Hit chance 25%

>hit chance 25%
>attack 4 times
>no hit

Other urls found in this thread:

i.4cdn.org/gif/1469943610098.webm
en.wikipedia.org/wiki/Bertrand's_box_paradox
twitter.com/SFWRedditVideos

>i dont know how probability works : the post

31% probability that you miss 4 times in a row at that hit rate. There's nothing unusual about that. Also this: Somebody ask me about probability.

what's the probability of getting a gf

0

>hit: 95% (max)
>miss

An urn contains 20 balls: 4 red and 16 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is 0.31

how many blue balls are in the second urn?

>making persuasion dialogue system where you can save and reload if you fail
>still bother giving you room for failure, even though they know you'll just reload
seriously, devs, if this is the case, don't bother capping off at a 95% success rate

So small, that we shouldn't even consider it a possibility.

10 blue balls.
Next question. I'll be here all week.

>hit: 5%
>hit

some guys have all the luck

>reread question
>both balls are the same color (not just blue)

Well. That's that. Gotta kill myself.

Not him but imma still try

Let x be the number of blue balls in the second urn.

Prob of drawing two red:
.2 * (16/(16+x))

Prob of drawing two blue:
.8 * (x/(16+x))


.2 * (16/(16+x)) + .8 * (x/(16+x)) = .31
Tell me if I'm on the right track with this so far

you got it just solve for x. note: i made this question up so i dont know if the number of balls will even be a whole number

Can you explain how you did that?

Don't worry. I didn't leave. I actually went to solve it after I posted that. You get 3.59 blue balls.

>10% chance to attack twice
>Procs twice in a row

i.4cdn.org/gif/1469943610098.webm

strange i just danced to this cover

Chance to miss =75%=.75
Missing 4 times in a row=.75×.75×.75×.75=.31=32%

.75 chance a miss will happen.

Two misses in a row is a .75 chance after a .75 chance. In other words, .75*.75, or .75^2.

Four misses in a row works the same way. Just raise it to the fourth power. .75^4 = .31 (approximately).

Not him but it's .75 x .75 x .75 x .75

Sauce on the song or the girls. Both or either would be nice.

what are the odds of getting tails 20 times in a row?

Huh that's pretty neato

>hit chance 25%
>decide the best course of action is to attack

So, you're interested in the probability that you would miss 4 times. There are NUMEROUS outcomes in this scenario, but you're only interested in one.

There's the probability that the first shot hits, the next three miss; the first shot misses, the second hits, the next two miss; the first two shots hit, the second two miss; ... etc. You're not interested in ANY of those. It would be a little more difficult to calculate the probability of any of those combinations. However, calculating successive procs of a chance is easy as dicks.

So, if he's got a 25% hit rate, that means he's got a 75% miss rate. We're interested in getting 4 misses in a row. So if we take swing once, that's:

0.75

Okay. If we were calculating for two swings, then it would be:

0.75 * 0.75

That's the probability of missing twice with a 25% hit chance. Now, you might ask, "Why are we multiplying instead of adding?" Any probability can never ACTUALLY exceed 1.0 (100%), because that doesn't make sense. In like Fire Emblem, you can have 150% hit rate, but that effectively becomes 100% because what the fuck does it mean to have more than a 100% chance? It's nonsense.

Anyway, so:

0.75 * 0.75 * 0.75 * 0.75

Or for a simpler approach:

0.75 ^ x

Where x is how many swings you take. So if you're playing DotA and a hero has a set probability of 0.16 of getting a crit, and they crit 6 times in a row, they only had a:

0.16 ^ 6

chance of that occurring, and you should report them because they use RNG as a crutch.

Oh, and after you calculate the probability of missing after that many hits, you can subtract that from 1 to get the probability that you would have hit AT LEAST once.

So he has a 31% chance of missing 4 times in a row. Alternatively, the probability of getting at least one hit was 69%. A 69% chance that he got some combination of hits on the enemy.

00.000009%

Does anyone here even like preacher?

...

Somebody explain Markov chains to me.

i close my eyes and pick up the third nearest object to me.
what are the odds that its this?

1 in 2^20

1 in 6.

Nice try, but it's actually 1 / (16 ^ 5)

i hope you're right, for my sake

...

Obligatory

Quick, what is the probability of rolling all 1s with three six-sided dice thrown at the same time?

100%

It's 50% because it doesn't remember what you picked last.

Either 50% or 0%.

