?

It's 1/3, not 1/2.

[/ QUOTE ]

Thanks for the reply (and all the other replies).

But this is still very unclear, for me, at least.

Let's take your example. You flip a coin twice. You write down the results (it's easier for me to look at it this way). Now, you tell me that one of them was heads, and you are now pointing with your finger at the *other* result. You ask me: "what is the other result, at which I point now?".

(The whole finger thing is to demonstrate the idea that the *other* result must be a specific one, for the question to make sense, IMO).

I cannot see how the prob. of the other result being heads is 1/3 and not 1/2, although I understand, of course, your reasoning.

What am I missing?

[Ghazban]

Sorry, I misunderstood the question. I thought you were saying that the first three flips were heads and the 4th was what you were betting on. If its 4 total flips and he can non-arbitrarily choose to report 3 of the results, you should think this way:

Of the 2^4 (16) possible results of 4 coin flips, there are 4 ways to get 3 heads and 1 tails (HHHT, HHTH, HTHH, and THHH) but only one way to get 4 heads (HHHH). Therefore, 80% of the time, that fourth flip was a tail (4 of the 5 possible outcomes) and the other 20% of the time it was a head so you should definitely say tails.

[karlson]

Right. So if I ask you what's the chance the second flip was heads, it's of course 1/2.

So you want to think of it as us playing this game, I flip twice, write it down, then give you one of the results. Fine. Say if it comes TT, we redo, if it comes HH, I pick one randomly, and if it comes HT or TH, I pick the one that came heads.

Then the chance that I'm referring to the first flip when I said that "One of them was heads" is 1/2(1/3)+1/3 = 1/2 which is good. Also notice that given that I'm referring to the first flip, the chances that the second one was tails is (1/3)/(1/2) = 2/3.

[PrayingMantis]

[ QUOTE ]
Sorry, I misunderstood the question. I thought you were saying that the first three flips were heads and the 4th was what you were betting on. If its 4 total flips and he can non-arbitrarily choose to report 3 of the results, you should think this way:

Of the 2^4 (16) possible results of 4 coin flips, there are 4 ways to get 3 heads and 1 tails (HHHT, HHTH, HTHH, and THHH) but only one way to get 4 heads (HHHH). Therefore, 80% of the time, that fourth flip was a tail (4 of the 5 possible outcomes) and the other 20% of the time it was a head so you should definitely say tails.


[/ QUOTE ]

Yes, this is pretty clear in general, but I suspect (as I wrote it the previous reply to Karlson) that there's a problem also with this reasoning.

You write: "If its 4 total flips and he can non-arbitrarily choose to report 3 of the results". What does "non-arbitrariliy" means here? He tells me that 3 results were heads, and asks me about the other result. I have to guess what was the result for this other flip.

To clearify more:

If he told me "3 of the results were heads. What is the probablity all of the results were heads?", then I see why it's 0.2. However, he's asking me about the other, "4th" result. I have to guess only this specific result.

I believe these 2 questions are not the same, and that's the crux of the problem, IMO.

[Ghazban]

By "non-arbitrarily", I meant that the 3 he told you about were not chosen at random. If, prior to the flips, he said he would tell you what the first 3 results were and that you would bet on the 4th, then it would be a 50/50 proposition. By giving you a sum total of the number of heads (outside of the last flip), the 4 coin flip events are no longer independent (because you now have information about them as a group, not as individuals-- namely that there were at least 3 heads total). I'm afraid I'm not explaining this well... been a few years since I took probability in college. Does that help at all?

[TomCollins]

How can you assume all four are equally as likely? This is a big assumption.

[bigpooch]

Suppose this friend would only ask the question if he had
flipped four heads? It wasn't stipulated at all that he
would always ask the question.

[kwyjibo00]

um, no it's not.

[brassnuts]

Making it more simple to understand by reducing the problem to 2 coins is a good idea. You flip the coin twice, here are the possible outcomes.

HH
HT
TH
TT

The probability of each one of these instances is exactly equal. So, out of the 4 possible cases, 3 of them had heads flip at least once. Out of these 3 instances, tails flipped twice. Hence, if there is no cheating and everything was done randomly, 2 out of the 3 times that one of the flips was heads, the other flip was tails. So the probability of the other flip being tails is 2/3 in this case, with 2 flips total.

