Suppose there are two face down cards each with a positive real number and with one twice the other. Each card has value equal to its number. You are given one of the cards (with value $x$) and after you have seen it, the dealer offers you an opportunity to swap without anyone having looked at the other card.

If you choose to swap, your expected value should be the same, as you still have a $50\%$ chance of getting the higher card and $50\%$ of getting the lower card.

However, the other card has a $50\%$ chance of being $0.5x$ and a $50\%$ chance of being $2x$. If we keep the card, our expected value is $x$, while if we swap it, then our expected value is: $$0.5(0.5x)+0.5(2x)=1.25x$$ so it seems like it is better to swap. Can anyone explain this apparent contradiction?

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Your question is very hard to understand. However, if I'm understanding you correctly, I fail to see any paradox. If I know the value of the other card is either double or half the value of my card (and I'm trying to maximize the value of the card in my hand) of course its beneficial for me to swap cards. – Ami Jul 21 '10 at 3:07
You never define what you mean by "value." After reading your question a few times I assumed that each player is trying to maximize "value." The math that you do is not well explained or justified. I don't know what you mean by: "so using expected value." Lastly, I don't see any paradox here. From a probabilistic standpoint, if the player switches cards he may lose or he may gain, but he stands to gain more than he stands to lose, thats what your math shows. Whats the problem? – Ami Jul 21 '10 at 3:23
@Ami: I wasn't very clear before. I think that it may be clearer now. – Casebash Jul 21 '10 at 3:24
This is exactly the two envelopes problem, if anyone wants to write up an explanation. – Larry Wang Jul 21 '10 at 3:49
An even more interesting problem (similar, but not the same) has been discussed to death on mathoverflow – BlueRaja - Danny Pflughoeft Jul 21 '10 at 4:05

This puzzle is known as the two envelope paradox. This paper contains a nice explanation of the two envelope paradox, and some references to further literature regarding the puzzle.

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in the interests of full disclosure, I know one of the authors of the paper, but know very little about the subject. it might well be that there is a much better review paper on the subject – Seamus Aug 4 '10 at 17:20
The article in the link seems to propose an axiomatization to explain the a priori indifference between the envelopes. It does not attempt to resolve the paradox itself. The paradox is discussed quite well in en.wikipedia.org/wiki/Two_envelopes_problem. – Ittay Weiss Sep 23 '12 at 7:12

This paradox has always interested me. Something to think about is that there does not exist a uniform probability distribution over the positive real numbers (since they are infinite). In arriving at your paradox, it seems you are assuming that any real number is equally likely, but this cannot be the case.

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There is a uniform distribution on the real numbers. I think you meant to say that the paradox is resolved when one realizes there is no uniform distribution on the natural numbers. However, that realization only solves one variant of the paradox. The paradox is recovered by using certain other distributions which still produce the paradox (see 'a new variant' in en.wikipedia.org/wiki/Two_envelopes_problem) – Ittay Weiss Sep 23 '12 at 7:10
@Ittay: How do you mean, "There is a uniform distribution on the real numbers"? There isn't in the usual sense of the term (see e.g. math.stackexchange.com/questions/14777). – joriki Jul 5 '13 at 6:47
@joriki bad formulation of what I meant to say. I meant to say that the infinitude of real numbers is not what prevents uniform distributions to exists (e.g., there are uniform distributions on finite intervals of real numbers). – Ittay Weiss Jul 5 '13 at 7:17
@Eric I was never satisfied by the common resolution. $x$ doesn't stand for two different things in the calculation (if you think about opening up the envelopes then the idea that $x$ is two different things becomes ridiculous). I do think there's something wrong with "suppose one is twice the other" - it probably subtly assumes that you can choose a random real number, as you suggest, but I'm not sure exactly where this subtlety is ... – Zubin Mukerjee Oct 9 '14 at 3:14

For the sake of argument, let's assume that the value of the lower envelope $V_L$ follows a uniform distribution from 0 to ∞. Let's call the probability function for that $P_L$. We define this to mean that values from equal ranges are equally likely. For example, for any constant $k$: $$P_L(0<V_L<k) = P_L(n<V_L<n+k)$$

Comparing this to the values of the higher envelope $V_H$, we get the probability distribution $P_H$, which when equated to $P_L$ is as follows:

$$P_L(n<V_L<n+k)=P_H(2n<V_H<2(n+k))$$

This means that the higher envelope's values are half as likely as the lower envelope's values over the same range.

For an example, let's say we opened the envelope and it's value $V$ ranged anywhere from 1 to 2 dollars inside it. The odds of the other envelope containing 2 to 4 are:

$$P_H(2<V_H<4) = P_L(1<V_L<2)$$

Similarly, the odds of the other envelope containing .5 to 1 are:

$$P_L(.5<V_L<1)$$

This is half as likely, since it covers half the range. Thus, the probability that we chose the lower-valued envelope for $0<V<∞$ is 2/3, not 1/2!

This may seem like another paradox, but it will all make sense in a moment.

Using this, we can now calculate the expected value of the other envelope:

$$2/3 * 2V + 1/3 * 1/2V = 1.5V$$

Thus, it is even higher than the $1.25V$ mentioned in the original post. Again, this seems like a paradox. How can we get a higher expected value for the other envelope across all cases?

To see why this is, we must count infinity, which helps us understand that We have NOT counted all the cases! First, we know that the possible values in our lower envelope range from 0 to ∞. However, this must mean that the values in the upper envelope range from 0 to 2∞. Thus, we know that one of the envelopes has greater than ∞ half the time since:

$$P_H(0<V_H<∞)=P_H(∞<V_H<2∞)$$

Since we'll pick that envelope half the time when it is available, we get that the probability that $V>∞$ is 1/4. In these cases, we should never switch since we have envelope with the bigger value.

For the other 3 out of 4 times, $0<V<∞$ will be true. We've already seen that the probability we've selected the lower valued envelope is 2/3 for this range. If we calculate the total probability that we selected the lower envelope over all values, it ends up being the expected 1/2, which resolves one of our "paradoxes":

$$2/3*3/4 + 1/4 * 0 = 1/2$$

Finally, to remove the other paradox, we calculate the average expected value of the other envelope over the entire range. For $∞<V<2∞$, which covers 1/4 of the cases, we know that the expected value is $.5V$. This averages out to be 3/4∞ since it is a uniform distribution from $.5∞<V<∞$. For $0<V<∞$, the expected value is $1.5V$, which also ends up being $3/4∞$ since it is a uniform distribution of values from $0<V<∞$, and $.5∞*1.5V=3/4V$. Thus, our average expected value of the other envelope is $3/4∞$, which equals the average of all possible values for both envelopes: $$∞/2 * 1/2 + 2∞/2 * 1/2 = 3/4∞$$

You can apply similar logic to any probability distribution you like and it will make cents ;)