Re: [R] Two envelopes problem

From: Greg Snow <Greg.Snow_at_imail.org>
Date: Wed, 27 Aug 2008 12:27:49 -0600

> -----Original Message-----
> From: Mario [mailto:mdosrei@nimr.mrc.ac.uk]
> Sent: Monday, August 25, 2008 5:41 PM
> To: Greg Snow
> Cc: r-help_at_r-project.org
> Subject: Re: [R] Two envelopes problem
>
> Dear Greg,
>
> The problem is that in your code you are creating a
> distribution where there are only 5-10 and 10-20 pairs. Yes,
> if I knew that there are only those two types of pairs, and
> if I new that the probability of each pair was .50, .50, then
> it is advantageous to switch, but that is because I have a
> priori information on the distribution of the pairs. Now lets
> assume, I opened the envelope, and I see 20, should I
> switch? Well, now it depends. Are you going to rewrite the
> simulation so that line 1 within tmpfun reads x <-
> sample(c(10,20), 1)? otherwise is not going to work. The
> question is, if I see 20, according to my friend's argument,
> I should switch, since there is a 50% chance of seeing 40 and
> a 50% chance of seeing 10. However, in your simulation, 40's
> are never seen, so under your simulator, switching every time
> you see a 20 is a sure loss. You see, that's the problem.
> You don't know the distribution of pairs, the fact that
> you've got a tenner note, does not give you any additional
> information. By the way, you could run your example with my
> own code (which is faster as I'm not using sample for the
> creation of the env pairs), just define the function
>
> r5.10 <- function(n) return (sample(c(5,10), n, rep=T))
>
> and now:
>
> env <- generateenv(r=2, r5.10, n=1e6)
> i10 <- which(env[,1]==10)
> mean(env[i10,1]) # Exactly 10
> mean(env[i10,2]) # ~ 12.50 you do get a definite advantage
> when switching
>
> But this is an example that was tailored to work with the
> actual value of 10.

Yes, conditional on you seeing 10, the only posibilities for x is 5 or 10. If you want a more general approach, try this:

> generateenv <- function (r, rdist, n, ...) {

+   env <- matrix(0, ncol=2, nrow=n)
+   env[,1] <- rdist(n, ...)  # first envelope has `x'
+   env[,2] <- r*env[,1]      # second envelope has `r*x'
+
+   # randomize de envelopes, so we don't know which one from
+   # the pair has `x' or `r*x'
+   i <- as.logical(rbinom(n, 1, 0.5))
+   renv <- env
+   renv[i,1] <- env[i,2]
+   renv[i,2] <- env[i,1]
+
+   return(renv)  # return the randomized envelopes
+ }

>
> TeachingDemos::char2seed('envelope')
> # same as
> # set.seed(5190728)
>
> env <- generateenv(r=2, rpois, n=1e6, lambda=10)
> # keep envelope # 1 strategy (no conditioning)
> mean(env[,1])

[1] 15.00811
>
> # given envelope #1 == 10, expected value of switching
> mean( env[ env[,1]==10, 2] )

[1] 16.54452
>

I generate data using your function and low and behold, given that the 1st envelope is 10, the switching strategy is the winner. However this is dependent on the distribution of x, change the distribution and the expected gain can either increase or decrease.

But the original puzzle does not say anything about the distribution of x (or implies uniform on 0,Infinity like Duncan mentions).

If you want this to work for any possible value of the 1st envelope, then the distribution of x needs to be the improper infinite uniform, but once you give a specific value to condition on, then the distribution of x only needs to be relatively flat in the region of the value of the 1st envelope and half that value (10 and 5 in this case):

> env <- generateenv(r=2, function(n,...) sample(3:52,n,replace=TRUE), n=1e6)
> mean( env[ env[,1]==10, 2] )

[1] 12.55169

(expected value still greater than 10)

Your original post says that you told your friend that he was talking nonsense, the real case is that you are working from a different set of assumptions than your friend, and answering a different question. So in a sense you were both correct, just working from different sets of assumptions of what was beyond the stated question. My intent (beyond continuing the discussion/contriversy) was to show that you were answering a different question than your friend and that your answer to the different question does not make your friend's answer to his question wrong.

