# Question on calculating a probability/expectation

I am pretty sure my mathematics correct and it is so simple but there is something bugging me like I’m missing something.

Suppose the expectation of an event E(x) = n. Now we add some variables. The first variable increases the probability of the event p(x) by 25%. The second variable decreases the probability of the event by 7%. What is the new E(x)?

I get E(x) = n x 1.25 x 0.93

Is that correct?

You said both “expectation” & “probability”. Are you using those as synonyms or do you have a more detailed understanding of the terms that you’d care to share?

You also want to be clear about what is meant by “increase the probability by 25%”. Does that mean adding 25 percentage points to the probability (e.g., 40% to 65%), or does it mean multiplying the probability by 1.25 (e.g., 40% to 50%)? I’ve seen it used both ways, in different contexts. If possible, ask your source to clarify.

Umm … the source is me. Increasing the probability by 25% means multiplying by 1.25 so that p(x)=0.5 increased by 25% means the new p(x) is 0.675

I don’t get LSLguy’s question. Let’s say I have a 20% chance to win \$500. p(x) is 0.2 and E(x) is \$40. Pretty standard definitions.

Oh, OK, then, you obviously know what you mean. If it’s important for anyone else to know, though, you might consider more careful phrasing.

If you can write expectation as as a function of the probability E(x) = f(p(x)) you only get what you want if f is linear map.

If f(p(x)) = c p(x) you’re golden.

If f(p(x)) = p(x) + 1 you’re not.

To be fair, the expectation being proportional to the probability only requires that the payoff be unaffected by the changes to the probability, which one would expect to be a fairly common situation.

It is indeed a fairly common situation and most likely the only situation the OP is considering. I only included my answer for completeness.

And that’s the difficulty I am having Chronos. For example (and greatly simplified)

p(insurance claim) = 0.1
Average value of repair/coverage = \$2500
E(claim) = \$250

But I have a fast expensive car so I have a 30% larger chance of filing a claim. And the value of my car is 50% more than the average. So is MY E(claim) = 250 x 1.3 x 1.5 = \$487.50

Yes given your formulation, that would seem correct.
To write it out, before your expectation was E = pV =0.12500=250

now E2=(p1.3)(V1.5)=pV1.31.5=E1.31.5.

This is different than your original question. This question has one factor that affects probability and another factor that affects value. So they would multiplied together as you show here.

But your OP asks about two factors that both affect probability. If they are independent then you should consider them separately:

E(x) = n x (0.25 - 0.07)

For example, if the probability of getting a heart attack if I am 50 is 0.08 and the chance of being hit by lightning if I live in Virginia is 0.001, then if I am a 50-year-old living in Virginia the chance of either having a heart attack or being hit by lightning is 0.081, the sum of the probabilities.

It is not clear from the formulation of your original question what calculation should be applied.

That’s all good logic. Which is why P(higher premiums for such a car) = 1.0

FYI for various folks above: a 50/50 raffle is probably the common situation where E is not a linear function of P. Buying two tickets doubles your chances, but also changes the prize amount.

Except buying another ticket doesn’t double your chances, either, since your chance also depends on the total number of tickets sold. In fact, your expectation in a 50-50 raffle is directly proportional to the number of tickets you buy.

Hmmmmm. D’oh! :smack: