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# Game show envelope conundrum

Posted on 2011-02-28
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so the game show gives you a choice of prizes, two sealed envelopes, all you are told is one envelope has twice the amount of money in it as the other, but you've no idea what it is.

So you take envelope A and it has \$100 in it - the host asks if you want to swap.  What is the value in swapping to envelope B?  It is 50% it has \$200 and 50% it has \$50 in it - so it is worth

.5 x 200 + .5 x 50 = \$125

so it has to be worth swapping - but you would therefore always decide to swap. or would you not swap?  Would you swap before seeing the envelopes contents?  It seems to be a contradiction
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Question by:deighton
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sdstuber earned 25 total points
ID: 34997066
This is similar to but different than the Monty Haul problem.
First, there are 3 choices in the Monty Haul
Second, Monty reveals one of the three choices which changes the information known so skews the results of the remaining two.

In a case starting with only 2 choices and no additional information is available there is no value in swapping
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Cluskitt earned 25 total points
ID: 34997098
Before seeing anything, you have a 50/50 chance of drawing either of them. After seeing one of them, and not knowing what the other one is, you still have a 50/50 chance of drawing either of them, whether you swap them or not. It's just your perceived interpretation that makes you think you have more chances when swapping, but statistically, the chances are ALWAYS 50/50.
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d-glitch earned 50 total points
ID: 34997436
Here is one analytical explanation.

http://www.maa.org/devlin/devlin_0708_04.html

The point is that you can not assume that just because you see \$100 in one envelope,
that \$50 and \$200 are equally likely for the other.
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ID: 34997837
And a better explanation at Wikipedia:

http://en.wikipedia.org/wiki/Two_envelopes_problem

The paradox comes from assuming that the second envelope contains either double or half the amount
found in the first envelope, irrespective of whether the smaller or larger amount was initially picked.
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ID: 34997840
d-glitch - however - in a on-off situation, the chances of picking the highest envelope is Bernoulli, with the amount seen in the open envelope not giving any guide as to whether it is the higher or lower value.   I still assert that to a contestant with no insider game show knowledge, the chances of doubling his prize has to be 50%

Only over a large number of trials would the contestant be able to assess an upper or lower bound to the prizes.  Obviously 1 cent is a lower bound, you would know the other envelope has 2c in it.

But that analysis given is a good try at solving the conundrum, maybe it does solve it!

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ID: 34997975
also take a look at 'A second problem' at wikipedia article, it lists a case where there is a probability distribution, although I think in that case you need a distribution for n would be the answer
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ID: 34997999
Once you know the value of the first envelope, the chance of doubling your prize is the same as the chance of halving it. 50/50. Just like the chance of picking the highest value is the same as picking the lowest. 50/50.
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ID: 34998182
Cluskitt - so you have a 50% chance of gaining \$100 and a 50% chance of losing \$50, making the swap more valuable.

the swap is worth 100 * .5 - 50 * .5 = 25
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ID: 34998261
No. The value doesn't matter at all. That is the flaw in the logic. You have a 50/50 chance of drawing either envelope. It doesn't matter which you open, because the chance is always the same. You can't multiply it like that. That assumes that both values exist on the other envelope, which they don't. You're calculating a statistic using a universe of data of 3 values for 2 envelopes, which is incorrect.
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ID: 34998505
In this specific example of double or half value does the classic 2-envelope problem apply?

There is no infinite distribution here.  The other envelope has one of exactly 2 known values.

And as such,  with the exception of the special case of 1 cent (or smallest denomination in whatever currency system is used)  there is no "extra" information objective or subjective to introduce that might push this problem into Bayes decision.

So,  50/50  switching doesn't help
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ID: 34998538
There is an infinite distribution here before you open the envelope (assuming the flawed logic was correct). That is, if swapping was better, then you would chose the other envelope instead (before opening). Which would make the unchosen envelope better, which would make a swap better which would make the original envelope better, etc.
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ID: 34998593
first, that doesn't help  :)

second,  that's not the case here.  We know the contents of the envelope we opened, so now we know exactly what choices we have for the other envelope.

