Envelope paradox / problem

[erdnase]

I have two identical envelopes - and all you know is that one of them contains exactly twice as much money as the other. You can open either of the envelopes, and then:

a) keep the money in that envelope

or

b) swap, and take the money in the other envelope.

Note that if you swap, you're stuck with your new choice. You can't swap back. I'll also add that there's no verbal trickery here, and that both envelopes contain a real amount of money. The question is - does swapping make sense, and if so why?

Note that there is no option C of beating me up and taking both envelopes

[ewenm]

No point swapping as knowing the amount in one envelope tells you nothing about the amount in the other. Unless that is, the envelope you've picked holds an odd value of money so can't be exactly double the other amount.

Of course if you're allowed to hold them both, pick the heavier one

Edited by ewenm on Monday 29th June 21:36

[voyds9]

Yes but don't know why. Carol Vorderman was on about it and even she had trouble understanding the maths.

[Rotary Madness]

depends on if its a cheque, or cash, and the cash is in the same denominaions. If its in pound coins, or 5 pounds notes, just give me a set of scales

It also depends on what amount we're talking. A fiver, id think what the hell and try the next envelope. If its something like 500k, then id be quite happy to stick with it!

Or punch you in the face and take both

[ewenm]

It's not the same as the 2 goats and a car problem.

[DrTre]

Always swap.

[tubbystu]

Isn't there some bizarre statistic that 2/3rds of the time the other one will have the higher amount in ? Dunno where I read that, and am ready to be if its tosh.

Wow, now where did that smilie come from ? Is there a secret set of coding for new smilies ?

[Pints]

This was in a movie, if I remember correctly. Can't remember the movie but I think the idea is to swap to increase your odds.

[sinizter]

Swapping works if its three envelopes/doors/whatever ... No benefit in two I think.

[grumbledoak]

With two envelopes it won't matter.

There is a puzzle with one prize behind one of three doors, and once you have picked one they open a losing door. In this circumstance you should always swap. You can try out all the combinations quite quickly in your head.

[ewenm]

The goats/car problem is this:

Three doors, behind 2 are goats, behind the last is a car. You get to pick one door. You are then shown one of the other doors with a goat behind it. Should you change your door choice to the other remaining door?

[DrTre]

I'd always swap just so that I knew and that it was MY choice. Couldn't abide not knowing. If I end up with a few extra spondoolicks then I'm happy.

[TheEnd]

the pick/shown a bad choice/ choice to swap is the one where you should swap.

With the 2 envelopes, there isn't anything fancy

[grumbledoak]

To add: The three door problem is the Monty Hall dilemma:

http://mathforum.org/dr.math/faq/faq.monty.hall.ht...

[Stevenj214]

You chose an envelope with value x inside. The other envelope has either 1/2x or 2x. i.e you are gambling half the money for a bet at odds of 2/1 where the odds that the other envelope holds more are Evens.

So you should swap. Always.

[erdnase]

This is different from the Monty Hall problem, which is fascinating in its own right. I'll not post any of the solution yet, but I'll clarify a few points that have been made.

There's no odd amounts in the envelopes, or weighing them, etc. Just two envelopes, one contains X and the other contains 2X.

[erdnase]

Stevenj214 said:

You chose an envelope with value x inside. The other envelope has either 1/2x or 2x. i.e you are gambling half the money for a bet at odds of 2/1 where the odds that the other envelope holds more are Evens.

So you should swap. Always.

Give that man a cigar - that's it on the nose.

You either lose 0.5x or gain 1x. Seems counter-intuitive, but there you have it. If anyone's interested, the equity/value of always swapping is 1.25x.

Head assplode!

[GreenV8S]

Stevenj214 said:

You chose an envelope with value x inside. The other envelope has either 1/2x or 2x. i.e you are gambling half the money for a bet at odds of 2/1 where the odds that the other envelope holds more are Evens.

So you should swap. Always.

That sounds convincing but something feels wrong about it. By this logic, if you're offered the chance to swap back (without opening the other envelope) you should do. Surely that can't be right?

[ewenm]

GreenV8S said:

Stevenj214 said:

You chose an envelope with value x inside. The other envelope has either 1/2x or 2x. i.e you are gambling half the money for a bet at odds of 2/1 where the odds that the other envelope holds more are Evens.

So you should swap. Always.

That sounds convincing but something feels wrong about it. By this logic, if you're offered the chance to swap back (without opening the other envelope) you should do. Surely that can't be right?

I'd be interested in the answer to that too. If you were allowed unlimited swaps, how would you decide when to stick?

If there is no way of deciding, doesn't that suggest it is just a 50:50 chance and you may as well stick with your first choice?

Edited by ewenm on Monday 29th June 22:25

[grumbledoak]

This doesn't sound right. The key to the Monty Hall problem is that his actions are not random, so his opening of a losing door changes your odds. But nothing like that is happening in this case.