One correction, in my prior post I talked about the n-envelope problem. The strategy that I mentioned maximizes the probability of getting the envelope with the highest value, not the expected value, any strategy for expected value would depend on the distribution of the values which is not known in that problem. Similar but different.

Is the idea that whichever envelope you choose, the strategy should be to switch is really that unbelievable? Just ask anyone with at least 2 children, more often than not when you give them something, they will all be convinced that someone else received more/better. My daughters once even counted the number of beans I put on their plate (their utility was the less was better), then the one with fewer pointed out that her beans were longer (I put a stop to it before they got out a ruler).

>
> system.time(env <- generateenv(r=2, r5.10, n=1e6)) # 0.500 0.020
> 0.521
> system.time(replicate(1e6, tmpfun())) #
> 38.211 0.148 38.364 which is about ~76 times slower
>
> Cheers,
> Mario.
>
>
>
> Greg Snow wrote:
> > I think it is like the girl problem, it is a matter of what
> you condition on (condition on at least 1 girl, or on a
> specific one being a girl).
> >
> > In your simulations (at least from what I can see), you
> generate an x from a distribution, randomize the order of x
> and 2x, then look at the average of env1 and the average of
> env2, but you never condition on the information you get from
> env1, so you are answering a different problem.
> >
> > Try this code:
> >
> > tmpfun <- function() {
> > x <- sample( c(5,10), 1 )
> > x2 <- 2*x
> > sample( c(x,x2) )
> > }
> >
> > out1 <- replicate( 100000, tmpfun() )
> > mean( out1[2, out1[1,] == 10] )
> > sum( out1[1,] == 10 )
> >
> > Here we first generate the data such that one of the
> envelopes will have 10, then only look at the cases where
> env1 is 10, the average of env 2 in this case is about 12.5.
> >
> > Now if we look at choosing env2 first:
> >
> > mean( out1[1, out1[2,] == 10] )
> >
> > We still have a mean of switching of about 12.5 (the paradox).
> >
> > The first envelope gives us information which changes the
> problem. You are looking at the problem from the start,
> without the information, not conditioning on that information.
> >
> >
> > Imagine a situation where you have 4 cards, 2 black and 2
> red. You shuffle the cards and randomly draw one out and
> place it face down in front of you (all backs are identical,
> probability of choosing each card is 1/4). Sitting at the
> table are 3 of your friends, you show friend #1 that one of
> the cards still in your hand (chosen randomly) is black, you
> show friend #2 that one of the cards in your hand (chosen
> randomly) is red, friend #3 sees none of your cards and they
> do not share any information about what they see. Now you
> ask them to each write down the probability that the face
> down card is black. Friend #1 will write down 1/3, #2 will
> write 2/3, and friend #3 will write 1/2. They are all
> looking at the same face down card, but each has different
> information to condition on (and if we simulate each friends
> case, we will find that their probabilities are correct for
> their knowledge).
> >
> > The 2 envelope case is a subset of a large problem. Assume
> that there are n envelopes each with a different value in
> them (you know nothing about the distribution of the values).
> You can open an envelope and see what is inside, then either
> keep that amount (and the game ends), or chose another
> envelope, but once you choose to open a new envelope you can
> never go back to a previous one. The goal is to determine a
> strategy that will maximize your expected winings. The best