It's "usually" going to be a choice of 2 values  (1/2  or 2)  so 50/50 in the normal case

One special case exists where we know the other value is double what we have.  100/0 in this one special case (the case of holding the smallest denomination)

I don't think this question qualifies as "The Two Envelope Problem"
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ID: 34998612
as for the second part,  would I swap before seeing?

No.  Since my odds for not seeing don't change, but for exactly one case my odds improve.  I would always look.

On the other hand, for that one special case, the reward for my cleverness is I win 2 cents instead of 1, so hardly worth it.
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ID: 34998651
Actually, it does. If you assume the flawed logic, swapping is always better. But, if swapping is always better, you might as well swap before you open the envelope. But if you swap before you open the envelope, that makes the other envelope a better choice. Basically, you fall into an infinite swap loop.

Not that it matter anyway, because the logic flaw is that you assume the other envelope may have double or half, when in fact it can't.
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ID: 34998693
>>> the logic flaw is that you assume the other envelope may have double or half, when in fact it can't.

I don't get that.

It not only can,  it must.  That's the what the question states.  one has twice the other.
So whichever you have, you know the other is double or half.

Unless you are throwing in another wrinkle that the game show host is also a liar.
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phoffric earned 25 total points
ID: 34998737
>> But if you swap before you open the envelope, that makes the other envelope a better choice. Basically, you fall into an infinite swap loop.

I skimmed the link, and I think the gist was that by using this logic, you could show a contradiction in ever increasing returns.

But, as with any recursive algorithm, there needs to be a way to stop the infinite recursion. To stop the recursion, you look at the value of the check. No longer does it make sense to keep swapping infinitely. You now know the other two possible values of the check, so a swap is beneficial.
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ID: 34998755
No. The flaw is exactly that. For example, imagine you have 2 envelopes, one with \$100 and one with \$200. Each has a 50/50 chance. If you draw the \$100, you're assuming the second envelope can have either \$50 or \$200. But it can't. It can only have \$200. You don't know which value it is, but it can only have that value. Likewise, you're assuming that, if you draw the \$200 envelope, the other can have either \$100 or \$400. So, you're assuming the possible values for both envelopes are \$50, \$100, \$200 and \$400, when in fact the only possible values are \$100 and \$200. It's a closed system which can't be infinitely regressed.
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ID: 34998838
ok, I wasn't understanding the argument to the argument.  The rephrasing help

I stand by my statement that the other envelope has either double or half.

Which is true,  but one of those is impossible given the starting values
Just as the other is one of  double, half, triple or third.
Even though 2 of those are clearly not possible, but still a true statement.

But I concede the point
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ID: 34998931
Basically, it's what the author refers to in the first link. Only he's using a value M for the value. But it's easier to visualize like that.
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ID: 34999202
the 'non-existent infinite probability' argument does not seem to work because of the alternate more complex game variant described at
http://en.wikipedia.org/wiki/Two_envelopes_problem#A_second_problem

This seems to describe the same paradox, but with a well behaved probability distribution

_________________________________________________________________________

BTW - would you agree that if the gameshow host offered to toss a coin, and you either doubled or halved your winnings based on it, then accepting the wager is a good bet?  If so, then why is the switch wager different?

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ID: 34999226
>>  imagine you have 2 envelopes, one with \$100 and one with \$200.

