I tell you that heads is worth £30. What is the probability that it landed on the more valuable side? 1/2. You knowing that it landed on heads and that heads is worth £30 makes no difference as you don't know which of heads and tails is the more valuable or how valuable the more valuable side is. — Michael
Open or don't open, with that framing, there's no gain from switching — fdrake
The puzzle is to find the flaw in the line of reasoning in the switching argument.
...
In particular, the puzzle is not solved by finding another way to calculate the probabilities that does not lead to a contradiction.
I don't think the problem with this formula is with its probability assignments. If I know that one envelope contains more money than the other, and if I pick one at random, then whether I open it or not the probability that I picked the envelope with the more money is — Michael
Probability assignments are done with respect to a space of events. — fdrake
I'm not sure what you mean. Perhaps you could answer the questions I posed earlier? — Michael
I do agree, given what you just said and your framing, that the calculation of gain is correct. What's wrong is the framing, not the calculation. — fdrake
The framing is the paradox. I pick one of two envelopes at random. One is twice the value of the other. Given that the probability that I picked the more valuable envelope, it is rational to switch. But also to then switch back. — Michael
Case A, you open it:
A3 ) If you open it and see 10, you don't know if your 10 is in the (5,10) pair or the (10,20) pair.
A4 ) Each of those is equally likely.
A5 ) Assume you're in the (5,10) pair, switching has 0 gain there under equal probability and expected loss.
A6 ) Assume you're in the (10,20) pair, switching has 0 gain under the same assumptions.
A7 ) The expected value of switching is 0. — fdrake
B3 ) If you don't open it, you don't know if your pair is the (5,10) pair or the (10,20) pair.
The reasoning is exactly the same. — fdrake
If there is a 50% chance that I am in (5, 10) and a 50% chance that I am in (10, 20), and if I have £10, then there is a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. According to — Michael
↪fdrake If I have a red ball, a blue ball, and a white ball, and if I pick two at random and put one in one package and another in another package, and if you open one and find a red ball, it is perfectly correct for you to say that the probability that the ball in the other package is blue is 50% and that the probability that the ball in the other package is white is 50%, even though the sample space, given that one ball is known to be red, is either (red, blue) or (red, white). — Michael
Exactly. You asked me to pick one, then treated that like drawing a ball from the bag. — fdrake
If I have a red ball, a blue ball, and a white ball, and if I pick two at random and put one in one package and another in another package, and if you open one and find a red ball, it is perfectly correct for you to say that the probability that the ball in the other package is blue is 50% and that the probability that the ball in the other package is white is 50%, even though the sample space, given that one ball is known to be red, is either (red, blue) or (red, white). — Michael
Case A, however, does not have the agent aware that the possible values in the other envelope are 5 or 20. — fdrake
They are aware. They are told before the experiment starts that one envelope contains twice as much as the other. They open their envelope to find £10. They know that the possible values in the other envelope are £5 or £20. — Michael
The puzzle is to find the flaw in the line of reasoning in the switching argument.
Do you understand the difference between case A and case C? — fdrake
Yes, and I think it is perfectly correct to say, in case A, that the probability that the other envelope contains £20 is 50%. — Michael
You both seem to be mixing up the participant's subjective assessment and some God's eye view objective assessment. — Michael
When you switch, you don't know if you're in case A subcase 1 or case A subcase 2. So you average the gain of switching over each of those. Which is 0. — fdrake
If you're in subcase 1 then you lose £5 by switching. If you're in subcase 2 then you gain £10 by switching. There is a 50% chance that you're in subcase 1 and a 50% chance that you're in subcase 2. So according to the E(z) calculation, it is rational to switch. — Michael
You don't know whether you're in Case A subcase 1 or case A subcase 2. Each of those has probability half. If you're in case A subcase 1, if you switch you gain 5 or lose 5. If you're in case A subcase 2, if you switch you gain 10 or lose 10. Each of those has 0. — fdrake
If you're in subcase 1 and you have £10 and you switch then you lose 5, if you're in subcase 2 and you have £10 then you gain £10. Each of these is equally likely, hence E(z) suggesting you should switch. — Michael
Nevertheless, if they observe n=10 in the first envelope, I still think there's a problem with assigning a probability distribution on the values (5, 20) in the other envelope. This is because that stipulates there being three possible values in the envelopes combined; (5, 10, 20); whereas the agent knows only two are possible. [...] — fdrake
cases=matrix(c(c(10,5),c(10,20)),nrow=2,ncol=2,byrow=FALSE) number_of_envelope_pairs=10000 resulting_envelope_value_C=rep(0,number_of_envelope_pairs) #Case_C for(i in 1:number_of_envelope_pairs){ my_envelope_pair=cases[,rbinom(1,1,0.5)+1] after_switch=my_envelope_pair[2] #this chooses "the other one than 10" resulting_envelope_value_C[i]=after_switch } case_C_gain=mean(resulting_envelope_value_C-10) case_C_gain #can see it's 2.5
#Something that sounds like Case_A my_envelopes=rep(0,number_of_envelope_pairs) my_envelopes_switch_gain=my_envelopes for(i in 1:number_of_envelope_pairs){ case_index=rbinom(1,1,0.5)+1 envelope_index=rbinom(1,1,0.5)+1 my_envelopes[i]=cases[envelope_index, case_index] my_envelopes_switch_gain[i]=cases[-envelope_index, case_index]-my_envelopes[i] } mean(my_envelopes_switch_gain) #approx 0, this is if you don't open the envelope #illustrating the conditioning, NB this is not the same as conditioning on #the pair being (5,10) or the pair being (10,20) #this is the gain given the chosen envelope is 10 and the #other envelope is known to be 5 or 20. mean(my_envelopes_switch_gain[my_envelopes==10]) #this gives you approx 2.5 as we saw in case C.
#Thing which is actually Case_A my_envelopes=rep(0,number_of_envelope_pairs) my_envelopes_switch_gain=my_envelopes for(i in 1:number_of_envelope_pairs){ what_case_am_i_in_given_i_have_10=rbinom(1,1,0.5)+1 #random assignment of case, 10 provides no knowledge of case five_ten_switch_gain_given_random_envelope=0 #conditioning on pair being 5,10, gain is known ten_twenty_switch_gain_given_random_envelope=0 #conditioning on pair being 10,20, gain is known case_gains=c(five_ten_switch_gain_given_random_envelope, ten_twenty_switch_gain_given_random_envelope) my_envelopes_switch_gain[i]=case_gains[what_case_am_i_in_given_i_have_10] } mean(my_envelopes_switch_gain)#this is just 0
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.