> strategy that I have heard (I saw the proof once, but don't
> remember the details) is to open about 1/3 of the envelopes
> (actually I think it was 1/e or maybe 1/pi, but 1/3 is a good
> approximation to both of those), then continue opening
> envelopes until the first one that is greater than the
> maximum of the 1st 1/3 and stop there (or stop at the last
> envelope if the maximum was in that first 1/3). As far as I
> know, nobody has come up with a better strategy yet (and some
> math people may have proven it is the best possible).
> Without any knowledge you don't know if the first envelope is
> high or low, but by opening the 1st 1/3 you get information
> that helps form the rest of the strategy. If n=2 (your 2
> envelope case), then 1/3 of 2 rounds off to 1, select the
> first envelope, then go on to the next (which in this case
> also happens to be the last). Always switching follows this
> same strategy.
> >
> > It is paradoxical, but in real life (and in your
> simulations) we generally have more information than what is
> in the puzzle.
> >
> >
> >
> >
> >
> >
> > --
> > Gregory (Greg) L. Snow Ph.D.
> > Statistical Data Center
> > Intermountain Healthcare
> > greg.snow_at_imail.org
> > (801) 408-8111
> >
> >
> >
> >
> >> -----Original Message-----
> >> From: Mario [mailto:mdosrei_at_nimr.mrc.ac.uk]
> >> Sent: Monday, August 25, 2008 2:51 PM
> >> To: Greg Snow; r-help_at_r-project.org
> >> Subject: Re: [R] Two envelopes problem
> >>
> >> No, no, no. I have solved the Monty Hall problem and the Girl's
> >> problem and this is quite different. Imagine this, I get
> the envelope
> >> and I open it and it has A (A=10 or any other amount it doesn't
> >> matter), a third friend gets the other envelope, he opens
> it, it has
> >> B, now B could be either 2A or A/2. He doesn't know
> what I have,
> >> he doesn't have any additional information. According to
> your logic,
> >> he should switch, as he has a 50% chance of having 2B and
> 50% chance
> >> of having B/2. But the same logic applies to me. In
> conclusion, its
> >> advantageous for both of us to switch. But this is a
> paradox, if I'm
> >> expected to make a profit, then surely he's expected to
> make a loss!
> >> This is why this problem is so famous. If you look at the
> last lines
> >> of my simulation, I get, conditional on the first envelope
> having had
> >> 10, that the second envelope has 5 approximatedly 62.6%
> of the time
> >> and 37.4% for the second envelope. In fact, it doesn't matter what
> >> the original distribution of money in the envelopes is,
> conditional
> >> on the first having 10, you should exactly see
> >> 2/3 of the second envelopes having 5 and 1/3 having 20. But I'm
> >> getting a slight deviation from this ratio, which is
> consistent, and
> >> I don't know why.
> >>
> >> Cheers,
> >> Mario.
> >>
> >> Greg Snow wrote:
> >>
> >>> You are simulating the answer to a different question.
> >>>
> >>> Once you know that one envelope contains 10, then you know
> >>>
> >> conditional on that information that either x=10 and the other
> >> envelope holds 20, or 2*x=10 and the other envelope holds
> 5. With no
> >> additional information and assuming random choice we can say that
> >> there is a 50% chance of each of those. A simple
> simulation (or the
> >> math) shows:
> >>
> >>>
> >>>> tmp <- sample( c(5,20), 100000, replace=TRUE )
> >>>> mean(tmp)
> >>>>
> >>>>
> >>> [1] 12.5123
> >>>
> >>> Which is pretty close to the math answer of 12.5.
> >>>
> >>> If you have additional information (you believe it unlikely
> >>>
> >> that there would be 20 in one of the envelopes, the envelope you
> >> opened has 15 in it and the other envelope can't have 7.5 (because
> >> you know there are no coins and there is no such thing as