1 in 216

I want to say 50%, but I feel like I might be misunderstanding the question. It's asking that if you have already received a box containing a gold ball, and you reach into the box again, what is the probability that you will obtain another gold ball? It should be 50% unless this is some kind of existential, nihilist, Schroedinger kind of question.

You mean, either 100% or 0%, right? Assuming this is the correct answer, then what I was saying about the philosophical bullshit was on point.

they have no ass what are they shaking

Seems like 50% to me too. Since if you get a gold ball you either picked box 1 or box 2 so if it's box 1 you will get a gold one next, and if it's box 2 you won't. Seems like 1/2 to me.

It's 66%

Explain.

1/6 per dice so it's probably that thing people used for OP's question. It doesn't change it whether you throw them all at once or one after another. So .46% I guess, if I'm right about you needing to do (1/6)^3.

it's 1/6^3

>FFXII
>50% chance of chest spawning
>10% chance of being what I want
>Get it first try

This better not be that stupid door again.

>accuracy: 90%
>miss 5 times in a row

FUCKING POKEMON

The important part to note is when you pick the first ball out of the box. When you reach into the left box, the probability you pick a gold is 100%, while if you reach into the center box, the probability you pick a gold is 50%. Because of this, as an observer, you can determine that the likelihood you've chosen the box with two golds is 2/3.

That's what I said, though I used parentheses because you need to ^3 what 1/6 is. You don't do 6^3 and then divide 1 by that, it's just less confusing to have parentheses.

50%

Once you draw the first ball though and get gold you have a 50% chance of having either of the first two because they're the only ones that have a gold ball. So if you choose another ball you have a 50% chance of getting another gold (that is, having chosen the first box) and 50% chance of getting a grey (or having chosen the second box).

im retarded and its 50%

No, it's not 50%. The first box has a higher chance, exactly twice as much as the second box, because it has twice the likelihood of having a gold ball picked out of it. The answer is 2/3.

>bait chance 1%
>post 1 time
>61 posts

This guy is correct. How the question was posed, it cannot be 2/3rds. As an example, when you ask, "What is the probability of a coin flip?", you are only concerned with the probability of this next action. You are not concerned with all that have occurred previous.

They are saying, "You got a gold ball. Given this situation, what is the probability that the next you pick up is gold?"

They are not saying, "What is the probability that you draw a gold ball from a selected box, and then draw a second?"

Basically, like many riddles, the question is poorly worded with the intention of miscommunication. Going back to the coinflip, you might then get a raging erection as you laugh as people say that it's 50% and then attempt to correct them saying that it's almost 0% given the probability of human life developing from the primordial soup and surviving and evolving as a species to develop minting technology.

The classic "Car and Two Goats" riddle would also be good here.

Nope. The question is perfectly worded. Both of those interpretations you described are the exact same; the second is always inherent to the first. Mathematically this can be easy to explain. The way you're seeing it, there are two possible scenarios. Either you pick a gold, and then another gold, or you pick a gold, then a silver. But in reality there's three possible scenarios. You could pick a gold and then another gold, or you could have picked that other gold first, then a gold, or a gold then a silver. There are three results whether you take the first or second interpretation, and in two of them you pick gold, while in only one of them you pick silver. 2/3.

>what is the law of large numbers

I think it's a case of "lowest common denominator" but I'm not an expert in probability. Once you pick a gold ball the chance that you picked box one is 50% and the chance you picked box 2 is 50%. So you draw again. If you know what is inside each box then if you chose box 1 you have a 100% chance of picking another gold ball. If you chose box 2 you have a 0% chance. But since the person in the problem doesn't know which box they picked, except that it's either 1 or 2, there's a 50% chance of having picked each, and so a 50% chance that the next ball drawn is also gold. And actually we as the people observing even though we know what is in all three boxes, we don't know which box the person chose. So the "lowest common denominator" is the 50% chance of having chosen 1 box or the other, and so that's the chance of choosing another gold ball.

You've already chosen one ball though. And the question asks about the next move. So you only have two choices now: that you chose box one and will choose another gold ball next, or that you chose box two and will draw a silver ball next. If the question was about your first move and you only had boxes 1 and 2 before you then your explanation might be right, but you have boxes 1 and 2 before you and one is already missing a gold ball, and that is what the problem is asking about.

