| Topic: |
Science > Physics |
| User: |
"ca314159" |
| Date: |
09 Feb 2006 07:16:41 AM |
| Object: |
Jailer's Paradox |
Here's a variation on the Monty Hall Problem called
the Jailer's Paradox[1] that's a little bit trickier:
A jailer knows which of three prisoners Adam, Bill, or Charlie
will be executed. Adam asks the jailor to reveal which one
of the other two prisoners will not be executed.
The jailor tells him that Bill will not be executed.
What is the probability now that Charlie will be executed?
This could be "solved" in the same way as the Monty Hall Problem:
Adam Bill Charlie Probability
Ia x + o 1/6
Ib x o + 1/6
II o x + 1/3
III o + x 1/3
x = executed (Monty's car)
+ = who jailor says will not be executed (Monty exposes a goat)
o = curtain
Cases Ia and Ib have a combined probability of 1/3
before the Jailor reveals who will not be executed
(before monty reveals a goat).
So, in the case where the jailer reveals that
Bill will not be executed, there are two possibilities:
Ia x + o 1/6
III o + x 1/3
III is twice as likely as Ia.
But this solution can lead to another paradox:
If Adam and Charlie both privately asked the jailer who
would not be executed and the jailor named Bill in both cases,
then Adam would think that Charlie was twice as likely to be
executed as he was, and Charlie would think Adam was twice as
likely to be executed as he was. Who would be right?
A problem with the previous solution is that Charlie's
chance of being executed seems to have gone from
1/3 to 2/3, or twice that of Adam based on new information
from the jailor. This is unintuitive and somewhat misleading.
A better solution might show Charlie as consistently having a
1/3rd prior probability while his 2/3 relative probability
of being executed comes from the 2/6th chance of _any_ prisoner
being in Charlie's position (that of not being the goat and not being Adam):
Label Prob. Description
x 1/3 chance of being executed
y 1/3 chance of being the person who asks the jailer for info (Monty's contestant)
z 2/3 not y (not the contestant)
p 2/6 not y and not x (goat; not executed)
q 2/6 not p and not y but possibly x
A = probability of being y and being executed = x * (1/3) = 1/9
B = probability of being q and being executed = x * (2/6) = 1/6
A = 2/3 B.
This might be interpreted as:
The probability of being the one out of three prisoners
to ask the jailor "who will not be executed" and being executed,
is 2/3 the probability of being executed and being the one of the
other two prisoners who was not named by the jailor.
In this way we don't say "Charlie is twice as likely to be
executed as Adam", which is misleading since they both have
an intrinsic 1/3rd chance of being executed which does not
change due to their situations.
We could say instead that "Someone in Charlie's situation
is 2/3 as likely to be executed as someone in Adam's situation."
or equivalently that "Someone in Charlie's situation is
twice as likely as Adam to be executed.". This would be
more revealing that any change in their initial 1/3rd
probability was derived from their situations.
[1] Randomness, Deborah J. Bennett, Harvard University Press, 1998
.
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| User: "Matthew T. Russotto" |
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| Title: Re: Jailer's Paradox |
09 Feb 2006 08:33:10 PM |
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In article <43EB40B9.76ACFCC0@bestweb.net>,
ca314159 <ca314159@bestweb.net> wrote:
Here's a variation on the Monty Hall Problem called
the Jailer's Paradox[1] that's a little bit trickier:
A jailer knows which of three prisoners Adam, Bill, or Charlie
will be executed. Adam asks the jailor to reveal which one
of the other two prisoners will not be executed.
The jailor tells him that Bill will not be executed.
What is the probability now that Charlie will be executed?
This could be "solved" in the same way as the Monty Hall Problem:
Adam Bill Charlie Probability
Ia x + o 1/6
Ib x o + 1/6
II o x + 1/3
III o + x 1/3
x = executed (Monty's car)
+ = who jailor says will not be executed (Monty exposes a goat)
o = curtain
Cases Ia and Ib have a combined probability of 1/3
before the Jailor reveals who will not be executed
(before monty reveals a goat).
So, in the case where the jailer reveals that
Bill will not be executed, there are two possibilities:
Ia x + o 1/6
III o + x 1/3
III is twice as likely as Ia.
But this solution can lead to another paradox:
Not surprising, because it's wrong. You're implicitly treating the
jailor's choice as independent of the person being executed when it is
not.
