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        riddles >> easy >> Easy: Two Coin flips
        
(Message started by: ootte on Jul 24^th, 2002, 2:29am)

Title: Easy: Two Coin flips
Post by ootte on Jul 24^th, 2002, 2:29am

The chance is still 50%, since it doesn't matter what the other coin come up with.

--
Oliver

Title: Re: Easy: Two Coin flips
Post by tealtoe on Jul 25^th, 2002, 6:11am

I don't want to give away the answer right on this forum (unless you ask me to), but let me tell you that surprisingly your answer of 50% is wrong; this one is more difficult than it first appears.

Email me if you'd like the answer.

Ken.

Title: Re: Easy: Two Coin flips
Post by ootte on Jul 25^th, 2002, 6:43am

on 07/25/02 at 06:11:03, tealtoe wrote:

but let me tell you that surprisingly your answer of 50% is wrong; this one is more difficult than it first appears.

If it is not some weird coin-throwing technique, and the coins are flipped one after another ... Umm. I don't know then.
Please post the solution or pm me, thanks.

--
Oliver

Title: Re: Easy: Two Coin flips
Post by tealtoe on Jul 25^th, 2002, 8:14am

I sent the solution to you via email.

Title: Re: Easy: Two Coin flips
Post by jmlyle on Jul 25^th, 2002, 8:28am

As I see it, it is either a 50% chance (being an independent event), or the problem is about the veracity of the statment "One of the coins came up heads."

It seems that another "EASY" problem depends on a similar statement not being exclusive (i.e. "One" not meaning "Not two"). If the statement is a complete description of the entire state of things, then there is one answer. But if it is just a simple truthful declaration about a single coin (which seems entirely reasonable), then it should be 50%.... I think.

Title: Re: Easy: Two Coin flips
Post by tealtoe on Jul 25^th, 2002, 9:14am

No, it certainly is not 50%, although you are on the right track--"One of the coins came up heads" is the key to the whole thing. If you don't get the answer, you can email me (kkitow "at sign" yahoo.com).

Title: Re: Easy: Two Coin flips
Post by Jonathan Wheaton on Jul 25^th, 2002, 3:55pm

As jmlyle noted, the statement "One of the coins came up heads" can be read as a description of the state of both coins or of only one. If the former, then it's more of a riddle; if the latter, then it's more of an exercise in probability. So which is it? I don't think there's a right answer to that one, although there's obviously only one right answer for each way of reading the statement. The riddle one is trivial, the probability one is harder (for me anyway).

In talking with one colleague, he looks at it this way. There are four states the two coins could be in. TT, HH, TH, HT. TT is out since you know one of them is heads. That leaves three possibilites, only one of which results in the second coin also being heads. Thus, 33% chance of the other coin being heads.

I'm not completely convinced, but it sounds reasonable. Thoughts?

Title: Re: Easy: Two Coin flips
Post by Mark Schnitzius on Jul 25^th, 2002, 4:39pm

Ha! Even if you consider "one of the coins came up heads"
to mean "AT LEAST one of the coins came up heads" (which
is how I believe it is meant), you are WRONG!

Here's how I responded to this question on another forum:

>June 15, 2001
>
>painfully easy aha:!
>
>i flip a penny and a dime and hide the result from you. "one >of the coins came up heads", i announce. what is the >chance that the other coin also came up heads?
>
>thanks tubby
>
>solution: easy
>
>
>Solved by David Grenier on June 17, 2001
>
>solution: painfully easy
>
>
>Assuming complete honesty on the part of the flipper, >wouldn't the solution be 33%?
>
>There are four possible scenarios:
>
>HH
>TH
>HT
>TT
>
>Obviously the TT possibility can be discounted because it >does not result in one of the two being heads.
>
>This lees us with three possibilities, only one of which has >the other coin also being heads.
>
>Therefore one third.
>
>I think.
>
>I usually get these wrong.

You called it, you are wrong.

This is one of my favorite problems, but I would dread
getting asked it in an interview, because stupid people
answer 50%, smart people answer 33%, and *really*
smart people KNOW the answer is 50%. ;-) And I bet
they'd be expecting to hear 33%.

Let me propose a similar question, the answer to which
*is* 33%:

i flip a penny and a dime and hide the result from
you. "Is at least one of the coins heads?" you ask.
"Yes" I truthfully reply. what is the chance that the
other coin also came up heads?

Do you see the subtle difference? You have to allow
for the possibility of one of the four cases to get
eliminated, if I had responded "No". The question, as
it is originally stated, does not allow for that.

To put it another way, suppose we do a number of
trials of the problem as it is originally stated. Since I
am volunteering the information about the state of
one of the coins, I decide, UNBEKNOWNST TO YOU, to
always tell you the state of the dime. "One of the
coins came up heads" I say if the dime came up heads,
and "One of the coins came up tails" I say if it
came up tails. Your reply essentially amounts to
guessing the state of the penny then. Are you telling
me that you can tell me you can do that with a better
than 50% accuracy?

If you knew that the flipper was eliminating all cases
of two tails being flipped, and only ever saying "At least
one of the coins is heads", then you could safely assume
the answer was 33%. But you don't know that the flipper
is doing that.

If this question ever comes up in an interview, I see
myself getting into a heated argument with the
interviewer and completely blowing my chances.
Of course, anyone here is welcome to analyze the
above at their leisure and send me a job offer if they
agree I'm right. ;-)

--Mark

Title: Re: Easy: Two Coin flips
Post by jmlyle on Jul 26^th, 2002, 8:19am

So..... The problem that is presented is actually only able to happen 75% of the time after the double coin toss.....

And therefore, the coin state is NOT an independent event. It's possible state is dependent on the fact that the tosser was able to say what he said.

So, I think I know, but I guess I'll go with tradition and not post what I think is the answer. Could someone send me the answer, please (spam@myborg.com)?

--jmlyle

Title: Re: Easy: Two Coin flips
Post by Peter Seebach on Jul 26^th, 2002, 6:37pm

The point is, *IF* he says "at least one is heads", then you're in the 3 cases where you had at least one heads. It's not independant.

Comparison: If you do a dime and a penny, and say "the dime came up heads", the chances that the penny came up heads are now 50%. Neat, huh?

Title: Re: Easy: Two Coin flips
Post by Ion Rush on Jul 27^th, 2002, 3:00am

tell me if this is going in the right direction

flip 2 coins, 4 possible outcomes

HH, HT, TH, TT

the fact that your buddy said one coin is heads means that TT must be removed, so you are down to 3 possible outcomes
HT TH HHs

now, if the question was, "what is the chance that BOTH coins are heads it would be 33%, however, the question is ,is the OTHER coin heads?

