# Idiot needs help

1. Jun 21, 2011

### WhiteSwan

Good afternoon all

I need help in settling a debate. If say I have a 1/5 chance of gaining promotion in a year, do the odds of gaining promotion increase as time progresses (all else being equal)?

Phrased negatively I can handle it i.e. chances of not gaining promotion after five years is 4/5^5 = 32%, 6years = 26% so it would seem yes that the odds increase (from a year 0 perspective)

But I get confused when I think of the chances of gaining promotion. Year 1 is 1/5 but how do I calculate all potential outcomes for year 2? It is still 1/5 for the 4 out of 5 times promotion isn't achieved in year 1 but how do I factor in that 1/5 chance it is got in year 1? Is it 4/20 in year 2 i.e. disregard the 1/5 from year 1 OR 9/25 i.e. If I get promotion in year 1 I will still have it in year 2 plus 1/5 * 4/5?

2. Jun 21, 2011

### tiny-tim

Welcome to PF!

Good afternoon, WhiteSwan! Welcome to PF!
Sorry, no.

It's like rolling a six …

no matter how many times you try, the chance next time is still one in six.

3. Jun 21, 2011

### WhiteSwan

Thank you for your response.

I'm understand chances each year are 1/5. However from a year zero perspective are the odds not increasing? With cumulative advantage I have more chance in gaining promotion over 5 years than 1 year alone. Like rolling a six, but with six attempts instead of one. Granted each individual throw is 1/6 but as a whole the odds of throwing a six are substantially better if I have six attempts. As soon as I throw the six game over and I stop trying, whether on first attempt or the last.

4. Jun 21, 2011

### tiny-tim

Hi WhiteSwan!
Yes, obviously.

The usual way to do it is P(promotion in the first n years) = 1 - P(no promotion in each of the first n years).

5. Jun 21, 2011

### cepheid

Staff Emeritus
Another way to do it without "phrasing it negatively", as you put it, is to consider all possible outcomes in which you get a promotion sometime in the first n years. (Let's consider n = 5). The outcomes can be labelled, Y1, Y2, Y3, Y4, and Y5, where Y1 = get a promotion in year 1, Y2 = get a promotion in year 2 etc. So what we're really looking to figure out is:

P(Y1 OR Y2 OR Y3 OR Y4 OR Y5) = ?

Now, let us further assume that the events are mutually exclusive i.e. you'll only get one promotion, so if you get a promotion in year n, the conditional probability that you'll get a promotion in any subsequent year is 0. For mutually exclusive events, the probability of one or another happening is just equal to the sum of the probabilities of the individual events. So the expression above becomes:

P(Y1) + P(Y2) + P(Y3) + P(Y4) + P(Y5) = ?

Now here's where it gets tricky. I've said that these events are mutually exclusive. That means that they cannot be independent. If one occurs, none of the others can occur, which means that the outcome of one very much affects the outcome of the others (they are not independent). However, if you don't get a promotion in year i, the probability of getting it in year i+1 is just yet again another 20-80 chance (regardless of what happened in the previous year). So, so long as you keep not getting promotions, the system behaves as though each year is an independent trial (just like each roll of the dice). This information is useful because:

P(Y1) = 1/5

P(Y2) = P((not Y1) AND Y2)

P(Y3) = P((not Y1) AND (not Y2) AND Y3)

P(Y4) = P((not Y1) AND (not Y2) AND (not Y3) and Y4)

P(Y5) = P((not Y1) AND (not Y2) AND (not Y3) and (not Y4) AND Y5)

For independent events, the probability of all of them happening is just the product of the individual probabilities, as you already know. That means we can write:

P((not Y1) AND Y2) = P(not Y1)*P(Y2) = (4/5)*(1/5).

Using this result, the probability we're looking for becomes:

P(Y1) + P(Y2) + P(Y3) + P(Y4) + P(Y5) = 1/5 + (4/5)*(1/5) + (4/5)2*(1/5) + (4/5)3*(1/5) + (4/5)4*1/5

= 0.67232.

Of course, tiny-tim's method where you just take one minus the complement of the outcome you're looking for is much easier!

1-(4/5)5 = 0.67232

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook