Free Money!!!

Daithio

You are a contestant in a gameshow, and the host presents you with two envelopes, each of which contains a cheque. You may choose only one of the two, and you get to keep whatever sum the cheque in the envelope you pick is made out for. Your host informs you that one of the envelopes is made out for a sum exactly twice as large as the other one, but he refuses to tell you which is which.

Having no means of telling which envelope contains the cheque with larger sum, you simply pick one by guessing, and after the host asks you to open your envelope, you discover that the cheque in it is made out to the sum of $50,000. The host now asks you whether you'd like to change your mind and pick the other envelope instead; do you take him up on the offer, or stick with what you have?

Shortstack

My heart tells me you have to change it as you can only lose $25k but could get an extra $50k for a 50:50 shot but something tells me that is wrong. I used to be good at maths

biteme

burn the money in the first envelope and take the second one. Pretty sure mike got the reasoning right too.

karlh

you DO know i just got my aces cracked by J9os, right!?

NickyOD

Obviously you should always switch in this instance since your equity in winning by switching is greater than winning by sticking

There are 4 possible outcomes.

You switch and get half (25,000)
You switch and double (100,000)

You stick and get the bigger amount (50.000)
You stick and get the half amount (50,000)

So by sticking you will on average win 50,000
By switching you will win 62,500 on average.

Daithio

Then why bother opening the envelope in the first place, seeing as you are always going to switch?

bohsman

So why bother looking at the first envelope at all? Just rip up the first one on the blind and take the second one? (I know the answer so I wont say more than that for the moment)

The Gecko

First of all I would never go on a game show without knowing the format. On this basis I would have seen at least one other episode. I would the calculate the maximum all contestents could have won which i recokon would give me the budget the show is working to there or there abouts.

Based on this if the average big pay out is €30-70K I would take the €50K

However if they are a particulary generous show, with big prizes etc say like Ant & Dec I would take the gamble. (Please no reference to the fact that I have watched Ant & Dec the loveable geordies)

bohsman

Daithio wrote:

Then why bother opening the envelope in the first place, seeing as you are always going to switch?

By that logic the second envelope would become the first envelope and you would have to switch to the original first

ocallagh

My Dad exlained this one to me a few years back. From memory these are my thoughts on why it doesn't make a difference if you switch. (When I looked up the web there ar some very strange solutions... )

Call the small envelope, envelope A
Call the large envelope, envelope B

The key to this is the value you assign to your first choice.

Afer your initial choice, you assume you have made your selection and now hold either envelope A or envelope B, whereas you're holding neither envelope. You still have the same initial decision to make, namely do you switch to Envelope A or Envelope B. So the only value you can assign to your first choice is the average of the two envelopes.

The average or expected value of your fist choice is the average between the two envelopes

Therefore, by switching you either gain or lose the difference between the 'average value' and the 'resultant value'. This gain or loss is the same so there is no point in switching.

A bit more detail:


Give the envelopes values
A is worth 1000
B is worth 2000

Pick an envelope

It's a 50/50 chance you picked 1000 or 2000. On average you end up with 1,500.

So, your average value is 1500.

If you switch and get Envelope B you get 2000, a gain of 500
If you switch and get Envelope A you get 1000, a lose of 500

Fatboydim

If it's on RTE - Stick

Happy Camper

NickyOD wrote:

Obviously you should always switch in this instance since your equity in winning by switching is greater than winning by sticking

There are 4 possible outcomes.

You switch and get half (25,000)
You switch and double (100,000)

You stick and get the bigger amount (50.000)
You stick and get the half amount (50,000)

So by sticking you will on average win 50,000
By switching you will win 62,500 on average.

Let's look at this problem more accurately:

There are 2 envelopes. Lets call the contents of each envelope "x" for the smaller prize and "2x" for the larger prize.

There are two possibilities: you either pick "x" or "2x" with equal probability.