Now, as I see it, one coin is totally out of play and should no longer be influencing the outcome, only the OTHER coin does

(for example, your buddy flips the coins one after the other while shielding them from you. he only actually checked the first flip, and yes it was heads. he then flipped the second, and has not checked it yet. )

Because one of the coin's results is know is irrelevant, we are taking about one single coin, the OTHER coin, so it should be 50%

is this correct?

Title: Re: Easy: Two Coin flips
Post by Wah on Jul 27^th, 2002, 5:42am

i don't think so
when he says one of the coins came up head, he could have meant either of the coins.
in the following 3 cases,his statement is true.
1st:T 2nd:H(the other coin is T)
1st:H 2nd:T(the other coin is T)
1st:H 2nd:H(the other coin is H)
therefor i think the chance are 1/3

Title: Re: Easy: Two Coin flips
Post by srowen on Jul 27^th, 2002, 7:09pm

No, if you believe that the flipper is telling you that "at least one of the two coins came up heads", then the answer is indeed 1/3, for reasons outlined previously. This is probably the intended interpretation and the riddle should probably include the words "at least".

If the flipper is actually reporting on the state of only one coin, then that is a different scenario (to which the answer is 1/2, yes). But this of course is not the same as the "at least one" interpretation.

And in an interview, your best bet is to demonstrate your total understanding of the subtleties by asking for clarification, since *that* is actually the really smart answer.

on 07/25/02 at 16:39:50, Mark Schnitzius wrote:

Ha! Even if you consider "one of the coins came up heads"
to mean "AT LEAST one of the coins came up heads" (which
is how I believe it is meant), you are WRONG!

...

To put it another way, suppose we do a number of
trials of the problem as it is originally stated. Since I
am volunteering the information about the state of
one of the coins, I decide, UNBEKNOWNST TO YOU, to
always tell you the state of the dime. "One of the
coins came up heads" I say if the dime came up heads,
and "One of the coins came up tails" I say if it
came up tails. Your reply essentially amounts to
guessing the state of the penny then. Are you telling
me that you can tell me you can do that with a better
than 50% accuracy?

If you knew that the flipper was eliminating all cases
of two tails being flipped, and only ever saying "At least
one of the coins is heads", then you could safely assume
the answer was 33%. But you don't know that the flipper
is doing that.

--Mark

Title: Re: Easy: Two Coin flips
Post by Mark Schnitzius on Jul 30^th, 2002, 8:25am

on 07/27/02 at 19:09:57, srowen wrote:

I disagree -- the answer is 1/2 in that case, UNLESS you make the assumption that the flipper keeps flipping until he gets at least one head, before telling you that at least one of the coins came up heads. That (to me) is a pretty big assumption.

on 07/27/02 at 19:09:57, srowen wrote:

If the flipper is actually reporting on the state of only one coin, then that is a different scenario (to which the answer is 1/2, yes). But this of course is not the same as the "at least one" interpretation.

The flipper is ALWAYS reporting the state of only one coin. He needn't choose the dime or the penny every time; he might always choose which coin settled first, or which was closest to him, or whatever, but it's always the state of one coin. Even if he's always saying "at least one". Or maybe I'm misunderstanding you. Please explain to me the difference between reporting that "at least one" coin came up a certain way, versus reporting on the state of one coin (to which the flipper could always tack on "at least one" without lying).

Title: Re: Easy: Two Coin flips
Post by chopotter on Jul 30^th, 2002, 9:32am

I've flipped both coins, and say "[at least] one of the coins is heads." If one of the coins is heads, there are only two possible states for the remaining coin, heads and tails, with heads having a 50% probability of being up.

There are not three possible outcomes, because the state of one of the coins has already been determined, i.e. you cannot use this coin to compute probabilites. There are only two possible outcomes. The remaining coin is heads or it is tails.

I'm quite ready to be disabused, but please do explain how this reasoning is faulty if you do.

Title: Re: Easy: Two Coin flips (spoiler)
Post by kenny on Jul 30^th, 2002, 11:34am

In my opinion, you're all overlooking something. Here's the puzzle:

"I flip a penny and a dime and hide the result from you. "one of the coins came up heads", i announce. what is the chance that the other coin also came up heads? "

It says nothing about the motivations of the flipper. Here are some possibilities:

1) Flip coins. If either or both comes up heads, say "one of the coins came up heads". Otherwise, say "one of the coins came up tails".

In this case, the probability of the other one being heads is 1/3.

2) Flip coins. Look at the dime. Say "one of the coins came up <facing on dime>".

In this case, the probability of the other one being heads is 1/2.

3) Flip coins. If either or both comes up tails, say "one of the coins came up tails". Otherwise, say "one of the coins came up heads."

In this case, the probability of the other one being heads is 1.

I think most arguments about the right answer hinge on assumptions about what the motivations of the tosser are.

-- Ken

Title: Re: Easy: Two Coin flips
Post by chopotter on Jul 30^th, 2002, 12:05pm

If no assumptions are to be made, the riddle has no answer. Actually, I guess it has an answer for each set of assumptions. How... messy. I'd prefer there to be only _one_ answer. :)

Title: Re: Easy: Two Coin flips
Post by josie on Aug 1^st, 2002, 3:26pm

I read it as only one coin came up heads. The question asks, "what is the chance that the other coin also came up heads?" which implies that the "other" coin is the one that didn't come up as heads. So, I think zero chance.

Title: Re: Easy: Two Coin flips
Post by jc on Aug 3^rd, 2002, 10:06pm

Yeah, I also got 1/3. Since it asks what the chances are that the "other" coin was also heads, it implies, "what's the chance that both were heads?". Since there are 4 possibilites, and you know tails/tails isn't one of them since it needs at least one heads, heads/heads is one of three possible answers. So 1/3. :P

It's much easier if you don't overcomplicate. ;)

Title: Re: Easy: Two Coin flips
Post by zoobiewa on Aug 4^th, 2002, 11:02am

I know the answer. First we should put the original question out there:

"i flip a penny and a dime and hide the result from you. "one of the coins came up heads", i announce. what is the chance that the other coin also came up heads?"

Then we need the table of all possible outcomes

tt th ht hh

The possibility of tt happening is impossible, so we can x that out. If you were randomly flipping to see if a pattern emerged, you would not be able to use any outcomes that came up with tt in them. This weighs the results, so now we have only three available options, which are: th ht and hh. With th and ht, they are the same thing, which means there are really only two options, one heads and the other tails, or both heads. That means there is a 50/50 chance.