> a .5 bill
> >> in the local currency), etc.) then that will change the
> >> probabilities, but the puzzle says you have no additional
> >> information.
> >>
> >>> Your friend is correct in that switching is the better strategy.
> >>>
> >>> Another similar puzzle that a lot of people get confused over is:
> >>>
> >>> "I have 2 children, one of them is a girl, what is the
> >>>
> >> probability that the other is also a girl?"
> >>
> >>> Or even the classic Monty Hall problem (which has many
> >>>
> >> answers depending on the motivation of Monty).
> >>
> >>> Hope this helps,
> >>>
> >>> (p.s., the above children puzzle is how I heard the puzzle,
> >>>
> >> I actually have 4 children (but the 1st 2 are girls, so it was
> >> accurate for me for a while).
> >>
> >>> --
> >>> Gregory (Greg) L. Snow Ph.D.
> >>> Statistical Data Center
> >>> Intermountain Healthcare
> >>> greg.snow_at_imail.org
> >>> (801) 408-8111
> >>>
> >>>
> >>>
> >>>
> >>>
> >>>> -----Original Message-----
> >>>> From: r-help-bounces_at_r-project.org
> >>>> [mailto:r-help-bounces_at_r-project.org] On Behalf Of Mario
> >>>> Sent: Monday, August 25, 2008 1:41 PM
> >>>> To: r-help_at_r-project.org
> >>>> Subject: [R] Two envelopes problem
> >>>>
> >>>> A friend of mine came to me with the two envelopes
> >>>>
> >> problem, I hadn't
> >>
> >>>> heard of this problem before and it goes like this:
> >>>> someone puts an amount `x' in an envelope and an amount `2x'
> >>>> in another. You choose one envelope randomly, you open it,
> >>>>
> >> and there
> >>
> >>>> are inside, say 10. Now, should you keep the 10 or swap
> >>>>
> >> envelopes
> >>
> >>>> and keep whatever is inside the other envelope? I told my
> >>>>
> >> friend that
> >>
> >>>> swapping is irrelevant since your expected earnings are
> >>>>
> >> 1.5x whether
> >>
> >>>> you swap or not. He said that you should swap, since if
> >>>>
> >> you have 10
> >>
> >>>> in your hands, then there's a 50% chance of the other
> >>>>
> >> envelope having
> >>
> >>>> 20 and 5% chance of it having 5, so your expected earnings are
> >>>> 12.5 which is more than 10 justifying the swap. I told
> my friend
> >>>> that he was talking non-sense. I then proceeded to write a
> >>>>
> >> simple R
> >>
> >>>> script (below) to simulate random money in the envelopes and it
> >>>> convinced me that the expected earnings are simply
> >>>> 1.5 * E(x) where E(x) is the expected value of x, a random
> >>>>
> >> variable
> >>
> >>>> whose distribution can be set arbitrarily. I later found
> out that
> >>>> this is quite an old and well understood problem, so I got
> >>>>
> >> back to my
> >>
> >>>> friend to explain to him why he was wrong, and then he
> >>>>
> >> insisted that
> >>
> >>>> in the definition of the problem he specifically said that you
> >>>> happened to have 10 and no other values, so is still
> >>>>
> >> better to swap.
> >>
> >>>> I thought that it would be simply to prove in my
> >>>>
> >> simulation that from
> >>
> >>>> those instances in which 10 happened to be the value
> seen in the
> >>>> first envelope, then the expected value in the second
> >>>>
> >> envelope would
> >>
> >>>> still be 10. I run the simulation and surprisingly, I'm
> getting a
> >>>> very slight edge when I swap, contrary to my intuition. I think
> >>>> something in my code might be wrong. I have attached it
> below for
> >>>> whoever wants to play with it. I'd be grateful for any feedback.
> >>>>
> >>>> # Envelopes simulation:
> >>>> #
> >>>> # There are two envelopes, one has certain amount of money
> >>>>
> >> `x', and
> >>
> >>>> the other an # amount `r*x', where `r' is a positive constant
> >>>> (usaully r=2 or r=0.5).
> >>>> You are
> >>>> # allowed to choose one of the envelopes and open it.