[well]

... then the probability of a fourth head would be .2

Of all the 2^4 ordered outcomes of four tosses, we're now in a situation of 5 equally likely outcomes:

HHHH
HHHT
HHTH
HTHH
THHH

Hence 1/5

[PrayingMantis]

[ QUOTE ]
... then the probability of a fourth head would be .2

Of all the 2^4 ordered outcomes of four tosses, we're now in a situation of 5 equally likely outcomes:

HHHH
HHHT
HHTH
HTHH
THHH

Hence 1/5

[/ QUOTE ]

Not necessarily. you can look at it this way:

HHHH - 4 ways to pick 3 H
HHHT - 1 way to pick 3 H
HHTH - 1 way to pick 3 H
HTHH - 1 way to pick 3 H
THHH - 1 way to pick 3 H

So, there are 4 ways for him to pick three H's if they were all H's (i.e, last one is H too), and 4 ways to pick 3 H's from all the cases where there were only 3 H's (i.e, last one isn't H). Therefore, it's 50-50, H or T for the other, "forth", flip.

[well]

[ QUOTE ]
Not necessarily. you can look at it this way:

HHHH - 4 ways to pick 3 H
HHHT - 1 way to pick 3 H
HHTH - 1 way to pick 3 H
HTHH - 1 way to pick 3 H
THHH - 1 way to pick 3 H

So, there are 4 ways for him to pick three H's if they were all H's [...]


[/ QUOTE ]

It's not about picking heads. It's about having thrown three or four of them, and then mentioning just the fact that this happened.

You see?

[PrayingMantis]

I will put it in another way:

If he only tells me that there were 3 heads, and doesn't tell me what the other flip was, than it's *equally* possible that the sequence was [3 heads, 1 tails (in any order)] or [4 heads], since for each case he has exactly 4 options for picking 3 heads. Thus, the other flip is 50-50 for heads or tails, according to this logic.

It gets more complicated if he's giving me the same information only at specific situations, according to his choice, as others have said here.

The truth is, although there were some very good replies here (and some very different ones too), I'm still not sure the answer for this problem is clear to me...

[ZeeJustin]

This is not a probability question that can be answered mathematically. It is psychological unless you know beforehand exactly what information he is going to give.

[AJo Go All In]

[ QUOTE ]
If he only tells me that there were 3 heads, and doesn't tell me what the other flip was, than it's *equally* possible that the sequence was [3 heads, 1 tails (in any order)] or [4 heads], since for each case he has exactly 4 options for picking 3 heads. Thus, the other flip is 50-50 for heads or tails, according to this logic.

[/ QUOTE ]

unfortunately this logic is incorrect. HHHH is a possible outcome. it happens 1/2*1/2*1/2*1/2 = 1/16 of the time. THHH (in that specific order) is also a possible outcome. this also happens (1/2)^4 = 1/16 of the time. same for HTHH, HHTH, and HHHT. your argument about "picking 3 heads" makes no sense assuming that he makes this same statement every time he tosses at least 3 heads. your argument is based on the notion that the results of his tosses are placed in a hat, and he only tells you he got 3 heads if he picks H out of the hat on his first 3 tries. in which case he would always offer you the bet if he had HHHH but would not always offer you the bet if he had only tossed 3 heads. so in this scenario, your argument is correct. however, there don't seem to be any specifications to the above in the original question.

[pzhon]

[ QUOTE ]
My friend has tossed a coin 4 times, and written down the results.

[/ QUOTE ]

This is simple enough.

[ QUOTE ]
Now he tells me: "In this sequence, I got three heads, and one other result, which is either heads or tails. Guess what it was. I'll give you 100$ if you guess it right!"

[/ QUOTE ]

This is not simple. Under what circumstances would he say that? Unless you know the answer to that, you can't give a reasonable probability that H is right.

It could be that your friend has decided to make that particular statement every time it is true. If so, then there are 5 equally likely sequences after which you get to choose, and in 4/5 the other toss was a T.

It could be that your friend would make that statement precisely when the first 3 tosses are heads. Then there are two equally likely sequences after which you get to choose, and in 1/2 the other toss was a T.

It could be that your friend will only make that statment after all tosses are heads. In that case, every time you get to choose, H is right.

Many people get the Monty Hall problem (http://rec-puzzles.org/new/sol.pl/decision/monty.hall) wrong when told that Monty Hall will always offer a chance to switch doors. However, in the actual show, Monty did not always allow the contestant a chance to switch. You have to know what Monty Hall is trying to do to decide whether switching wins 0%, 50%, 2/3, or 100%. The same issue arises here, and this problem is underspecified.

[Dov]

Flip a coin

[DeadRed]

Your explanation is only valid in a situation where your friend decides (either before or after the flips) that he will tell you about three of the specific flips (e.g. the first, third, and fourth). If this were the case, then 1/2 is correct. But I think that the way the original quetion is posed, your friend just makes the general observation "I see three heads" (meaning AT LEAST 3) without other influence, then you get the .8T/.2H probability.

