Probability of coin flip

[StopSign]

This question was brought up in class back in high school and I'm still thinking about it. The question is:

If you flip two coins and know the outcome of one of them, what is the probability both coins landed on the same side?

My teacher argued that it's 1/3 because the possibilities are:
Known/Unknown
Unknown/Known
Known/Known

I argued that it's 1/2 because if you know the outcome of one of them, then the chance that the other one is the same is 1/2. It shouldn't matter which coin you know because it doesn't affect the result of the other. Example: I flip two coins and one is H. What's the probability that the other coin also lands on H?

...so who's right here?

 

[chcarnage]

4 cases of equal probability:

1) You flip the allways-head coin first: Of course, you get a head
a) The normal coin flip results in head, too
b) The normal coin flip results in tails

2) You flip the normal coin first:
a) It's a head. The always-head coin is another head
b) You get tails. The always-head coin is a head

If you first have a head, it's twice as probable that the second coin is a head too than that it is a tails (1a, 2a vs 1b, thus 67%)

But the probability is 50% if you don't know if the manipulated coin always is head or tails.

 

[BrownTown]

The answer is 1/3 since if you know that for example there is one head than it cannot be the case that there are 2 tails, so there are only 3 possible outcomes of which only one has them on the same said (where both are heads). That should be the end of the thread.

Having said that, since this is AT is likely a bunch of people will try to argue semantics and say its 1/2, however they (and you) are not answering the question as it is stated and meant. The answer is supposed to be 1/3, however there is always some ambiguity in the question which will result in some people misreading it, and at AT people are generally to stubborn to admit they are wrong (see .999... =1 and airplane treadmill). I would make the suggestion concerning mathmetical problems/riddles that they are gennerally designed so that the "obvious" answer is in fact not the correct one since this allows the teacher to see if you really understand the concepts, or are simply making the obvious guess.

 

[TuxDave]

Originally posted by: chcarnage
4 cases of equal probability:

1) You flip the allways-head coin first: Of course, you get a head
a) The normal coin flip results in head, too
b) The normal coin flip results in tails

2) You flip the normal coin first:
a) It's a head. The always-head coin is another head
b) You get tails. The always-head coin is a head

If you first have a head, it's twice as probable that the second coin is a head too than that it is a tails (1a, 2a vs 1b, thus 67%)

But the probability is 50% if you don't know if the manipulated coin always is head or tails.

I don't think you're supposed to deal with 1 specific coin that always comes up heads. Just flip the 2 coins and ideally you you should get 25% HH, 25% TT, 50% HT. Say I say given that one of them is heads, so you throw away the 25% TT results, what's the probability that the other is heads?

You end up with 1/3rd.

 

[Mark R]

The answer is 1/3. You know the result of one of the tosses - this could be either the first, or the second.

There are 4 possible combinations - HH, HT, TH and TT. However, if you are told that there is an H, then the TT can no longer apply. As each of the remaining probabilities are equal, they must equal 1/3 each.

This can be expressed in terms of Bayes' theorem:

Prob (Getting 2 heads given you know you have 1 head) = Prob (getting 1 head when you know you have 2) x Prob (2 heads) / Prob (at least 1 head)

P (2 heads given 1 head) = 1 [if you've got 2 heads, you've got 1 as well] x 1/4 / 3/4 [3 possibilities]
= 1/4 / 3/4 = 1/3

 

[Aluvus]

The problem is that "know the outcome of one of them" is an ambiguous statement.

It could mean, as OP describes, that you flip one coin, know that result, and then flip another. The probability of the second flip matching the first one is 50%.

It could also mean, as some other posts have assumed, that two coins are flipped simultaneously and you know that at least one of them came up with a particular result. This means you can eliminate the case where both coins come up with the opposite result. Of the remaining possible outcomes, one of the three (each of which is equally probable if you assume differentiable coins) indicates the coins match. So you have a 1/3 probability that they match.

My personal opinion is that the first interpretation is more significant, and the second one is mainly useful as an "oh that is kind of neat" exercise.