Your first chart shows the probability of the 'x' being executed AND
the '+' being chosen by the jailor. What we want to know is the
probability of the 'x' being executed given the '+' being chosen by
the jailor.
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| User: "Alan Morgan" |
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| Title: Re: Jailer's Paradox |
10 Feb 2006 10:45:40 AM |
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In article <43EB40B9.76ACFCC0@bestweb.net>,
ca314159 <ca314159@bestweb.net> wrote:
Here's a variation on the Monty Hall Problem called
the Jailer's Paradox[1] that's a little bit trickier:
A jailer knows which of three prisoners Adam, Bill, or Charlie
will be executed. Adam asks the jailor to reveal which one
of the other two prisoners will not be executed.
The jailor tells him that Bill will not be executed.
What is the probability now that Charlie will be executed?
[snip]
But this solution can lead to another paradox:
If Adam and Charlie both privately asked the jailer who
would not be executed and the jailor named Bill in both cases,
then Adam would think that Charlie was twice as likely to be
executed as he was, and Charlie would think Adam was twice as
likely to be executed as he was. Who would be right?
Ah, this is a very different situation. In the first case we
know, for a fact, that the Jailer can always name someone and
we can, arbitrarily, assign that person the name Bill. In the
second case, it might not be the case that the jailer can give
the same name to both people. If, for example, Bill was going
to be the one executed then the jailer would have to give the
Adam's name to Charlie and Charlie's name to Adam, which gives
a completely different problem (in the first case, if Bill were
the one being executed the jailer could just give the name
Charlie, which does *not* give a completely different problem).
In short: Adam and Charlie are equally likely to be executed
in this scenario.
Alan
--
Defendit numerus
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| User: "Nick Wedd" |
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| Title: Re: Jailer's Paradox |
09 Feb 2006 03:08:15 PM |
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In message <43EB40B9.76ACFCC0@bestweb.net>, ca314159
<ca314159@bestweb.net> writes
Here's a variation on the Monty Hall Problem called
the Jailer's Paradox[1] that's a little bit trickier:
A jailer knows which of three prisoners Adam, Bill, or Charlie
will be executed. Adam asks the jailor to reveal which one
of the other two prisoners will not be executed.
The jailor tells him that Bill will not be executed.
What is the probability now that Charlie will be executed?
2/3
This could be "solved" in the same way as the Monty Hall Problem:
Adam Bill Charlie Probability
Ia x + o 1/6
Ib x o + 1/6
II o x + 1/3
III o + x 1/3
x = executed (Monty's car)
+ = who jailor says will not be executed (Monty exposes a goat)
o = curtain
Cases Ia and Ib have a combined probability of 1/3
before the Jailor reveals who will not be executed
(before monty reveals a goat).
So, in the case where the jailer reveals that
Bill will not be executed, there are two possibilities:
Ia x + o 1/6
III o + x 1/3
III is twice as likely as Ia.
But this solution can lead to another paradox:
I'm not sure what you think "the solution" was.
If Adam and Charlie both privately asked the jailer who
would not be executed and the jailor named Bill in both cases,
then Adam would think that Charlie was twice as likely to be
executed as he was, and Charlie would think Adam was twice as
likely to be executed as he was. Who would be right?
The jailor was lying. When I answered before, it was on the assumption
that he was honest.
If the jailor can lie, his statements tell nothing.
A problem with the previous solution is that Charlie's
chance of being executed seems to have gone from
1/3 to 2/3, or twice that of Adam based on new information
from the jailor. This is unintuitive and somewhat misleading.
A better solution might show Charlie as consistently having a
1/3rd prior probability while his 2/3 relative probability
of being executed comes from the 2/6th chance of _any_ prisoner
being in Charlie's position (that of not being the goat and not being Adam):
Label Prob. Description
x 1/3 chance of being executed
y 1/3 chance of being the person who asks the jailer for info
(Monty's contestant)
z 2/3 not y (not the contestant)
p 2/6 not y and not x (goat; not executed)
q 2/6 not p and not y but possibly x
A = probability of being y and being executed = x * (1/3) = 1/9
B = probability of being q and being executed = x * (2/6) = 1/6
A = 2/3 B.
This might be interpreted as:
The probability of being the one out of three prisoners
to ask the jailor "who will not be executed" and being executed,
is 2/3 the probability of being executed and being the one of the
other two prisoners who was not named by the jailor.