I tried this by flipping a penny and a dime 35 times, and tossing out any touble tails. After 35 throws, they were almost neck and neck, with ht winning by one point. It is 50%

Title: Re: Easy: Two Coin flips
Post by jc on Aug 4^th, 2002, 7:03pm

on 08/04/02 at 11:02:38, zoobiewa wrote:

With th and ht, they are the same thing, which means there are really only two options, one heads and the other tails, or both heads. That means there is a 50/50 chance.

Nope, sorry. They're not the same thing. He doesn't tell you which coin already landed heads. th would mean, say, that the penny is tails and the dime is heads, whereas ht would mean the penny is heads and the dime is tails. They're both viable possibilities.

Title: Re: Easy: Two Coin flips
Post by srowen on Aug 5^th, 2002, 6:19am

on 07/30/02 at 08:25:25, Mark Schnitzius wrote:

Sure, you are given in this riddle that it happens that at least one of the two coins came up heads. If you want to say that as "the flipper kept flipping until getting a head", OK.

on 07/30/02 at 08:25:25, Mark Schnitzius wrote:

I am saying the difference between the two interpretations is this:

A) "At least one of the coins is heads."
Possible outcomes were: HT TH HH

B) "I only looked at one coin (chosen according to my whim) and it is heads."
Possible outcomes were: HT HH

The difference in what the interpretation tells you about the possible outcomes of the flips makes the difference.

Or to put it in the terms you used, it is the difference between flipping two coins until at least one is heads, and flipping two coins, selecting one to look at, and continuing until that selected one is heads.

You're right that it doesn't matter how the flipper selects which coin to report on, under interpretation B. But it does matter if she is only giving you information about one coin, because it affects your information about the outcomes.

Title: Re: Easy: Two Coin flips
Post by anshil on Aug 5^th, 2002, 7:40am

The answer is 50%. No I did not think about it, I just wrote a little pyhton program.


from random import *

t = 0; # number of cases the condition was true
f = 0; # number of cases the condition was false

while 1:
    a = choice(range(2))  # flip the first coin
    b = choice(range(2))  # flip the second coin

    if a == 0 and b == 0: # if both are tail ignore and reflip
                continue;

    if a == 1 and b == 1: # if both are head the condition is true
                t = t + 1
    else:                 # otherwise the other coin is not head
                f = f + 1

    if f > 0:    # just don't device by zero
                         print "t=", t, " f=", f, " ratio=", 1.0 * t / f

#end of program

It will print out 0.5 very fast, every time I run. So the answer is 50% as long I've not coded something wrong. Now theorize so long until you get the answer the expirment showed. (science, believe the expermient, not the theory)

Title: Re: Easy: Two Coin flips
Post by anshil on Aug 5^th, 2002, 7:45am

Damm am I stupid, of course the answer is not 50%.

2/3 and 1/3 are actually true, if I devide false by truth I get 0.5, so truth is twice as often as false. so 33% is correct!

Title: Re: Easy: Two Coin flips
Post by peter johnson on Aug 6^th, 2002, 3:00pm

>i flip a penny and a dime and hide the result from you. "one >of the coins came up heads", i announce. what is the >chance that the other coin also came up heads?

The chance of the second coming up heads is zero, 0%. ONE of the coins came up heads !

Title: Re: Easy: Two Coin flips
Post by B on Aug 8^th, 2002, 12:15pm

Possible results:
HpHd, TpHd, HpTd, TpTd

"one of the coins came up heads" == "either P or D or both came up heads", which means TpTd is not possible. The other three are equally possible, and in only one case are both coins heads. 1/3.

Title: Re: Easy: Two Coin flips
Post by JW on Aug 9^th, 2002, 7:13am

To all of those people saying that 'one of the coins came up heads' equates to the other coming up tails I think that is a misinterpertation. If both coins came up heads then the statement 'one of the coins came up heads' would still be true since one of the coins did indeed come up heads, the fact that two of the coins came up heads does not change this.

Now as for all of the people questioning the motive of the flipper I say that just makes your conclusion less accurate. Since we do not know the motives of the flipper (if he/she is looking only at the dime while making the statement) then to venture a guess as to the motives is foolish. The best answer is 1/3 because the only information given to you is that ONE of the coins came up heads. Given this and the fact that there are three possible scenarios involving ONE of the coins coming up heads (HT, TH, HH) the chance of both coins coming up heads is 1/3.

Title: Re: Easy: Two Coin flips
Post by Wanjoon on Aug 18^th, 2002, 1:06pm

Oops, didn't see this thread.

But seriously guys, the answer is really 33 1/3 %. This is an elementary probability problem used in statistics and discrete mathematics courses.

TH and HT are NOT the same, precisely because there are two different ways to get a heads and tails combination, whereas there is only one way to get HH.

If you had a six-sided die that replaced the 5 side with another 6 side, would you still say that the probability of getting a 6 was only 1/6? Of course not.

Title: Re: Easy: Two Coin flips
Post by crawfish on Aug 23^rd, 2002, 11:00am

The answer is 50%.

If I asked the question as follows: "If I flipped two coins and one came up heads, what is the probability the other also came up heads", the answer would definitely be 33%. But in this case the coins have already been flipped, and we are given the state of one of the coins. Thus, we can remove one coin from consideration and evaluate the second coin alone as a random occurence of one coin flip.

At least, I think so. ;)

Title: Re: Easy: Two Coin flips
Post by S. Owen on Aug 23^rd, 2002, 11:32am

The question is whether "one came up heads" refers only to the state of one of the two coins (in which case the answer is 50%), or to both - that is, "at least one of the two came up heads." (in which case it's 33%).

It makes no difference whether the coins were already flipped, or whether you are talking about some hypothetical situation. That is, the answer is 33% even if the flipper has already flipped the coins, provided the flipper is reporting on the state of both coins ("at least one...").

Title: Re: Easy: Two Coin flips
Post by Dustin D. on Aug 23^rd, 2002, 9:34pm

The question isn't clear. Depending on how you interpret it, the answer is either 25% or 50%. The outcome of the first coin toss does not effect the outcome of the second. So if you're strictly talking about what the chance is of the second coin being heads, its outcome is independent of the first coin, which means 50%. However, as I read the question, I see a lot of emphasis on the word "also," thus implying you're looking at the probability of both coins coming up heads. This means we're looking at the 4 possible cases, HH, TT, HT, and TH. Obviously HH has a 25% chance.

The 33% argument is invalid, because you can't discount a possible outcome. You don't erase the odds of getting heads or tails just by finding out information about the outcome after it happened. Probability deals with what might happen before it happens, so you can't judge probability after the fact.

