Three people called Alf, Beth, and Gemma, sit together in the same room.
One of them always tells the truth.
One of them always tells a lie.
The other one tells truth or lies at random.
In each of the following situations, your task is determine how each person acts.
Suppose that Alf says “I always tell lies” and Beth says “Yes, that’s true, Alf always tells lies”.
Who always tells the truth? Who always lies? Briefly explain your answer.
First, if Alf is telling the truth, we have a contradiction. So Alf must be, on this occasion, lying. So either Alf always lies, or he lies at random.
If Alf always lies, he’s telling the truth on this occasion, another contradiction. So by process of elimination, Alf is the random liar.
Now Beth says that Alf always tells lies—but we’ve just figured out that this isn’t true, so Beth is lying on this occasion, and must therefore always tell lies since Alf is the random liar.
We get: Alf lies at random, Beth always lies, and Gemma always tells the truth.
There are 6 possible combinations, we could present this reasoning as a table considering each statement in turn against these. We place a cross wherever a contradiction arises.
In these tables, R stands for random-truth-and-lie-teller, L for liar, and T for truth-teller.
|
A is B is G is |
T R L |
T L R |
R T L |
R L T |
L T R |
L R T |
| A says A=L | x | x | \(\checkmark\) | \(\checkmark\) | x | x |
| B says A=L | – | – | x (know A=R) | \(\checkmark\) | – | – |
Showing the result we argued above.
Suppose instead that Gemma says “Beth always tells the truth” and Beth says “That’s wrong.”
Who always tells the truth? Who always lies? Briefly explain your answer.
Consider Gemma’s statement, “Beth always tells the truth.” If it’s true then Gemma has spoken a truth and can’t be the liar. If it’s false then both Beth and Gemma lie at least some of the time. So either Beth is the truth-teller and Gemma is random, or Beth is random and Gemma the liar.
Beth then contradicts Gemma. If Beth is the truth-teller then Gemma’s statement is false so Beth is not the truth-teller—a contradiction.
We are left with only one option: Beth is random, Gemma is the liar and Alf the truth-teller.
|
A is B is G is |
T R L |
T L R |
R T L |
R L T |
L T R |
L R T |
| G says B=T | \(\checkmark\)(G lie) | \(\checkmark\) | x | x | \(\checkmark\)(G true) | x |
| B says B\(\neq\)T | \(\checkmark\)(B true) | x | – | – | x | – |
Suppose instead that Alf says “Beth is the one who behaves randomly” and Gemma says “Alf always lies”. Then Beth says “You have heard enough to determine who always tells the truth”.
Who always tells the truth? Who always lies? Briefly explain your answer.
Using the earlier conventions we can fill in a table as before:
|
A is B is G is |
T R L |
T L R |
R T L |
R L T |
L T R |
L R T |
| A says B=R | \(\checkmark\) | x | \(\checkmark\)(lie) | \(\checkmark\)(lie) | \(\checkmark\) | x |
| G says A=L | \(\checkmark\) | – | \(\checkmark\) | x | \(\checkmark\) | – |
| B states one solution only | \(\checkmark\)(lie: still 3) | – | x | – | x | – |
Just before Beth speaks, we have three possible combinations, so what she says is a lie. She cannot therefore be the truth-teller and this eliminates two options. Once she has spoken it, her statement in a sense becomes true!
We conclude that Alf always tells the truth, Beth lies at random and Gemma always lies.
