Card doubling paradox

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.

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.

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 ;)

Paradox resolved!