## Which Way?

Posted by Karl Sharman on May 13, 2011 – 11:15 am

You are lost in the woods. You come to a T-junction and are wondering which way to go. You have a funny feeling that one way will get you out, the other will mean certain death. Weird, eh?

Wizard of Oz approaches you. “A left turn will get you out of the woods,” he advises. Nearby, Chris hears this and calls out, “Beware of him. He lies!” “Exactly 4 times out of 10, I do!” Wiz retorts, and disappears.

You know from experience that Chris always tell the truth. You also know that throughout his life, Wiz tells the truth a certain number of sentences out of ten, determined by chance at birth. You also know this number is an integer. Which way should you turn to maximise your chance of escape?

Chris won’t sleep tonight!

May 13th, 2011 at 5:54 pm

I’m only ever 40% right with my answers. Chris is 60% right to call me a liar!

May 13th, 2011 at 9:33 pm

We don’t know whether the Wiz lied about the number of times he lied so it is impossible to get a numeric value. Your best bet is to shout out to Chris and ask him. Happy trails!

May 14th, 2011 at 1:49 am

I don’t think so. Between two evils ’tis not worth choosing

May 14th, 2011 at 9:22 am

Its just luck. I would choose the right path, because it would be more probable to get out of the woods, but there is a 40% chance you would die. Or you could be completely wrong about the situation and both paths could lead to freedom or to death.

May 14th, 2011 at 4:19 pm

i say go straight down the middle, don’t take either path and at least you’ll be halfway right.

May 14th, 2011 at 5:08 pm

. . . or go back the way you came from.

May 14th, 2011 at 9:18 pm

I think the extra info is trying to throw us off on this one. Here are the important parts:

1. Wiz says: “A left turn will get you out of the woods”

2. Chris says: “He lies!”

3. Chris always tells the truth

Therefore, statement 1 is a lie and you should go right. 100% chance of escape.

May 14th, 2011 at 9:34 pm

Does statement 2 in post # 7 mean “He ALWAYS lies” or “He SOMETIMES lies”? If it’s the latter case then we’re back where we started, i.e. having to second guess Wiz (that’s me, but I don’t know whether I’m lying or not).

I still say go back the way you came, then find another way round. It may take longer but it’s safer.

May 14th, 2011 at 11:37 pm

Since I posted the problem, I have had some new information – a hoard of man-eating, rapacious and predatory pedantic pygmies (purely for the alliteration!) are approaching from behind. There is no going back! Yes – they are hungry, and they eat women too!

A quick search on google maps also shows that the woods surrounding the paths are impenetrable.

SP – The statement is “Beware of him. He lies!” Chris would be telling the truth whether Wiz’s following statement “Exactly 4 out of 10 times, I do!” is a lie or the truth. Chris’s statement is conditional that Wiz lies – not necessarily all the time though!

May 15th, 2011 at 12:16 am

praying, that God will show me the way its 100% without failing lol

May 15th, 2011 at 4:22 pm

All we know for sure is that I, Wiz, tell the truth an integer number of times out of 10. It can’be 10 out of 10 as that would contradict my statement that I lie 4 times out of 10, also it would make Chris a liar, and we all know him to be a man of impeccable integrity, don’t we?

My self assessment of 4 lies out of 10 obviously can’t be relied on. So, with no other basis to make a judgment, all we can do is to assign an equal probability to each integer from zero (always lying) to nine. (Yes, zero is an integer – would I lie to you?). The average of the integers from 0 to 9 is 4.5, so there’s a 45% chance that I’m telling the truth when I say turn left, and a 55% chance that I’m lying. So the odds favour turning right.

But if I originally said turn left, and I now say turn right, why should you believe anything I say?

May 15th, 2011 at 5:21 pm

I’m struggling with trying to decide on how to use the fact that you said “4 out of 10″ to assess the probability that you actually spoke the truth (or even if it actually cotains information at all. Before you said that, then each of 1-10 would be equally likely. But after you said it, I fancy that 4 out of 10 is no longer an equally likely option.

And just to make it more complicated, why did I give the caution? Why did I say anything at all? Was I really saying “turn right” (but then why didn’t I simply say “Wiz lied” or “turn right”? Ah!, puzzleland rules.

May 16th, 2011 at 12:23 am

“A left turn will get you out of the woods” can also mean you’ll be dead.

