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[Andres]

First time posting on this thread. I need help with chance. Here is my question: Say there's a one in ten chance of Scenario A occurring. What would be the chance of Scenario A occurring if you multiply the chance of it happening by 11.6? I know it's not 100% or more but I don't know the equation to work out the right answer.

[TheBiggerFish]

.1 * 11.6?

No wait.

Yeah, I have no clue.

[frostshotgg]

I believe you would take the chance of it not happening, eg 0.9, and divide that by 11.6 to get 0.0776 as the chance of it not happening, then find the chance of it happening from that, so 0.9224. I think.

I hate statistics.

[TheBiggerFish]

That makes some amount of sense.

I know nothing of stat though.

[Amperzand]

I did some basic stuff over the summer, but yeah that sounds weird.

[Reelya]

First time posting on this thread. I need help with chance. Here is my question: Say there's a one in ten chance of Scenario A occurring. What would be the chance of Scenario A occurring if you multiply the chance of it happening by 11.6? I know it's not 100% or more but I don't know the equation to work out the right answer.

It's not a meaningful statement. You have to break down what exactly you mean by "scenario A occurring", since that's completely undefined here. For a single event any multiplier is just multiplied, and if you get an answer over 100% it just means you inputted meaningless data. A 20% chance is twice as likely as a 10% chance, 30% is three times as likely as 10%, etc up to 100% which is 10 times as likely as 10%. Is any multiplier over 10 even meaningful here?

Putting things in words helps. Consider this statement "there is a 10% chance of it raining on tuesday" and "there is a 20% chance of it raining on wednesday" so here you'd say "it is twice as likely to rain on wednesday than tuesday" and that makes perfect sense. If friday had a 100% chance of rain (big storm coming), then that's 10 times as likely as tuesday. But if you go over 10 times tueday's chance it no longer makes sense: "it is 11.6 times as likely to rain on thursday than tuesday" isn't meaningful. You can't have a chance that's over 100%.

[chaotic skies]

So I currently have no use for this thread, as school has yet to start in on many subjects, but I'm going to PTW so I don't forget this thread.

[Andres]

First time posting on this thread. I need help with chance. Here is my question: Say there's a one in ten chance of Scenario A occurring. What would be the chance of Scenario A occurring if you multiply the chance of it happening by 11.6? I know it's not 100% or more but I don't know the equation to work out the right answer.

It's not a meaningful statement. You have to break down what exactly you mean by "scenario A occurring", since that's completely undefined here. For a single event any multiplier is just multiplied, and if you get an answer over 100% it just means you inputted meaningless data. A 20% chance is twice as likely as a 10% chance, 30% is three times as likely as 10%, etc up to 100% which is 10 times as likely as 10%. Is any multiplier over 10 even meaningful here?

Putting things in words helps. Consider this statement "there is a 10% chance of it raining on tuesday" and "there is a 20% chance of it raining on wednesday" so here you'd say "it is twice as likely to rain on wednesday than tuesday" and that makes perfect sense. If friday had a 100% chance of rain (big storm coming), then that's 10 times as likely as tuesday. But if you go over 10 times tueday's chance it no longer makes sense: "it is 11.6 times as likely to rain on thursday than tuesday" isn't meaningful. You can't have a chance that's over 100%.

Ok, let's get a scenario with a few made up numbers.
1 in 1000 humans become mutants.
For people who live near radioactive waste, 11.6 in 1000 humans become mutants, so you're 11.6 times as likely to become a mutant due to living near radioactive waste.
There is a family where a family member has a 1 in 10 chance of becoming a mutant.
If the family were to live near radioactive waste, the amount of family members who'd become mutant would multiply by 11.6.
What are the final odds of a family member living near radioactive waste becoming a mutant?

[Bauglir]

It'd be additive, not multiplicative.

1/10+116/10000=.1116=11.16%.

that doesn't seem right

you double count cases where a person is a mutant by family and by radioactive waste that way, don't you?

[Reelya]

Let's step back and get our semantics right first. Back to the original wording:

What would be the chance of Scenario A occurring if you multiply the chance of it happening by 11.6?