In this way we don't say "Charlie is twice as likely to be
executed as Adam", which is misleading since they both have
an intrinsic 1/3rd chance of being executed which does not
change due to their situations.
We could say instead that "Someone in Charlie's situation
is 2/3 as likely to be executed as someone in Adam's situation."
or equivalently that "Someone in Charlie's situation is
twice as likely as Adam to be executed.". This would be
more revealing that any change in their initial 1/3rd
probability was derived from their situations.
[1] Randomness, Deborah J. Bennett, Harvard University Press, 1998
Nick
--
Nick Wedd
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| User: "draccarlawpet" |
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| Title: Re: Jailer's Paradox |
10 Feb 2006 11:33:12 AM |
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Isn't the answer 50%?
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| User: "Frank Ferrese" |
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| Title: Re: Jailer's Paradox |
10 Feb 2006 02:15:12 PM |
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How about this? Since Adam knows that one of the prisoners will be
executed, he knows that one of Charlie or Bill will not be executed
before even asking the jailer. There is no need to ask the question
because it doesn't matter what the answer is if you are Adam. Your
odds of being executed are 1/2 from the very beginning based on the
knowledge that you are one of three and one of three will be executed.
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| User: "Nick Atty" |
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| Title: Re: Jailer's Paradox |
11 Feb 2006 03:20:55 AM |
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On 10 Feb 2006 12:15:12 -0800, "Frank Ferrese" <frank.ferrese@navy.mil>
wrote:
How about this? Since Adam knows that one of the prisoners will be
executed, he knows that one of Charlie or Bill will not be executed
before even asking the jailer. There is no need to ask the question
because it doesn't matter what the answer is if you are Adam. Your
odds of being executed are 1/2 from the very beginning based on the
knowledge that you are one of three and one of three will be executed.
Yeah, that's been bothering me since I read it yesterday. It bothers me
in two ways
a) it is clearly right
b) when I first heard the Monty Hall problem I argued in exactly the
same way (he can always do that, so it makes no difference).
I think it probably is relevant in this case, unlike in MH, because
there is no "swapping" stage. In MH the information he gives you (that
this one of the two that doesn't have a car behind) is information you
can use - you can swap to the other. So which one of the two it is
matters. In this problem you can't take any action, so knowing which of
B or C it is doesn't change your circumstances.
Does it?
--
On-line canal route planner: http://www.canalplan.org.uk
(Waterways World site of the month, April 2001)
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| User: "" |
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| Title: Re: Jailer's Paradox |
11 Feb 2006 07:42:17 AM |
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Dont be bothered by it. It is a hoax. A mathematical practical joke.
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| User: "George Weinberg" |
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| Title: Re: Jailer's Paradox |
11 Feb 2006 11:57:23 AM |
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On Sat, 11 Feb 2006 09:20:55 +0000, Nick Atty
<nospam@nandj.freeserve.co.uk> wrote:
On 10 Feb 2006 12:15:12 -0800, "Frank Ferrese" <frank.ferrese@navy.mil>
wrote:
How about this? Since Adam knows that one of the prisoners will be
executed, he knows that one of Charlie or Bill will not be executed
before even asking the jailer. There is no need to ask the question
because it doesn't matter what the answer is if you are Adam. Your
odds of being executed are 1/2 from the very beginning based on the
knowledge that you are one of three and one of three will be executed.
Yeah, that's been bothering me since I read it yesterday. It bothers me
in two ways
a) it is clearly right
I think you mean "clearly wrong". Since exactly one of the
three will be executed, your chance of being the one is initially 1/3.
Saying "Bill won't be executed" in response to your question doesn't
change that.
b) when I first heard the Monty Hall problem I argued in exactly the
same way (he can always do that, so it makes no difference).
Thar depends on what you mean by "do that". MH can always reveal
a goat, but he can only reveal a goat behind the door on the left if
there is in fact a goat behind the door on the left. If he opens the
door on the right, the probability that you initially picked the car
door doesn't change, but the probability that the car was behinf te
door on the right has dropped to zero. In 2 cases out of 3 it's
because he was compelled to open the door on the right by his rules,
and in the third case he could have opened either door but chose to
open the one on the left.
