Famous and Challenging Logic Puzzles
Three of the hardest and most famous logic problems ever – can you solve them?
Famous Logic Puzzles
There have been many puzzles throughout history that have stumped both average man and mathematician. Can you solve these famous logic problems? (answers on the bottom)
Knights and Knaves
We’ve all heard this one before, in some variation or other. It was developed by Ramond Smullyan.
Problem: On a fictional island, all inhabitants are either knights, who always tell the truth, or knaves, who always lie. The puzzles involve a visitor to the island who meets small groups of inhabitants. Three inhabitants named A, B, and C approach you. You ask A what type he is, but do not hear A’s answer. B says “A said that he is a knave” and C says “Don’t believe B, he’s lying!” What type of inhabitant is B and C?
The Monty Hall Problem
This is a challenging riddle that has stumped many mathematicians.
Problem: “Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?”
The Hardest Logic Puzzle Ever
This self-proclaimed king of logic puzzles is quite confusing and challenging. It was thought up of by George Boolos in 1992.
Problem: “Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which. t could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question).
Liked it
The first two are nice, but the last one is messy and confusing! 
Besides, you’ve messed up #1 and #2 in “Assume ‘ja’ means no and ‘da’ means yes”
I am terrible at puzzles but I do like to try. When I look at the answers I see the logic. Thanks for a good brain work-out.
The answer to the Monty Hall problem (3 doors) is bogus. At the point the host asks if you wish to switch or not the problem changes and you have a choice of 2 doors and therefore the probability of success is .5, or 50/50.
Re: Knights and Knaves
I disagree with the conclusion that you cannot determine A’s status because of lack of sufficient information.
You correctly noted that A cannot say he is a “knave” so it follows that B is a knave and C is a knight. A has to be a knight.
Jonathan (above) – the Monty Hall problem has stumped everyone. The answer is not bogus – focus on this distinction: if Monty Hall did not know which door had a car behind it then you would be right – - it would be 50/50; however, since Monty Hall knows where the car is, that changes the probabilities.
I’ve read the mounty hall problem in a book…
The mysterious incident of the dog in the night time?
For those confused of the rationale behind the Monty Hall solution, exaggerate the conditions to see the strategy behind always making the switch:
“There are one million doors. You are told to pick one. After making your pick, 999,998 other doors are opened, each revealing a goat. Only two doors remain unopened: your original pick, and one other. You are asked if you would like to switch to the other unopened door. Is it in your best interests to switch?”
In this scenario the question becomes simple. The chances you initially picked a door behind which a goat stood among the 1,000,000 doors is 999,999/1. Thus, there was an extremely, extremely good probability that the other unopened door was instead the door behind which a car sat.
Reducing the severity upon which that situation exists, in the case of the original problem involving three doors there is a 2/1 chance the initial door you picked was the goat door. As such, 66% of the time, making the switch results in choosing the car door.
In addition (referring to Dan’s observation of the Knights & Knaves problem), stipulating that A must be a knight given the fact he (supposedly) declared his identity is incorrect (he gave an answer, which does not necessarily equate to an either/or identity posulation of knight or knave, e.g. “I didn’t hear you, could you repeat?,” which could either be a lie or the truth). While a knave would not admit to being a knave, he could admit to being a knight while adhering to his falsehood. Thus, a knight declaring (truthfully) to be a knight would be indistinguishable from a knave declaring to be a knight without external referential.
Logic, yes you are right. I see my error.
Let me make it simple: You have a big room with 3 doors on one of its walls. Somewhere inside the room is a reward. You divide that room in 2 parts, one taking 2/3 of space and 2 doors, and the other one taking 1/3 of space and 1 door. Where is the greater chance for the reward to be? – In the bigger part of course. That’s what it makes it a good choice. It’s actually a rather simple puzzle. You just gotta have a good approach.
Monty Hall. The problem says the host knows where the prize, or car, is located.
The stated (or math) solution sucks. Here’s why:
Initially there were three choices and one prize. The player had a 1/3 possibility of winning. After the host opens a door containing a goat, regardless of whether it was the original choice of the player, the odds now are: two choices, one prize. This means the player now has a 50-50 chance of winning. So the only thing that has really changed is that now you have better odds. But the choice is still random- regardless. So if you switch or not, your chances are still going to be 50-50. Which means there is no clear benefit to switching. However, the fact that the host knows where the prize is, and that this is specifically stated in the problem, means it is more than a mere question of statistics. Which is why you should take the host’s intentions into consideration.
First case scenario: the host does not want you to win.
If your first choice would have been wrong, he would have opened it and achieved his purpose. Which means your first choice was right. So you do not switch.
Second case scenario: the host wants you to win (highly unlikely- if he goes out of his way to help you win he will certainly be fired).
If your first choice was the correct choice he would have opened it. Since he did not, your first choice was wrong. You should switch. Again this scenario is highly unlikely.
Third case scenario: the host is indifferent. This scenario is impossible. If he were indifferent, he would have opened the door you told him to open. Period.
Given all three options the most likely is the first. So good logic says you should not switch your choice.
