Example 4.29 Suppose that we have two envelopes in front of us, and that one envelope contains twice the amount of money as the other (both amounts are positive integers). We are given one of the envelopes, and asked if we would like to switch.

page 182. 4.3-4&5

  1. Using the notation introduced in Example 4.29, show that in the example of Brams and Kilgour, if x is a positive power of 2, then
\[\begin{equation} \frac{p_{x/2}}{p_{x/2}+p_x}=\frac{3}{5} \end{equation}\]

Solution:

If x is a positive power of 2

\[\begin{equation} p_x= \begin{cases} \frac{1}{3}(\frac{2}{3})^(k-1), & \text{if}\ x=2^k \\ 0, & \text{otherwise} \end{cases} \end{equation}\]

\(x/2 = 2^{k-1}\)

\(p_{x/2}=\frac{1}{3}(\frac{2}{3})^{k-2}\)

\(\frac{p_{x/2}}{p_{x/2}+p_x} = \frac{\frac{1}{3}(\frac{2}{3})^{k-2}}{\frac{1}{3}(\frac{2}{3})^{k-2}+\frac{1}{3}(\frac{2}{3})^{k-1}}\)

\(\frac{p_{x/2}}{p_{x/2}+p_x}=\frac{(\frac{2}{3})^(k-2)}{(\frac{2}{3})^(k-2)(1+\frac{2}{3})}\)

\(\frac{p_{x/2}}{p_{x/2}+p_x}=\frac{3}{5}\)

5 Using the notation introduced in Example 4.29, let \[\begin{equation} p_x= \begin{cases} \frac{1}{3}(\frac{2}{3})^(k-1), & \text{if}\ x=2^k \\ 0, & \text{otherwise} \end{cases} \end{equation}\]

Show that there is exactly one value of x such that if your envelope contains x, then you should switch.

If

\(x = 2^k\), \(\frac{p_{x/2}}{p_{x/2}+p_x}=\frac{3}{5}\),\(\frac{p_{x}}{p_{x/2}+p_x}=\frac{2}{5}\)

\(\frac{p_{x/2}}{p_{x/2}+p_x}\frac{x}{2} +\frac{p_{x}}{p_{x/2}+p_x}2x >x\)

\(\frac{3}{5}\times \frac{x}{2} + \frac{2}{5}\times2x = \frac{11}{10}\times x\)

\(\frac{11}{10}\times x > x\)

So when x is a positive power of 2, we should switch.