I think it probably is relevant in this case, unlike in MH, because
there is no "swapping" stage. In MH the information he gives you (that
this one of the two that doesn't have a car behind) is information you
can use - you can swap to the other. So which one of the two it is
matters. In this problem you can't take any action, so knowing which of
B or C it is doesn't change your circumstances.
Does it?
Right, it doesn't.
George
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| User: "" |
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| Title: Re: Jailer's Paradox |
11 Feb 2006 01:08:49 PM |
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Revealing information before the problem is completely solved is the
equivalent of posing a completely different question.
What do you supose would happen if you were calculating acceleration
due to gravity where you drop a rock from a tall building, but before
the rack hits ground, the Earth is transformed into planet Jupiter ?
The original question is just plain stupid.
You do NOT change the parameters of a question in mid stream and expect
everything else to remain the same. That goofy Monty Hall problem is
pure idiocy.
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| User: "ca314159" |
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| Title: Quantum Monty Hall Problem |
09 Feb 2006 10:35:27 PM |
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Monty Hall Problem:
http://en.wikipedia.org/wiki/Monty_Hall_problem
Quantum Monty Hall Problem:
Paper:
http://xxx.lanl.gov/abs/quant-ph/0202120
Java Applet:
http://www.imaph.tu-bs.de/qi/monty/
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| User: "" |
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| Title: Re: Quantum Monty Hall Problem |
10 Feb 2006 12:05:09 AM |
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Who on Earth would post such a thing on Wikipedia ?
That's the best mathematical joke I ever read. It's almost like a
running cartoon. A work of pure brilliance.
What's the probability of flipping 5 consecutive tails ?
Answer : 1/2. It's a trick coin. It has 2 headed, or 2 tailed, but you
dont know which.
Comical.
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| User: "Proginoskes" |
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| Title: Re: Quantum Monty Hall Problem |
10 Feb 2006 01:51:11 AM |
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wrote:
Who on Earth would post such a thing on Wikipedia ?
Well, take the number of people who have an Internet connection,
subtract 2, and that's how many people _could_ post such a think on
Wikipedia.
One of the reasons why Wikipedia should not be used as an official
source.
--- Christopher Heckman
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| User: "" |
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| Title: Re: Quantum Monty Hall Problem |
10 Feb 2006 08:13:36 AM |
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I must say though that I did really enjoy reading it.
Next time somebody asks me, I'll just scratch my chin and say "Ahhhhh
yes, the Monty Hall problem...hmmmmmm."
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| User: "ca314159" |
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| Title: The Two Envelopes Paradox |
13 Feb 2006 09:17:59 AM |
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The Two Envelopes Paradox
Monty offers you two envelopes, each containing a check.
You may choose one, keeping the money it contains.
Monty tells you that one envelope contains exactly
twice as much as the other, but does not tell you which is which.
Since you have no way of knowing which envelope contains the
larger sum, you pick one at random.
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
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| User: "Helmut Wabnig" |
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| Title: Re: The Two Envelopes Paradox |
13 Feb 2006 09:55:44 AM |
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On Mon, 13 Feb 2006 10:17:59 -0500, ca314159 <ca314159@bestweb.net>
wrote:
The Two Envelopes Paradox
Monty offers you two envelopes, each containing a check.
You may choose one, keeping the money it contains.
Monty tells you that one envelope contains exactly
twice as much as the other, but does not tell you which is which.
Since you have no way of knowing which envelope contains the
larger sum, you pick one at random.
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
Three possibilities, probability 1/3 each
1 I am at maximum, do not swap,
merit function value= 1/3 * (40000) = 1,3 E3
2 I will get another 40000 (I have already 40000)
if I swap, merit 1/3 * 40000, = 1,3 E3
3 I will lose and have 20000 if I swap merit = 1/3 * 20000 = 0,67 E3
it tells me not to swap because 1 and 2 yield the same merit function
value, and the merit value cannot be increased.
(The win sum can be increased, but only with a certain risk,
and risk has to be excluded).
w.
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| User: "Helmut Wabnig" |
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| Title: Re: The Two Envelopes Paradox |
13 Feb 2006 10:22:02 AM |
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On Mon, 13 Feb 2006 16:55:44 +0100, Helmut Wabnig
<...._.--_.-_-..._-._.._--.@.-_---_-._*_.-_-> wrote:
On Mon, 13 Feb 2006 10:17:59 -0500, ca314159 <ca314159@bestweb.net>
wrote:
The Two Envelopes Paradox
Monty offers you two envelopes, each containing a check.
