[Brain Teaser] Does this improve your chances?

[Tasoth]

Considering I am allowed to change my choice, I would choose this as the opertune moment to shuffle back and forth between the doors making as much noise as I can without looking suspicious. If the lion really is hungry, he's going to get antsy and give away his position.

Or else I get killed by a centurion.

[Formless]

Spoiler

True, but I find that fate generally hates me in real life with all the bizarre shit that I seem to stumble on when I'm just minding my own business. So why take the chance?

Spoiler

But you have no way of knowing either way that the emperor is going to screw you. And if he is intent on making you dead, you have no way of changing the fact. So you might as well assume that he isn't going to fuck with you, and take the option that you know will give you better odds mathematically.

[Kuroneko]

Spoiler

As mentioned before, even with all psychological considerations removed, there are two reasonable interpretations consistent with the OP as stated that lead to different answers to the title question. The wording is simply too ambiguous. However, it can be said that under either scenario, switching does not worsen one's chances, so the poll has a correct answer of 'yes'. (Surlethe's being a bit too tricky here--I was thinking of the question in the title when I made my previous post.)

[Napoleon the Clown]

Spoiler

Well, there's initially a 1 in 3 chance of living. The Emperor gets rid of one of the lions, but definitely not the one in front of the door you just chose. So now there's a 1 in 2 chance of survival. This looks like better odds, but keep in mind that if you change your answer, you're still going to have the same odds of survival as if you stayed with the same door. Your odds improve by definition.

However, changing doors does nothing for your new odds. It's the same fifty percent chance however you choose. The door you picked initially had a 1 in 3 chance of being right. But it's now a 1 in 2 chance of being right, just like the other door. Changing doors does nothing for your odds of survival.

[Gerald Tarrant]

Spoiler

Yes, change your door. 2/3rds of the time your guess will be a door with a lion. So 2/3rds of the time the unpicked door will be the slave maiden. This is a more killy variation of some old gameshow final round, where there was a goat behind 2 doors and a car behind the last. The odds that you're playing are that you chose a deadly door during your first guess.

[The Grim Squeaker]

Ah, a real braintwister. When I was 9 years old and first heard it. Spoiler

Switch, since your odds are improved mathematically by doing so, instead of 1 in 3, you're using 1 in 2. It's a load of statistical waffery, but mathematically sound.

[Garlak]

Well... if you *really* want to "stick it to The Man," then instead of going with your first choice to spite him... commit suicide.

Bite your tongue and suffocate on it.

I mean... if your objective is to deny the Emperor satisfaction... how do you know whether the jerk would laugh more at you switching, or keeping to your first choice? What I'm saying is: if you stick with your first choice just to spite the Emperor... to YOU, it might be spiting him. But the Emperor may just laugh anyway.

But if you commit suicide, well, you're making it a lot more obvious that you will NOT play along with the Emperor's game...

[CaptainChewbacca]

Spoiler

Well, there's initially a 1 in 3 chance of living. The Emperor gets rid of one of the lions, but definitely not the one in front of the door you just chose. So now there's a 1 in 2 chance of survival. This looks like better odds, but keep in mind that if you change your answer, you're still going to have the same odds of survival as if you stayed with the same door. Your odds improve by definition.

However, changing doors does nothing for your new odds. It's the same fifty percent chance however you choose. The door you picked initially had a 1 in 3 chance of being right. But it's now a 1 in 2 chance of being right, just like the other door. Changing doors does nothing for your odds of survival.

Spoiler

You are incorrect, because they are linked events. As was explained previously, you only had a 1/3 chance of being right. If there were 100 doors, and only 1 was safe, and after you picked one, 98 were eliminated, there is a 99/100 chance that you did NOT pick the safe door. The fact that 98 doors are now gone is irrelevant.

Similarly, with only 3 doors, the fact that a door was removed after you picked is irrelevant to what the odds were WHEN you picked. In the 100-door scenario, the sexy lady will be behind the door you didn't pick 99 times out of 100. In the 3-door scenario, the lady will behind the door you didn't pick two out of three times. It isn't a 50/50 choice, because the two picks are linked events.

For EXTRA fun thinking, suppose after you're offered this choice, another prisoner comes out, and he's told to pick one of two doors, and he can't know which one you picked originally. Does he have the same chance of survival as you?

[B5B7]

I've posted on a few other sites about the Monty Hall problem.
On this site mtgsalvation, in pos # 87 user Cornflake Pirate gave a very elegant solution. I am RebelmageX on that site, and posted a few solutions in that thread.

[Fingolfin_Noldor]

Spoiler

I remember doing a probability problem like this... Gosh I forgot how to do it, but apparently, you can improve your chances by changing your mind, if I recall correctly.