Title: Re: Easy: Two Coin flips
Post by S. Owen on Aug 24^th, 2002, 7:00am

So you are saying that there is a nonzero probability that there were two tails, given that "one came up heads"? Whoa!

You can reason about the probability of an outcome, even if it has already happened. Maybe you would find the question more acceptable if it said "if I were to flip two coins, and one were to come up heads, what would be the probability that the other is heads?" But this is the same thing.

If you are still not convinced, write a computer program that simulates this situation (your program does have to discard TT flips, yes), and you will see that it is 33%.

Title: Re: Easy: Two Coin flips
Post by flapp on Aug 25^th, 2002, 11:37pm

The key is not the statement "One of the coins came up heads", but rather "what is the chance that the other coin <B>also</B> came up heads?"

Title: Re: Easy: Two Coin flips
Post by S. Owen on Aug 26^th, 2002, 6:59am

Can you elaborate?

It was argued above that "also came up heads" means that we are talking about the probability that both came up heads (and then, that the answer is 25%, which is wrong).

There is a potential ambiguity in "one of the coins came up heads" but it is not resolved by "also"... in fact, remove the word and the riddle is the same.

Title: Re: Easy: Two Coin flips
Post by ggkrause on Aug 26^th, 2002, 10:50am

This riddle is an example of conditional probability. What you are trying to determine here is the probability that (A) BOTH coins have landed heads-up GIVEN that (B) AT LEAST one coin has landed heads-up. The formula is:

P(A|B) = P(A <intersect> B)/P(B) [sorry, can't find the upside-down 'U']

The intersection of A and B is the case HH. Its probability is 1/4.
The probability of B is 3/4 (all cases except TT).

(1/4)/(3/4) = 1/3

Title: Re: Easy: Two Coin flips
Post by Emprod on Aug 31^st, 2002, 4:46pm

Don't overthink it.

Plug the exact facts given by this problem into a machine, and it will tell you 0%.

0%.

Title: Re: Easy: Two Coin flips
Post by Jonathan_the_Red on Sep 1^st, 2002, 4:30pm

Kenny has it right here: the solution cannot be determined without knowing the algorithm used by the flipper to decide what to say after flipping. In this, it is kind of similar to the Monty Hall problem, which cannot be solved unless you are assured that Monty will always open a door to reveal a goat before giving you the option of switching.

Title: Re: Easy: Two Coin flips
Post by Alex on Sep 12^th, 2002, 8:11pm

we are analyzing the question, not what it could say... so i'm thinking the question is "what is the chance that the other coin also came up heads?"

not what is the chance that they BOTH came up heads.

so the chance that the <b>other</b> also came up heads in 1/2 not 1/4 or 1/3!

Title: Re: Easy: Two Coin flips
Post by David Balmain on Sep 19^th, 2002, 9:51pm

Lets take the motivation of the flipper out of the equation.

Consider this. The flipper flips two identical coins and by mistake reveals that one of the coins is a head.

If anyone thinks that the probability of the other coin being a head is not 50%, please tell me why? If not, please explain how this is different to the original problem?

Title: Re: Easy: Two Coin flips
Post by S. Owen on Sep 20^th, 2002, 5:45am

If you mean that, for example, the flipper accidentally gave you a peek at one of the coins, and it was heads, and asks what the chance is that the other is heads, then yes it's definitely 1/2. This is equivalent to one interpretation of the original problem.

However if the flipper somehow lets it slip that "at least one of the two coins came up heads" and then asks what the chance is that the other is heads, the answer is 1/3. This is equivalent to the other interpretation, probably the intended interpretation, of the original problem.

Title: Re: Easy: Two Coin flips
Post by Zaius on Nov 16^th, 2002, 1:11am

I think it's 50%. The problem with the HT, TH, HH theory is that in this particular case HT and TH are identical, since we do not care which coin was tails and which was heads. In other words, the only 2 possibilities are a combination of tails and heads (in whatever order) and a combination of heads and heads. Only 2 possibilities: the other coin is tails, or the other coin is heads. And either possibilities are equal in value (2 sides to a coin).

This is, of course, assuming that the riddler does not mean "one and only one of the coins came up heads" when saying "one of the coins came up heads". In that case, the other coin would be tails, and the answer would be 0.

Title: Re: Easy: Two Coin flips
Post by S. Owen on Nov 16^th, 2002, 10:04am

While HT and TH are in some sense indistinguishable to you, it's not valid to count them as one outcome. One heads and one tails happens twice as often as two heads. If HT and TH were the same thing, then if you repeatedly flipped two coins, you would see a heads and tails 1/3 of the time. You can confirm however that you actually get it 1/2 the time.

For this problem, try repeatedly flipping two coins and discarding two-tails outcomes - you will find that two heads come up once for every two times a head and tail come up, on average - that is, 1/3 of the time.

Title: Re: Easy: Two Coin flips
Post by babe72724 on Nov 22^nd, 2002, 4:58pm

if there are two coins and one was heads then the other one would have 33% chance of bein heads wouldn't it? because the heads side is heavier than the tails side therefore making the tails side show more. then the heads side would only show once every three tosses. isn't the weight of one a factor of the other?

Title: Re: Easy: Two Coin flips
Post by KAuss on Nov 25^th, 2002, 11:53pm

Here is my crack at this given ONE assumption...

1) The flip of the coin in question is a fair coin and gaining a head or a tail from a result of the flip is a fair 50%

there is actually 2 answers to this question.

BEFORE the flip, your chnaces are 50% Reason being is because no matter how many million times you flipped a coin before you flipped this one, no matter how many times you hit heads, or how many times you hit heads in a row, your NEXT flip IS, HAS been, and forever WILL be 50% because when you flip a fair coin, you'll only have either a head or a tails.

So before you flip, your chances are 50%

AFTER you flip with results in hand, AND if you ask the question of the coin's possibility to land heads, then your chances are STILL 50% due to physical abilities of the coin. BUT, if you are asking what are the chances you get 2 heads in a ROW, then your chances are 1/4 (if the first coin is not revealed). The reason being is that the very same reason the coin is just as easy to fall as tails on every flip, to consecatively land on heads in a row will be increasingly harder as you flip more heads. As you get more biased and biased, you're defying the 50% probable which we all understandingly know.

And to put this in betting terms and to stretch the situation to the extreme...

If you were to bet someone after flipping 1 million heads in a row if their NEXT FLIP will be a head, your chances are 50% < you cannot argue this if you're judging only a single flip no matter what they flipped before hand.