The third problem is nice.The difficult and brilliant part was concieveing that type of question that would get you the thruthfull answer while not knowing the foreign language.
Excellently put, Marcos, the problem doesn’t become a simple one involving statistics anymore once you actually take into account the game show host knows where the car is. I like to point out one more thing abt this 2out of 3 and 1out 2 stuff, Everyone seem to say that before the probability was 1/3 and now it is 1/2 for the door, but once that door with the goat is opened both doors stand at the samelevel and the question “Would you change your choice?” cannot involve mathematical logic.
Monty Hall question. While reading this a was amused at the total lack of mathmatic reasoning behind this answer. I agree in the original choice you have a 1 in 3 chance to pick the door with the care, however they go and breakdown the possible situations that could have occured when he opened the door, however, in order to prove their logic, they lumped 2 situations together and made it one. The correct breakdown would be as follows:
1.The player has picked the door with the car. The host has shown GOAT A (1/4 chance)
2.The player has picked the door with the car. The host has picked GOAT B (1/4 chance)
(note..these are NOT the same. They are 2 separate scenarios.)3.The player originally picked the door hiding Goat A. The game host shows Goat B. (1/4 chance)
4.The player originally picked the door hiding Goat B. The game shows Goat A. (1/4 chance)
Therefore, that leaves you with a 2 out of 4 chance to have picked the car. It leaves you with the choice do you change? It really doesn’t matter. Taking the “intentions” of the host out of the equation. You have a 2/4 chance or 50% chance of having the correct door. There is a 1/4 chance that the goat behind the other door is goat A and a 1/4 chance that the goat behind the other door is gaot B.
The two situations ARE the same. The host always chooses a goat when he opens the door, that’s part of the game, you have three choices at the start, Youre chances of picking the car are 1 in 3. The presenter will pick a goat, he will not have a biased decision as it makes no difference which goat. There is a 1 in 2 chance of the presenter picking goat a and the same chance of him picking goat b. This results in a chance of 1/3 x 1/2. This is a sixth not a quarter. Check with a calculator if you don’t believe me:) I’m 12 and I knew better than you. Lol.
Monty Hall Problem, easy explanation (there’s no way you can read this and still be confused):
People are ignoring the original question “Is it better to switch?” and the fact that the host always reveal a goat. Let’s see what happens in the two scenarios when you switch:
1. 66% chance that you pick a goat. Host reveals the other goat. Therefore, when you switch, you will get the car. 66% chance you will end up with the car.
2. 33% chance that you pick the car. Host reveals one of the goats. Therefore, when you switch, you will get a goat. 33% chance you will end up with a goat.
Now let’s see what happens in the two scenarios when you don’t switch (it should be obvious by now):
1. 66% chance that you pick a goat. Host reveals the other goat. Therefore, when you don’t switch, you will keep your goat. 66% chance you will end up with a goat.
2. 33% chance that you pick the car. Host reveals one of the goats. Therefore, when you don’t switch, you will keep your car. 33% chance you will end up with the car.
Those are all the scenarios, and you can clearly see that your odds are better of getting the car when you switch, due to the fact that the host will reveal a goat every time.
no lol. all of u are looking into it too hard. it is a new question, with a new number of doors, and a new probability. 50% end of story
In his puzze, each letter represents a different number. Can you work out what number (0-9) each letter stands for? The leftmost letter can not be zero in any word.
Logic’s explanation of the Monty Hall problem by exaggerating the figures to larger numbers is the most logically sound as yet….although common sense sort of says that it should be a 50-50
i can explain better. Say there are 100 doors. 1 car and 99 donkeys. you chose 1 door. the host then opens 98 of the remaining doors showing all donkeys. what are the odds you picked the car in the first place? 1 in 100. what are the odds you are left with the car if u switch? 99 out of 100. think about it. the host is opening the remaining doors and they will all be donkeys every time. thats the way the game works. if he opens the door to the car u will know u get a donkey anyway.
The Monty Hall problem is NOT bogus. It is NOT a 50/50 chance, even though that seems counterintuitive. The fact that the host revealed one goat DOES have an effect on the odds. Here is an obvious way of looking at it:
Imagine if we changed the situation a little bit. Let’s say there were 15 doors lined up in a row: 14 doors with goats behind them, 1 door with a car. You get to pick one door, and then the host will reveal 13 goats behind 13 doors, leaving only your original door and one other door closed.
So let’s say you pick door #1. The host says, “Ok, here’s where 13 goats are hidden.” He goes down the row of doors opens every single one except #1 (because you picked it), and #9.
Tell me, standing there with that entire row of opened doors, 13 goats staring you in the face… would you stick with your original choice? Or would you switch to door #9, which is the only door in that long row that has conspicuously been left shut?
Well, OF COURSE you would switch to door #9!! At this number of doors, it’s completely intuitive. You KNOW there’s a better than 50% chance that #9 conceals the car. Because unless you picked the right door when you said door #1, it MUST be door #9 that has the car. And you know that you only had a 1-in-15 chance of picking the right door at the start.