The expected gain from a switch is: P(Pick 2x).E(2x and switch) + P(Pick x).E(x and switch) = (0.5)(-x) + (0.5).(x) = 0. Thus, it doesn't matter whether you switch or not.

I think your problem is that you don't fully understand conditional probabilities.

NickyOD

Happy Camper wrote:

The expected gain from a switch is: P(Pick 2x).E(2x and switch) + P(Pick x).E(x and switch) = (0.5)(-x) + (0.5).(x) = 0. Thus, it doesn't matter whether you switch or not.

Should this not be " = ((.5)(-x) + (x)(2))/2 " ?

okidoki987

Should this not be " = ((.5)(-x) + (x)(2))/2 " ?

Can't really see RTE getting their heads around that one

Iago

could have sworn I was in the Poker and not the physics/advanced mathematics forum..

Do we need a probability sub-forum perhaps? either way my head hurts, I think it's best if I avoid reading these threads in future

Ste05

Iago wrote:

could have sworn I was in the Poker and not the physics/advanced mathematics forum..

Do we need a probability sub-forum perhaps? either way my head hurts, I think it's best if I avoid reading these threads in future

I actually laughed out loud at that!! ..... and I'm in work, so try explaining the joke to the non poker playing people !!!

This is exactly what I've been thinking lately.
You have my vote for a physics/advanced mathematics sub-forum anyway...

bohsman

The easiest way to explain it: There is only 2 envelopes so it will either be bigger or smaller its 50 50.

Most people calculate as if there is 3 envelopes, The one handed to you, one smaller and one bigger. Hence if there were 3 envelopes and you were handed the middle amount you should pick another one

NickyOD

bohsman wrote:

The easiest way to explain it: There is only 2 envelopes so it will either be bigger or smaller its 50 50.

Most people calculate as if there is 3 envelopes, The one handed to you, one smaller and one bigger. Hence if there were 3 envelopes and you were handed the middle amount you should pick another one

The thing is although there are only 2 envelopes you don't get told the possible outcome of the second envelope until after you've chosen so there are 3 possible values.

25K 50K and 100K.

By switching you have an equal chance of increasing your winnings by 100% as you have to lose 50%.

If you tihnk of it in terms of pot odds. If you are getting 1:1 to risk 50% of your stack to double up you should always call.

AmarilloFats

I don't get what happy camper was saying. whats E?Happy camper, you sound clever seek clarity not obscurity.
there is a third option.Kick Marty Whelan in the knackers... and perhaps cleveland steam his head.

delrodge

NickyOD wrote:

Should this not be " = ((.5)(-x) + (x)(2))/2 " ?

No It shouldn't. Happy Camper is right from the maths standpoint. -(i'm a maths student )
The GAIN you would get from switching if you chose the smaller amount is x and not 2x -(which is what your formula is saying, having chosen the smaller envelope you alrady have x, which you double, to 2x)

Daithio

Delrodge and HappyCamper are both right, just thought it was an interesting problem as it seems to go against your natural intuitions. Most people assume the same as Nicky did, and that to switch is the right option, but what's the point in even looking at the first envelope if you are simply always going to switch. It doesn't make a difference, it's still 50-50.

NickyOD

Daithio wrote:

Delrodge and HappyCamper are both right, just thought it was an interesting problem as it seems to go against your natural intuitions. Most people assume the same as Nicky did, and that to switch is the right option, but what's the point in even looking at the first envelope if you are simply always going to switch. It doesn't make a difference, it's still 50-50.

But you're not told you can switch until after you make your choice which changes the probability. Its a completely diferent decision from your initial choice of 2 envelopes because there are now 3 possible values instead of 2. You now have 50,000. you are risking 25K to win 50K on an even money shot. On average the second envelope will be worth 62,500. I must be missing somthing that is distorting the numbers. i.e. the 50/50 that your initial coice is the higher figure.

