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Let's assume on given year you have a 3% chance of being fired , 97% of not being fired

What are the odd after 4 years you are still employed?

This is hard because you cannot be fired twice

Denote N=not fired, F=fired

for four years, we have the possible outcomes

NNNN

FNNN

NFNN

NNFN

NNNF

But once you're fired employment ceases.

y=years

here is my attempt at the problem:

$\frac{N^y}{Y*F*N^{(Y-1)}+N^Y}$

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Considering that the probability of staying employed in a given year is independent of what happened on the previous years, the probability of staying employed after 4 years equals

$$ 0.97 ^ 4 = 0.88529281 \approx 88,5 \%. $$

Since you've asked for the odds of staying employed, we must divide this probability by its complementary, which gives us

$$ \frac{0.88529281}{1 - 0.88529281} \approx 7.71785, $$

meaning you're almost 8 times more likely to stay employed after 4 years than getting fired before that.

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    $\begingroup$ This seems to be the obvious answer that I would have given if I believed that is what the OP is asking. But I don't think he is expressing the problem correctly. The strange formula and the set of events that he lists. also notice the statement that the problem is hard because you can't be firede twice. That would not be a factor for your solution. I think he is confused and needs ti explain the problem better. $\endgroup$ – Michael Chernick Jan 3 '17 at 4:10

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