[Singular Intellect]

For EXTRA fun thinking, suppose after you're offered this choice, another prisoner comes out, and he's told to pick one of two doors, and he can't know which one you picked originally. Does he have the same chance of survival as you?

Spoiler

Nope. He has a fifty fifty chance of being right or wrong.. You have a sixty seven percent of being right, assuming you had changed your guess. If not, you have a 67% chance of being wrong.

[Acidburns]

Spoiler

What happens if after he reveals one of the lions, you then flip a coin, heads being the door on the left and tails being the door on the right. That must be 50/50 (two doors, one good, one bad) but you might end up picking the same door as last time. Which according many people, has a higher chance of being a lion?

[Kuroneko]

Spoiler

What's different is that, assuming the Emperor's strategy is to reveal a random lion that's not behind the player's first pick, is the Emperor's choices are constrained by the player's--he no choice at all about which door to reveal if the player picked a lion. Thus, the Emperor's behavior tells you more about what's behind the doors than just the fact that it's one of the two remaining ones. If your strategy is to switch with probability p, then your chances of winning are (2/3)p + (1/3)(1-p) = (1+p)/3.

On the other hand, if the Emperor simply picks a random lion and it just happens to be not the player's first choice, then switching doesn't improve your odds. The OP is ambiguous as to which strategy the Emperor employs, even if we assume that the question is intended to be entirely probabilistic (i.e., the Emperor is not actively leading you to a particular outcome once the game is underway).

[Acidburns]

Thank you very much Kuroneko, I understand now.

[RedImperator]

I won't bother trying to answer because others already have (I accept the math, but it's really counterintuitive; hey, surprise, the universe doesn't conform to human intuition! Who knew?!). But I will say that this would be a much more interesting television program than Let's Make a Deal.

[Adrian Laguna]

This scenario has an interesting deviation from the actual Monty Hall Problem as it appeared in Let's Make a Deal. In the TV show you could take a third option: open no door and walk away with $5000 in cash. In an article I read on the issue some time ago, Monty Hall himself said the best choice is to take the cash, and I am inclined to agree, because then you are 100% certain to walk away with something.

[Nieztchean Uber-Amoeba]

Yes, of course.

[B5B7]

This scenario has an interesting deviation from the actual Monty Hall Problem as it appeared in Let's Make a Deal. In the TV show you could take a third option: open no door and walk away with $5000 in cash. In an article I read on the issue some time ago, Monty Hall himself said the best choice is to take the cash, and I am inclined to agree, because then you are 100% certain to walk away with something.

Well, in that case, utility theory would come into play, and you would need the door prize to be above $7,500 before considering it.

[Ted C]

Spoiler

I originally had a one-in-three chance of being correct, meaning there was a two-in-three chance I was wrong. By changing my guess, I switch from the one-in-three chance to the two-in-three chance. Explainers of the "Monty Hall" puzzle often suggest that you imagine it as 100 doors. You choose one, then Monty opens 98 of the others to reveal booby-prizes. Would you want to change your guess to the one other unopened door at that point?

I have seen this puzzle explained before.

[Ariphaos]

Spoiler

How is the filtering that fate provides not the equivalent to intent?

You have a 2/3rd's chance of choosing a lion. One of the lions is revealed, the chances are either 50/50 that it is yours, or 0% (if you chose the slave). Or, 1/3rd. That fate or human intent remove that 1/3rd chance should make no difference.

[Kuroneko]

Spoiler

If the Emperor's choices are not affected by the player's, then the result will be a uniform probability distribution, except for some events of vanishing probability. That's not at all what happens if the the Emperor cannot pick the player's first guess, in which he is sometimes forced by the player. In that case, the Emperor's choice gives you more information.

Spoiler

Consider the following possible strategies for the Emperor:
A. Reveal a random losing door that's not the player's first choice.
B. Reveal a random losing door.
C. Reveal a random door.
The first two are compatible with the OP as stated, while the third is not. There are other strategies compatible with the OP, but those are the simplest. A further assumption is that the Emperor picks with a uniform probability out of the set of doors consistent with the individual strategies.

Because the situation is initially completely symmetric, we can always label the player's first pick as door 0, and the other two in arbitrary order. Call the winning door as n, the emperor's pick as m, and write P(n,m) for the joint probability that the winning door is n and the emperor's pick is m.

If the Emperor's strategy is C, then it doesn't matter--he picks all doors with equal probability, with nine possibilities. B is very similar, except that the case n = m is disallowed, with six possibilities left.

Strategy A, however, is very different. If the winning door is 0, then the Emperor has 1/2 chance to pick 1 and 1/2 chance to pick 2. Since the probability that 0 is the winner is 1/3, the joint probabilities are therefore both (1/3)(1/2) = 1/6 in that case. On the other hand, if the winning door is 1, he Emperor picks door 2 with probability 1. Hence, the joint probability is (1/3)(1) = 1/3, and similarity for the case of n = 2.

