You are led into a room where everyone will be shot if two dice come up double sixes, but a sign says at...
probability/shooting.room New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
You are led into a room where everyone will be shot if two dice come up double sixes, but a sign says at least 90% of all who ever enter will be shot. How worried should you be? You awake from a drugged sleep, and along with a group of other people (but you don't know how many) you are led into a large room. You are told that a pair of regular dice will be rolled, and if they come up double sixes, everyone in the room will be shot. Otherwise you will all be set free. You are worried, but not *too* worried. Then you notice a sign on the wall. It says, "At least 90% of all people who ever enter this room will be shot." Your captors explain, and by their explanation you are satisfied that the sign is true. What's going on? Your captors are playing the following morbid game. They lead a single person into the room, and roll the dice. If the dice come up double sixes, they kill the person, and the game is over. Otherwise, the person is set free, and they lead nine other people into the room. Then they roll again. Double sixes, shoot 'em all. Otherwise, set them free and lead another 90 people into the room. Roll again.... And so on, at each subsequent stage n leading another 9*10^n people into the room until the dice come up double sixes. Clearly the sign is true. Now, what are your chances of surviving? How worried should you be? Everyone I've asked says that your chances of surviving are good, your chance of being shot only 1/36. (But shouldn't the same reasoning apply to each and every person who enters? Yet 90% of them will be shot!) But now suppose that the morning after the game ends, an account is printed in the newspaper, and a complete list of everyone who was led into the room is published. Your mother is reading the story, and she sees your name on the list (so here is what she knows: that you were led into the room). How worried should *she* be? Everyone I've asked says that she should be very worried, she should take your chance of being shot to be 90%. So here's the question. Is it really correct for you and your mother to have different views of the probability of your being (or having been) shot? If so, what is the difference in the information you each have? (This paradox is due to John Leslie.) -- Jamie Dreier <pl436000@brownvm.brown.edu> [ Editor: the first paragraph is the editor's summary. ]
Show solution
(ring ring)
ME: Hello?
MA: Melvin! Thank goodness you're alive!
ME: Why shouldn't I be alive?
MA: I saw your name in the paper! You were in that Shooting Room thing,
weren't you?
ME: Oh yes. But fortunately I survived. The odds weren't too bad.
PA (on another extension): Son, at least nine-tenths of the people whose
names were printed in that list are dead.
ME: What? I estimated my chance of survival as 35/36.
PA: How can that be? (Checks himself) Obviously, we had different
information.
ME: Right. Let Ei be the event that I was led into the room at step i,
and S be the event that I was shot. Clearly, Pr(S|Ei)=1/36. When I
got led into the room, I looked around. There were around 90,000
people in the room, so E5 obtained. Hence my estimate of Pr(S) was
1/36.
PA: Yes, that's right. Now let Fj be the disjunction of all Ei with
i<=j. In the paper we read that F(infinity) held for you; but
Pr(S|F(infinity)) >= 9/10.
But you didn't have to condition your estimate on E5. For any n, if
you had used En, your answer would have been the same.
ME: That's true. I could also have used any Fn, for n finite. Since
the paper said the game lasted 17 stages, you really had the
information F17; but you knew when the game ended, which I did not.
If I had been in the room at the fateful 17th stage, I would still
have estimated my probability of survival at 35/36, but this would
have been optimistic.
PA: It's strange, though, that Pr(S) should be different when
conditioned on any Ei and on their (disjoint) disjunction.
MA (annoyed): It's just countable infinity rearing its head again,
Patel!
PA: I'm sorry?
MA: In a countably additive probability space, if the Ei were disjoint
positive probability events and Pr(S|Ei)=p for all i [Editor: "for
all p" in the original], we would necessarily also have
Pr(S|(union Ei))=p. But the Shooting Room assumes a countably
infinite set Q of people and some sort of random choice from Q.
And, as we all know---
MA+PA: ---You can't put a countably additive measure on a countably
infinite set.
PA: We can restore the Kolmogorov axioms by looking at the situation
overall and not from one person's point of view. Let xi be the
number of people led into the room at stage i (or 0 if the game
terminates before then), yi the number shot at stage i, Xi the sum
of all xj, where j<=i, and Yi the sum of all yj, for j<=i.
Now E(yi) = 1/36 E(xi) for all i, and therefore E(Yi) = 1/36 E(Xi)
for all finite i, but even in the usual sort of probability space
there is no problem in making Y(infinity) >= 9/10 X(infinity) with
probability 1. Of course, the Monotone Convergence Theorem then
requires that we have E(Y(infinity)) = E(X(infinity)) = infinity,
and this is clearly true in our scenario. In fact, we have just
reinvented something like the St. Petersburg paradox.
MA: I wonder how those Shooting Room guys knew they wouldn't run out of
people?
ME: They just had to be able to carry out their threat eventually.
Remember President Asimov's election slogan?
PA: I didn't vote for him.
ME: But you remember the slogan, surely.
PA (disapprovingly): "An all-human galaxy and population growth without
bound."
ME: There you are.
PA: You know, I think I remember reading about this before. It was when
I was taking a course in pre-Diaspora philosophers.
ME: Do you remember anything else?
PA: One of those guys came up with a very strange argument. He thought
that the human race would either die off quickly or expand
throughout the galaxy. Then he viewed himself as a random sample
from the set of all humans that had lived or would ever live. But
he was pre-expansion, and the chance of such a sample being
pre-expansion was much larger with an early death for humanity. It
followed that there was a large Bayes factor in favor of a quick
die-off.
MA: What an idiotic argument!
ME: How so?
MA: Isn't all science dependent on one's circumstances? Obviously, any
estimate he might make of the probability of die-off had already
implicitly incorporated the facts about when he was born. It would
be double counting to use it again.
PA: I suppose you're right, Martha.
MA: Always!
Well, we'd better say good-bye now.
MA+PA: Good-bye.
ME: Good-bye.
-- David Moews <dmoews@xraysgi.ims.uconn.edu>
And in a follow-up:
I forgot to say that a _uniform_ countably additive measure (i.e., one
that assigns each singleton set the same value) is what cannot exist.
You can easily construct non-uniform measures.
-- David Moews
Jamie Dreier wrote:
>
> It might be interesting to see how things would be different if there
> were an upper limit on the number of people available. Let it be very
> very high, say 10^100, so that the chance of the dice coming up double
> six some time in the sequence is very high.
Interestingly enough, even with 10^100 people, everyone escapes death
nearly 6% of the time!
If there is 1 person, the chance everyone survives is 35/36
If there are 10 people, the chance everyone survives is (35/36)^2
If there are 10^n people, the chance everyone survives is (35/36)^(n+1)
If there are 10^100 people, the chance everyone survives is:
(35/36)^(101) = .058
Obviously, the small chance that such a huge number of people survive
balances out the fact that usually most of the people who enter the room
get shot, and _on average_ (say if we perform this experiment many
times), 1 out of every 36 who enter the room will be shot. Which is of
course correct... Everyone who enters the room has a 35/36 chance of
survival, so given finite people, 35/36 of them will survive on average.
Duh. :) (Obvious after I thought about it for a couple days, but it
wasn't at all clear to me what was going on when I first heard the
puzzle. Great paradox!)
-- Alex Zamanian <azamania@bbn.com>