If you had picked door #2 instead of #1, doesn’t it seem pretty likely that door #9 would STILL be the one the host left shut? It seems like a 1-in-15 chance to me.
It’s not 50/50. If you don’t believe me, run an experiment with a friend. Have a friend be the “host,” and set up some situation where you have three pieces of paper to pick from — one that says “car” and two that say “host”. You get the idea. Switch your choice every time. Run this experiment like 40 times and keep track of how many times you get the car. It will be about two thirds of the time.
Regarding the monty hall problem, assume 100 doors instead of 3 (as many have suggested/posted previously), but ALSO consider this scenario:
* instead of JUST you and the game show host, it is the you, your neighbor and the game show host.
* either you or your neighbor gets to pick FIRST.
* the person that picks FIRST, gets to pick ONE door, the other person gets the remaining 99.
q1. at this point in time, would you rather pick first or would your rather have your neighbor pick first?
# note: if you’d rather pick first and get only 1 door, you can stop reading now. 
* the host now exposes 98 doors (of the person selecting second) as goats.
q2. there are only two doors left now, would you rather
be the person that selected first (1 door) or second (originally 99, now only 1)?
q3. if he answer to q1 is: “i’d rather pick second”
and
if the answer to q2 is: “it doesn’t matter”
then, q3 is:
* what changed that would make you change your mind?
* at what point in time did the probability change of the first person being right change?
* at what point in time did the probability change of the second person being right change?
Now the same logic holds when there are just 3 doors, right?
This is a fun, interesting and counter-intuitive puzzle.
(i didn’t get it the first time either.)
Every year there are more people born that will also be “tricked”, hopefully more get it sooner rather than later.
(and hopefully, i’m not “submitting” with many typos)
cheers and thanks for the web content!
After you select a door, and opt to change your choice, your odds of finding the prize are then 50/50, as only 2 doors remain. 50/50 is much better than your original odds when you were faced with three doors, or a 33% chance of winning which is 1/3. ~ciao
Knowing, of course, that you probably chose the wrong door initially because only one of the three doors has the prize.
I think I have solved the third problem.
It is most definitely one of the hardest of these types of questions I have ever seen.
You start by asking god A “Answer yes if you would say both of these statements are correct and Answer yes if you would say both these statements are false “da means yes” and “God B does not answer randomly”. Otherwise answer no.”
If the response is da then God B does not answer randomly, if it is ja the God C does not answer randomly. If A is random then it doesn’t matter which one you ask the second question too.
You then ask the God you know not to answer questions randomly this “Answer yes if you would say both of these statements are correct and Answer yes if you would say both these statements are false “da means yes” and “1+1=2”.Otherwise answer no.”
If the God answers yes then it tells the truth if not is lies. Then Ask the same god if god A is a random answering god. If it turns out he is a random answering god then the last god must be ether a liar if the god you just asked the question to is a truth teller and vise versa if not.
If the god answers no then the other god is a random answering god and god A is the opposite of the god that just answered this question.
From this you then should know the states of all the gods.
Sorry if my answer is confusing but i tried my best to make it hard confuse the wording of the questions.
The real reason why monty hall is 2/3, not 50/50 is this:
The host will -never- open the door with the car behind it.
Now, here are the two options:
1. You picked the door with a goat first. We can all agree this happens 2/3 of the time, right? Then, the two other doors have a goat and car. He will only reveal that other goat. So, the remaining door must be the car. So, 2/3 of the time, you should switch.
2. You picked the door with a car first. This happens 1/3 of the time. Both of the other doors have a goat. You switch, you lose, sorry.
Anyway, the point is that he’s not randomly opening -any- door, he’s only opening a door without a car.
Another example:
You’re rolling a die, and say you want to get a six. You roll, and then after you roll, someone takes the die and magically removes a number (that you know -won’t- be 6) from the die, and asks if you’d like to roll again. Originally, you had a 1/6 chance of winning. Now, you know you have a 1/5 chance. Roll again!
i felt this site answered the Monty hall question rather well. also, it has a table at the end which unarguably shows the answer of switching is more beneficial:
http://coastaltech.com/monty.htm
Esk Iom – excellent explanation
For those who don’t get it. If you don’t get it after this you are clueless and shouldn’t be attempting logic puzzles.
There are only 3 possible scenarios.
1. Car is behind A
2. Car is behinc B
3. Car is behind C
Lets do this 3 times, always pick A and always keep our original pick.
1. Pick A. You are right. Host shows B or C. You stay with A and win.
2. Pick A. You are wrong. (car is in B). Host shows C. You stay and lose.
3. Pick A You are wrong. (car is in C). Host shows B. You stay and lose.
You won 1/3.
Lets do it again and always switch.
1. Pick A. You are right. Host shows B or C. You switch and lose.
2. Pick A. You are wrong. (car in B). Host shows C. You switch and win.
3. Pick A. You are wrong. (car in C). Host shows B. You switch and win.
You won 2/3.
Conclusion. If you switch, your odds are 2/3 vs. 1/3 if you do not switch. (not 50/50)