 

[CycloWizard]

Originally posted by: Aluvus
The problem is that "know the outcome of one of them" is an ambiguous statement.

I think this is the problem with most riddles that are posted here. One can't solve the problem if one doesn't know what the problem is in the first place.

 

[BrownTown]

No, the problem is that people solve it wrong and then try to cover it up be saying the question is worded wrong instead of admitting that they were wrong.

 

[Aluvus]

Originally posted by: BrownTown
No, the problem is that people solve it wrong and then try to cover it up be saying the question is worded wrong instead of admitting that they were wrong.

Well at least you didn't get all antagonistic about it.

 

[CycloWizard]

Originally posted by: BrownTown
No, the problem is that people solve it wrong and then try to cover it up be saying the question is worded wrong instead of admitting that they were wrong.

No, the problem is that the question is phrased poorly. The original question is demonstrably ambiguous (per Aluvus' post). Someone needs to feed you a slice of humble pie before you put an eye out swinging your e-penis around like a weapon.

 

[chcarnage]

Also, the probability that the thread starter has left this topic for good is about 0.9

 

[PowerEngineer]

Originally posted by: Mark R
The answer is 1/3. You know the result of one of the tosses - this could be either the first, or the second.

There are 4 possible combinations - HH, HT, TH and TT. However, if you are told that there is an H, then the TT can no longer apply. As each of the remaining probabilities are equal, they must equal 1/3 each.

This can be expressed in terms of Bayes' theorem:

Prob (Getting 2 heads given you know you have 1 head) = Prob (getting 1 head when you know you have 2) x Prob (2 heads) / Prob (at least 1 head)

P (2 heads given 1 head) = 1 [if you've got 2 heads, you've got 1 as well] x 1/4 / 3/4 [3 possibilities]
= 1/4 / 3/4 = 1/3

Wait a minute... If you are told that the first coin flip is "heads", then both the TT and TH combinations are taken off the table. You only have the HT and HH combinations left, and therefore the chances for a HH outcome must be 1/2.

Or so it seems to me...

Edit: Aha... I should read more carefully. It is very much like the old Monty Hall problem if it isn't necessarily the first coin flip that you know the results of. My mistake,

 

[CycloWizard]

Originally posted by: PowerEngineer
Wait a minute... If you are told that the first coin flip is "heads", then both the TT and TH combinations are taken off the table. You only have the HT and HH combinations left, and therefore the chances for a HH outcome must be 1/2.

Or so it seems to me...

I think the teacher's argument is that you don't know the first one or the second one - you only know that one of those two is heads, but not which one.

 

[PhatoseAlpha]

Depends entirely on what "know the outcome of one of them means".
Does it mean that you know:
(Coin A is heads)
(Coin B is heads)
or
or does it mean you know:
Coin (A or B) is heads.

Not reality a probability question at all, but a semantic one.


Edit: Ah, I see. It's the old Monty Haul Problem.
He's going by the last interpretation.
Thing to remember here is that there is a non-random actor at work here. For you to
know that (A or B) is heads means that the coins have both already been flipped, and are already decided. Then you're being given partial information. The 50/50 you would expect in a genuinely random situation isn't there, because it's not truly random anymore.

 

[StopSign]

Originally posted by: chcarnage
Also, the probability that the thread starter has left this topic for good is about 0.9

Huh? Who said I left? Not everyone stays and posts on this forum 24/7, especially the HT board.

You guys bring up good points. I guess I was also caught up in the wording of the question. Like I said, this was a while ago so I don't remember the exact wording of the question. Based on how I think I interpreted it originally, I think the coins were not unique. I believe I was thinking that when you flip the coins, they're either going to be the same or not the same.

I think it makes sense to me that it is 1/3 because "same or not the same" is different from "knowing one is heads or tails."

 

[chcarnage]

Sorry StopSign, this wasn't meant as a personal attack. Sometimes people drop questions in which they aren't that interested. Other topics go horribly off topic and the OPs leave it, tearing at their hair. My remark wasn't appropriate in ATHT.