If you were to bet someone to FLIP 1 MILLION heads in a row, your chances are 50%^n or 1/2^n where n = the number of times in a row in this case 1 Million Your chances to flip 2 heads in a row is actually 1/4 but since you already got 1 head, you do 2 - 1/4 which gives you 1/2 since you only got 2 flip probables left.

But if you are to bet someone 1 MILLION in a row, but some how know they just flipped 999,999 heads in a row, then your chances are back to 50% because the flipper no longer has to worry about achieving 999,999 more heads after the next one. It'll all boil down to ONE last flip in which a fair coin will give you 50%... (The reason for this is because you actually do a count down effect. Where after the first head, you only need 999,999 more which means your chances are 1/2^999,999 and you keep going down till you're 1/2^1 = 1/2, initially it'll be a good bet to bet against someone but as they land more and more, their chances actually get easier and easier till it ultimately boils down back to 1/2)

^ If that didn't do it, I must suck in expressing the english language... (Well my test scores already show me I do but anywho ;D)

Title: Re: Easy: Two Coin flips
Post by Icarus on Nov 26^th, 2002, 4:31pm

KAuss, to apply your analysis to this question, you would have to change the question to "Flip a coin twice. The first time it is heads. What is the probability it will be heads a second time?" Everyone agrees that the answer to this question is 50%. But that question is not equivalent to the one asked here. Try this: Flip two coins several times. Count the number of occurences where at least one of the coins was heads. Count the number of occurences where both coins are heads. You will find the ratio of (both heads) to (at least one head) is approximately ~1/3. (The more flips, the closer you will get.)

Title: Re: Easy: Two Coin flips
Post by KAuss on Nov 27^th, 2002, 7:24pm

But it's like many of the people here have stated, there is a logical mathmatical way to figure out probability....

If you want to find out what the probability is to flip 2 heads in a row, it's 1 / 2^n where n is 2 in this case for 2 heads...

so the answer is 1/4 of the time you should get heads since there are 4 probables..

c1 c2
H T
H H
T T
T H

as you can clearly see, there are four outcomes to flipping a fair coin 2 times... and H, H is one out of the 4 options... so it only makes sense to say 1 / 4...

My point was, that each time you hit heads, you subtract the n... which means that if you bet someone who wants to flip 3 heads in a row, which is 1 / 8 chance of happening, after the first head, it'll be a 1 / 4 chance...

and reguardless of the # of flips in a row they need, a billion or a trillion, if they got those many heads previous to the last flip, the last flip will ALWAYS be 50%...

I think you're saying what I'm saying, if so then forget this post =P

Title: Re: Easy: Two Coin flips
Post by Icarus on Nov 29^th, 2002, 11:08am

Unfortunately, we are not saying the same thing. My point is, you are misunderstanding the situation. The experiment was to bring to your attention where the misunderstanding was. It is true you can figure out the probabilities here by logic. But only if you start from the correct point. Pythagorus very logically deduced extremely stupid concepts about how the world worked, because he was believed it beneath him to ever check himself with experiments. The answer to the puzzle is 1/3. The reasons for this have been well-explained already in the thread, and the experiment will confirm the result. The problem with your logic is that you are examining a situation which is slightly different than the one in the puzzle. In particular, in relation to what you said in your last post, note that the TT case has been precluded because you are told that did not happen. But that is all you have been told. The other 3 cases are still equally possible, so 1/3 is probability of 2 heads.

Title: Re: Easy: Two Coin flips
Post by redPEPPER on Dec 16^th, 2002, 10:21am

The answers 1/2 and 1/3 are arguably both valid. It boils down to why (how) the person said that one of the coins was a head.

We flip two coins, there are 4 possible occurences: HH, HT, TH, TT.

We receive an additional information. Either this information means that one of the coins, arbitrary chosen (say, the first one), happens to be a head, then that removes TH and TT, and leave us with a 1/2 probability. Or the information means that both coins were checked to see if one (any one) of them is a head. It happens that one is. That only removes TT, and the probability is then 1/3.

After reading this thread I'm guessing that the original intent was to go for a 1/3 answer, but the riddle would certainly benefit from a more accurate wording. I liked Mark Schnitzius' idea of a question: "- Is at least one of them a head? -Yes."

Title: Re: Easy: Two Coin flips
Post by Zaius on Jan 31^st, 2003, 11:45pm

First we need to understand that the flipper kept flipping his coins until at least one coin came up heads, since having both coins come up tails doesn't work with this question. Thus, he can only come up with 3 possibilities, all equally likely:

TH
HT
HH

Second, we assume that the flipper really says: "At least one of the two coins came up heads, what is the chance that they are both heads?" If the flipper instead meant "One and only one of the two coins came up heads," then we would KNOW the other would be tails and the experiment would be invalid. In other words, I don't see how the flipper could be interpreted ANY other way that with my interpretation. There is no reason to think that the flipper was only looking at one coin before announcing that he saw a head.

In his list of possibilities, again all equally likely: TH, HT, HH, only one of them has two heads, and it only happens 1/3 of the time.

Therefore, yes it's 1/3.

*Humility Disclaimer*: I may be wrong

Title: Re: Easy: Two Coin flips
Post by Chronos on Feb 3^rd, 2003, 4:04pm

Quote:

First we need to understand that the flipper kept flipping his coins until at least one coin came up heads, since having both coins come up tails doesn't work with this question.

But there are other ways to handle this. We don't know what the flipper is doing. Maybe he flips two coins, picks one at random to look at, and then says either "At least one is heads" or "At least one is tails". In this case, the 50-50 answer is correct. Maybe he keeps on flipping until at least one is heads, reflipping all cases of TT, and then says "At least one is heads". In this case, there's a 1 in 3 chance that both are heads. Heck, maybe he flips them only once, looks at both coins, and says one of three things: Either "They both came up tails" for TT, "They both came up heads" for HH, and "At least one came up heads" iff it's HT or TH. In this case, givent hat he said that, there's no chance of HH.

Without knowing what he's doing, no meaningful answer is possible.

Title: Re: Easy: Two Coin flips
Post by Logan Manning on Mar 8^th, 2003, 12:50am

I learned probability a different way. the probability of the first coin landing heads is (1/2), and the second landing heads is also (1/2). therefore the chances of both coins landing heads is (1/2)*(1/2)= 1/4 or .25 . just a different perspective

Title: Re: Easy: Two Coin flips
Post by Icarus on Mar 8^th, 2003, 5:16am

Logan - There is nothing different about the way you are calculating probabilities than anyone else - except that you are ignoring some of the information you have.