The difference between the two games is this:

If your friend flips four coins and tells you about three of the faces that he sees, whether they are the three he decides to mention, the three you ask about, the three closest together, the three shiniest (I could go on... /images/graemlins/wink.gif , and this criteria can change from flip to flip) then the identity of the fourth is 50/50.

If instead he flips four coins repeatedly, only stopping to tell you when he sees (at least) three heads, then you're 80/20 for tails on the fourth.

So there are not two right answers, but two different GAMES, each having it's own right answer.

[PrayingMantis]

[ QUOTE ]
Your explanation is only valid in a situation where your friend decides (either before or after the flips) that he will tell you about three of the specific flips (e.g. the first, third, and fourth). If this were the case, then 1/2 is correct. But I think that the way the original quetion is posed, your friend just makes the general observation "I see three heads" (meaning AT LEAST 3) without other influence, then you get the .8T/.2H probability.

The difference between the two games is this:

If your friend flips four coins and tells you about three of the faces that he sees, whether they are the three he decides to mention, the three you ask about, the three closest together, the three shiniest (I could go on... , and this criteria can change from flip to flip) then the identity of the fourth is 50/50.

If instead he flips four coins repeatedly, only stopping to tell you when he sees (at least) three heads, then you're 80/20 for tails on the fourth.

So there are not two right answers, but two different GAMES, each having it's own right answer.

[/ QUOTE ]

Your reply has got me thinking. I understand what you mean about the 2 different games, however - according to the information he gives me, as I see it, I do not know what game it is. In other words: it is possible he will make his offer only when he gets at least three heads, or that he's simply telling me he got 3 heads NOW (these could be the first three he sees, or whatever).

So, my thinking here is: If I do not know what game it is, and I know that the the prob. for heads is either 0.5 or 0.2, for each different game, shouldn't I try to go for somewhere in between these two, i.e, put heads, for instance, on 0.35 prob, for this unspecified game? Does it make any sense?

It looks pretty clear, that even if I'm not sure which game we're playing, there is still a greater chance that the "forth" result is tails. The question is, how much greater.

[pzhon]

[ QUOTE ]
It looks pretty clear, that even if I'm not sure which game we're playing, there is still a greater chance that the "forth" result is tails. The question is, how much greater.

[/ QUOTE ]

No, your friend could decide to make the statement only when all 4 tosses were heads. The problem is underspecified.

The economist Knight (http://cepa.newschool.edu/het/profiles/knight.htm) distinguished between risk and uncertainty. Risks have knowable probabilities. Uncertainties may not. You don't always have enough information to assign probabilities to outcomes in a meaningful/consistent fashion. Imagine we play rock-paper-scissors, but I get to make my choice after yours. You can't determine a probability distribution for my choice, then make an optimal play based on it.

[PrayingMantis]

[ QUOTE ]
No, your friend could decide to make the statement only when all 4 tosses were heads. The problem is underspecified.

The economist Knight distinguished between risk and uncertainty. Risks have knowable probabilities. Uncertainties may not. You don't always have enough information to assign probabilities to outcomes in a meaningful/consistent fashion. Imagine we play rock-paper-scissors, but I get to make my choice after yours. You can't determine a probability distribution for my choice, then make an optimal play based on it.


[/ QUOTE ]

Thanks, the link for Knight's biography and writings is really helpful. I should read some of it.

Anyway, I think your reasoning here leads to a somewhat strange conclusion.

If the problem is indeed underspecified as you say, then if my friend offers me money on my bet here, I have no information to decide what is a better choice: heads or tails. Therefore, when guessing at it, I am practically looking at it as a 50-50 bet (as in your rock-paper-scissors example, where it's 1/3-1/3-1/3).

So, at the bottom-line, according to this logic, I'm treating this as a the simplest coin-tossing problem. To make it more clear: if I'm asked to make a bet on a problem like this many times (by different people, say, so I don't know anything about any of them), without having any information about their "strategy", my best EV move is to toss a coin myself, every time, to make a choice.

This IS a solution to the original problem, underspecified or not.

Do you agree?

[PrayingMantis]

After a little more thought, I understand why 50-50 is not better than any other way of solving it.

It is basically the same as betting on the come of coin-tosses. Your EV will not be higher if you always bet heads, or always tails or 87-13 or 50-50 or whatever. This is what you mean by underspecified, if I understand it correctly.