3 Prisoners

Ah, but there is a subtle difference with the Monty Hall puzzle. There the door is open so the information is verified as true. Now, what if the guard lied about B having to be executed? Now we have two possible situations:
1) The guard said the truth (i.e. B is to be executed)
2) The guard lied (i.e. B is to live) = A is to die.
Assuming that they are equally possible we now have a completely different problem.

Originally posted by PawnCurry
To BLR - proof of the Monty Hall problem.

Assume there are three safes labelled A, B, C. There are only three possible outcomes each with P=1/3.

P(A) = P(B) = P(C) = 1/3 where P(A) is the probability of the prize being in safe A, etc

The three possible outcomes are shown below:-

A B C
1 O O X
2 O X O
3 X O ...[text shortened]... Phew! Do we get a prize for contributing to one of the longest threads on RHP?

🙂

Originally posted by ilywrin
Ah, but there is a subtle difference with the Monty Hall puzzle. There the door is open so the information is verified as [b]true. Now, what if the guard lied about B having to be executed? Now we have two possible situations:
1) The guard said the truth (i.e. B is to be executed)
2) The guard lied (i.e. B is to live) = A is to die.
Assuming that they are equally possible we now have a completely different problem.[/b]

Difficult for me to argue in English (I'm German) but let me put it
this way: with the MH-problem the question is not whether the price
is in A or B after the moderator has opend C. The question is whether
the candidate improves his chances by changing his first selection.

That's exactly the same situation as if the candidate had selected
door A and then the moderator asks "now, should I open that door
A for you or would you prefer me to open doors B AND C

You may empirically "prove" this problem by using a simple
EXCEL-sheet:

Let Col-A denote where the car is; let Col-B give the canditate' random
selection; so Cols A and B both should contain the formula
Randbetween(1;3)
Col-C is titeled "candidate does not switch" and we know that in this
case he will win if his first selection was correct; thus formula in Col-C
should read
if(A1=B1";"win";"lose&quot😉

OK?
Now copy/paste this line 5000 times - and the let EXCel count the
numer of "win"s

Convinced???

But --- is the prisoner's problem REALLY the same as the MH-problem?
Not sure about this ...

Originally posted by BLReid
Now that's just silly.

Originally posted by ilywrin
Would you care to explain why?



because the problem implicitely assumes that the guard tells the truth;
if you say" the guard may or may not be honest" you create a totally
different problem.

Thankyou PawnCurry, very good explanation... the penny finally dropped when I realised that the guard specifying that one of the others will die does not exclude any of the three outcomes defined initially for C, since if B was going to die, the guard would have said B and if A would have died the guard would have said A. Hence the guards response to C's question would change depending on which of the three outcomes were chosen initially and hence would make no difference to C's probability of dying. However, if B is stated to die, from this point on A would know that the outcome where B lives is excluded from his future and hence his chances of living increases.

Sheesh, even after getting it, it still baffels the mind a little, since for A and B you can now exclude one of the outcomes for the future but for C you can't.

Originally posted by ilywrin
Would you care to explain why?

Originally posted by BLReid
As EdSchwartz stated above, the truth of the guard's statement is implied in the spirit of the puzzle. But you already know that don't you?🙂

I understand your point, but I think that a fair amount of assumption must be made as far as the guard's veracity. Without that, we would all be spinning our wheels here (not that we aren't just wasting time anyway 😀).

*edit: Important to note that I am the guy who was wrong for a fair portion of this thread, so what do I know anyway?!🙂

I think with a lot of these puzzles you need to make a certain amont of assumptions, such as being told true information etc, otherwise you could reduce things down to absurd proportions.

You could introduce plainly ridiculous possibilities such as being broken out of prison - how would you factor that in? What if one of the prisoner's knows he is going to die and hangs himself - does that mean the other 2 will be executed? I know the examples I've given are extreme (and a tad ridiculous!), but in order for their to be a problem worth taxing our brain-cells, you do need to make a few basic assumptions that aren't explicitly stated in the problem.

In this case, I personally would make the assumption that I was being told the truth. Monty Hall, is a no brainer, as he PROVES which safe is empty by opening it.

I think what all of this shows, is that whoever coined the phrase "lies, damned lies and statistics" was spot on! A lot of these probability questions require us to ditch our immediate presumption as to the obvious, and go right back to basics.

I remember being shown the equivalent of the "3 Prisoners" problem while at school (ooh, 20 years ago?), and struggled to come to terms with it then - when my brain functioned a lot more cleraly than it does now!

Even now, KNOWING the answer, AND being able to prove it (at least myself), the "natural" answer would seem to be 50/50. I still have to perform some mental gymnastics and the odd leap of faith to get the real answer.

Anyone got any more??!!

🙂

Before asking the guard A's chances of surviving is 1/3. As the guard explains that either B or C may be executed 2 morrow, this does not exclude the possibility that it could still be A, as A said "tell me nothing of himself". Chances that A will be executed remain the same, 1/3.....

How to minimize the probability that there's a bomb on your flight?

Take a bomb with you yourself, because the probability for TWO
bombs on one flight is very small

🙂

Originally posted by EdSchwartz
How to minimize the probability that there's a bomb on your flight?

Take a bomb with you yourself, because the probability for TWO
bombs on one flight is very small

🙂

This thread will go on forever. 😛