Yes, if you flip two coins, the probability of two heads is .25. But in the puzzle there is a restriction on the outcomes, and that restriction changes the probabilities. In particular, for a straight flipping of two coins, the probability of 2 tails is also .25, but in this situation, the probability is 0. Since the probability of all events must add up to 1, the probability lost by the 2 tails situation must be picked up by the other outcomes.

Consider this: the probability of getting a 6 when rolling a die is 1/6. Now suppose that someone else rolls the die and tells you that the result was not 1. What is the probability of it being 6?
In the original case there are six possible outcomes, all equally likely, so the probability of each is 1/6. In the second case there are only 5 possible outcomes, again all equally likely, so the probability of each is 1/5.

The big disagreement in this thread is over how to interpret the puzzle. By one interpretation, one coin's result is completely decided, and the probability of the second being heads is 1/2. By the other (more common) interpretation, we don't know which of the two coins was refered to, and the three combinations other than two tails are all equally likely, so the probability of two heads is 1/3.

Title: Re: Easy: Two Coin flips
Post by william wu on Mar 9^th, 2003, 5:42am

on 03/08/03 at 05:16:35, Icarus wrote:

The big disagreement in this thread is over how to interpret the puzzle. By one interpretation, one coin's result is completely decided, and the probability of the second being heads is 1/2. By the other (more common) interpretation, we don't know which of the two coins was refered to, and the three combinations other than two tails are all equally likely, so the probability of two heads is 1/3.

Oy, sorry about this guys. The phrasing problem didn't come to my attention until someone e-mailed me about it recently. As the puzzle was stated, both 1/2 and 1/3 were equally defensible. I have reworded the puzzle as follows:

i flip a penny and a dime and hide the result from you. "at least one of the coins came up heads", i announce. what is the chance that both coins came up heads?

contrast this with the old phrasing:

i flip a penny and a dime and hide the result from you. "one of the coins came up heads", i announce. what is the chance that the other coin also came up heads?

Title: Re: Easy: Two Coin flips
Post by Chronos on Mar 11^th, 2003, 10:28pm

I'm afraid that that still doesn't fix the puzzle, Willy. That does remove the argument for 0 chance (if one interprets the statement to mean "exactly one came up heads", then there's no chance of two heads), but it does not remove the 1/2 - 1/3 ambiguity. We still don't know what the announcer's "system" is. The simplest way to fix the problem is to make the announcement the answer to a question. So: You flip, and look. I ask, "Is there at least one head?". You say "Yes".

In this case, the answer is 1 in 3 chance that both are heads.

Title: Re: Easy: Two Coin flips
Post by william wu on Mar 11^th, 2003, 11:18pm

Oh I see. I admit I didn't read this thread before posting; If looks like this problem suffers from the same flaws that my original Monty Hall phrasing suffered from -- when you do not know the motivations of the announcer / game show host, it's not clear what's the right answer.

How about if we just take the pesky humans out of the picture:

Two coins are flipped. Given that one of them came up heads, what's the probability that both came up heads?

What do you think?

Title: Re: Easy: Two Coin flips
Post by Drowko on Mar 13^th, 2003, 4:53pm

ok here is a guess

2 coins

1 came up Heads
which was a 3/4 chance
the second could be could be both heads and tails and had a 1/3 for chance.

so you muiltiply 1/3 * 3/4 and you get 3/12 which is 1/4

or
it could be 1 came up heads
1/2 chance
another is fliped which also haves a one 1/2 chance

The chance of having them both heads is 1/4 again

or
1 is heads.
3/4 chance of having this happen

2 is fliped
1/2chance of head

3/4 * 1/2 which is 3/8

there are a few ways

Title: Re: Easy: Two Coin flips
Post by Icarus on Mar 13^th, 2003, 5:32pm

??? Did you forget to take your pills today? ???

Title: Re: Easy: Two Coin flips
Post by rmsgrey on Apr 14^th, 2003, 9:06am

on 03/11/03 at 23:18:41, william wu wrote:

Still open to semantic quibbles. "Two coins are flipped. Given that at least one of them came up heads, what's the probability that both came up heads" works. Otherwise you've got the problem of the use of "one" to label as in: "I have two coins which total 6 cents. Given that one of them is not a dime, what is the other?"

Alternatively, "I flip two coins. 'is at least one of them heads?' you ask. 'Yes,' I reply. What is the probability of them both being heads?" keeps the human interest.

Title: Re: Easy: Two Coin flips
Post by James G on Sep 18^th, 2003, 9:20am

The answer's definately 33.3%

If we assume the tosser is telling the truth, then since we know that one coin came up heads, and want to know if the other one did too, there are 3 possible cases:
Either only one coin came up heads, so the tosser is talking about that coin, and the other one is tails,
Or both coins came up heads, and the tosser is talking about the penny, in which case the dime is also heads,
Or both coins came up heads, and the tosser is talking about the dime, in which case the penny is also heads,
So the probability would be 2/3.
But first there are the 2 possibilities that the tosser might either be telling the truth or lying,
If the tosser is lying, then neither came up heads,
If they are telling the truth, then the probability is 2/3,
Which makes the actual probability 0/2 + 2/3/2 = 1/3 or 33.3%

Gold star for me. ;-)

Title: Re: Easy: Two Coin flips
Post by James G on Sep 18^th, 2003, 10:26am

Actually, I think the old wording was slightly better, (and more interesting).

With the new wording, we could (if we want to be picky) not assume the tosser is truthful, in which case there is a 1/4 chance (that's easy).

With the old wording, it seems to me we actually do know the tosser is truthful (thanks to that "also" that someone pointed out).

If we had the old wording, and the "also" wasn't there, and if we don't assume the tosser is truthful, I think there's a 1/2 chance (the tosser is still talking about one of the coins, even though they're not necessiraly being truthful about it). But the "also" was there, so I'm not going to worry too much about this. (Besides, it's just too weird.)

If we do assume the tosser is truthful (one way or the other), then we must completely disregard their motivation, for exactly the same reason that we would completely disregard their statement if we didn't assume they were truthful, i.e. although we could guess at their motivation, it would be just that, only a guess. Hell, their procedure might be that they flip 2 coins, and if they both come up heads, they hit you in the face. If we assume they are telling the truth, then all we know for sure is that their statement is true, i.e. that one of the coins came up heads, and hence we can discard the TT case.

I think perhaps I haven't actually said anything new here, but hopefully I've put it all together in a more understandable way.