> >>>>
> >> After you know
> >>
> >>>> the amount # of money inside the envelope you are given
> >>>>
> >> two options:
> >>
> >>>> keep the money from # the current envelope or switch
> envelopes and
> >>>> keep the money from the second # envelope. What's the best
> >>>>
> >> strategy?
> >>
> >>>> To switch or not to switch?
> >>>> #
> >>>> # Naive explanation: imagine r=2, then you should switch
> >>>>
> >> since there
> >>
> >>>> is a 50% # chance for the other envelope having 2x and 50% of it
> >>>> having x/2, then your # expected earnings are E = 0.5*2x +
> >>>>
> >> 0.5x/2 =
> >>
> >>>> 1.25x, since 1.25x > x you # should switch! But, is this
> >>>>
> >> explanation
> >>
> >>>> right?
> >>>> #
> >>>> # August 2008, Mario dos Reis
> >>>>
> >>>> # Function to generate the envelopes and their money # r:
> >>>> constant, so that x is the amount of money in one envelop
> >>>>
> >> and r*x is
> >>
> >>>> the
> >>>> # amount of money in the second envelope
> >>>> # rdist: a random distribution for the amount x # n: number of
> >>>> envelope pairs to generate # ...: additional parameters for the
> >>>> random distribution # The function returns a 2xn matrix
> containing
> >>>> the (randomized) pairs # of envelopes generateenv <-
> function (r,
> >>>> rdist, n, ...) {
> >>>> env <- matrix(0, ncol=2, nrow=n)
> >>>> env[,1] <- rdist(n, ...) # first envelope has `x'
> >>>> env[,2] <- r*env[,1] # second envelope has `r*x'
> >>>>
> >>>> # randomize de envelopes, so we don't know which one from
> >>>> # the pair has `x' or `r*x'
> >>>> i <- as.logical(rbinom(n, 1, 0.5))
> >>>> renv <- env
> >>>> renv[i,1] <- env[i,2]
> >>>> renv[i,2] <- env[i,1]
> >>>>
> >>>> return(renv) # return the randomized envelopes }
> >>>>
> >>>> # example, `x' follows an exponential distribution with
> >>>>
> >> E(x) = 10 #
> >>
> >>>> we do one million simulations n=1e6) env <-
> generateenv(r=2, rexp,
> >>>> n=1e6, rate=1/10)
> >>>> mean(env[,1]) # you keep the randomly assigned first envelope
> >>>> mean(env[,2]) # you always switch and keep the second
> >>>>
> >>>> # example, `x' follows a gamma distributin, r=0.5 env <-
> >>>> generateenv(r=.5, rgamma, n=1e6, shape=1, rate=1/20)
> >>>> mean(env[,1]) # you keep the randomly assigned first envelope
> >>>> mean(env[,2]) # you always switch and keep the second
> >>>>
> >>>> # example, a positive 'normal' distribution # First
> write your won
> >>>> function:
> >>>> rposnorm <- function (n, ...)
> >>>> {
> >>>> return(abs(rnorm(n, ...)))
> >>>> }
> >>>> env <- generateenv(r=2, rposnorm, n=1e6, mean=20, sd=10)
> >>>> mean(env[,1]) # you keep the randomly assigned first envelope
> >>>> mean(env[,2]) # you always switch and keep the second
> >>>>
> >>>> # example, exponential approximated as an integer rintexp <-
> >>>> function(n, ...) return (ceiling(rexp(n, ...))) # we use
> >>>>
> >> ceiling as
> >>
> >>>> we don't want zeroes env <- generateenv(r=2, rintexp, n=1e6,
> >>>> rate=1/10)
> >>>> mean(env[,1]) # you keep the randomly assigned first envelope
> >>>> mean(env[,2]) # you always switch and keep the second i10 <-
> >>>> which(env[,1]==10)
> >>>> mean(env[i10,1]) # Exactly 10
> >>>> mean(env[i10,2]) # ~ 10.58 - 10.69 after several trials
> >>>>
> >>>> ______________________________________________
> >>>> R-help_at_r-project.org mailing list
> >>>> https://stat.ethz.ch/mailman/listinfo/r-help
> >>>> PLEASE do read the posting guide
> >>>> http://www.R-project.org/posting-guide.html
> >>>> and provide commented, minimal, self-contained,
> reproducible code.
> >>>>
> >>>>
> >>>>
> >>>
> >>
> >
> >
>
>
>

--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow_at_imail.org
(801) 408-8111

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