May 16th, 2011 at 4:11 am

Assuming that Wiz could only say “I lie on N occasions in 10″, where N = 0 to 10 (because he’s bound by Puzzleland rules), then I can analyse it as follows:

Before Wiz spoke for the second time it could be that he actually lies on 1-10 times out of 10. Also I guess that we have to assume that each of those possibilities is equally likely.

If Wiz actually lies on N in 10 occasions (i.e. he tells the tuth on 10-N out of 10 occasions) he’d say “N” (100-10N) times out of 100 times and each of the other 10 numbers N times each in each 100. In particular, if he lies on N in 10 occasions he’d say “4″ N times if N ≠ 4, and he’d say “4″ 60 times out of 100 if N = 4. Altogether in the order N = 1,2,…10, he’d say “4″, 1,2,3,60,5,6,7,8,9,10 times out of 100. So over what was the 10 equally likely cases, we now have 1+2+3+60+5+6+7+8+9+10 = 111 ways (out of 1000) that he could have said “4″. So Given that he said “4″, the probability that he actually lies N in 10 is N/111 if N ≠ 4 and 60/111 if N = 4. So the probability Wiz tells the truth more often than he lies is (1+2+3+60)/111 = 66/111, and the probability that he lies more often than he tells the truth is (6+7+8+9+10)/111 = 40/111. The probability that he lies half the time is 5/111, which I shall ignore.

So you’re best to take Wiz’s advice and take the left turn. Now I’ll wait for slavy and/or Karl to shoot that down

May 16th, 2011 at 4:46 am

Chris – with my prob -ability – I doubt I will be the one to shoot you down!

However, the solution to this problem is found through conditional probability. I hinted at this above. With zero being an integer, there are 11 possibilities for the number of times out of 10 Wiz lies…..

Counter-intuitive, or just a trick?

May 16th, 2011 at 4:58 am

Hi Karl. That was a conditional probability analysis. The condition being that Wiz said “4″.

But I guess your comment means that I’ve got it wrong

May 16th, 2011 at 8:03 am

go left since wizard is not lieing, because he did not tell the way in a sentance

May 16th, 2011 at 2:26 pm

So the wiz lies some percentage of the time.

(10%, 20%, … 90%)

This gives an average of 50%.

IF he is telling the truth, that he lies 40% of the time, then that makes us lean toward going to the Left. (60% chance he’s telling the truth)

If he is lieing about the 40%, then we average the remaining possible lie percentages (10%, 20%, 30%, 50%, … 90%) for a 51.25% change he’s lieing about going Left. (48.75% chance he’s telling the truth about going left)

So it’s either a 60% chance he truthful or a 48.75% chance.

This averages to a 54.375% chance he is truthful about going left.

I’m going left …. and crossing my fingers.

May 16th, 2011 at 2:49 pm

Hi cazayoux. Wiz could lie 100% of the time.

May 16th, 2011 at 3:11 pm

Hi Karl. I just re-read your comment #15. Chris said that Wiz lies, and you have (quite rightly ) suggested that Chris can be trusted. So it cannot be that Wiz lies 0 out of 10 times. So there are only 10 possibilities, not 11. But I have assumed that Wiz could have given 11 retorts. Also, as Wiz said that he lies 4 out of 10 times, he can’t always tell the truth, as saying 4 out of 10 would be a lie.

May 17th, 2011 at 7:33 am

Ahh … Thanks, Chris.

Well … that pushes my calculation further from trusting Wiz, of course.

So it’s either a 60% chance he truthful or a 43.33% chance.

This averages to a 51.67% chance he is truthful about going left.

Still heading left … still crossing my fingers.

May 18th, 2011 at 7:29 am

Thinking again … the 43.33% is a average of all options other than if Wiz is truthful about the 4 in 10, so I cannot average these two numbers as equals. The 60% chance of truthfulness is not a 50/50, but a 1 in 10.

So I need to give teh 43.33% an average (9 in 10) chance.

So that combined view is now a 45% chance of truthfulness. I must go RIGHT!