Title: Re: Easy: Two Coin flips
Post by James G on Sep 18^th, 2003, 10:40am

Whoops, sprinkle "at least" through-out my previous message, where appropriate, and all that. Sorry, it's 5:48am here...
:-[

Title: Re: Easy: Two Coin flips
Post by James G on Sep 18^th, 2003, 11:14am

Or perhaps that should be "there is a coin that is heads"?
??? Speaking of heads, mine hurts.
Right, I'm off to sleep.

Title: Re: Easy: Two Coin flips
Post by S. Owen on Sep 18^th, 2003, 11:24am

You know, I've come to the conclusion that this is actually a lousy riddle! It is almost entirely a question of interpreting the problem statement (or if you want to get tricky, deciding whether the flipper is even telling the truth!) -- once you decide what you think the statement is saying, then the probability is trivial.

I've heard this riddle repeated to non-math types as an example of how tricky probability can be, many of whom quite reasonably interpret the statement in one way and give the "wrong" answer of 50%, even though their probability skills are fine.

I think the Monty Hall riddle is a much better "cute" probability riddle for the uninitiated.

Title: Re: Easy: Two Coin flips
Post by Ouroborus on Jan 13^th, 2005, 1:30pm

'i flip a penny and a dime and hide the result from you. "at least one of the coins came up heads", i announce. what is the chance that both coins came up heads?'

The question:
- What is the probability that both coins came up heads?

The assumptions:
- There are two coins, different enough to distinguish them from each other, but otherwise behave the same.
- Each coin has two (and only two) sides. (Yes, a real coin has more than two sides and the other side(s) have infintismal chances of showing. Ignoring this is part of this assumption.)
- The probability of any specific side of any specific coin showing is 50%.
- The coin flipper is honest. (The flipper's statement is true and they're not going to cheat.)

Now, IANA Statistition, but this is how I understand the puzzle. The probability of any two coins both landing heads is 1/4 (one condition we are looking at out of four possible conditions) or 25%. It remains at this fraction until something further becomes known about the state of the coins. When it is announced that at least one of (same as: one or more of) the coins is indeed heads, the probility of both coins being heads suddenly changes. The condition in which both coins are tails is eliminated leaving only three possible conditions left. Since we're still only interested in one of these conditions the probability changed to 1/3 (one condition we're looking at out of three possible conditions) or approximately 33%.

It would only be 50% if you knew that a *specific* coin was heads. In that case, the known coin could be eliminated leaving you with 1/2 (one condition we're looking at out of two possible conditions).

In my opinion, this is the sort of mind bending logic that makes quantum physics so hard to grasp... At least for me...

Title: Re: Easy: Two Coin flips
Post by Grimbal on Jan 13^th, 2005, 3:09pm

on 01/13/05 at 13:30:27, Ouroborus wrote:

'i flip a penny and a dime and hide the result from you. "at least one of the coins came up heads", i announce. what is the chance that both coins came up heads?'

One assumption is that the probability of all 4 outcomes are equal. But when you claim that there is at least one heads, either you have a 1/4 chance of lying, or there is something that happened or would have happened that make the probability of having 2 tails zero. If you don't know what happened, you have no way to compute any probability. If you know, it becomes trivial.

Example.
1. whenever you get 2 tails, you replay.
=> one case is effectively removed, the probability is 1/3 to have 2 heads.
2. you just announce one of the coins without preference, you could as well have announced there is at least one tails.
=> There are 8 cases (2 coins outcome and choice of the coin). Half of these match the situation where you announce one heads, and 50% of this half has 2 heads.
3. you always announce one heads, unless there are 2 tails, then you announce one tails.
=> here, among the 3 cases where you announce one heads, 1 has two heads. That makes 1/3 probability.
4. you always announce one tails, unless there are 2 heads, then you announce one heads.
=> 100% of the cases where you announce heads are 2 heads.
5. you pick the biggest coin and announce that one.
=> there are 4 cases, but 2 only where you announce heads. It is true that 3 cases have at least one heads, but in one of them you annouce tails. So in the end, 2 cases remain where 50% have 2 heads.
6. you always say there is at least one heads regardless of the outcomes.
=> the probability of having 2 heads is only 25%.
7. you lie. When you announce at least one heads, there are none and vice-versa.
=> the probability of having 2 heads is zero.

Title: Re: Easy: Two Coin flips
Post by Niko Popovic on Feb 21^st, 2005, 8:49pm

First of all I think it's amazing how we are able to spend this much time analyzing the problem.

Here's my view: The statement 'at least one of the coins is heads' has two possibilities. Either the nickle is heads and we have a 50% chance for the dime to be heads OR the dime is heads and we have a 50% chance for the nickle to be heads.
0.5*0.5 + 0.5*0.5 = 0.5 = 50% as my final answer.

Any comments?

Title: Re: Easy: Two Coin flips
Post by towr on Feb 21^st, 2005, 11:18pm

on 02/21/05 at 20:49:11, Niko Popovic wrote:

Any comments?

You can't simply add the probabilities, since they are not independant.
(HH is in both sets)

Title: Re: Easy: Two Coin flips
Post by Niko on Feb 22^nd, 2005, 2:31pm

The statement 'at least one of the coins is heads' presents a STATE one of the coins is IN. All you need is the probability of the other coin to be heads which is obviously 50%.
I think it is not the case that there are 4 potential situations, but 3:
-tails/tails
-heads/tails no matter which is which
-heads/heads

Seems so easy, but this one is worth thinking about, I'm still not sure about it. Somebody prove me wrong...

Title: Re: Easy: Two Coin flips
Post by towr on Feb 23^rd, 2005, 12:56am

but tthe state with HT is twice as likely to occur.
And also you're definitely not in state TT, but in either of the other two (in which at least one is heads).

Title: Re: Easy: Two Coin flips
Post by Niko on Feb 23^rd, 2005, 11:58am

I see my mistake, it is 33.33%.