Code: Select all

(n)|m->  A     |     B     |     C     | Mutual information:
Win| 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | A = 2/3     ≅ 0.6667 bits
---+---+---+---+---+---+---+---+---+---+ B = lg(3)-1 ≅ 0.5850 bits
.0.| 0 |1/6|1/6| 0 |1/6|1/6|1/9|1/9|1/9| C = 0       = 0.0000 bits
.1.| 0 | 0 |1/3|1/6| 0 |1/6|1/9|1/9|1/9|
.2.| 0 |1/3| 0 |1/6|1/6| 0 |1/9|1/9|1/9|

Let's impose the criterion that the revealed door was not the winning one. Neither A nor B change in that case, since they never reveal the prize. However, C changes, and we have the conditional
P(n,m|n≠m) = P(n,m,n≠m)/P(n≠m) = (1/9) / (6/9) = 1/6,
since every individual case has probability 1/9, but only six of them have n≠m. Thus, C will now look like B.

There is also the requirement that the emperor's door may not be the player's first pick. Again, this does not change the strategy A, but B and C are affected. For n≠m:
P(n,m|m≠0) = P(n,m,m≠0)/P(m≠0) = (1/6) / (4/6) = 1/4,
since all cases where m≠0 and n≠m have probability 1/6 and four of them have n≠m.

The result:

Code: Select all

(n)|m->  A     |     B     |     C     | Mutual information:
Win| 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | A = 2/3     ≅ 0.6667 bits
---+---+---+---+---+---+---+---+---+---+ B = 1/2     = 0.5000 bits
.0.| 0 |1/6|1/6| 0 |1/4|1/4| 0 |1/4|1/4| C = 1/2     = 0.5000 bits
.1.| 0 | 0 |1/3| 0 | 0 |1/4| 0 | 0 |1/4|
.2.| 0 |1/3| 0 | 0 |1/4| 0 | 0 |1/4| 0 |

Sanity check: For A,
P(n=1|m=2) = P(1,2)/( P(0,2)+P(1,2)+P(2,2) ) = (1/3)/(1/2) = 2/3 = ... = P(n=2|m=1)
Thus, switching--i.e., picking the door that was not your first (0) and not the Emperors--gives probability of winning as 2/3. For the other two strategies, switching or keeping your first guess has not effect.

[Fingolfin_Noldor]

^Kuroneko: Damn, thanks for reminding me how the homework problem I solved was done. Thank you sir.

[Ariphaos]

Spoiler

Gwuh. : /
http://beta.anenris.com/rome.php

Code: Select all

<?php

  $CHOICEMAX = 2;

  $emperorchosen = 0;
  $lionchosen = 0;
  $slavechosen = 0;

  for ($i = 0; $i < 100000; $i++)
    {
      $cage = array ();
      for ($j = 0; $j <= $CHOICEMAX; $j++)
        {
          $cage[$j] = 'Lion';
        }

      $cage[mt_rand(0,$CHOICEMAX)] = 'Slave'; // Slave is placed.

      $choice = mt_rand (0,$CHOICEMAX); // Person makes choice.

      $emperorb = mt_rand (0, $CHOICEMAX);

      while ($cage[$emperorb] == 'Slave')
        {
          $emperorb = mt_rand (0, $CHOICEMAX);
        }

      if ($choice == $emperorb)
        {
          $emperorchosen++;
        }
      elseif ($cage[$choice] == 'Slave')
        {
          $slavechosen++;
        }
      else
        {
          $lionchosen++;
        }
    }

  echo 'Emperor reveals player\'s door to be a Lion: '.$emperorchosen.'<br />';
  echo 'Emperor reveals non-player door, player chose slave: '.$slavechosen.'<br />';
  echo 'Emperor reveals non-player door, player chose lion: '.$lionchosen.'<br />';

Kuroneko is quite right.

[Surlethe]

By the way, to answer the semantics question: the OP was intended to be a Monty Hall problem. I always have lost points for imprecision in proofs ... .

[Kuroneko]

I accept the math, but it's really counterintuitive; ...

It's possible to rephrase the problem so that it is fairly intuitive. Spoiler

One way you can think of this is that you're allowed a choice between
1) Sticking with your first choice--the Emperor's information is completely ignored and thus does nothing.
2) Picking the other two doors at the same time, with the Emperor agreeing to get rid of one the losing doors from your dual pick.
If the Emperor's strategy is to pick a losing door that's not the player's, then (2) is completely equivalent to what actually happens. However, if the Emperor could pick any losing door, even outside your "dual pick", it is no longer equivalent, and thus it should not be too surprising that 'filtering by fate' gives a different answer.

I hope that makes a bit more sense.