Simple coin tossing problem?

[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 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... [img]/images/graemlins/wink.gif[/img] , 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.