I think I like math more than video games

You're missing a huge key in probability: you're not factoring in the possibility of having NOT picked a gold from either box. Remember, even if you pick one of them, in the center box there is still a 50% chance you pick a silver on the first try, ending the experiment. However, in the left box there is a 100% chance you pick a gold. Knowing this, but not knowing the order of the boxes, any observer or the tester could say with ease that the next ball will more likely be gold than not. Like I said, the first choice does matter. Each possibility in the second choice branches off of that.

Their soulless husk

>There's three possible scenarios
Oooh, I see what you're saying. When you describe it like that, I understand perfectly. So the crux of the argument then becomes "Does order matter?" which is what is arguing.

I think that guy is correct, however. Order should not matter if they're providing that you have already acquired a gold ball.

So in Box 1, we have Ball 1 and Ball 2. In Box 2, we have Gold and Silver.

You argue that the options are:

Ball 1 -> Ball 2 [x]
Ball 2 -> Ball 1 [x]
Gold -> Silver [ ]

Right? Which makes it 2/3rds. However, I would argue that order does not matter because it is given you already have in your possession one ball, which limits your options to:

Gold [x]
or
Silver [ ]

The lynchpin of this situation is whether or not you think order matters, which given the description, I don't think it does.

>Knowing this, but not knowing the order of the boxes, any observer or the tester could say with ease that the next ball will more likely be gold than not
I don't understand why you think this. All an observer knows is that the person picked a gold ball. And actually an observer that isn't us wouldn't even know the contents of any of the boxes. But we do. So the person chooses a gold ball, meaning that he has either box one or box two. So we have a gold ball in our hands so our next choice will either be another gold ball, or a grey ball. Yeah, we could have picked either of the gold balls in box one if we have box one but I don't see anything in the problem that has anything to do with that. We have a gold ball, it doesn't matter which one, we reach into our box and we either get another gold ball (if we have box one) or a silver ball (if we have box two). Whether we picked the left gold ball or the right gold ball (if we have box one) doesn't matter since both are gold and the problem doesn't specify a specific gold ball, nor does it tell us which gold ball the person picked.

>either you win the lottery, or you don't; 50%

That's not how it works user. Order always matters.

...

The comparison is wrong because one side is weighted more heavily than the other. That is not what I'm saying. Order does not matter because we are not concerned with what preceded our current situation. It's like saying, "You have won the lottery. You are now flipping a coin. What's the probability of getting a heads?"

You do NOT factor in the probability of winning the lottery because it is already given that you have won the lottery. The problem gives extraneous information such as the box with two silvers. Given the current information, that is totally irrelevant to the actual outcome of the current situation. Regardless of whether you picked Ball 1 or Ball 2, it doesn't matter. Order does not matter because it is given that you got a gold. There were three possibilities for initial golds, but that does not matter. You already HAVE a gold. It is now asking what the likelihood of each outcome will be following that.

>Yeah, we could have picked either of the gold balls in box one if we have box one but I don't see anything in the problem that has anything to do with that
>We have a gold ball, it doesn't matter which one
>Whether we picked the left gold ball or the right gold ball (if we have box one) doesn't matter since both are gold

No, that's an overlook. Just because both are gold and lead to the same result doesn't mean they aren't two separate and individual situations. If you can't understand that then you can't understand probability.

en.wikipedia.org/wiki/Bertrand's_box_paradox

but that's not how it works, user!

>accuracy 90%
>miss three times in a row

>hit chance 99%
>miss

thanks xcom

2/3

Since there's only 3 gold balls, and you can't pick the box with only silver balls, there are 3 possible outcomes:

>you pick gold ball 1, box one, success
>you pick gold ball 2, box one, success
>you pick gold ball 3, box two, failure

Assuming you have an equal chance of selecting any one gold ball, you're more likely to select box one. Thus the 2/3 chance.

They may be different situations but the problem doesn't require us to care about that. The balls aren't numbered in any way and the problem doesn't ask about the probability of us drawing gold ball number 2 next after drawing number 1. Just the probability of drawing another gold ball. What you're talking about may matter in some problems but it doesn't in this one.

See this is what I'm talking about. The problem doesn't ask for the probability of picking a particular gold ball, just the probability of picking another gold ball.

Although reading some of your posts I am starting to think I'm missing something but I still don't think which gold ball you pick matters since the problem only asks about picking another gold ball and not about picking a specific gold ball next.

I actually do see where you guys are coming from now though.

Nigger, read the page. The image posted is literally the Bertrand Box Problem, and it has a full and total explanation as to why you're looking at the problem totally wrong. In fact, if you were to do the problem described in the image, the answer would turn out as 2/3. It's been done.