You are on a game show. The host says he has hidden money under two cups,
one amount being double the other one. You pick one at random, he turns it
over, it's $100. He gives you a second chance (that he would do so was
determined before the game and not dependent on which cup you'd turn over):
You can switch, or take the $100.

You figure: If I switch, I will either get $50 or $200 with equal
probability. Expected return is thus $50*1/2 + $200*1/2 = $125, which is
larger than the $100 I'd spend to "play" this second game. Looks like a good
deal. Is it?

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somebody wrote:
> You are on a game show. The host says he has hidden money under two cups,

Did you initially hear this problem phrased as money under cups, or did 
you just change it to that so it's be harder go Google for the answer?

At any rate, it's a very neat problem.

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"Kevin Wampler" <wam### [at] uwashingtonedu> wrote in message
news:4b284efd@news.povray.org...

> or did  you just change it to that so it's be harder go Google for the
answer?

Good catch.

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somebody wrote:
> You are on a game show. The host says he has hidden money under two cups,
> one amount being double the other one. You pick one at random, he turns it
> over, it's $100. He gives you a second chance (that he would do so was
> determined before the game and not dependent on which cup you'd turn over):
> You can switch, or take the $100.
> 
> You figure: If I switch, I will either get $50 or $200 with equal
> probability. Expected return is thus $50*1/2 + $200*1/2 = $125, which is
> larger than the $100 I'd spend to "play" this second game. Looks like a good
> deal. Is it?
> 
> 

Unless I am vastly misreading this, it seems there is a simple way to
set up this problem. Two cups, arbitrarily labeled A and B.

A contains $X, B contains $2X
or
A contains $2X, B contains $X

I don't see an obvious way to turn this into the Boy or Girl, or Monty
Hall, paradox, so I would have to guess that yes, it would be a good bet
to take the switch.

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> Unless I am vastly misreading this, it seems there is a simple way to
> set up this problem. Two cups, arbitrarily labeled A and B.
>
> A contains $X, B contains $2X
> or
> A contains $2X, B contains $X
>
> I don't see an obvious way to turn this into the Boy or Girl, or Monty
> Hall, paradox, so I would have to guess that yes, it would be a good bet
> to take the switch.

How can a tactic where you *always* switch the cup be more profitable than 
never switching?  In both cases, in the long run you're going to end up 50% 
of the time choosing the higher amount and 50% the lower amount.

I'm still trying to figure out where the mistake is in the OP's reasoning...

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Le 16/12/2009 09:55, scott nous fit lire :
>> Unless I am vastly misreading this, it seems there is a simple way to
>> set up this problem. Two cups, arbitrarily labeled A and B.
>>
>> A contains $X, B contains $2X
>> or
>> A contains $2X, B contains $X
>>
>> I don't see an obvious way to turn this into the Boy or Girl, or Monty
>> Hall, paradox, so I would have to guess that yes, it would be a good bet
>> to take the switch.
> 
> How can a tactic where you *always* switch the cup be more profitable
> than never switching?  In both cases, in the long run you're going to
> end up 50% of the time choosing the higher amount and 50% the lower amount.
> 
> I'm still trying to figure out where the mistake is in the OP's
> reasoning...

finding 100$, you do not know if it is the low or high amount.

The computation is NOT $50*1/2+ $200*1/2 ( = $125)
but either :

 $50*1/2 + $100*1/2 = $75
or
 $100*1/2 + $200*1/2 = $150

Which turn out $75*1/2 + $150*1/2 around $112.5... if that was random.

Finding $100, you have no idea yet if you have a good value or a bad one.
It's a one time game, no long run (once you know the best value... that
you learn at the end of the choice), and girl & boy might be coming in.

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> finding 100$, you do not know if it is the low or high amount.

Indeed, it's a 50/50 chance whether you pick the high or low amount.

> The computation is NOT $50*1/2+ $200*1/2 ( = $125)
> but either :
>
> $50*1/2 + $100*1/2 = $75
> or
> $100*1/2 + $200*1/2 = $150
>
> Which turn out $75*1/2 + $150*1/2 around $112.5... if that was random.

So you'd always change and take the other box?  Why even bother opening the 
1st box, just leave it shut and take the 2nd one.  Oh but hang on, then when 
you look inside the 2nd one you might want to change back to the 1st...

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scott wrote:

> So you'd always change and take the other box?  Why even bother opening 
> the 1st box, just leave it shut and take the 2nd one.  Oh but hang on, 
> then when you look inside the 2nd one you might want to change back to 
> the 1st...

Isn't this basically the entire premise of Deal Or No Deal?

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Le 16/12/2009 14:39, scott nous fit lire :
>> finding 100$, you do not know if it is the low or high amount.
> 
> Indeed, it's a 50/50 chance whether you pick the high or low amount.
> 
>> The computation is NOT $50*1/2+ $200*1/2 ( = $125)
>> but either :
>>
>> $50*1/2 + $100*1/2 = $75
>> or
>> $100*1/2 + $200*1/2 = $150
>>
>> Which turn out $75*1/2 + $150*1/2 around $112.5... if that was random.
> 
> So you'd always change and take the other box?  Why even bother opening
> the 1st box, just leave it shut and take the 2nd one.  Oh but hang on,
> then when you look inside the 2nd one you might want to change back to
> the 1st...
> 
> 
For optimising my happiness' function, I will take the first box and say
good bye, not wanting to know which case I was playing.

Playing otherwise might diminish my happiness' value, and that's not
something I would risk.

That's not rational, but I'm human, I have the right to make error (as
in 'to err is human') and to seek happiness rather than optimal
solution. (seeking happiness, nothing about getting it!)

P.S.: tuning the numbers, what if the amount ratio instead of 1:2 is 2:3 ?
Found 300€ under your choice, swap for other ? (450€ or 200€)

And for 4:5 ? (found £100, £80 or £125)

and what about 1:100 ? (100, 1 or 10000)

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somebody wrote:
>> or did  you just change it to that so it's be harder go Google for the
> answer?
> 
> Good catch.

Well, it was easy since I've always heard this problem with a different 
phrasing.  I think I actually like the way you stated it better though 
-- it works nicely in the gameshow context.

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On 12/16/09 11:08, Le_Forgeron wrote:
> For optimising my happiness' function, I will take the first box and say
> good bye, not wanting to know which case I was playing.
>
> Playing otherwise might diminish my happiness' value, and that's not
> something I would risk.
>
> That's not rational, but I'm human, I have the right to make error (as

	Nope - that's rational.

	The economists use "rational" in terms of choosing the path that gives 
you the most money. Unfortunately, too many people tend to use the same 
definition outside the sphere of economics.


-- 
Would the capacity of a Palaeozoic Hard Dive be measured in Trilobites?

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"Le_Forgeron" <jgr### [at] freefr> wrote in message
news:4b2913f1$1@news.povray.org...

> That's not rational, but I'm human, I have the right to make error (as
> in 'to err is human') and to seek happiness rather than optimal
> solution.

For some, happiness *is* finding the solution.

> P.S.: tuning the numbers, what if the amount ratio instead of 1:2 is 2:3 ?
> Found 300? under your choice, swap for other ? (450? or 200?)
>
> And for 4:5 ? (found £100, £80 or £125)
>
> and what about 1:100 ? (100, 1 or 10000)

You are onto something here.

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somebody wrote:
> You are on a game show. The host says he has hidden money under two cups,
> one amount being double the other one. You pick one at random, he turns it
> over, it's $100. He gives you a second chance (that he would do so was
> determined before the game and not dependent on which cup you'd turn over):
> You can switch, or take the $100.
> 
> You figure: If I switch, I will either get $50 or $200 with equal
> probability. Expected return is thus $50*1/2 + $200*1/2 = $125, which is
> larger than the $100 I'd spend to "play" this second game. Looks like a good
> deal. Is it?
> 
> 

There are actually four cups. Two cups in the $300 total universe and 
two cups in the $150 total universe. No amount of switching will change 
the universe in which you exist and that universe was selected before 
you chose a $100 cup.

  -Shay

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Shay wrote:

> 
> There are actually four cups. Two cups in the $300 total universe and 
> two cups in the $150 total universe. No amount of switching will change 
> the universe in which you exist and that universe was selected before 
> you chose a $100 cup.
> 
>  -Shay

Yes! You've got it. :-)

-- 

Best Regards,
	Stephen

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Stephen wrote:
> Shay wrote:
> 
>>
>> There are actually four cups. Two cups in the $300 total universe and 
>> two cups in the $150 total universe. No amount of switching will 
>> change the universe in which you exist and that universe was selected 
>> before you chose a $100 cup.
>>
>>  -Shay
> 
> Yes! You've got it. :-)
> 

Here's how the math really works out, as well as a better explanation of 
where Andrew got it wrong.

Andrew counted the built-in profit of the system as "winnings" he showed 
for switching.

Four cups, right?
{$50 | $100} ------ {$100 - $200}

I'll change the scenario just a bit to make it clearer. Contestant is 
allowed to select one set of two cups ($50-$100 and $100-200) and to 
then select one of the two cups from his selected set. His expectation 
from this game? ($50 + $100 + $100 + $200) / 4 = $112.50. That's what 
he's being *given* for playing the game, so that amount must be 
subtracted from his winnings to reveal the profits of his strategy.

So, subtract $112.50 from each of the four cups to get
{-$62.50 | -$12.50} ------ {-$12.50 | $87.5}

Now, back to Andrew's premise. The contestant turns over a cup to find 
-$12.50. Should he switch?
Possibility 1: He's in the low set[1] - expectation = -$37.50
Possibility 2: He's in the high set - expectation = +$37.50
A 50% probability either way.

That's the honest math.

  -Shay

[1] Actually, not *the* but *a* high set. There are two ways that could 
go, but it doesn't change the problem.

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somebody wrote:
> You are on a game show. The host says he has hidden money under two cups,
> one amount being double the other one. You pick one at random, he turns it
> over, it's $100. He gives you a second chance (that he would do so was
> determined before the game and not dependent on which cup you'd turn over):
> You can switch, or take the $100.
> 
> You figure: If I switch, I will either get $50 or $200 with equal
> probability. Expected return is thus $50*1/2 + $200*1/2 = $125, which is
> larger than the $100 I'd spend to "play" this second game. Looks like a good
> deal. Is it?

Yes.

Regards,
John

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