"Chance" is the percentage chance.
"Multiply" is just the multiplication operation.

This is just two numbers being multiplied together. e.g. 10% is 5% multiplied by 2. Another way to word this is "Thing A is X times as likely as Thing B". With this definition, nothing can be over 10 times as likely as 10%. And if 11.6 times as likely as 10% is less than 100%, then that makes it less likely than 10 times as likely, which is clearly nonsense.

Given what you wrote in the mutant example, 100% of the family become mutants. In gaming terms, the family rolled a D10 and if they roll a 0 they become a mutant (1/10). If the chance is multiplied by 10, then it becomes 10/10. Any higher than this chance is nonsense: 11.6/10 means that for each 1 person you get 1.16 mutants. It's meaningful mathematically but we know it's nonsense because our semantic knowledge is that you can't get more mutants than the original number of people.

You'd get a different answer if you looked at multiple independent probability events. e.g. "the chance of rain is 10% per day, what is the chance of it raining this week". In that sense, you're taking "10%" and working out the chance of it ever happening over 7 days. But you can't do that operation with "11.6", you need to specify a whole number of events.

[Reelya]

Separate post, but I wanted to look at the radiation question as a separate maths problem. Suppose that we're not talking linear multiplication of percentage chances, let's look at it as an aggregate of smaller chances. Each "unit" of radiation would have a chance of mutating you. You can't just linearly add up the chances either, since you're double-counting people who were mutated "twice" by different exposures.

So, the high-mutation family is more susceptible to the radiation, but it's not 11.6 times as likely, it's 11.6 times the exposure. The actual chance is an aggregate sum of the effects of smaller exposures. But, if we're making this assumption for the high-mutation family, we also need to make the same assumption for the low-mutations families: the chance is not a linear multiple. So first, let's work out how many multiples of the "background" radiation each person living near the radioactive waste is exposed to, just to be consistent.

The way you work this out is to work out the chance of NOT being mutated by a single exposure, and put this number to the power of the number of exposures. Then subtract the result from 1, and there's your probability (p) of being mutated at least once. This way, we can work out how many "background level" exposures you'd need to have to be mutated at least once, if you live at the waste dump:

p = 1 - (999/1000)^n

p = 0.0116 and we want to solve for 'n'

0.0116 = 1 - (999/1000)^n
(999/1000)^n = 1 - 0.0116
log((999/1000)^n) = log(1 - 0.0116)
n log(999/1000) = log(1 - 0.0116)
n = log(1 - 0.0116) / log(999/1000)

n = 11.66... units of radiation. A little bit more than 11.6 times the radiation.

But the "susceptible" family has a base rate of 0.1 mutations per unit of radiation. So, their chance of not being mutated by a single dose is 9/10. We need to work out 'p' for this family:

p = 1 - (9/10)^11.66
p = 1 - 00.29266953
p = 0.7073 or 70.73% chance of being mutated.

But of course, this is not the same thing as where we said "multiply the chance". The amount of exposure given here multiplies the chance of mutation by 7, not 11.6.

[Andres]

Thanks for the help.

[Spehss _]

What's the derivative and anti-derivative of e^x/2? It's not just e^x/2 is it?

[TheBiggerFish]

Derivative of e^(x/2).
Generic letter terms, which I just do because it's easier to work out...
Becomes: f^g
Take the derivative:
Chain Rule
f'(g)g'
Plug in everything:
(e^x >> e^x, x/2 >> 1/2)
(e^(x/2))/2
So no.

I don't know how to do antiderivatives though.

E:And in the couple minutes of Googling done, it doesn't look like I'll know in the immediate future.  I will ask, if it doesn't actually come up, though.

[frostshotgg]

What's the derivative and anti-derivative of e^x/2? It's not just e^x/2 is it?

dy/dx(e^(x/2)) = e^(x/2) * dy/dx(x/2) = 1/2 * e^(x/2)
Basic chain rule stuff.

integral of e^(x/2) = 2 * e^(x/2)
Not really sure how to explain that one beyond "recognize that e^(x/2) will stay in the integral and then whatever is the derivative of the ^ term needs to be divided out.