Hey fellas. I was walking home with some friends and we argued

>>8335435
0.5%

I don't understand. Why would it be 5 out of 10 chance if each attack is independent of each other?

>>8335435
1-0,9^5=0,40951; -> ~41%

>>8335442
0,9 is the chance of the attack not landing
0,9^5 is the chance of the attack not landing 5 times in a row
1-0,9^5 is the chance of the attack not not landing 5 times in a row - therefore, the chance of the attack landing in one out of five times

>>8335435
32.805%

>>8335435
Landing at least once?
I don't see why the probability wouldn't be 10%.

>>8335435
Exactly once or at least once?

Either way take a gander at the binomial distribution.

[eqn] {5 \choose 1} \cdot 0.9^4 \cdot 0.1^1 [/eqn]

>>8335462
>Exactly once or at least once?
80% of the time exactly once IS at least once.

Probability of landing the first attack and missing the subsequent 4:

[math]0.1 \times 0.9^{4} = 0.06561[/math]

There's 5 possible ways of landing a hit.

[math]5 \times 0.06561 = 0.32805[/math]

Might've forgotten how to latex

>>8335448
>1-0,9^5 is the chance of the attack not not landing 5 times in a row - therefore, the chance of the attack landing in one out of five times

[math] 9^5 [/math] is the chance of missing 5 attacks in a row.

[math] 1 - 9^5 [/math] is the chance of NOT missing 5 attacks in a row.

You know what's not missing 5 attacks in a row?

>Landing 1 hit out of 5
>Landing 2 hits out of 5
>Landing 3 hits out of 5
etc.

>>8335472
In this case OP has a 65.13% chance of hitting at least once but a 38.7% chance of hitting exactly once, I'd say that's an important distinction to get.

0 hits = 34.86784%
1 hit = 38.74205%
2 hits = 19.37102%
3 hits = 5.739563%
4 hits = 1.116026%
5 hits = 0.1488035%
6 hits = 0.0137781%
etc. just muh binomial distribution

>>8335485
I don't know where you are getting those numbers from. I got >>8335442 For at least 1 hit (which implies ~59% for 0 hits, not ~34%). And for exactly 1 hit, I got >>8335458. The 80% just came from dividing one by the other, it was an observation not a part of the problem.

Again, assuming the criteria is "at least once", how is it not 10%?

In the real world with a complex spectrum of factors at work, I'd loosely take the odds as cumulative (and be torn apart and treated like braindead gutter trash for it). In a simple question like this I'd just treat them all as unrelated events. The chance every time being 10%. Thus 10% chance of getting at least one hit.

Only greater than, less than, or exactly criteria where n!=1 differ.

>>8335515
You need to think about every possible situation. I like to make a table. Lets simplify this problem to only 2 attacks instead of 5 for clarity. The first row will be the first attack and the second row will be the second attack. 1 signifies a hit and a 0 signifies a miss. Here are all the possibilities:
0011
0101
Now let's calculate the probability that each happens. A miss has a 90% chance of happening while a hit has 10% chance. We need to multiply the probability of the first attack by the probability of the second attack to find the total probability for that situation. So our table then becomes:
0 0 1 1
0 1 0 1
(.9^2) (.9*.1) (.1*.9) (.1^2)
Now we can check our work by adding up the probabilities. The total should be 1, because these are all the possible scenarios. It should be obvious to you that "hitting at least once" or even just "hitting once and only once" is not 10%. Hitting at least once is the sum of probabilities that don't include the (0, 0) column. Hitting once and only once is the sum of probabilities not including the (0, 0) column OR the (1, 1) column.

>>8335513
it is [(0.9)^(10-hits)]*[(0.1)^(hits)]*[10 choose (hits)]

>>8335547
That doesn't explain anything to me. [10 choose (hits)] is meaningless to me. I'm sorry I don't know the math lingo but I haven't taken probability yet. What's so hard about actually putting the formula for [n choose m] instead of writing [n choose m]? It shows you have no idea what's actually happening and you just memorized or copy pasted that formula. Also where is that 10 even coming from? I did this method >>8335528 but for all 5 attacks rather than 2 attacks, and my answer is consistent with everyone else in the thread. The fact that you are getting 34% for 0 hits should be indication enough that your method is incorrect.

File: 67271825.jpg (108KB, 400x400px) Image search: [Google]

108KB, 400x400px

>>8335575
I was just posting the formula that guy was using. His computations were for 10 attacks, at 10% chance per attack (that is obvious since he put 6 hits in his percentage table). If you want it to be for 5 attacks, then you do [(0.9)^(5-hits)]*[(0.1)^(hits)]*[5 choose (hits)]. This is called a binominal distribution. You get a binominal distribution when doing this because a sequence like Hit,Miss,Hit,Miss,Miss is distinct from a sequence like Miss,Miss,Miss,Hit,Hit; if you count up all of the distinct sequences of 5 attacks, but 2 hits, then you get 5 choose 2, because you have 5 positions in the sequence to choose from to put your hit in, and you choose 2 of those positions to put your hit in. However, your calculation (1 - (0.9)^5) is for hitting at least once, rather than exactly once, because it is the complement of the probability of not hitting for all 5 attacks.

>>8335600
>if you count up all of the distinct sequences of 5 attacks, but 2 hits, then you get 5 choose 2, because you have 5 positions in the sequence to choose from to put your hit in, and you choose 2 of those positions to put your hit in.
I understand that the two situations are distinct, what I was salty about is that [5 choose 2] is the notation used. There has to be some formula for [n choose m], so why not post the formula instead of using the notation [n choose m]. It makes it more difficult for people who haven't formally learned probability to follow what you're doing.

>However, your calculation (1 - (0.9)^5) is for hitting at least once, rather than exactly once, because it is the complement of the probability of not hitting for all 5 attacks.
I know, in my original post I said that ratio of hitting once and only once to hitting at least once was 80%. I did the calculation for both, but I just don't understand how this simple statement was interpreted to mean 10 attacks in a row by that other anon. That just came out of nowhere.

>>8335435
>If an attack has a 10% of landing, and you attack 5 times, what percent chance do you have of landing once?

You have a 10% chance of landing once on each attack because each event is independent of the other.

Your question is ambiguous, however.

You could also mean "what is the chance of landing exactly once in a series of five attempts?" you must solve this with the binomial probability formula. Which would result in 0.32805.

Surely you can see that (1 out of 5 attempts) and (1 independent attempt) differs.

>>8335435
At least one hit: 0.40951
Exactly one hit: 0.32805

c(1,5) * 0.1 * 0.9^4