Math question for you guys!

[idiotec]

Alright, I have seen some pretty smart math guys in here, so I thought I would throw a question at you.

Scenario: A company has a contract with a 5 year life. The contract has an infinite number of renewal options. That is to say, at the end of the first five years, the company has an option to renew the contract for another 5 years. At the end of that 5 years, the company has another option to renew for 5 more years. And so on and so on. The Company has determined that there is an 80% probability that they will renew the contract at the end of the first 5 years. And if renewed, there would again be an 80% probability of them renewing again. And so on, and so on.

How long do you expect the contract to last?

That is the jist of it. Any factors I am leaving out that would be necessary, let me know. I have been out of school for far too long to remember this ****

Edit: Forgot the actual question

[hilowe]

I'm going to admit right now that probability and statistics are not my favorite things to work on. That being said, I'll give this one a shot.

My answer is that the contract will be renewed 3 times. The way I'm working the math is that it's an 80% chance the contract will be renewed once, a 64% (.8*.8) chance the contract will be renewed a second time, a 51% (.8*.8*.8) the contract will be renewed a third time, and a 41% (.8*.8*.8*.8) the contract would be renewed a fourth time. At that point the chances are less than 50-50, so it likely would not be renewed.

[Nubius]

I iz nawt that shmart. I dun nevr gawt that fawr

[idiotec]

Thanks for the help Nubius

hilowe, I think that makes some sense. It does seem to be problem of conditional probability. And I do remember multiplying probabilities in that case, but I don't know if it's that simple or not. I should of saved my college text books

[RadRacer]

OK. If the company likes the contract and is happy with it, the probablility may go up, but if they don't, it will go down. Since you stated it will 80% each time, then it wouldn't be lowering each time they renew. So, they may renew it an infinite amount of times, since the percent of them renewing is constantly 80%. You didn't say that next time they renew, they're chance of renewing may be 80% of their last chance, you just said that it would be 80%. That gives the impression that it will always be 80%.


What do you guys think?

[Nikkon]

~105 years.

The initial equation is (if I understand the problem correctly) .8^x = .009 (something less than 1%).

To simplify you take the natural log of both sides -> ln (.8^x) = ln .009 which simplifies to xln.8 = ln .009

Solve for x which gives you 21.xxxxxx. Round it off multiply by 5.

[Chankama]

I feel sleepy, so there might be mistakes here. At least it'll give you guys something to check and criticize if I made a mistake ..

Notice how each "value" is possible after successes (0.8) in previous times and a fail (0.2) at the end.

Expectation = sum {value_i x p(value_i) }
5 * 0.2 + 10 * 0.8 * 0.2 + ... + (5*n * 0.8^(n-1)*0.2) + ..., as n-> inf
= sum {n*0.8^(n-1)}, n=1...inf

Now we know that sum{k^n} = k/(1-k}, when n=1->inf

Take the derivative of both sides.

sum{n*k^(n-1)} = -k(-1)(1-k)^(-2) +(1-k)^(-1), n=1->inf

Sub in this formula to our problem:
k=0.8

Therefore, expectation = 0.8*(0.2)^(-2)+(0.2)^(-1) = 25 years

hmm... well.. I guess it "could" be reasonable. I will check back later after my nap ..

[idiotec]

Chankama, I think you might own!

The person asking me this question thought that it should work to take 1/(1-.8) X 5. That equation works out to 25. The problem was, no one new if that equation actually worked, or why it worked.

With you coming up with 25 years as well, I think it does!

[killians45]

3 days

[Chankama]

Originally posted by idiotec
Chankama, I think you might own!

Yeay! .. BOOM HEADSHOT??!

Originally posted by idiotec
The person asking me this question thought that it should work to take 1/(1-.8) X 5. That equation works out to 25. The problem was, no one new if that equation actually worked, or why it worked.

With you coming up with 25 years as well, I think it does!

Cool . Then I don't have to recheck it then..

I did the derivative of that thing on the fly, so hopefully there isn't a mistake there!

btw, that k/(1-k) formula I used to derive the second one is just the sum of a geometric series - which is very easy to prove.