In the chain letter problem (see Example 10.14) find your expected profit if
\(p_0\) = 1/2, \(p_1\) = 0, and \(p_2\) = 1/2.
\(p_0\) = 1/6, \(p_1\) = 1/2, and \(p_2\) = 1/3.
Also, show that if \(p_0\) > 1/2, you cannot expect to make a profit.
Chain letter problem from Example 10.14: The chain required a participant to buy a letter containing a list of 12 names for 100 dollars. The buyer gives 50 dollars to the person from whom the letter was purchased and then sends 50 dollars to the person whose name is at the top of the list. The buyer then crosses off the name at the top of the list and adds her own name at the bottom in each letter before it is sold again.
The equations from the textbook:
\(50m + 50m^{12} - 100\)
m = \(p_1 + 2p_2\)
m = \(p_1 + 2p_2\)
m = 0 + 2 * (1/2) = 1
\(50(1 + 1^{12}) - 100 = 50(2) - 100 = 100 - 100 = \$0\)
m = \(p_1 + 2p_2\)
m = (1/2) + 2 * (1/3) = (1/2) + (2/3) = (7/6)
\(50(\frac {7} {6} + \frac {7} {6}^{12}) - 100 = 50(\frac {7} {6} + 6.35859956) - 100 = 50(7.41836615) - 100 = 370.918308 - 100 = \$270.92\)
According to the description of the problem from the textbook: “But this will be true if and only if m > 1. We have seen that this will occur in the quadratic case if and only if p2 > p0.”
Therefore, to make a profit, \(p_2\) must be greater than \(p_0\). In the problems above, (a) \(p_2\) = 1/2 and (b) \(p_2\) = 1/3. If \(p_0\) > 1/2, then \(p_0\) > \(p_2\) in both cases, so you will not make a profit. Additionally, when \(p_0\) is more than 1/2, this means the probability of the buyer losing money is greater than the probability of the buyer receiving a profit. Since they have to spend $100 but there’s more than a 50% chance they will lose money, then they will lost more money than they will gain. Even if they do earn some profit, the losses will overwhelm the wins due to that 50% chance of losing money.