You may choose one, keeping the money it contains.
Monty tells you that one envelope contains exactly
twice as much as the other, but does not tell you which is which.
Since you have no way of knowing which envelope contains the
larger sum, you pick one at random.
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
Three possibilities, probability 1/3 each
1 I am at maximum, do not swap,
merit function value= 1/3 * (40000) = 1,3 E3
2 I will get another 40000 (I have already 40000)
if I swap, merit 1/3 * 40000, = 1,3 E3
3 I will lose and have 20000 if I swap merit = 1/3 * 20000 = 0,67 E3
it tells me not to swap because 1 and 2 yield the same merit function
value, and the merit value cannot be increased.
(The win sum can be increased, but only with a certain risk,
and risk has to be excluded).
(Forgot the finale):
Actually we have to calculate two decision values
because it's a yes/no 2 state decision
a) no change, equals top 1 in above, merit = 1,3 E3
b) equals a combination of 1/2 top 2 and 1/2 top 3
that is ; 1/2 * (1,3 + 0,67) = 0,985 merit
Decision a > b which tells me not to swap.
Question: above example doubles / halves the sum.
is there a factor say triple / third or higher which
will make us swap the envelopes?
:-)
w.
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| User: "dgates" |
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| Title: Re: The Two Envelopes Paradox |
14 Feb 2006 12:40:04 AM |
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On Mon, 13 Feb 2006 10:17:59 -0500, ca314159 <ca314159@bestweb.net>
wrote:
The Two Envelopes Paradox
Monty offers you two envelopes, each containing a check.
You may choose one, keeping the money it contains.
Monty tells you that one envelope contains exactly
twice as much as the other, but does not tell you which is which.
Since you have no way of knowing which envelope contains the
larger sum, you pick one at random.
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
What's the most I've ever seen him give out? What's the least? Have
I watched enough times that I have a sense of what the average amount
is?
If I feel that I'm a little low of his average, I'm gonna have to
switch. If I feel that I'm right about at his average, I'll have to
think about how far he tends to veer from his average amount.
You've hit a tough number for me. If it were $4k, I'd switch for
sure. If it were $1 million, there's no way I would switch. But
$40,000...
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| User: "Gareth Owen" |
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| Title: Re: The Two Envelopes Paradox |
13 Feb 2006 09:56:15 AM |
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ca314159 <ca314159@bestweb.net> writes:
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
Absolutely. From your complete ignorance of the distribution of the
money, your can only assume that if you swap you have a 50% chance of
$80k and a 50% chance of $20k.
That gives you an "expected return" of $50k.
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| User: "Alec McKenzie" |
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| Title: Re: The Two Envelopes Paradox |
13 Feb 2006 11:40:32 AM |
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Gareth Owen <usenet@gwowen.freeserve.co.uk> wrote:
ca314159 <ca314159@bestweb.net> writes:
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
Absolutely. From your complete ignorance of the distribution of the
money, your can only assume that if you swap you have a 50% chance of
$80k and a 50% chance of $20k.
That gives you an "expected return" of $50k.
I'm afraid not. If there were any merit in such reasoning it
would apply just as well to swapping envelopes without even
opening the first. And again by the same reasoning to swap them
again, and again . . .
--
Alec McKenzie
un2312.e.armck@xoxy.net
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| User: "The Qurqirish Dragon" |
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| Title: Re: The Two Envelopes Paradox |
13 Feb 2006 12:03:18 PM |
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Gareth Owen wrote:
ca314159 <ca314159@bestweb.net> writes:
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
Absolutely. From your complete ignorance of the distribution of the
money, your can only assume that if you swap you have a 50% chance of
$80k and a 50% chance of $20k.
That gives you an "expected return" of $50k.
If you want to maximize your expected value, yes. However, I would
probably say, before even looking at the envelope, "If there is at
least $X in here, but not more than Y, I am keeping it"
For example, I would probably keep the $40,000 in the example, since
for me the difference between $40,000 and $20,000 is too big.
If I opened it and there was $5, I wouldn't really care if I lost
$2.50, so I'd switch.
If I opened it, and there was $30,000,000, then since I have no idea
what I would do with the $15,000,000 if I lost, these values are
effectively the same for me, and I may decide to switch, since I would
be more than happy with the smaller amount anyway.
