# The Amex EveryDay Bonus: A Stochastic Valuation Model

January 21, 2019

The EveryDay credit cards from American Express feature an unusual complication in their reward structures: in addition to awarding the cardholder points in proportion to spending (e.g., 2 points per dollar on grocery purchases, 1 point per dollar on everything else), cardholders who use their EveryDay card a certain number of times during a billing period receive a bonus at the end of the month, equal to some percentage of the total number of points earned during that month:

EveryDay Credit CardEveryDay Preferred Card
Required purchases / month2030
EveryDay point bonus+20%+ 50%

For example, if you earned 1,000 points on the EveryDay Preferred credit card during August, and made exactly 30 purchases during August, you would receive a bonus of 500 points, for a total award of 1,500 points.

But if you made only 29 purchases with that card during August, you would receive no bonus — just the 1,000 points earned from regular spending.

A potential cardholder who makes approximately 20 or 30 purchases each month — sometimes more, sometimes less — may not know precisely how to value this bonus structure. In this article, I will develop a simplified model for valuing the EveryDay purchase bonus, under the following assumptions:

1. On average, the cardholder makes $$R$$ purchases per month
2. The average purchase price is $$D$$ dollars
3. Purchases are distributed randomly throughout the year
4. A bonus worth some multiple $$M$$ of the points earned in the previous month is awarded if at least $$B$$ purchases are made during the month
5. The cardholder makes no specific effort to earn the monthly bonus

For the present undertaking, I will ignore the tiered point system that awards two or three points for spending in specific categories, and assume that all spending earns (pre-bonus) 1 point per dollar.

To fix ideas, the regular card awards a 20% bonus when 20 purchases are made; that is, $$M=0.2$$ and $$B=20$$. For the Preferred card, $$M=0.5$$ and $$B=30$$ (50% bonus if 30 purchases are made).

Assumptions 1 and 3 let us describe purchases as random arrivals, and therefore model the probability of making exactly $$n$$ purchases during a given month using a Poisson distribution that has parameter $$\lambda = R$$:

$P(x=n) = \frac{e^{-\lambda} \lambda^n}{n!} = \frac{e^{-R} R^n}{n!}$

The probability of making at least $$B$$ purchases is then:

$P(x\ge B) = \sum_{n=B}^\infty \frac{e^{-R} R^n}{n!}$

First we model the value of the pre-bonus spending in terms of points. The expected number of points earned during the month is given by the Poisson probability of each number of purchases occurring, multiplied by that same number of purchases, multiplied by $$D$$, the average purchase price in dollars (since one dollar yields one point):

$E[points] = \sum_{n=0}^\infty \frac{e^{-R} R^n}{n!} \times n D$

Evaluating this summation is not especially interesting, as it just yields the well-known mean of a Poisson distribution, $$RD$$. However, this sum formulation is useful for properly valuing the end-of-month bonus.

How much is the bonus worth? If a bonus were always awarded, it would be worth on average $$MRD$$, the bonus multiple times the average number of points earned. However, because the bonus is only awarded when a certain number of purchases are made, the size of the bonus is conditional on the purchase count (more purchases yield a larger bonus). The average awarded bonus will therefore be larger than $$MRD$$.

The bonus in a month where exactly $$n$$ purchases are made is worth $$MnD$$. Thus the expected value of the bonus is given by:

$E[bonus] = \sum_{n=B}^{\infty} \frac{e^{-R} R^n}{n!} \times MnD$

To simplify the expression, we can add and subtract the expected value of all the nonexistent bonus points where $$n \lt B$$:

$E[bonus] = \sum_{n=B}^{\infty} \frac{e^{-R} R^n}{n!} \times MnD + \sum_{n=0}^{B-1} \frac{e^{-R} R^n}{n!} \times MnD - \sum_{n=0}^{B-1} \frac{e^{-R} R^n}{n!} \times MnD$

The first two summations collapse:

$E[bonus] = \sum_{n=0}^{\infty} \frac{e^{-R} R^n}{n!} \times MnD - \sum_{n=0}^{B-1} \frac{e^{-R} R^n}{n!} \times MnD$

The collapsed summation evaluates to the mean of a Poisson distribution, $$MRD$$:

$E[bonus] = MRD - \sum_{n=0}^{B-1} \frac{e^{-R} R^n}{n!} \times MnD$

Factoring out the $$MRD$$, the expression becomes:

$E[bonus] = MRD \left(1 - 1/R \sum_{n=0}^{B-1} \frac{e^{-R} R^n}{n!} \times n\right)$

Which lets us rewrite the summation as1:

$E[bonus] = MRD \left(1 - \sum_{n=0}^{B-2} \frac{e^{-R} R^n}{n!} \right)$

Plugging in some values, a cardholder who makes an average of $$R=25$$ purchases of \$40 each month can expected to earn $$RD=1000$$ points from regular spending. The expected value of the bonus for the regular card ($$B=20$$, $$M=0.2$$) works out to 181.6 points per month, with the bonus earned 86.6% of the time. The expected value of the bonus for the Preferred card ($$B=30$$, $$M=0.5$$) works out to 118.3 points, with the bonus earned 18.2% of the time.

Other things being equal (i.e. ignoring the annual fee and tiered point structures), this hypothetical cardholder can expect to earn 73.3 more bonus points per month using the regular card.

We can also solve for the value of $$R$$ such that the expected bonus from both cards are equal, that is, solve for $$R$$ such that2:

$0.2 RD\left(1 - \sum_{n=0}^{18}\frac{e^{-R} R^n}{n!}\right) = 0.5 RD\left(1-\sum_{n=0}^{28}\frac{e^{-R} R^n}{n!}\right)$

Canceling $$D$$ and solving numerically with a linear search, this equality can be seen to hold when $$R\approx 27.1$$.

Therefore the naive advice to choose the Preferred card only when $$R\ge 30$$ deserves a slight modification. In the model developed here, the Preferred card is a better option whenever $$R > 27.1$$.

If you want an more precise answer that takes into account both the annual fees and the tiered point system, I’ve developed an interactive calculator, hereby christened Evan’s Ultimate Amex EveryDay Bonus Calculator.

Alternatively, you could just plan to use the card once a day, and expect to earn $$MRD$$ bonus points every month.

But where’s the fun in that?

### Notes

1. Interestingly, the term in parentheses is equal to $$P(x\ge B-1)$$, that is, the upper CDF of the Poisson distribution starting at $$B-1$$, rather than $$B$$. The expected bonus can thus be written more succinctly as:

$E[bonus] = MRD \sum_{n=B-1}^{\infty} \frac{e^{-R} R^n}{n!}$

Or even:

$E[bonus] = MRD \times P(B-1, R)$

Where $$P(x, R)$$ is the regularized incomplete gamma function.

We can therefore quantify the expected difference between doing the exact calculation with the bonus size conditioned on the previous month’s purchases, and an approximate calculation where the bonus is assumed to equal $$MRD$$. The difference is simply the first term in the summation above; that is, it is equal to the Poisson probability of making exactly $$B-1$$ purchases in a given month. If, for example, $$B=R=20$$, the difference works out to $$P(x=19) = 8.88\%$$ of $$MRD$$.

2. Using the gamma function relation in the previous note, an even more succinct version of this equation is:

$P(19, R) = \frac{5}{2} P(29, R)$

Solving for $$R$$ in this representation may not be so easy unless standard math libraries improve their support for special functions.

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