13
$\begingroup$

I have a doubt because I think that once the result of the first coin is obtained, just simply await the outcome of the second, which is completely independent of the previous one, and then we have a chance of $\frac12$ to get a head again.

But someone else tells me that as the possible events are:

  • Head - Head
  • Head - Tail
  • Tail - Head
  • Tail - Tail

then when we get a head we restrict ourselves to the first three cases, so the probability would be $\frac13$.

What is the right way?


I know there's a difference between saying "first came head" to say "one of the two came head", but if we have the first fact, aren't we supposed to know which one is that came head?

$\endgroup$
  • $\begingroup$ His (or her) way........ $\endgroup$ – barak manos May 30 '16 at 4:47
  • 1
    $\begingroup$ 1/3 is the right ans. since once you get head on the first, your event becomes conditional. $\endgroup$ – chandresh May 30 '16 at 4:53
  • 17
    $\begingroup$ I am not happy with the wording "one gets head" since it can be interpreted in several ways. If it said "at least one gets head," then the calculation that leads to $1/3$ would be the only reasonable one. $\endgroup$ – André Nicolas May 30 '16 at 4:57
  • 4
    $\begingroup$ Closely related is this question. $\endgroup$ – drhab May 30 '16 at 7:36
  • 2
    $\begingroup$ Um, "at least one gets head" can also be "interpreted multiple ways", if you know what I mean... Usually one says of a coin that it "comes up heads". $\endgroup$ – Rahul May 31 '16 at 3:09
23
$\begingroup$

I think this is one of the cases where logic/mathematics totally goes overboard on a trivial problem, and you (as well as your friend) are overthinking it. But it's fun anyway, so...

There are at least 3 possible answers which are equally correct, depending on how pedantically one tries to twist the wording one way or the other. But for practical purposes it does not make a lot of sense because there is only exactly one solution that is correct (and immediately obvious) in each situation.

The first obvious interpretation is that you toss two coins into the air at the same time (which is not what you describe!). There are 4 possible outcomes. One of these outcomes has two heads, and one has no heads at all. You have set up the precondition that one coin gets head, which rules out the "no heads" outcome, leaving 3 possible outcomes. Only one of the three has two heads in it, thus: 1/3. This is a dependent probability. It is also an example of a Monty Hall Problem.

The second obvious interpretation is you toss one coin, and it comes up head. That's the precondition. You could just as well not have tossed the first coin at all. You now toss the second coin. Alternatively, you can toss the two coins together, but ignore all cases where the precondition that the first one gets head isn't fulfilled.
Assuming the second coin is not weighted or a trick coin with two heads or such, the chance is, of course, 1/2. From the point of view of the second coin, the first coin doesn't exist at all. This is a single (independent) probability.

The third obvious solution is zero. If two coins are flipped and exactly one coin gets head, the probability of both coins getting head is zero. This is a smart ass probability.

$\endgroup$
  • $\begingroup$ If one coin lands the number of remaining outcomes becomes 2, not 3. $\endgroup$ – Euri Pinhollow May 31 '16 at 5:24
  • $\begingroup$ Regarding third point: if you claim that something has already happened you'd better not say that it has probability '1'. Probability is a mathematical concept. $\endgroup$ – Euri Pinhollow May 31 '16 at 5:28
  • 2
    $\begingroup$ @EuriPinhollow: It is absolutely correct to say that something that has happened has probability $1$. You can mathematically define any probability space you like that has that event as an element, and you will then find that for the probability space to match what you know about the real world you need that event to be assigned probability one. $\endgroup$ – user21820 May 31 '16 at 6:29
  • $\begingroup$ Just for the record, I consider the third possibility a very remote one for a native English speaker. I believe most of them would not think of it on hearing or reading the question. But certainly such ambiguity is much more present for larger numbers, such as "If ten coins are flipped and five come up heads"... $\endgroup$ – user21820 May 31 '16 at 10:38
38
$\begingroup$

The problem here is that your phrase "one gets head" is not precise enough.

If it means "at least one of the coins comes up heads", then indeed there are 3 equally likely possibilities (HT,TH,HH) out of which exactly 1 has both coming up heads. This means that the desired probability is $\frac13$.

If it means "out of the two coins A,B that were flipped, A comes up heads", then B is equally likely to come up heads or tails, and so the desired probability is $\frac12$.

$\endgroup$
  • $\begingroup$ I don't understand how the phrase "one gets head" is not precise at all. If it meant "at least one of the coins comes up heads" it would have been written "at least one of the coins comes up heads". No to say that if at least one of the coins comes up head it's still one granted head. Not to say that nowhere in the question is made mention of a wanted order of outcome, so "HT" and "TH" are the same output. $\endgroup$ – motoDrizzt May 30 '16 at 10:48
  • 13
    $\begingroup$ @motoDrizzt: My answer already shows the two possible interpretations. I'm a native English speaker, and can affirm that other native speakers will see both interpretations and probably ask for clarification. Secondly, the question is about probability, and so whether or not you treat HT and TH as the same output does not change the probability of "exactly one head coming up". $\endgroup$ – user21820 May 30 '16 at 11:11
  • $\begingroup$ @motoDrizzt: See also the answer at math.stackexchange.com/a/1801261/21820 explaining pretty much the same thing, in more words. $\endgroup$ – user21820 May 30 '16 at 11:17
  • $\begingroup$ When simple probabilty questions like these are posted, 99% of the time any disagreement is about imprecise wording. Worst when "a very famous mathematician" said something that isn't exactly what is claimed they said and therefore becomes totally incorrect. $\endgroup$ – gnasher729 May 30 '16 at 16:32
  • 9
    $\begingroup$ And if it means "out of the two coins, exactly one was heads" then the answer is 0 :) $\endgroup$ – hobbs May 30 '16 at 19:05
4
$\begingroup$

It's called conditional probability:

Let $A$ denote the event of getting exactly $2$ heads.

