Compound interest, mortgages and why the casino wins
Tonight's idea
Three pieces of mathematics run most of your financial life: a geometric series, a limit, and a weighted average.
Compounding. Put \(P\) in an account paying annual rate \(r\), compounded \(n\) times a year. Each period multiplies the balance by \(1 + r/n\), so after \(t\) years \(A = P\left(1 + \frac{r}{n}\right)^{nt}\). As \(n \to \infty\) this tends to \(Pe^{rt}\), with \(e = 2.71828\ldots\) from The number e.
The mortgage formula. Borrow \(P\) at monthly rate \(r\), repay a fixed \(M\) every month for \(n\) months. Each month the balance grows by the factor \(1 + r\) and drops by \(M\):
| after month | balance owed |
|---|---|
| 0 | \(P\) |
| 1 | \(P(1+r) - M\) |
| 2 | \(P(1+r)^2 - M(1+r) - M\) |
After \(n\) months it must be zero:
\[P(1+r)^n - M\left[(1+r)^{n-1} + \cdots + (1+r) + 1\right] = 0.\]
The bracket is a geometric series with sum \(\dfrac{(1+r)^n - 1}{r}\). Solve for \(M\):
\(M = P \cdot \dfrac{r(1+r)^n}{(1+r)^n - 1}\). Every mortgage and car loan is this one line.
Expected value. A bet pays \(x_i\) with probability \(p_i\). Its expected value \(E = \sum p_i x_i\) (add up each outcome times its probability) is the average result per play in the long run. Casinos are built on bets with \(E\) slightly negative for you. A European roulette wheel has 37 pockets (0 to 36); an American wheel adds a 00, making 38. Same payouts, different denominator.
Watch one: a $300,000 mortgage at 6% over 30 years
Monthly rate \(r = 0.06/12 = 0.005\), \(n = 360\) months, \((1.005)^{360} = 6.0226\). So
\[M = 300{,}000 \times \frac{0.005 \times 6.0226}{6.0226 - 1} = \$1{,}798.65 \text{ per month.}\]
Total paid: \(360 \times 1798.65 = \$647{,}514\). Interest: \(\$347{,}514\), more than the house. That is what \((1.005)^{360} \approx 6\) does over thirty years.
One spin of American roulette, $1 on a single number paying 35 to 1: \(E = \frac{1}{38}(35) + \frac{37}{38}(-1) = -\frac{2}{38} = -5.26\%\). You can win tonight; you cannot win on average.
Your turn
- $1,000 at 5% for 10 years. Find the balance with yearly, monthly and continuous compounding (\(e^{0.5}\)). How much does "compounding forever" add over monthly?
- $200,000 at 4.5% annual, repaid over 15 years (\(n = 180\)). Find the monthly payment and the total interest paid.
- A $2 lottery ticket wins $5,000,000 with probability 1 in 10,000,000 and $5 with probability 1 in 100. Find the expected value of the ticket and the expected loss per ticket. If 50 million tickets are sold, how much does the lottery keep on average?
- Find the expected value of a $10 bet on red on a European wheel (18 red, 18 black, 1 green zero; even money), and of a $10 bet on a single number on an American wheel (35 to 1, 38 pockets). Which is the better bet for the player, and by how much per spin?
Stretch: a pension is a mortgage run backwards
A pot earning 4% a year must pay out $40,000 at the end of each year. How big a pot pays forever (a perpetuity)? Now show that a pot lasting exactly 25 years is \(40{,}000 \times \dfrac{1 - 1.04^{-25}}{0.04}\) by summing a geometric series, and compute it.
Answers
- Yearly: \(1000 \times 1.05^{10} = \$1{,}628.89\). Monthly: \(1000 \times (1 + 0.05/12)^{120} = \$1{,}647.01\). Continuous: \(1000e^{0.5} = \$1{,}648.72\). Continuous adds only $1.71 over monthly.
- \(r = 0.00375\), \((1.00375)^{180} = 1.9616\). \(M = 200{,}000 \times \dfrac{0.00375 \times 1.9616}{0.9616} \approx \$1{,}530\); to the cent, \(\$1{,}529.99\). Total paid \(180 \times 1529.99 = \$275{,}398\), so interest \(\$75{,}398\).
- \(E = 5{,}000{,}000/10^7 + 5/100 = 0.50 + 0.05 = \$0.55\). Expected loss \(= 2 - 0.55 = \$1.45\) per ticket. 50 million tickets: \(\$72.5\) million kept on average.
- Red, European: \(10 \times \frac{18 - 19}{37} = -\$0.27\) per spin. Single number, American: \(10 \times \frac{35 - 37}{38} = -\$0.53\) per spin. Red on the European wheel loses about half as fast; it is the better of two bad bets.
Stretch. Forever: the interest alone must cover the payout, so \(0.04 \times PV = 40{,}000\) and \(PV = \$1{,}000{,}000\). For 25 years each payment due in \(k\) years is worth \(40{,}000/1.04^k\) today; summing \(k = 1\) to 25 gives \(40{,}000 \cdot \dfrac{1}{1.04} \cdot \dfrac{1 - 1.04^{-25}}{1 - 1/1.04} = 40{,}000 \cdot \dfrac{1 - 1.04^{-25}}{0.04} = \$624{,}883\). Let 25 become infinity and \(1.04^{-n} \to 0\): the million dollars again. The mortgage formula is this sum solved for the payment.