Milage reimbursement calculation

[Bonesy777]

I'm trying to come up with a conclusion expressed in cents/mile for the following scenario:
For the first 99 miles I drive, I get reimbursed nothing. Once I drive 100 miles, I get reimbursed 25 cents for all 100 miles ($25). Once I drive anything over 100 miles but less than 200 miles, I still just get paid for 100 miles. Once I drive 200 miles, I get reimbursed 25 cents for all 200 miles ($50). If I drive anything over 200 miles, but less than 300 miles, I still just get paid for 200 miles. This goes on for every 100 mile increment.
Examples:
I drive 99 miles, I get $0
I drive 137 miles, I get $25
I drive 367 miles, I get $75
I drive 780 miles, I get $175
If I drive an infinite number of trips and the probability is random for the miles driven each trip, what am I being reimbursed per mile in cents/mile?

 

[ksdhart2]

Well, if you're assuming an infinite number of trips, then does that mean there's no cap on the number of miles you might drive? That is to say, is it theoretically possible for you have to driven a trillion miles? Assuming that's the case, then you assign a variable, say n, to be the maximum number of miles it's possible for you to drive, and look at the behavior as n grows infinitely large. Now, think about how much you get paid. Out of the n possible choices for how many miles you drive, 100 of those you get paid $0, 100 of those you get paid $25, etc. So, you've got a probability of (100/n) of making 25k dollars, where k is a variable that can take on only integer values. You'd then need to sum up n of these values for when k=0, k=1 ... k=n

Brushing up on sigma notation may be helpful. (try here or here). Now you try continuing from here. Where does this all lead?

 

[Bonesy777]

Sorry if I'm a little rusty here. It's been since 1989 that I've had a math class. I think it was calc for business majors. Lol!
Anyhoo, I determined that the maximum reasonable number of miles a person could travel in my situation is 1,680. That should go on the top of the sigma. Zero should go on the bottom of the sigma (I think). To the right of the sigma is 100/K. Then, I would sum up the values when K is 0, 1, 2, 3, up to 1680. My problem here is the first number, 0. 100/0 is undefined. Where this is going? I'm not sure. Lol! Just thinking out loud and graphically, it makes me think this is a curve of a parabola. Am I on the wrong path?

 

[ksdhart2]

Well, you're on the right track, but not quite there yet. The formula I was attempting to describe, and I may have described it poorly, was the expected value formula. It says that the expected value of a random variable X is:

\(\displaystyle \displaystyle E(X)=\sum_{k=0}^{N} \: x_k \cdot p_k\)

where N is the maximum value that X can take on, x_k is the value (i.e. how much you get paid) when the variable has value k, and p_k is the probability of X taking on the value k. Your follow up post indicates that the maximum number of miles is 1680, so we know that N = 1680. I'm assuming that you're equally likely to drive any number of miles, so the p_k will always be 1/N =1/1680. That refines the formula to:

\(\displaystyle \displaystyle E(X)=\sum_{k=0}^{1680} \: x_k \cdot \frac{1}{1680}\)

Now, it's just a matter of figuring out what x_k is. From the given information, we know that you get paid $0 for driving 0-99 miles, $25 for driving 100-199 miles, etc. This is why I suggested dividing the values of X into chunks of 100 and saying the payout is 25k dollars. We can also treat the 1/1680 as a constant and "pull it out" as per the standard rules of sums (I think the pages I linked to in my last post have information on this):

\(\displaystyle \displaystyle E(X)= \left( \frac{100}{1680} \cdot \sum_{k=0}^{16} \: 25k \right) + \frac{80}{1680} \cdot \text{???}\)

To finish up, you just need to find the last expression for how much money you get paid if you drive 1600-1680 miles (hint: it won't be in terms of k any longer, but will be a constant value, because this part is outside the sum and k "doesn't exist" outside the sum). You can also see that you can simplify the formula a bit if you're willing to sacrifice some accuracy and say the maximum number of miles is 1700.

 

[Bonesy777]

So, let's say the maximum number of miles is 1700.

100/1700 times summation sign with 17 on the top and k=0 on the bottom and 25k on the right side.

Then I end up with1/17 times 3,400 which equals 200.

However, I'm stumped as to what 200 represents besides expected value. Is it $20/100miles?

 

[ksdhart2]

I think you did the math a teeny bit wrong, as the actual sum should be 3825:

\(\displaystyle \displaystyle \sum^{16}_{k=0} \: 25k=3400\), but \(\displaystyle \displaystyle \sum^{17}_{k=0} \: 25k=3825\)

That aside, 1/17 of 3825 is 225, so that's your expected value. It represents the "average" amount you'd expect to be paid. You can see that if you just pick a random amount of miles driven (i.e. a random value of X), you probably won't be close to $225. But, if you pick another random value and average the two, you'll get closer to $225. Pick a third, and take the average of all three, even closer. And so on. If we let F(n) be the average payout after we've n picked and averaged n payouts, we can see that:

\(\displaystyle \displaystyle \lim _{n\to \infty }F(n)=\$225\)

Or, in other words, if we were to theoretically be able to take infinity values and average them together, we'd get $225. In much the same way, if you flip some finite number of fair coins, you'll get an unbalanced number of heads/tails, but the limit as the number of coins approaches infinity is a 50/50 split.

 

[Bonesy777]

Now things are coming back. For individual instances, it may be way different than $225 but the more instances you do it, the average will be closer and closer to $225.

This reminds me of a intro to finance class I took probably in 1989. Approximately 300 students, we all flip a coin. If you get heads, you keep flipping, tails you stop. Although we all knew there was a 50/50 chance of getting heads or tails, there is going to be some freak that keeps getting heads for a while since you had so many flippers. I was that freak, Lol! 7 heads in a row. Suprised me as well as everyone else. Good thing it didn't turn me into a compulsive gambler. Lol!

Thanks for all your help!