# Three coins are tossed. If one of them shows a tails, what is the probability that all three coins show tails? [duplicate]

To solve the following problem

Three coins are tossed. If one of them shows a tails, what is the probability that all three coins show tails?

I tried $$1\cdot\frac12\cdot\frac12$$ where $$1$$ is the probability for the first coin that shows tails, and $$\frac12$$ is the probability for the other two coins that can be heads or tails.

This answer does not match the expected one. Where am I wrong?

## marked as duplicate by quid♦Jul 6 at 13:45

• A priori all $8$ possible outcomes are equally likely. Your condition rules out one of them. Therefore... – lulu Aug 21 '18 at 13:21
• Is this a statistics question: e.g. all coins are crooked in the same way to give tails with some as yet unknown probability $p$ and the fact that the first lands on tails gives you some information about what $p$ is? Or is this a mathematics question where it is given on forehand that all three coins are fair? – Vincent Aug 21 '18 at 13:21
• Rather like the "two sons/daughters" problem, the answer depends on how you find out one coin shows tails: do you see the coin, or something else – Henry Aug 21 '18 at 13:21
• In other words, it all depends on whether you know which coin is tails. – Kyle Delaney Aug 23 '18 at 22:45

The contrast between your interpretation and that in the other answers points out that there is an ambiguity in the wording of the question, in particular in the use of "one". You interpret it as "the first". The other answerers interpret it as "at least one". Someone might interpret it as "exactly one" in which case the answer would be 0.

To elaborate on Henry's comment: suppose the three coins are tossed but I have placed cups over them so you cannot see them.

• If you are allowed to pick one cup and lift it, and after doing that you see the coin under that cup was tails, we are in your interpretation of the problem.
• If I am allowed to look under the cups (but you aren't) and I tell you afterwards: 'at least one coin shows tails' then we are in the other posters answer to the problem.
• If you pick a cup but are not allowed to look under it, and I (knowing the outcome of all three coins) lift a different cup, showing that the coin under that one is tails (and offer you to change cups even if that is irrelevant for the current question, but hey, perhaps we are in a game-show, who knows?), then you can only compute the probability of three tails under some assumptions about what is going on in my head (e.g. would I always show you heads if I had the chance?).

The important thing is that all three scenario's are consistent with the given that 'one of them shows a tails'. So in other words: the answer depends on how you interpret this given and you should go complain to the person who gave you this riddle.

• Amazing. Thanks. – user3508140 Aug 21 '18 at 14:40
• Why isn't it "25%"? (Since -- after knowing the value of one coin -- we're left with only two unknown coins, and the odds of two coins being tails is 25%.) – RonJohn Aug 21 '18 at 15:50
• That's just the point, @RonJohn: "knowing the value of one coin" is the first alternative presented in this answer, but not the other two. In the second alternative, you know only that there is at least one tails, which is different. The third can be equivalent to either of the first two, or to something else, depending, as the answer says, on how the coin whose value is shown to you is chosen. – John Bollinger Aug 21 '18 at 18:06
• This description suggests a variation on the Monty Hall Problem. The fact that Monty knows things you don't (heh) is central to understanding why the naïve "1/4" is incorrect. He has the freedom to choose which Let's Make A Deal door to reveal, and @Vincent has claimed the same freedom with these coins. – Monty Harder Aug 21 '18 at 18:21
• @user3508140 In Wildcard's example, the probability is very close to 100% that it will come up heads. The probability that the flips were randomly coming up heads every time is close to zero. The coin is likely a double-headed coin. – InterstellarProbe Aug 23 '18 at 13:43

Here is the chart of every possible outcome where there is at least one tails:

$$\begin{array}{c|c|c}1 & 2 & 3 \\ \hline H & H & T \\ H & T & H \\ H & T & T \\ T & H & H \\ T & H & T \\ T & T & H \\ T & T & T\end{array}$$

There are 7 possibilities (each equally likely) where at least one coin is a tails. Only one of them is all tails. The probability is $$\dfrac{1}{7}$$

Let E $=$ all three coins show tail.

Let F $=$ one of the coins show tail.

$E=\{TTT\}$

$F=\{HHT, HTH, THH, HTT, THT, TTH, TTT\}$

$$P(E|F)=\dfrac{P(E\cap F)}{P(F)}$$ where $P(F)=\dfrac78$ and $P(E\cap F)=\dfrac18$

Therefore, the probability that all the three coins show tails is $$P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{\dfrac18}{\dfrac78}=\dfrac18\times\dfrac87=\dfrac17$$

• Right, almost. If you were using only a sequence of 2 events. Even so, what's being asked is something that's more seen in reality, not an abstract concept $P(E|F)$ (the Conditional Probability), which is $P(E \cap F)$, the actual probability of $TTT$ – Dehbop Aug 22 '18 at 7:42

The issue is, you don't know which coin shows tails. So, let's count all the possible 3 coin tosses that include tails. In fact, there is only one combination, i.e., HHH, where there are no tails.

So, the number of ways of getting 3 tails (1) is divided by the new total number of possibilities (8-1). So the answer is $\frac17$.

Answer concerning the scenario "at least one coin shows a tail".

Let $T$ denote the number of tails that are thrown.

Then you are looking for:$$P(T=3\mid T\geq1)=\frac{P(T=3\wedge T\geq1)}{P(T\geq1)}=\frac{P(T=3)}{1-P(T=0)}=\frac{\frac12\frac12\frac12}{1-\frac12\frac12\frac12}=\frac17$$

Answer concerning the scenario "exactly one coin shows a tail".

Let $T$ denote the number of tails that are thrown.

Then you are looking for:$$P(T=3\mid T=1)=\frac{P(T=3\wedge T=1)}{P(T=1)}=\frac{P(\varnothing)}{P(T=1)}=\frac{0}{P(T=1)}=0$$

Answer concerning the scenario "the first coin shows a tail".

Here I refer to your own answer. The probability then equals indeed $1\cdot\frac12\frac12=\frac14$ and your reasoning is correct if this is the aimed scenario (I have serious doubts that it is).

Let me work through your reasoning and demonstrate exactly why it doesn't work.

Just now, I asked my friend to flip 3 coins, one at a time. I then asked her if at least one of the coins came up tails. She said yes.

What's the probability that all three coins came up tails?

Define $A$ as the first coin which came up tails. Define $B$ as the first coin which is not $A$, and then define $C$ has the remaining coin.

If I make your reasoning more explicit, it says something like:

The probability that $A$ is tails, $P(A = T)$, is $1$. The remaining two coins are equally likely to be heads or tails, so $P(B = T) = 1/2$ and $P(C = T) = 1/2$. Finally, the three coins are independent, so the probability that all three coins are tails is the product of these: $P(A = T, B = T, C = T) = 1 \cdot 1/2 \cdot 1/2 = 1/4$.

However, this reasoning isn't correct. We have defined $B$ in an "unfair" way, so that $B$ is actually more likely to be heads than tails.

(What's so unfair about the definition of $B$? If the first coin comes up heads, then $B$ comes up heads, because, according to our definition, $B$ is the first coin. But if the first coin comes up tails, $B$ may come up heads anyway.)

We can see the actual probabilities of $B = T$ and $C = T$ by writing out the seven possible outcomes (all of which are equally likely):

• $HHT$: $B = H$, $C = H$
• $HTH$: $B = H$, $C = H$
• $HTT$: $B = H$, $C = T$
• $THH$: $B = H$, $C = H$
• $THT$: $B = H$, $C = T$
• $TTH$: $B = T$, $C = H$
• $TTT$: $B = T$, $C = T$

By looking at this table, we can see that $P(B = T) = 2/7$ and $P(C = T) = 3/7$. The two events aren't independent, either: $P(B = T)P(C = T) = 6/49$, whereas $P(B = T, C = T) = 1/7 = 7/49$.