Illegal Alien

This is completely different to the other thread with 3 evelopes you know.

Because the last two in that example isn't a 50/50 shot statisticly. It's 66/33

Culchie

delrodge wrote:

No It shouldn't. Happy Camper is right from the maths standpoint. -(i'm a maths student )
The GAIN you would get from switching if you chose the smaller amount is x and not 2x -(which is what your formula is saying, having chosen the smaller envelope you alrady have x, which you double, to 2x)

Still confused, have we an answer?

It reminds me of a puzzle I heard before in a pub ..... the answer is always to pick the one with the biggest boobs. ..... easy one that :rolleyes:

Tourneque

ocallagh wrote:

My Dad exlained this one to me a few years back. From memory these are my thoughts on why it doesn't make a difference if you switch. (When I looked up the web there ar some very strange solutions... )

Call the small envelope, envelope A
Call the large envelope, envelope B

The key to this is the value you assign to your first choice.

Afer your initial choice, you assume you have made your selection and now hold either envelope A or envelope B, whereas you're holding neither envelope. You still have the same initial decision to make, namely do you switch to Envelope A or Envelope B. So the only value you can assign to your first choice is the average of the two envelopes.

The average or expected value of your fist choice is the average between the two envelopes

Therefore, by switching you either gain or lose the difference between the 'average value' and the 'resultant value'. This gain or loss is the same so there is no point in switching.

A bit more detail:


Give the envelopes values
A is worth 1000
B is worth 2000

Pick an envelope

It's a 50/50 chance you picked 1000 or 2000. On average you end up with 1,500.

So, your average value is 1500.

If you switch and get Envelope B you get 2000, a gain of 500
If you switch and get Envelope A you get 1000, a lose of 500

I thought this had explained it ?

There's no gain to be had by switching.

DubTony

Iago wrote:

could have sworn I was in the Poker and not the physics/advanced mathematics forum..

Do we need a probability sub-forum perhaps? either way my head hurts, I think it's best if I avoid reading these threads in future

Ste05 wrote:

I actually laughed out loud at that!! ..... and I'm in work, so try explaining the joke to the non poker playing people !!!

This is exactly what I've been thinking lately.
You have my vote for a physics/advanced mathematics sub-forum anyway...

I'm with Iago and Steve on this one. But to avoid confusion could it be called the Doors / Cars / Envelopes / Goats sub-forum?

ocallagh

NickyOD wrote:

But you're not told you can switch until after you make your choice which changes the probability. Its a completely diferent decision from your initial choice of 2 envelopes because there are now 3 possible values instead of 2. You now have 50,000. you are risking 25K to win 50K on an even money shot. On average the second envelope will be worth 62,500. I must be missing somthing that is distorting the numbers. i.e. the 50/50 that your initial coice is the higher figure.

So you are saying you gain an advantage by switching?

Expand the problem. Imagine you are given the chance to switch envelopes 1,000 times. Do the envelopes grow in value each time you switch?

Because when you switch, you might have traded the larger one for the smaller one.. and applying your logic you must switch again, and again, and again!!

Rodge

Daithio wrote:

made out to the sum of $50,000. The host now asks you whether you'd like to change your mind and pick the other envelope instead; do you take him up on the offer, or stick with what you have?

My answer is that I stick with what I have. $50,000 is a lot of money to me and I dont want to risk it.

I cant see how anyone can say that there is a definitive right answer or wrong answer. This problem has been around for years. There is the maths argument and the philosphical argument (which we havent seen yet).

I always thought that the probability side of it would only apply over a large sample space, seeing as we have only one choice on one occasion then probability doesnt apply.

NickyOD

Tourneque wrote:

I thought this had explained it ?

There's no gain to be had by switching.

From the point at which you ask the question "do you want to switch envelopes" - in a 100% random and honest game - the EV of switching is 1.25x the amount in the first envelope...