Can we assume instead an equal probability distribution of check pairs (50/100) and (100/200):
C1) 200 100
C2)  50 100

We have two drawers ~ C1 and C2
Pick a drawer at random and then pick a check which happens to be \$100.
P(C1) = P(C2) = .5

Z = picked   100
X = other is 200
Y = other is  50

P(X|Z & C1) = 1
P(Y|Z & C1) = 0
P(X|Z & C2) = 0
P(Y|Z & C2) = 1

P(C1) * P(X|Z & C1) 200 + P(C1) * P(Y|Z & C1) 50 +
P(C2) * P(X|Z & C2) 200 + P(C2) * P(Y|Z & C2) 50 =
.5 * ( 1*200 + 0*50 + 0*200 + 1*50) = .5( 200 + 50 ) = .5(250) = 125
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ID: 34999233
Yes, because with a coin toss, you will then have both values possible: double and half, whereas with the envelopes you don't (you just think you do).
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ID: 34999271
>>Can we assume instead an equal probability distribution of check pairs (50/100) and (100/200):
C1) 200 100
C2)  50 100

You may not assume this, because this gives you information about the maximum
return available in the game.
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ID: 34999306
The whole point of the paradox is that there's a closed system of 2 values, but the way the problem is presented leads the reader into interpreting a 3 value system.
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ID: 34999327
Why do you say 'you just think you do'.

But I sort of agree, if you spring the game on numerous  un-connected people, always with 50 and 100 (but they don't know that) the switchers will have an average of Â£75, and non-switchers again \$75 average.

so it only applies if the \$100 & \$200 game is somehow just as likely as \$50 & \$100 - which is shown to be difficult to achieve, but not impossible with the wikipedia version
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ID: 34999395
The thing is, you're assuming both are possible. Once you open one envelope, the chances aren't 50/50 for the other envelope. They're 100/0. The fact that you don't know which is which is irrelevant. Knowledge doesn't affect statistics.
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ID: 34999430
>>Can we assume instead an equal probability distribution of check pairs (50/100) and (100/200):
C1) 200 100
C2)  50 100

In the problem statement, \$100 is either the max or min payout.
With the above assumption, it is neither.
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ID: 34999436
>>> Knowledge doesn't affect statistics.

Sort of depends on what you're looking at.

Check the Monty Haul problem.

The odds of winning with your original guess don't change with added knowledge.
The odds of the "other" door double with added knowledge.

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ID: 34999506
Given two envelopes, with different amounts of money in them, and two contestants.

They flip a coin.  A wins, chooses an envelope and opens it to find \$100.
B takes the other envelope and doesn't open it.

Is there any reason to trade?  How can ignorance make B's envelope more valuable?
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ID: 34999520
What is the goal of this question?  To "solve" the general 2 Envelope problem?  If so, wow, that's optimistic for a bunch of part-time volunteers.

In the long run, it doesn't matter if you switch or keep

You can verify simply by trying it.  Of course Monte Carlo  simulation, doesn't "prove" anything, but it helps support the supposition.

By my random number generator, if you keep your \$100 envelope several million times you'll net about a penny.

But only about half the time.  :)
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ID: 34999726
You can do a Monte Carlo simulation without assigning a priori probabilities to the possible initial conditions:  [50, 100] and [100, 200].
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ID: 35000226
Sorry!!!

You CAN'T do a Monte Carlo simulation without assigning a priori probabilities to the possible initial conditions:  [50, 100] and [100, 200].
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ID: 35000605
right, I figured it was a typo.

and you're right, I used 50/50, which is an assumption on my part.

If there is a 1/10000000 chance the \$100 will be doubled and all other times the second envelope will be halved then I'll want to keep my envelope when I see the result.

If the doubling/halving chance is reversed then I'll want to swap after seeing my result.

Of course that assumes I both know the choices are 50/100 or 100/200 and I also know the probability of each.

If I don't, and all I have to go on is the scenario as described above, I have no reason to believe it's "rigged" one way or the other. So I assume 50/50.

That doesn't make my 50/50 guess more correct than any other guess.
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ID: 35000665
>>Can we assume instead an equal probability distribution of check pairs (50/100) and (100/200):
>> In the problem statement, \$100 is either the max or min payout.
>> With the above assumption, it is neither.
Maybe I'm not getting the definition of "payout".
Could you elaborate why it is neither.
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ID: 35000998
This is a game with two envelopes.  One has X dollars, the other has 2X.