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| User: "Oliver Wong" |
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| Title: Re: The Two Envelopes Paradox |
13 Feb 2006 01:00:03 PM |
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"The Qurqirish Dragon" <qurqirishd@aol.com> wrote in message
news:1139851196.794275.191190@g14g2000cwa.googlegroups.com...
Gareth Owen wrote:
ca314159 <ca314159@bestweb.net> writes:
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
[...]
If I opened it, and there was $30,000,000, then since I have no idea
what I would do with the $15,000,000 if I lost, these values are
effectively the same for me, and I may decide to switch, since I would
be more than happy with the smaller amount anyway.
If I were playing this game in real life, and I opened first the
envelope and found $30 million, I'd probably get incredibly excited and
unable to think straight. Therefore, the best strategy for me would be to
keep that money, less in my excitement I misunderstood some part of the
rules and somehow lose all the money.
- Oliver
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| User: "dgates" |
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| Title: Re: The Two Envelopes Paradox |
14 Feb 2006 12:44:25 AM |
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On Mon, 13 Feb 2006 19:00:03 GMT, "Oliver Wong" <owong@castortech.com>
wrote:
"The Qurqirish Dragon" <qurqirishd@aol.com> wrote in message
news:1139851196.794275.191190@g14g2000cwa.googlegroups.com...
Gareth Owen wrote:
ca314159 <ca314159@bestweb.net> writes:
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
[...]
If I opened it, and there was $30,000,000, then since I have no idea
what I would do with the $15,000,000 if I lost, these values are
effectively the same for me, and I may decide to switch, since I would
be more than happy with the smaller amount anyway.
If I were playing this game in real life, and I opened first the
envelope and found $30 million, I'd probably get incredibly excited and
unable to think straight. Therefore, the best strategy for me would be to
keep that money, less in my excitement I misunderstood some part of the
rules and somehow lose all the money.
Ahh, yes, misunderstanding the rules is a huge issue here. Unless
I've got lawyers going over the exact wording, I'd be very afraid that
I had missed some subtlety, and that would make me more prone to stay
with a sure amount.
Oh, also, if I would never find out what's in the other envelope, I'd
be more likely to keep the money. But if he were going to show me
regardless of whether or not I switch, then I'd be more likely to
switch.
I'd rather kick myself for having lost $20k than for having lost $40k!
:-)
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| User: "George Weinberg" |
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| Title: Re: The Two Envelopes Paradox |
13 Feb 2006 11:27:41 AM |
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On 13 Feb 2006 15:56:15 +0000, Gareth Owen
<usenet@gwowen.freeserve.co.uk> wrote:
ca314159 <ca314159@bestweb.net> writes:
Monty asks you to open the envelope. You do so and take out a
check for $40,000.
Monty now says you have a chance to change your mind and
choose the other envelope. Should you swap?
Absolutely. From your complete ignorance of the distribution of the
money, your can only assume that if you swap you have a 50% chance of
$80k and a 50% chance of $20k.
That gives you an "expected return" of $50k.
You're kidding, right?
To the OP: This one has been done to death.
George
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| User: "Mike Williams" |
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| Title: Re: Jailer's Paradox |
09 Feb 2006 07:46:10 AM |
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Wasn't it ca314159 who wrote:
Here's a variation on the Monty Hall Problem called
the Jailer's Paradox[1] that's a little bit trickier:
A jailer knows which of three prisoners Adam, Bill, or Charlie
will be executed. Adam asks the jailor to reveal which one
of the other two prisoners will not be executed.
The jailor tells him that Bill will not be executed.
What is the probability now that Charlie will be executed?
The only added complexity over the normal Monty Hall situation is the
potential for confusion about the frame of reference.
In Monty Hall, the frame of reference is always that of the contestant.
The only other possible frame of reference is that of Monty himself, and
he knows which door has the car. In Monty's frame of reference the
probability of the car being behind that door is one and the probability
of it being behind either of the other two doors is zero.
In the Jailer's Paradox, we now have four frames of reference to
consider. That of Adam, Charlie and the Jailer, but also the frame of
reference of someone (ourselves) who has been told all the information
available to both Adam and Charlie.
The conditional probabilities of the executions conditional on the
information revealed to *either* Adam *or* Charlie are the same as in
the Monty Hall situation.
The conditional probabilities of the executions conditional on the
information revealed both Adam *and* Charlie are 1/2 to 1/2. Only two of
the original possibilities remain, and they are equally likely.