If the puzzle was - "There are two envelopes - one contains 1/2 the cash of the other one...after you pick the first one - you will be given the option to switch..." Under THAT scenario - then there is no advantage one way or the other...but since the switching aspect wasn't mentioned as part of the problem until after the first choice - I don't see not switching as an option from a strictly EV standpoint

The_Daddy_H

NickyOD wrote:

From the point at which you ask the question "do you want to switch envelopes" - in a 100% random and honest game - the EV of switching is 1.25x the amount in the first envelope...

If the puzzle was - "There are two envelopes - one contains 1/2 the cash of the other one...after you pick the first one - you will be given the option to switch..." Under THAT scenario - then there is no advantage one way or the other...but since the switching aspect wasn't mentioned as part of the problem until after the first choice - I don't see not switching as an option from a strictly EV standpoint

Its irrelevent whether or not you're told before hand if you can switch or not, just as its irrelevant whether or not you actually know the amount of money in the envelope you first pick. The mistake you're making is calculating the expectation of the switching action after you've open the first envelope and in effect treating it as a completely seperate event. You you need to do is look at the situation in its entirity and calculate whats known as the conditional expectation.

NickyOD

The_Daddy_H wrote:

Its irrelevent whether or not you're told before hand if you can switch or not, just as its irrelevant whether or not you actually know the amount of money in the envelope you first pick. The mistake you're making is calculating the expectation of the switching action after you've open the first envelope and in effect treating it as a completely seperate event. You you need to do is look at the situation in its entirity and calculate whats known as the conditional expectation.

Because you only find out about the switching AFTER you choose an envelope, it in fact IS a separate event. Once something actually occurs, the probability of the remaining steps changes and becomes different from the original probability from the beginning. Think of it like this:

You have 2 coins. 1 is a normal heads/tails, the other is double-sided heads. We know that after flipping both coins, the odds of 2 heads showing are 50%. Now say you only flip the 1st coin, and it shows heads. What are the odds that the 2nd coin will also show heads? The answer is 75%.

OR, a typical roulette situation (for sake of argument and math simplicity, lets say there are no 0's). The odds of Red coming 3 spins in a row are 12.5%. However, after 2 Red spins, the odds of a 3rd are 50%.

My examples aren't direct correleations, but are to demonstrate that you cannot base on assumptions, that only facts known at the time are relevant.

Sometimes it's difficult to calculate conditional probabilities at different points of an event, when YOU in fact know what the different conditions will be in the future steps.

delrodge

NickyOD wrote:

Because you only find out about the switching AFTER you choose an envelope, it in fact IS a separate event. Once something actually occurs, the probability of the remaining steps changes and becomes different from the original probability from the beginning. Think of it like this:

You have 2 coins. 1 is a normal heads/tails, the other is double-sided heads. We know that after flipping both coins, the odds of 2 heads showing are 50%. Now say you only flip the 1st coin, and it shows heads. What are the odds that the 2nd coin will also show heads? The answer is 75%.


My examples aren't direct correleations, but are to demonstrate that you cannot base on assumptions, that only facts known at the time are relevant.
.

You're example aren't just not direct correlations, they're completely irrelevant. You are mixing up conditional expectations, as was said above
your expected gain from switching is your
expected gain if in the bigger envelope times gain if in bigger envelope+
expected gain if in the smaller envelope times gain if in smaller envelope=
-(you accept that this is the way to go about the problem, and below formula is correct?)
* (0.5)(-x)+(0.5)(x)=0

So i assume you query one of the figures- the 0.5s are self evident,
Now the -x and +x -(not +2x as you think) are conditional expectations,
i.e. my gain if i switch condtioned on the fact that i'm in smaller envelope.
if i'm in the small envelope i have x, if i switch i get 2x gaining x
if i'm in the big envelope i have 2x if i switch i get x gaining -x.

So we accept that "before we start" the e.v. of switching is 0- as you said above.

