Sorry for a meme math thread but the one on /r9k/ is getting out of hand

This is correct

>pic
>assuming

Never assume. If you don't know, then you can't solve the problem correctly. You may as well assume fluffy bunnies for all it matters.

Ambiguous problem specification. See "boy or girl paradox" on e.g. wikipedia.

Basically, it's a language puzzle, not a probability puzzle. To turn it into a probability puzzle you'd need to specify the problem formally, at which point the ambiguity disappears and the answer (either 50% or 33% depending upon formulation) is obvious.

The issue is *why* the case of no crits is excluded. You can't simply assert it /a priori/ then pretend that the probabilities are well defined.

In essence, the problem as specified is self-contradictory. It says "assuming 50% crit chance". If the crit chance was 50% (without any qualification), then the probability of no crits is 25%. But it also says that "at least one of the hits is a crit", which implies that the probability of no crits is 0%. These cannot both be true.

>/r9k/
don't bring your cesspool over here... fucking plebs

Physics student here. Here's how I think the problems should be solved:

If we had no information and the question "what is the probability both hits are crits?" knowing only that each hit has a 50% chance to crit, then the answer would be 0.5*0.5=0.25 or 25%

Knowing that one one of the hits must be a guaranteed crit we have 1*0.5=0.5*1=0.5 or 50% (the 1 indicates the 100% guaranteed crit either on the first or second hit)

I'm not sure how you get 33%, but in any case the other possible way of reasoning with this is that you have 3 out of 4 (ruling out the case in which neither of the hits are crits) cases in which at least one of the hits is a crit, so that makes it a 75%. BUT this would be a problem involving permutations by which the order in which the crit lands is relevant, which is not something stated by the problem

Crack your teeth on this one, lads.

Once you eliminate the last expression, the probability of having two crits is 1/(1+2+2) or 1/5. The other guy's code backs this up.

Ambiguous.

It's totally consistent with this wording that you have an ability that makes every second hit a critical strike. Therefore, one of the two hits is a crit and the other is not, therefore the probability for both being crits is zero.

To elaborate, once you assert that "at least one of the hits is a crit", a statement such as "assuming a 50% crit chance" becomes meaningless gibberish. Crits cannot then be independent events (if they were, the chance of zero crits would be 50%, not 0%), so you cannot specify an unconditional probability for them.

Instead, you have to specify conditional probabilities; either:

a) the probability that the first hit is a crit, and the probability that the second hit is a crit given that the first hit isn't a crit.

b) the probability that the second hit is a crit, and the probability that the first hit is a crit given that the second hit isn't a crit. Or.

c) the probabilities of some additional events, and the probability of each hit being a crit for each possible combination of those events.

If you do one of those three, "the probability both hits are crits" can be determined. Otherwise, it can't.

Ambiguous.

It doesn't say that anyone has won, it just says that they stop playing.

There are 4 rounds. Each round albert has a chance of 0.6 to win, so after the 4 rounds his chance of having won is 0.6*0.6*0.6*0.6=0.1296 or 12.96%

25 %

there are 3 outcomes which end on Albert winning 3 games after the 4th game:
aaba
abaa
baaa
each of these has chance (0.6)^3*0.4=0,0864
multiplied by 3 gives 0,2592, which is the answer

Because at first you have 50 % chance to choose y or n, after n must be y so that takes 50 %, than 50 % is left between yy and yn, so you get 25 %

Really? Do you think Albert has a lower chance of winning?

no?
there is no rule saying "If Albert doesn't win after 4 games, Bertha wins unconditionally"

That's an actual probability question, which doesn't really belong in this thread. But anyhow, it's 9/13.

There are 3 ways in which Albert can win after 4 moves, each with probability (3/5)^3*(2/5), for a total of 3*(3/5)^3*(2/5).

There are 3 ways in which Bertha can win after 4 moves, each with probability (2/5)^3*(3/5), for a total of 3*(2/5)^3*(3/5).

The probability of the game ending after 4 moves is the sum of these, 3*(2/5)*(3/5)*((2/5)^2+(3/5)^2).