Title: Re: Easy: Two Coin flips
Post by baH on Dec 27^th, 2005, 5:10am

Very old topic, no posts in a very long time, but I stumbled upon this page, and the answer is flawed to this riddle
1. data from the perl program shows that in a true life situation it is 50%
2. when you eliminate the tt option, then the model changes

penny first:::::::::dime first
hh::::::::::::::::::::hh
ht::::::::::::::::::::ht
th::::::::::::::::::::th
tt eliminated:::::::::tt eliminated

the odds that the penny is heads is 2/3
the odds that the penny is tails is 1/3
the odds of the dime being heads is 2/3
the odds of the dime being tails is 1/3
if the penny is tails, then the odds that the dime is tails is 0 and vise versa
given this a new chart is needed
h h t
h hh hh ht
h hh hh ht
t th th tt<<impossible

4 hh 2 ht 2 th, 50% chance
easy as genetics :P

Title: Re: Easy: Two Coin flips
Post by towr on Dec 27^th, 2005, 5:34am

I don't get how you get 50%
from 4 hh 2 ht 2 th, you have 8 h and 4 t, so it's 2/3 and 1/3

Title: Re: Easy: Two Coin flips
Post by BaH on Dec 27^th, 2005, 2:28pm

public class hh {
/**
* throw 2 coins, eliminate the tail tail combo,
* and calculate the chance of a remaining
* head head combo
*/
public static void main(String[] args) {
int total=0;
int aa=0;
int all=0;
int r1=10;
int r2=10;
for (int i=0;i<10000;i++){
r1 = (int) (Math.random()*2);
r2 = (int) (Math.random()*2);
System.out.println(r1 + " " + r2);
if (r1+r2!=0){
total++;
if (r1+r2==2) {
aa++;
}
}
all++;
}
System.out.println(all +" coin tosses " + aa + " hh " + total + " valid flips = " + (100*aa/total) + "%");
}
}

well, the numbers don't lie, I thought the previous entry with the python code was correct, but I didn't check it exactly.. this is a java program I jammed together real quick to check, and I was wrong

10000 coin tosses 2476 hh 7471 valid flips = 33%

ph - dh yes
\ dt no

pt - dh no
\ dt <-disallowed

dh - ph yes
\ pt no

dt - ph no
\ pt <- disallowed

75% of all the coin tosses generate a condition where at least one is heads, but of those, only one third generates 2 heads.

This problem last night gave me such a headache, much clearer to see when the coins are actually tossed

in case anyone who reads this is interested on how to execute java on your comp, first dl a jdk and install it,make sure that the jdk bin is in your path, then copy and paste the code above into a file named hh.java then type javac hh.java then type java hh

Title: Re: Easy: Two Coin flips
Post by notariddle on Feb 17^th, 2010, 9:13am

I'm 8 years too late to the party but oh well:

Nobody has pointed out where anshil's original python program went wrong. It was in this line:

Code:

print "t=", t, " f=", f, " ratio=", 1.0 * t / f

It should be:

Code:

print "t=", t, " f=", f, " ratio=", 1.0 * t / (t + f)

His experimental result of 50% implies that in his original code f = 2t. Thus with the corrected code of t / (t + f) we can deduce that the correct result must have been t / (t + f) = t / (t + 2t) = 1/3, or 33%.

An experiment is only any good if you design it correctly :)

That being said I really don't understand people's concern with the tosser's "algorithm". The riddle presents a particular scenario and that's all you should be reasoning from. It's not the same as the Monty Hall problem because that problem specifically presents a different scenario, one where the host explains an algorithm. This scenario doesn't present an algorithm so you can only take it at face value, and in that case the specific answer for the specific scenario is 33%.

Title: Re: Easy: Two Coin flips
Post by tanmaya on Feb 27^th, 2010, 11:35am

well, the problem can be simply thought like this...

the possible cases are 3 where we get at least one head ( HH, HT, TH)
out of these, the favorable case is where we get both heads which is one in number.
so the required probability is 1/3

Title: Re: Easy: Two Coin flips
Post by rmsgrey on Mar 1^st, 2010, 8:04am

on 02/27/10 at 11:35:39, tanmaya wrote:

It depends whether each of those three cases is equally likely...

If they are, then if you are told instead that "there is at least one tail", the probability that they are both tails is 1 - if it were HT or TH, you'd have been told "there is at least one head" instead...

Title: Re: Easy: Two Coin flips
Post by tanmaya on Mar 1^st, 2010, 8:24am

I didn't actually understand what you mean. Can you please elaborate. And are you saying that what i have proposed is wrong?

Title: Re: Easy: Two Coin flips
Post by rmsgrey on Mar 1^st, 2010, 9:04am

on 03/01/10 at 08:24:32, tanmaya wrote:

I didn't actually understand what you mean. Can you please elaborate. And are you saying that what i have proposed is wrong?

What happens if I flip the same two coins, look at them, and then tell you "at least one of the coins came up tails"?

If you apply the same logic as to the original question, then there are again three possibilities, TH, HT and TT, one of which has both coins T.

The trouble is, if you assume I always say either that at least one came up heads, or that at least one came up tails, then there are 6 possibilities:
HH (at least one H)
HT (at least one H)
HT (at least one T)
TH (at least one H)
TH (at least one T)
TT (at least one T)

Since HH, HT, TH and TT are all equally probable, the six cases can't all be equally probable - if HH, HT("H") and TH("H") are equally probable, then HT("T") and TH("T") each have to have probability zero, so, of the three cases where I say there's at least one T, the only possibility is TT...

[edit]

What you said is the originally intended answer, but it's not the only sensible answer...

Title: Re: Easy: Two Coin flips
Post by Grimbal on Mar 1^st, 2010, 9:45am

If you want to give a probability, you need to know the whole story.

Obviously, you need to know exactly how the coins behave. If they are not balanced, you need to know the probability of each side. Or at least, you need to know what is the probability of each possible imbalance. If you don't have that, you can't give a probability for any outcome.

The same applies to what the thrower announces. There are 4 outcomes. HH, HT, TH, TT. If the thrower announces "there is at least on H", you need to know what was the probability of that particular declaration for each outcome. If you don't know that, again you can't give a probability for the outcome HH. Depending on how the thrower decides what to declare, the probability of having HH can be anything from 0 (he only says there is a H if there is only one) to 1/3 (he always says there is a H if there one) or 1/2 (he chooses a coin randomly and gives its state) or 1 (he only says there is a H if there are two).

In general, to give a final probability in any situation, you need to know exactly the whole process, what could have happened, what where the chances that what happened did happen, and why the things that did not happen didn't.

The problem is generally worded in a way loose enough to allow different interpretations, and different assumptions of how the whole process works. So it is likely that people who argue are both right, except that they have slightly different assumptions.

Title: Re: Easy: Two Coin flips
Post by littstles on Mar 4^th, 2010, 9:54pm

So is it 50% or 33% or something else? I am more confused now. I didn't read every reply since there are too many of them but I read the whole first page and didn't get a solid answer.

My first guess would be 50% but I can see how it would be 33% and I am not smart enough at things like that to know for sure.

Title: Re: Easy: Two Coin flips
Post by Grimbal on Mar 5^th, 2010, 12:32am

As I explained, it depends on the assumptions on how you came to the situation described.

What would have happened if the flip resulted in 2 tails? And if the result is a heads and a tails, would it always be "one heads" that were announced, or sometimes "one tails"? The answers to these questions are necessary to give the correct probability.