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There's are many people saying either one of two answers for this:

If you flip a coin twice, what's the chance that both will be heads given 
that at least one will be heads?

The more "educated" answer is that the three possible outcomes are H/H, H/T, T/H so the chance is 33.33% The more obvious answer is that since one is heads, you should only count it as a 1 coin toss so it's a 50% chance.

I think 50% is correct even though the above explanation is wrong.

So, one of the coin flips is a guaranteed heads flip..

Case 1: The first flip is is the guaranteed heads: Outcomes: H/T, H/H

Case 2: The second flip is the guaranteed heads: Outcomes: T/H, H/H

H/H should be counted as 2 outcomes, like T/H and H/T are. They're both the same in quantity (i.e T/H, H/T is 1 heads 1 tails, H/H, H/H is 2 heads)

The thing that differs is the ORDER, in which case (guaranteed)H/H and H/(guaranteed)H should be counted as two different outcomes, therefore making it 50%. Am I wrong here?

Joe Defill
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  • The argument is not right. It is certainly true, however, that given the first is (guaranteed to be) head then the probability of $2$ heads is $1/2$, and the same holds if the roles of first and second are interchanged. If you doubt the $1/3$, toss a coin twice $10000$ times. Let $A$ be the number of situations where we got at least one head, and $T$ be the number where we got two. Calculate $A/T$. The experiment is not much fun to do by hand, but we can do a computer simulation. – André Nicolas May 29 '15 at 18:18
  • Nevermind, deleted – Joe Defill May 29 '15 at 18:23

4 Answers4

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Let $E_{HH}$ be the event that both coin tosses are heads, $E_{HT}$ (resp. $E_{TH}$) resp. that the first is Heads and the second Tails (resp. the converse). Letting $E$ be the event that at least one coin flip is Heads, you get exactly $$E=E_{HH}\cup E_{HT}\cup E_{TH}$$ and this is a disjoint union.

You want to compute $\Pr[E_{HH}\mid E]$, which can be written $$ \Pr[E_{HH}\mid E] = \frac{\Pr[E_{HH}\cap E]}{\Pr[E]} = \frac{\Pr[E_{HH}]}{\Pr[E_{HH}]+\Pr[E_{HT}]+\Pr[E_{TH}]} $$ as $E_{HH}\cap E = E_{HH}$. As the coins are fair and independent, this is $\Pr[E_{HH}\mid E] = \frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}} = \frac{1}{3}.$

Clement C.
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  • I know why people people say 1/3, I know your reasoning. But I'm saying why use this equation on an answer that says "at least one will be heads" which gives us 2 cases (either first flip is heads, or the second flip is heads). Where each case has two outcomes. There are 4 outcomes, not 3 – Joe Defill May 29 '15 at 18:19
  • But if you do that, you are double-counting one event -- the one where both are Heads. I.e., your two events are not disjoint. – Clement C. May 29 '15 at 18:20
  • But it's not the same event, because a guaranteed heads is different from heads that you acquire from a 50/50 chance. The quantity of heads in each is the same, like T/H and H/T but the order of where the guaranteed heads is is different (in both) – Joe Defill May 29 '15 at 18:27
  • It's not quite true, although it may seem intuitive. Try to write down fully and rigorously your argument using e.g. conditional probabilities, and you will see something does not go through (basically, again, it boils down to counting separately the probabilities of events that are not disjoint), (As a followup comment: either you think something is wrong with your argument, or that something is wrong with mine, as both cannot be correct at the same time. I'm pointing out a flaw in your reasoning -- if you discard this, then it's your turn to explain what is wrong in the argument above.) – Clement C. May 29 '15 at 18:29
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Your analysis is correct, but for a different question: Suppose you flip two coins. You pick one of the coins at random, and examine it. It proves to be heads. The probability that the other coin is heads also is $1/2$, for the reasons you provide.

If you represent the two coin flips as binary variables $A$ and $B$, with heads being $1$ and tails being $0$, then the difference between the original question, and the question you really answered, is that the original question is asking for the probability

$$ P(A \land B \mid A \lor B) = \frac{1}{3} $$

where $\land$ and $\lor$ are logical AND and OR, respectively, while the question you actually answered asks for the expression

$$ P(B \mid A) \cdot P(A) + P(A \mid B) \cdot P(B) = \frac{1}{2} $$

Brian Tung
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I think, in your case, order counts: $H, T \ne T,H$ since they are two different events, and they have their own possibilities. Task only says that one $H$ is guaranteed, but doesn't say anything about, when is it guaranteed, therefore the answer is indeed $33,3$%.

Atvin
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  • That's what I said in my last sentence. (guaranteed)H/H and H/(guaranteed)H are two different H/H outcomes and therefore should be counted as 2 just like T/H and H/T. Making it 50% – Joe Defill May 29 '15 at 18:09
  • But you can't differ $H,H$ and $H,H$, since you can't decide which is the guaranteed one. – Atvin May 29 '15 at 18:10
  • One of them has the first heads as guaranteed, the other has the second heads as guaranteed. Regardless of which is which. We can't see it with our eyes obviously, but I'm talking about logically – Joe Defill May 29 '15 at 18:13
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The point is that there are more ways to get exactly one head if you have two coins and don't assume anything, compared to the number of ways to get exactly one head if you assume that the first coin (or second coin) is heads or tails. This is in essence why the probability is indeed $1/3$ and not $1/2$.

If you want a rigorous argument, just take the 4 equally likely and disjoint possibilities $TT, HT, TH, HH$. The law of conditional probability says that the probability of getting $HH$, assuming you have at least 1 head, is $P(HH)/(P(TH) + P(HT) + P(HH))$. So it's 1/3.

user2566092
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  • My argument is that there's more than one way to get heads / heads. Whether the first or the second is the guaranteed heads means that there are TWO orders of H/H. – Joe Defill May 29 '15 at 18:12
  • @JoeDefill See update giving conditional probability analysis. You can look up the event-based law of conditional probability to verify this. – user2566092 May 29 '15 at 18:17
  • I know, I gave the same simplified explanation in my post. My point is why is H/H not two different events? It has the guaranteed heads in a different position in both. – Joe Defill May 29 '15 at 18:29