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| User: "Matthew T. Russotto" |
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| Title: Re: Jailer's Paradox |
09 Feb 2006 08:35:30 PM |
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In article <fTNULCAie06DFwkX@econym.demon.co.uk>,
Mike Williams <nospam@econym.demon.co.uk> wrote:
Wasn't it ca314159 who wrote:
Here's a variation on the Monty Hall Problem called
the Jailer's Paradox[1] that's a little bit trickier:
A jailer knows which of three prisoners Adam, Bill, or Charlie
will be executed. Adam asks the jailor to reveal which one
of the other two prisoners will not be executed.
The jailor tells him that Bill will not be executed.
What is the probability now that Charlie will be executed?
[...]
The conditional probabilities of the executions conditional on the
information revealed to *either* Adam *or* Charlie are the same as in
the Monty Hall situation.
The conditional probabilities of the executions conditional on the
information revealed both Adam *and* Charlie are 1/2 to 1/2. Only two of
the original possibilities remain, and they are equally likely.
On the contrary: Adam's chance of getting executed is 1/3, and
Charlie's is 2/3. (Bill, of course, is 0).
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| User: "Al Zenner" |
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| Title: Re: Jailer's Paradox |
09 Feb 2006 09:26:08 PM |
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ca314159 <ca314159@bestweb.net> wrote in
news:43EB40B9.76ACFCC0@bestweb.net:
Here's a variation on the Monty Hall Problem called
the Jailer's Paradox[1] that's a little bit trickier:
A jailer knows which of three prisoners Adam, Bill, or Charlie
will be executed. Adam asks the jailor to reveal which one
of the other two prisoners will not be executed.
The jailor tells him that Bill will not be executed.
What is the probability now that Charlie will be executed?
This could be "solved" in the same way as the Monty Hall Problem:
<snip>
This is *not* a Monty Hall Problem variation. In Monty Hall problems
there is no new information transmitted. In this example new
information is introduced. In Monty Hall the contestant makes a
selection without having any information other than the fact that
one choice will bring a reward. Allowing a change in the choice
is merely an excitement enhancer which doesn't alter the conditions
of the original problem.
In this problem:
Before information is exchanged:
Jailer knows who will be executed. His belief is 1/1
Adam, Bill, and Charlie all have a belief of 1/3
Then information is exchanged:
Jailer knows who will be executed. His belief is 1/1
Adam - knows Bill not executed. His belief is 1/2
Adam is the only one of the 3 to believe his chances
got worse for having the knowledge, though it has been
predetermined who is to be executed, so only his belief
changes, the facts do not.
Bill knows nothing new. His belief is 1/3
Charlie knows nothing new. His belief is 1/3
In this problem the discussion is not about the actual chance
of execution which has previously been determined, but it is
only about the belief of each prisoner as to what the chances
are that he has been selected for execution. Changes in the belief
of any prisoner do not affect the belief of any other prisoner
because there is no causality involved. As the problem is written
there is no exchange of information among prisoners nor can the
other prisoners overhear the discussion between Adam and the jailer.
Throwing Monty Hall into the discussion serves only to confuse.
There is no paradox involved in this problem which yields politely
to an ordinary frontal assault style analysis.
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| User: "" |
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| Title: Re: Jailer's Paradox |
09 Feb 2006 10:02:47 PM |
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Prior to the information being revealed to Adam, each has equal
probability of 1/3 that he will be executed.
When it is revealed that Bill will not be executed, the probability for
Charlie becomes 1 3rd in 2 3rds, and the probability for Adam is also 1
3rd in 2 3rds.
Also written as 1/3 : 2/3 which happens to equal 50 : 50.
I happen to have firsthand knowledge of this paradox and the confusion
it causes to prisoners awaiting execution. But there really is a
contradiction in all of this - and that would be just how in the heck
people who cant even count are supposed to reckon probabilities in the
n'st place ?
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| User: "Ben Rudiak-Gould" |
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| Title: Re: Jailer's Paradox |
09 Feb 2006 01:35:12 PM |
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ca314159 wrote:
A jailer knows which of three prisoners Adam, Bill, or Charlie
will be executed. Adam asks the jailor to reveal which one
of the other two prisoners will not be executed.
The jailor tells him that Bill will not be executed.
What is the probability now that Charlie will be executed?