You choose one envelope at random.  You have a 50/50 chance of selecting max/min.

You open you envelope and see \$100.  This seems like a lot of information.
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ID: 35005084
Actually, a simple rephrasing should clarify things:

You have envelope A and envelope B. One has a higher value than the other. You have a 50/50 chance of choosing either. Once you open one envelope, what are the odds that it's the higher value? Still 50/50. What are the odds that swapping will increase the value? Still 50/50. The amount by which one is higher than the other is irrelevant. You have one that is higher, one that is lower, and both have 50% chance of being drawn.

sdstuber: Knowledge does NOT affect statistics. In Monty's, the chance of choosing the right door is based on two separate statistics (whether you defend that the chance is 1/2 or 2/3, seeing as none is proven). The first statistic is 1/3. After you gain knowledge of another door, you then have another separate statistic.
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ID: 35005477
Cluskitt - I agree with that, however from the point of view of the contestant holding \$100, there is a 50% chance that he has the lower amount, therefore he has a 50% chance of gaining \$100 and a 50% chance of losing just \$50 - so it is like an evens bet with half the stake back if you lose.

This contradiction seems to have led some people to conclude the game can't exist, 'proved' by the impossibility of an infinite  uniform probability distribution.  However the modified game at wikipedia shows a case where there is a plausible probability distribution, although it has to allow for unbounded prizes, which can not exist in real money, but if it was Monopoly money, they could exist.
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ID: 35005496
The point of view of the contestant is skewed, though. There isn't a 50% chance for both values. There's a 100% chance of one and 0% of the other. You not knowing which is which doesn't affect the probability.
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ID: 35006063
Cluskitt - not really, because you can devise any number of situations where the outcome is pre-determined, but un-known to the decider - in those cases surely they can only be guided by probability.  In this scenario they will double their money 50% of the time, would you not agree?

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ID: 35006197
They will double their money by choosing the lowest amount and halve it by choosing the lowest amount. Or, to say it another way, they will win +\$X when they choose the lowest amount and lose -\$X when they chose the highest. You're getting lost in both semantics and perceived reality (which is different from "real" reality). Despite the fact that you're saying "double" and "half", the difference amount is always the same. You just apply double and half depending on your first choice being lower or higher amount.

When there are many ways to phrase a problem, you have to reduce it to its simplest form:
You have 2 envelopes with different amounts. You open one. What is the chance that swapping them will increase your wins?

The fact that it's double/half, or square/square root or any other similar operation becomes irrelevant. 2 options: 50% chance. Playing on semantics and most especially on perceived reality is how most games of chance always favour the "house".
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ID: 35006218
And I can give you another paradox example:

Imagine you have 3 types of cards: One has spades on both sides, one has hearts on both sides, one has spades on one side and hearts on the other. You're asked to chose one at random. What are the odds that you'll chose one with different suits? Obviously it's 1/3.

Now, imagine that, after you've chosen your card, and before you see it, I reveal one of the other cards. What is the chance? It's not 50%. It's still the same. 1/3. Because your choice was made before seeing any card, seeing one after that makes no difference to your odds. Revealing one card won't change anything at all.
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ID: 35006288
You should read this book: "Aha! Gotcha: Paradoxes to Puzzle and Delight" by Martin Gardner. There are many similar examples of perceived reality statistical differences leading to paradoxes. Although there are more books about the subject, I found that this book was an excellent gateway into these issues, being both accessible, so you don't have to be a math guru to understand it, and challenging, meaning it's not written for babies. :)
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ID: 35006949
Cluskitt,

>>> you then have another separate statistic.