May 18th, 2011 at 8:03 am

Hi cazayoux. I couldn’t agree less If Wiz hadn’t made a statement about his honesty, then you best guess is that it’s equally likely that he lies on 1,2,3,4,5,6,7,8,9 or 10 out of 10 times. So 1,2,3,4 you’d trust him, and for 6,7,8,9,10 you wouldn’t. For 5 out of 10, it doesn’t matter if you trust him or not. So on that basis it’s 5/4 = 1.25 times more likely that he’s predominantly a liar, and so you shouldn’t trust him. However, that completely ignores the fact that Wiz said that he lies 4 times out of 10.

Although Karl has implied that I’ve got it wrong, I think it is most likely that he tells the truth more often than not, for the reasons I gave in post 14.

May 18th, 2011 at 2:21 pm

this is not a math question. Chris heard the wiz say turn left. Chris told you he lies. Chris is telling you to go right. No matter what the wiz says he is lying.

May 18th, 2011 at 3:20 pm

Even though he lies, there is a possibility that the two statements were actually true. Even if he lies 9 out of 10 times, these could be his 2 true statements out of 20. On the flip side, even if he lies only 1 out of 10 times, these could be his two lies out of 20. I think if you are to run probability calcs, we should consider using 20 statements.

Gut feeling is that we are told ‘he lies’ from a realiable source for a reason, so I’m headed Right. Who’s with me? JG? SP? man…what’s up with all of these initials anyway?

May 18th, 2011 at 8:19 pm

Hi Karl. Have I misunderstood the question? The only thing I can think of is that I’ve incorrectly guessed the scope of the possible retorts that Wiz gave.

May 19th, 2011 at 5:32 am

@we shouldnt ignore d fact dat if we decide upon our own

we have 50%probability of surviving

May 19th, 2011 at 7:02 am

Conditional probability says that the probability of Wiz lying X number of times out of 10 in his life is equal to the probability of him being able to make the statement (X/10, unless X is 4) divided by the sum of probabilities for all the cases.

So, for our example, the probability that Wiz tells lies 7 of 10 times is: (7/10) / (0/10 + 1/10 + 2/10 + 3/10 + 6/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 10/10) = 7/57

Similarly, the probability that Wiz lies X of 10 times is: X/57, for X not equal to 4, 6/57, for X equal to 4.

Now we can find the probability of Wiz lying in his first statement.

There is a X/57 chance (or 6/57 for the 4 of 10 case) that Wiz lies X of 10 times, and for each of these cases there is a X/10 chance that Wiz lied about the path. To find the odds that Wiz lied, simply add the products of probabilities for each case:

0/570 + 1/570 + 4/570 + 9/570 + 24/570 + 25/570 + 36/570 + 49/570 + 64/570 + 81/570 + 100/570 = 393/570 = 68.947%.

So, the probability Wiz lied is 68.947%, and you should go right.

So, Chris, as this is my answer to a probability question, I would accept that the probability it is right is damn near to 0. So, your input at this stage will be invaluable.

Is now a good time to state that this is the answer as I remember it, rather than was written down at the time?

I felt the question was a nice little slant on the old “Fork in the Road – one tells the truth, one lies” type of question.

May 20th, 2011 at 2:02 pm

Hi Karl. I bet you were beginning to think that I wasn’t going to respond to your official answer. LOL, here goes. Your first paragraph is wrong. It starts off with an incorrect statement, and then without explanation, picks on “4″ as being special.

If Wiz actually lies N in 10, then he’d say so (10-N) in 10 times, and one of the other phrases N in 10 times. As there are 10 others, and as I have no reason to believe that Wiz has any preference for them, he’d say each of them on N in 100 occasions.

If N is 1, Wiz would say “4 in 10″ 1 in 100 times.

If N is 2, Wiz would say “4 in 10″ 2 in 100 times.

If N is 3, Wiz would say “4 in 10″ 3 in 100 times.

If N is 4, Wiz would say “4 in 10″ 60 in 100 times.

If N is 5, Wiz would say “4 in 10″ 5 in 100 times.

If N is 6, Wiz would say “4 in 10″ 6 in 100 times.

If N is 7, Wiz would say “4 in 10″ 7 in 100 times.

If N is 8, Wiz would say “4 in 10″ 8 in 100 times.

If N is 9, Wiz would say “4 in 10″ 9 in 100 times.

If N is 10, Wiz would say “4 in 10″ 10 in 100 times.

I’m sticking with my previous post (#14), but will admit to not being completely satisfied with it.