Here's another puzzle that I hope will elucidate the correct answer to yours.
Problem: On one side of a card is the sentence "the sentence on the other
side of this card is false"; on the other side is the sentence "the sentence
on the other side of this card is true". How is this possible?
Solution: It's possible because there's no law of physics preventing someone
writing anything they want on a piece of paper. If you believe either
sentence then you're inconsistent, but the problem didn't tell you to
believe them, nor give any reason why you should.
It's the same with your jailor paradox. Without some knowledge of the
jailor's motivations, the jailor's statement tells me nothing at all, not
even that Bill won't be executed. To make the problem solvable you have to
turn the jailor into an automaton following a known algorithm, and you have
to give a prior probability of execution for each prisoner. Otherwise
there's no correct answer, and trying to find one is futile.
This is also true of the Monty Hall problem. When correctly stated it has
the oft-quoted surprising answer, but it's usually not correctly stated.
-- Ben
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| User: "Ben Rudiak-Gould" |
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| Title: Re: Jailer's Paradox |
11 Feb 2006 07:51:26 PM |
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I wrote:
To make the problem solvable you have to turn the jailor into an
automaton following a known algorithm, and you have to give a prior
probability of execution for each prisoner. Otherwise there's no correct
answer, and trying to find one is futile.
There are now quite a lot of wrong answers to the problem in this thread, so
I think I'll elaborate a bit on what I said.
Suppose that exactly one prisoner will be executed (which wasn't explicit in
the problem statement) and let the prior probability of execution for each
be 1/3 (which doesn't follow from the problem statement, and is not
obviously correct).
Now suppose that the jailor behaves as follows: when asked who from a given
set won't be executed, she answers truthfully; if there's more than one
truthful answer, she prefers Adam over Bill and Bill over Charlie.
Then there are three equally probable outcomes:
Adam will be executed; jailor answers Bill.
Bill will be executed; jailor answers Charlie.
Charlie will be executed; jailor answers Bill.
The actual answer, Bill, means we can eliminate the second option. Thus
Adam, if he's a good Bayesian, will assign a probability of 1/2 to his own
execution and 1/2 to Charlie's, after hearing the jailor's answer.
Now suppose we modify the jailor's behavior: instead of disambiguating in
favor of particular prisoners, she flips a coin where necessary to decide
who to name. Then we have four outcomes, two of which are half as likely as
the others because there's a coin flip involved:
Adam will be executed; jailor answers Bill. (1/2)
Adam will be executed; jailor answers Charlie. (1/2)
Bill will be executed; jailor answers Charlie. (1)
Charlie will be executed; jailor answers Bill. (1)
This time the answer allows Adam to eliminate the two middle possibilities,
which leaves
Adam will be executed; jailor answers Bill. (1/2)
Charlie will be executed; jailor answers Bill. (1)
and so Adam should assign a probability of 1/3 to his own execution and 2/3
to Charlie's. Everyone who's given this answer to the problem has been
assuming this prior for the prisoners and this behavior for the jailor,
whether they realize it or not.
The same analysis applies to the Monty Hall problem. Suppose that when Monty
has a choice of two doors to open, he always chooses the one on the left,
rather than flipping a coin. Then if you choose the leftmost door and he
opens the middle one, there's no advantage (nor disadvantage) to switching
to the rightmost one.
-- Ben
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| User: "Phil Carmody" |
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| Title: Re: Jailer's Paradox |
12 Feb 2006 03:27:44 AM |
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Ben Rudiak-Gould <br276deleteme@cam.ac.uk> writes:
Now suppose we modify the jailor's behavior: instead of disambiguating
in favor of particular prisoners, she flips a coin where necessary to
decide who to name. Then we have four outcomes, two of which are half
as likely as the others because there's a coin flip involved:
Adam will be executed; jailor answers Bill. (1/2)
Adam will be executed; jailor answers Charlie. (1/2)
Bill will be executed; jailor answers Charlie. (1)
Charlie will be executed; jailor answers Bill. (1)
This time the answer allows Adam to eliminate the two middle
possibilities, which leaves
Sorry if I'm being thick, but what happened to
Bill will be executed; jailor answers Adam.
Charlie will be executed; jailor answers Adam.
?
Phil
--
What is it: is man only a blunder of God, or God only a blunder of man?
-- Friedrich Nietzsche (1844-1900), The Twilight of the Gods
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