Let $B$ denote the event of getting at least $1$ head.

Then the probability that $A$ will occur given that $B$ has occurred is:

$$\frac{P(A\cap B)}{P(B)}=\frac{1/4}{3/4}=\frac13$$


EDIT:

Please note that in this specific case, $A\cap B=A$.

Therefore, one could simply calculate $\frac{P(A)}{P(B)}$.

However, this is not always the case.

For example, consider the following question:

A fair $6$-sided die shows a number larger than $3$.

What is the probability that this number is even?

Let $A$ denote the event of the die showing an even number.

Let $B$ denote the event of the die showing a number larger than $3$.

In this example:

  • $A$ denotes the event of the die showing $2$ or $4$ or $6$
  • $B$ denotes the event of the die showing $4$ or $5$ or $6$
  • $A\cap B$ denotes the event of the die showing $4$ or $6$
$\endgroup$
4
$\begingroup$

Think of it this way:

If you toss the first coin until it comes head and only then toss the second, the probability of the second being head is 1/2. Because it is independent of the first coin.

Now, toss the first coin again and no matter the result, toss the second coin. Obviously, there is a difference. You do not know beforehand that you will get a head at all. Out of four equally likely events, you are interested in three, which contains a head. And out of these 3, only one contains two heads.

$\endgroup$
2
$\begingroup$

Your doubt is valid because it hinges on how the "one gets heads" report is being generated.   What are the possible alternative responses?

Is it a choice between "one gets heads", and "neither gets heads", or is "both get heads" also a possibility?

Or is it a choice between "one gets heads", and "one gets tails", with the reporter choosing between them when the coins show different faces?   Can you guarantee that when given such a choice, the reporter makes an unbiased call?

$\endgroup$
2
$\begingroup$

Yes the reasoning is correct.

You can solve the problem by assuming that the coins are distinguishable, and you get the following counts:

$$\begin{align}HH&: 1,\\ HT&: 1,\\ TH&: 1,\\ TT&:1.\end{align}$$ This is a uniform distribution. Then the requested probability is

$$\frac{HH}{HH+HT+TH}=\frac13.$$

Or you can consider that they are undistinguishable, and just working with the number of heads

$$\begin{align}HH&:1,\\ H&: 2,\\ -&: 1.\end{align}$$ This is a binomial distribution. The probability is still

$$\frac{HH}{HH+H}=\frac13.$$

$\endgroup$
0
$\begingroup$

You are totally right.

The premise of your question is "If two coins are flipped and one gets head": this means that on the two coins is guaranteed to be head, there is no doubt about it. Then it doesn't matter if we launch them at the same time or one after another or if we give them an order, a name, or if we toss coins from two different countries, there is no chance to get the "4) Tail - Tail" option.

$\endgroup$
0
$\begingroup$

The probability of a coin getting Head or Tail is not influenced by how many other coins were tossed.

The probability of your coin getting a Head is always 1/2.

When you look at the probability of getting Heads on 2 or more different coins, then the other logic comes in.

$\endgroup$
0
$\begingroup$

I have now thought about this a bit more, and I think the way we get the information "one gets head" is important.

In the question, I put that the possible cases were:

  • Head - Head
  • Head - Tail
  • Tail - Head
  • Tail - Tail

Each tends to happen $25$% of the time.

If we see the first to fall comes up head, then we restrict to cases $1$ and $2$, and so the probabilities are $1/2$. Only in $50$% of the time we can see the first one coming up head, and in its half both coins end having head.

If the information was gotten from another person who sees the outcomes while you have not, and that person is forced to tell you that one coin came up head always that it happens, then that person will do it $75$% of the time, of which its $1/3$ is when both coins get head. So, once we know we are in those $75$%, we have $1/3$ probability.

If that another person was free to say you either "one coin came up head" or "one coin came up tail", then the probabilities are $1/2$ for both coins coming up head. The reason is that in the two cases we have got a head and a tail, that person has $1/2$ probability to report "one coin came up head".

$1)$ $25$% of the time we get two heads, and in all this cases that person tells you that one coin came up head.

$2)$ $50$% of the time we get a head and a tail, but only in half of its cases ($25$%), he tells you that one of them came up head. In the other half, he tells you that one of them came up tail.

$3)$ $25$% of the time we get two tails, and in all this cases that person tells you that one coin came up tail.

So, only in $25$% $+ 25$% $= 50$% he says you that one coin came up head, of which its half is when you have both coins head.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.