You keep thinking about it as what would happen if you play repeatedly. I'm just talking about playing a single game. — Michael
Every amount has a 50% chance of winning for a 2:1 payout. — Michael
(2) Michael and @andrewk are right about the expected value of the other envelope, and this is a genuine paradox. (What they're wrong about is thinking swapping works, since it's really obvious that it does not and cannot.) — Srap Tasmaner
(1) There is a flaw somewhere in the way Michael and @andrewk recommend we should estimate the value of the remaining envelope. If there is, it's not exactly obvious. — Srap Tasmaner
I'm just doing this:
1. We pick an envelope at random
2. There's a 50% chance that my envelope is the X envelope and a 50% chance that my envelope is the 2X envelope.
3. I open my envelope and see £10
4. From 2 and 3, there's a 50% chance that my £10 envelope is the X envelope and a 50% chance that my £10 envelope is the 2X envelope.
5. From 4, there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5.
We seem to agree on 1-3, but you disagree with 4 and/or 5? — Michael
I have numbered the steps from 0 to 4. Which step(s) do you not believe? — andrewk
Expected value in B = 1/2 ( (Expected value in B, given A is larger than B) + (Expected value in B, given A is smaller than B) )
If we then take the sum in one envelope to be x and the sum in the other to be 2x the expected value calculations becomes:
Expected value in B = 1/2 (x + 2x)
(3) Michael and @andrewk have part of the right analysis, but something more is needed. — Srap Tasmaner
I agree with both of these. The point about the maximum possible amount has been made before, but has been lost in the length of the thread. I mentioned it somewhere early on, and in the interests of simplicity, declined to probe further because (1) I thought digging into it would have required some fancy notation like integral signs, and I couldn't remember how to invoke latex on this site (Edit: I see from the above that Michael has mastered it) and (2) I was pretty confident it would not make a material difference.I disagree that the 50% chance can be assumed. If the sample space is finite, then there is a non-zero chance that the $10 amount is the maximum amount. — Andrew M
The point about the maximum possible amount has been made before, but has been lost in the length of the thread. I mentioned it somewhere early on, and in the interests of simplicity, declined to probe further because (1) I thought digging into it would have required some fancy notation like integral signs, and I couldn't remember how to invoke latex on this site and (2) I was pretty confident it would not make a material difference. — andrewk
This can't be written as B = 1/2 (x + 2x) without conflating different values of x. — Michael
If A and B are sets, then A is called a subset of B if, and only if, every element of A is also an element of B.
A sample space is the set of all possible outcomes of a random process or experiment. An event is a subset of a sample space.
You just write it as .
is whatever's in the envelope in your hand. — Srap Tasmaner
As I have said many times over, the sample space of the other envelope is [X,2X]. — Jeremiah
Your sample space becomes:
[[10=X,10=2X],[10=X,10=2X]]. — Jeremiah
Therefore your sample space could never be [5,20], as it does not follow the form of [X, 2X]. Your sample space could be [5,10] or [10,20] but there is no way you can mathematically justify a sample space of [5,20] as it is not consistent with [X,2X]. — Jeremiah
Intuitively this seems like new information which would call for you to update your prior; however, it still does not tell you if are you in case R or case S. You don't know if that is 10=X or 10=2X. So in truth you can't really update your prior, as your prior was based on the uncertainty of being in case R or case S. — Jeremiah
Ya, jumped the gun there and made a mistake myself (unlike you I am able to admit that). However, if you going to use my argument to justify your position, my argument, that you referenced there, still completely counters your statement of [5,20] and proves you 100% wrong. By using my argument as proof you have just admitted that you are wrong. — Jeremiah
Let's review some definitions here:
If A and B are sets, then A is called a subset of B if, and only if, every element of A is also an element of B.
A sample space is the set of all possible outcomes of a random process or experiment. An event is a subset of a sample space.
Discrete Mathematics, An Introduction to Mathematical Reasoning, By Susanna S. Epp
Do you understand that these are not my definitions? They are established definitions that may be used in formal proofs.
Let's call the two envelopes A and B. Now envelope A could have X or A could have 2X and likewise B could have X or B could have 2X. Those are all the possible outcomes so by the definition of a sample space our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X], which means our sample space could also be written as [[X,2X],[X,2X]].
Did you follow that? They are sets in a set. In this case the elements of our sample space are the set [X, 2X]. So that means since your proposed solution does not follow the format of [X,2X] then, by the definition of a subset, it cannot be a subset of either A or B. — Jeremiah
Are you sure that was me? It doesn't sound familiar, but maybe I've just forgotten. Can you link the exchange?And I asked you last night about the minimum possible value of X. Your response then was that what matters is not what might or might not be put in the envelope, but whether you the player have reason to exclude a given value. — Srap Tasmaner
Are you sure that was me? It doesn't sound familiar, but maybe I've just forgotten. Can you link the exchange? — andrewk
What in the rules of the game or in the knowledge you have acquired (i.e., that Y = 10) assures you that the games mistress includes both 5 and 10 in the sample space for X? Maybe a tenner is the smallest bill she has. — Srap Tasmaner
It's not about the games mistress's sample space. She doesn't have one. She has two amounts that she has always intended to put into the envelopes. It's about my sample space, which consists of the events that I cannot rule out based on my knowledge to date.
That knowledge - based on seeing 10 in my envelope - allows me to rule out every other event (an event being a particular sum being in the other envelope) except for 5 and 20. So that's my sample space - just two events. — andrewk
Let's call the two envelopes A and B. Now envelope A could have X or A could have 2X and likewise B could have X or B could have 2X. Those are all the possible outcomes so by the definition of a sample space our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X], which means our sample space could also be written as [[X,2X],[X,2X]]. — Jeremiah
we can also model the other envelope using the sample space [X, 2X], despite the fact that in doing so we're positing two different possible values of X (either 5 or 10). — Michael
Get involved in philosophical discussions about knowledge, truth, language, consciousness, science, politics, religion, logic and mathematics, art, history, and lots more. No ads, no clutter, and very little agreement — just fascinating conversations.