I think we're in agreement on the concept, just maybe not agreeing on the phrasing.
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ID: 35006994
That is usually the source of these paradoxes :)
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ID: 35007098
Look at the psychology and the value of information:

Given two envelopes, with different amounts of money in them, and two contestants.

I.  They flip a coin.  A wins, chooses an envelope and opens it to find \$100.
Now he doesn't want it anymore.

II.  They flip a coin.  A wins, chooses an envelope.  Before he can open it,
B opens his to find \$50, and offers a trade.  A refuses because he thinks
he may have \$100.

III.  They flip a coin.  A wins, chooses an envelope.  Before he can open it,
B opens his to find \$200, and offers a trade.  A refuses because he thinks
he may have \$400.

IV.  They flip a coin.  A wins, chooses an envelope.  Before he can open it,
B opens his to find \$2000.  [There was a mistake at the bank, and they
put in twenty \$100 bills instead of twenty \$10's.]  B and offers a trade,
but A refuses because he thinks he may have \$4000.
B is annoyed, but only briefly.

The initial choice of envelopes gives A a clear, fair 50/50 chance at the max amount.

Information about the contents of any one envelope provides absolutely no
additional information.  This should be clear since there is no dollar threshold
that would convince an irrational contestant to hold on to an open envelope.

Since knowing the content of a single envelope provides no information, it provides no reason to second guess an initial choice.
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ID: 35007291
Given two envelopes, with different amounts of money in them, and two contestants.

The envelopes are placed on the left and right sides of a computerized sensor table.
There is an advanced computerized currency scanner on the left side that reads the
microfibers in the bills through the envelope, and calculates the total.  As soon as a
contestant chooses an envelope, the contents of the left side envelope are displayed to
the studio audience and the world.

You win the coin toss.  You always choose the right side, because you will know how much
money is in the other envelope before you open yours, and yours may have more money.
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ID: 35007367
>> You always choose the right side, because you will know how much
money is in the other envelope before you open yours, and yours may have more money.

Why would this be? The chance of the right side having more money is exactly the same as the left side. All you would gain is knowing what amount you didn't chose, which doesn't change anything at all. As you said before: Since knowing the content of a single envelope provides no information, it provides no reason to second guess an initial choice.
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Accepted Solution

d-glitch earned 50 total points
ID: 35007387
Sorry.  I was doing reductio ad absurdum
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ID: 35008558

Given two envelopes with different amounts of money in them.

After you choose an envelope and before you open it, you may ask the host to tell you the
value of either envelope.

You choose either A or B, but then you must ask about the value of the other envelope because
then yours could be worth twice as much.

You would never ask about your own envelope, becsause then  the other envelope might be  twice yours.  And we wouldn't want that.

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ID: 35009357
clusskit - I like that car one, I agree that you probability remains at 1/3, the third card is therefore 2/3
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ID: 35011002
yes, that's the Monty Haul problem.
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ID: 35028025
A parting shot from the Always-Swap Contingent:

Given two envelopes, with different amounts of money in them, and two contestants.  But this time, the difference in the amounts is a factor of 100.

You win the coin toss.  You open your envelope to find \$100.

The possible values of the other envelope are  \$1  and  \$10,000.
But what is the expected value?

I have been arguing that it isn't   \$5000.50

Would you offer to buy the unopened envelope for \$5K?
How about for the bargain price of  \$2.5K?
Would you pay a lot of cash out of pocket for something you could have had for free just a
few seconds earlier?

Maybe.  Opening your envelope does gives you a some information.  You may not be able to use it mathematically, but there is also real-world context to consider.

It the contest is being run by the professor of your Intro to Game Theory Class you would
certainly hold on to the \$100.  There is no chance that he's put  \$10K in the other envelope.

But if Monty Hall  or Regis Philbin is running the contest in prime time during sweeps week, you would really want the unopened envelope.  You might even want it if you found \$10K in the first one.  \$10K x 100 is only \$1M.  They they put millions on the line every week.
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