From: Deepayan Sarkar <deepayan.sarkar_at_gmail.com>

Date: Mon, 25 Aug 2008 14:35:15 -0700

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. Received on Mon 25 Aug 2008 - 21:40:12 GMT

Date: Mon, 25 Aug 2008 14:35:15 -0700

On Mon, Aug 25, 2008 at 1:51 PM, Mario <mdosrei_at_nimr.mrc.ac.uk> wrote:

> 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.
*

That has nothing to do with switching, and everything to do with properties of regression. Try:

> env <- generateenv(r=2, rintexp, n=1e6, rate=1/10)

*> colMeans(env)
*

[1] 15.76573 15.75436

> mean(env[which(env[,1] == 10) , 2]) # greater than 10

[1] 10.64476

> mean(env[which(env[,1] == 20) , 2]) # less than 20

[1] 18.10053

And now with a distribution with mean less than 10

> env <- generateenv(r=.5, rintexp, n=1e6, rate=1/10)

> colMeans(env) # mean less than 10

[1] 7.881469 7.873117

> mean(env[which(env[,1] == 10) , 2])

[1] 8.966886

-Deepayan

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