Probability Section 4.3 Exercise 5 Page 182

Using the notation introduced in Example 4.29,

let \[{p_{ x }= \begin{cases} { \frac { 2 }{ 3 } \left( \frac { 1 }{ 3 } \right) ^{ k } } , & if\quad x=2^{ k } \\ 0 , & otherwise \end{cases}}\]

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

If the envelope we are given contains two dollars or more:

\(k \ge 1\) so \(x = 2^k \ge 2\) dollars.

The amount in the envelope is either the smaller amount or the larger amount (but, the decision-maker does not yet know which.)

The probability that this is actually the smaller amount is:

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

If we switch, we receive the envelope containing \(2x\) and the contribution to long-run probability becomes:

\[\frac {p_{x}}{p_{\frac{x}{2} }+p_{ x }} = \frac { \frac { 2 }{ 3 } \left( \frac { 1 }{ 3 } \right) ^{ k } } { \frac { 2 }{ 3 } \left( \frac { 1 }{ 3 } \right) ^{ k-1 } + { \frac { 2 }{ 3 } \left( \frac { 1 }{ 3 } \right) ^{ k } }} = \frac { \left( \frac { 1 }{ 3 } \right) ^{ k } } { \left( \frac { 1 }{ 3 } \right) ^{ k-1 } + { \left( \frac { 1 }{ 3 } \right) ^{ k } }} = \frac{1}{3+1} = \frac{1}{4} = 0.25 \]

so the contribution to expected value is \(2x \cdot \frac{1}{4} = \frac{x}{2}\) .

On the other hand,

if we were already holding the envelope containing the larger sum, and we switch, we get the envelope containing \(\frac{x}{2}\) and the contribution to long-run probability becomes:

\[\frac {p_{\frac{x}{2} }}{p_{\frac{x}{2} }+p_{ x }} = \frac { \frac { 2 }{ 3 } \left( \frac { 1 }{ 3 } \right) ^{ k-1 } } { \frac { 2 }{ 3 } \left( \frac { 1 }{ 3 } \right) ^{ k-1 } + { \frac { 2 }{ 3 } \left( \frac { 1 }{ 3 } \right) ^{ k } }} = \frac { \left( \frac { 1 }{ 3 } \right) ^{ k-1 } } { \left( \frac { 1 }{ 3 } \right) ^{ k-1 } + { \left( \frac { 1 }{ 3 } \right) ^{ k } }} = \frac{1}{1+\frac{1}{3}} = \frac{1}{\frac{4}{3}} = \frac{3}{4}=0.75 \]

so the contribution to expected value is \(\frac{x}{2} \cdot \frac{3}{4} = \frac{3x}{8}\) .

Adding the two parts, the expected value from switching becomes \(\frac{x}{2} + \frac{3x}{8} = \frac{7x}{8} = 0.875x\) .

So, if the envelope we are given contains 2 dollars or more, the expected value of switching results in a loss of \(\frac{1}{8}\) of whatever amount we currently hold.

So, donโ€™t switch.

Alternatively:

if we are given an envelope containing exactly one dollar:

\(k = 0\) so \(x = 2^0 =1\) dollar .

In this case, we know that this must be the smaller amount, because it is not possible for the other envelope to contain 50 cents.

Therefore, the other envelope must contain 2 dollars.

This is the only case in which it makes sense to switch.

Conclusion:

If the envelope we are given contains 1 dollar, then it makes sense to switch because we know the other envelope must contain 2 dollars.

If the envelope we are given contains 2 dollars or more, then the expected value of switching would result in losing money.