Poll of the Day > How do you math a thing when you can't think the thing?

(+) Yeah! (+)

I used to be great at math, then I took an arrow to the knee

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So, you got a 1/9 chance of something happening, and you need to get at least 1... the quick'n'dirty value is 9, you need 9 attempts, that's just the ideal point. Likewise you'd need 18 attempts if you must score a 2, or 36 attempts or a 4 and so on. , But that's not the actual odds, that's just a ballpark

It's actually something like 1-(8/9)^n or something... but I can't remember, I haven't used stuff like this in forever.

What if you have the probabiliy instead (75% or something) and need to know the number of attempts (n) needed to get that? it's certainly not 75% of 9, that'd be silly.

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Actually if you NEED a certain number you can never have a 100% chance of getting it no matter how many attempts you make, the chance just constantly approaches 100% (but never gets there.

So if you roll a 9 sided dice (which would be a really weird dice btw), you could actually roll 10,000 times and never once get a 9. The chance of that is incredibly small, but it's possible.

Now if you just wanted to exceed a certain % chance of getting a 9, that you can work out pretty easy.

As may people who have tried to use the always bet on black method (martingale method? I forget) can attest to, you can eventually have a string of reds that'll bankrupt you (20+ even!!). The odds are low, but when it happens, you lose it all.

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Lokarin posted...

What if you have the probabiliy instead (75% or something) and need to know the number of attempts (n) needed to get that? it's certainly not 75% of 9, that'd be silly.

The quick and dirty estimate would be 9 divided by 75%, which is 12.

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Let's treat this like 9 cards from a deck. There are 8 spades and 1 diamond. The question is now what to do when you draw a spade. Does it get reshuffled or discarded? If you reshuffle then you always have an 8 in 9 chance of drawing not diamond. However if you discard then consecutive drawings have increasingly better chances of drawing not spade.

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If the odds of something happening are X, and you're looking for the odds of it happening in N tries, the formula for that is 1-((1-X)^N). 1-X gives the odds of it not happening, then (1-X^N) gives the odds of it never happening in N tries, and subtracting that from 1 gives you the odds of it happening at least once.

Example: X=25%, N=6
1-.25 = 75% chance of not happening each time
.75^6 = 17.80% chance of never happening
1-0.1780 = 82.20% chance of happening at least once

If you're looking to calculate how many tries you need to have a given chance of seeing it happen, you need to rearrange that formula (Y=1-((1-X)^N)) to solve for N instead of Y, which means using a logarithm (the inverse of an exponential function).

Y=1-((1-X)^N)
1-Y=(1-X)^N
N=log(1-Y)/log(1-X)

Example: X=25% chance to happen, y=83% chance to happen overall

log(1-.83)/log(1-.25)=6.16 attempts to have an 83% chance.

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As a video gamer, I find the only thing I really need to know is that there's a 50/50 chance of getting a drop that I want after I try 69% of the "1 in ___" times. (Which is a specific usage of the equation in the above post)

For example, if something has one of those old 1 in 256 drop rates I'll have a 50/50 chance of getting what I want after killing 178 of them. (I wish I didn't know that off the top of my head)

For 1 in 9 odds you'll need to try 6.25 times ON AVERAGE. Of course, you might get it on your first try...or 50th.


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wpot posted...

As a video gamer, I find the only thing I really need to know is that there's a 50/50 chance of getting a drop that I want after I try 69% of the "1 in ___" times. (Which is a specific usage of the equation in the above post)

For example, if something has one of those old 1 in 256 drop rates I'll have a 50/50 chance of getting what I want after killing 178 of them. (I wish I didn't know that off the top of my head)

For 1 in 9 odds you'll need to try 6.25 times ON AVERAGE. Of course, you might get it on your first try...or 50th.

This is pretty much how I apply it, too. It's never a sure thing, by the nature of random chance, but this is all you really need to do to be able to say "it'll probably take me about this long." If you want to be really thorough about that, you can figure out the standard deviation on that and get an even better idea of how long it'll probably take (it follows a normal distribution, so you've got a 68% chance of being within one SD of the 50% mark and a 95% chance of being within two), but all you really need is to evaluate whether or not farming a given drop is going to be worthwhile, and you don't need really detailed probabilities for that.

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adjl posted...

This is pretty much how I apply it, too. It's never a sure thing, by the nature of random chance, but this is all you really need to do to be able to say "it'll probably take me about this long." If you want to be really thorough about that, you can figure out the standard deviation on that and get an even better idea of how long it'll probably take (it follows a normal distribution, so you've got a 68% chance of being within one SD of the 50% mark and a 95% chance of being within two), but all you really need is to evaluate whether or not farming a given drop is going to be worthwhile, and you don't need really detailed probabilities for that.

Right. Of course my focus on 50/50 tricks my brain into expecting to get a drop around 178 tries for my 1/256 odds and makes me mad when I hit 400 without seeing anything, even though 400 tries is far from rare for 1/256. To calm myself down (and out of boredom) I multiply it out:

75% odds after 178 x 2 = 356 tries.
87.5% odds after 178 x 3 = 534 tries, etc etc.
...which is more or less the same as standard deviations without using the formal terms.

Sometimes that makes me feel better. Other times... :)

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wpot posted...

Right. Of course my focus on 50/50 tricks my brain into expecting to get a drop around 178 tries for my 1/256 odds and makes me mad when I hit 400 without seeing anything, even though 400 tries is far from rare for 1/256. To calm myself down (and out of boredom) I multiply it out:

75% odds after 178 x 2 = 356 tries.
87.5% odds after 178 x 3 = 534 tries, etc etc.
...which is more or less the same as standard deviations without using the formal terms.

Sometimes that makes me feel better. Other times... :)

Yeah, most of the time when I calculate the odds of not having seen it yet, it's because I want to determine just how unlucky I've been. It doesn't change anything, but it can be nice to realize that I'm not actually that unlucky (or if I am, that my frustration is justified).

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