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I have a scenario where there is a 29.9% chance of a 235% increase to a certain value, and want to determine what the average percent increase of that value will be.

I'm fairly certain that it's not (.299)(2.35), but I'm not sure what I should be doing.

Apologies if this has been asked before; the number of times "percent" appears in my title question gives an idea of how poorly this question works with search engines.

EDIT: I ended up brute-forcing it by the following:

$$\frac{\frac{299(100*2.35)+701*100}{1000}}{100} = 1.40365$$

or an average percent increase of 40.365%.

It's worth noting that I'm 100% certain this is the answer, as it's very close to another known answer (30% chance of 235% increase is 40.5% average increase).

I'm now interested in a less batshit-crazy more elegant method.

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Let $X$ be the random variable expressing the increase in %. We are interested in $\mathbb{E}[X]$. There are only two possible events - $E$ = 'there is increase' or $F$ = 'there is no increase'. We have $$X(E)=2.35,\; P(E)=0.299;\qquad X(F)=0,P(F)=1-P(E)=0.701$$ Therefore $$\mathbb{E}[X]=P(E)X(E)+P(F)X(F)=0.299\cdot 2.35+0.701\cdot 0 =0.299\cdot 2.35 $$ so your intuition was actually correct.

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If you want to increase something by $235$% you have to multiply it by $(1+\frac{235}{100})=3.35$

So your expected increase factor is $0.299\times3.35=1.00165$ and this means an expected percentage increase of $0.165$%

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