mjhjmtu

I tried $1cdotfrac12cdotfrac12$ 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.

Where am I wrong?

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.

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

$$beginarrayc1 & 2 & 3 \ hline H & H & T \ H & T & H \ H & T & T \ T & H & H \ T & H & T \ T & T & H \ T & T & Tendarray$$

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 $$dfrac17$$

Let E $=$ all three coins show tail.

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

$E=TTT$

$F=HHT, HTH, THH, HTT, THT, TTH, TTT$

$$P(E|F)=dfracP(Ecap F)P(F)$$ where $P(F)=dfrac78$ and $P(Ecap F)=dfrac18$

Therefore, the probability that all the three coins show tails is $$P(E|F)=dfracP(Ecap F)P(F)=dfracdfrac18dfrac78=dfrac18timesdfrac87=dfrac17$$

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:$$P(T=3mid Tgeq1)=fracP(T=3wedge Tgeq1)P(Tgeq1)=fracP(T=3)1-P(T=0)=fracfrac12frac12frac121-frac12frac12frac12=frac17$$

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:$$P(T=3mid T=1)=fracP(T=3wedge T=1)P(T=1)=fracP(varnothing)P(T=1)=frac0P(T=1)=0$$

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

Here I refer to your own answer. The probability then equals indeed $1cdotfrac12frac12=frac14$ 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 $P(B = T, C = T) = 1/7 = 7/49$.

First, as opposed to everyone else, I will show the possibilities with only varying the last 2 toss:

A (listed here just for the sake of demonstration):

$HHH$

$HHT$

$HTH$

$HTT$

B:

$THH$

$THT$

$TTH$

$TTT$

We only consider B because the first toss is DETERMINED & as you can see the possibilities are only 4, not 8; leading to $P(TTT) = 1/4$. It agrees with the initial intuition that, despite the efforts of these other users, is correct.

Now, some people are invoking Conditional Probability as what's being asked. I would have to say that what's being asked is something more concrete as $P(TTT) = 1/4$

& the confusion of Conditional Probability being asked for is because of the shortcomings of natural languages.

EDIT:

Another simple comment. I just realized this, but even asking for a Conditional Probability (I still assert is not what's being asked) CAN virtually be the same answer. $fracP(X1 = T cap TTT)P(X1 = T)$ where X1 is the first toss, as long as the conditional is $P(X1 = T)$, seeing that $P(X1 = T) = 1$. This is STILL because all possibilities in consideration is B. Won't work if the conditional is $P(TTT)$

I'd say these are Math & Actuarial Science TAs that are posting these questions & answers, as they are VERY familiar with what I'd call "verbal/logical disconnect", specially seen in Math & Stats.