So the question is does that change?
All opening the envelope does is give you a figure, how does this change the above formula?
beyond this figure
you gain no information, yeah? - The fact it's 50thousand doesn't change the problem,
-(please enlighten me as to how you think it does)
What it does do, is give you an estimate of x -(37500 = 0.5x50k + 0.5x25k)
which you can use in above formula, to give you an e.v of .... 0


You are bastardising the above formula * by changing which x you use midway through the sum-(assuming you think the formula is right)
why are you suddenly allowed to do this when you're given a figure?

What you are saying is always switch, regardless of which envelope you choose, so say you always choose the envelope in the presenter's left hand -(since it doens't matter which one you pick)?
You then always switch, So you always choose the one in his right hand.

roryc

My head hurts....

NickyOD

Allright. Chill out delrodge. Its just a discussion. Of course your formula is correct.

The reason it appears to be positive EV is because the fact that we get to simply open an envelope is +EV. In other words, before we open the first envelope, we are guaranteed at least 25% of the maximum we "believe" we can win. This is where the 25%+ EV is coming from. Although it appears by switching we are picking up this EV, we actually picked it up by simply being able to open an envelope. Therefore, you have to subtract that amount from the EV of the second move of swapping. This causes there to be no difference of EV between keeping the original envelope and swapping.

hotspur

Guys this is a famous paradox, hence the arguing. The possible solutions to the paradox are waaaay more complicated than any of you have come close to, but I'll be buggered if i'm gonna spend 15 mins trying to explain it in an unsatisfactory way. All I'll say is that there is no way it's gonna be resolved on a bloody poker forum lol, and I don't say that disrespectfully, it's just that it rages among philosphers unsatisfactorily resolved. Pot odds to this is like a mole hill to Everest, shuffle up and deal!

Crumbs

hotspur wrote:

Guys this is a famous paradox, hence the arguing. The possible solutions to the paradox are waaaay more complicated than any of you have come close to, but I'll be buggered if i'm gonna spend 15 mins trying to explain it in an unsatisfactory way.

That's also why I didn't respond to this thread (errr...until now). I feared that discussing infinite probability distributions and the like would lead to a lynch mob.

One thing that I don't think was mentioned - if the amount in the first envelope is an odd number (like €999.99), then switching is +EV.

The_Daddy_H

Crumbs wrote:

That's also why I didn't respond to this thread (errr...until now). I feared that discussing infinite probability distributions and the like would lead to a lynch mob.

This is not a famous paradox, its a brainteaser they give to kids in school. What you're thinking of is a different problem. The expected value calculation is, as has been clearly demonstrated, simply flawed.

Even without resorting to any mathematics whatsoever and mearly appealing to common sense it should be obvious, you cant make money out of thin air despite what any of these so-called philosophers may say.

delrodge

Now we could leave things as they stand, However just to completely screw with your heads I feel compelled to mention the following:

When we first play the game we know we will gain-(from the Entire game) x with probability 0.5 and 2x with probability 0.5 so our expected gain is 1.5x
(remember x is the smaller of the two amounts)

So when we see 50,000 in the envelope we get we know x is either 25,000 or 50,000 with probabilities
0.5, so we have an expected value of x as 37500. ok?

But then we can estimate our gain from the game as 1.5x= (1.5)(37500)=

:eek: 56250!!


(So if you were asked would you like to give back the 50k cheque, reseal the envelope, shuffle them up and then choose one randomly and stick to it, would you?
Is this the same (effectively) as switching?

Does this muddy up the waters again?

NickyOD

Just so I never make the same mistake again.

How about substiuting the $ values into the formula below where one envelope is 100 times the other. Is there still no gain in switching? Theoretically it should make no difference whether one is 5,10 or 100 times the other amount, right?

P(Pick 2x).E(2x and switch) + P(Pick x).E(x and switch) = (0.5)(-x) + (0.5).(x) = 0