Given that the game ends after 4 moves, the probability that it was won by Albert is
3*(3/5)^3*(2/5) / 3*(2/5)*(3/5)*((2/5)^2+(3/5)^2)
= (3/5)^2 / ((2/5)^2+(3/5)^2)
= (3^2) / (2^2+3^2)
= 9 / (4+9)
= 9 / 13

Its not a must for one to win

> That's an actual probability question
I'm not so sure of that.

It says that "they stop playing after four rounds". It doesn't say whether they stop playing because one side won or because they got bored with the game or whatever.

So it looks like another ambiguous question (i.e. bait). There's

a) the probability of Albert winning given that one of the two players won on the fourth round (9/13), and

b) the probability of Albert winning given that the game wasn't won in three rounds and stopped after the fourth round due to either a win or abandonment (9/25).

Same old stupid ambiguous "you meet a woman with one child, at least one child is a boy" argument-provoker.

It causes confusion because it states it as a concrete case: in this example, "You hit an enemy twice." Then it adds information, "At least one of the hits is a crit" without stating how you got that information.

The most natural interpretation, in the context of a single case, is that you saw the results of one of the hits, and the question is about the probability of the other result. In other words, interpreting "At least one of the hits is a crit... what is the probability that both hits are crits?" as "The first hit is a crit... what is the probability that the other hit is a crit?"

There is absolutely nothing wrong with this interpretation. On the other hand, if you assume that through some contrived means, you actually got the information, "Both hits have already happened, one or two may be crits, but not zero. You were certain to receive this and only this information, or its negation, no matter what the outcome was." the answer is different, and while that interpretation is strange because it's hard to imagine a scenario in which you'd receive this precise amount of information, it's not wrong.

There is no single correct answer, because there is no single correct interpretation. Anyone who insists there is, particularly someone who insists that "The first hit is a crit... what's the probability that the other is as well?" is an invalid interpretation, is either stupid or trolling.

3/4*1/3 = 1/4 = 25%

Lets enumerate the possibilities:

AAAA 81/625
AAAB 54/625
BBBA 24/625
BBBB 16/625

AABA 54/625
ABAA 54/625
BAAA 54/625

ABBB 24/625
BABB 24/625
BBAB 24/625

AABB 36/625
ABAB 36/625
ABBA 36/625
BAAB 36/625
BABA 36/625
BBAA 36/625

The four can't happen; they result in one of the players winning after 3 rounds, so the game can't end after four rounds. Total probability

The second 3 are a win for Albert; total probability 162/625.

The third 3 are a win for Bertha; total probability 72/625.

The last 6 indicate that no-one won after 3 rounds; total probability 216/625.

If "they stop playing after four rounds" implies that one player won, then the probability that it was Albert is 162/(162+72)=9/13.

If it allows for the possibility that they just stopped playing, the probability that Albert won is 162/(162+72+216) = 9/25.

> The four can't happen
The FIRST four can't happen

Thank you!

I clearly know the answer is 1/3 and can work this out using a probability tree.

Though how come I don't get the result 1/3 when I apply what you used there with the 50% stated in OP picture question?

50% = 1/2

3/4 * 1/2 = 3/8 = 0.375

Because it's wrong

In fact, if the first is not a crit, the event that the second one is 100%, but the fact that the first is not a crit is only 50%, so it is 25%

That's his chance of winning 4 rounds familio

Very probable

What is wrong about it?

Is this wrong?

Your friend tells you he flipped 2 coins and one of them came up heads. What is the probability that were both heads.

I see you have multiplied 1/2 and 1/2 together to get 0.25 is right.

However, this question seems to be a is a conditional probability question there is more to it.

It is based on the condition of a critical hit being made we don't know if it happened first or second time out of two hits.

To do this

P(C) P(CC) P(CN) P(NC)
0.25 / (0.25 + 0.25 + 0.25) = 1/3

You have to include the case of no crit hit then the second time of being a critical hit. Why? Because this an event that can still occur.

Whereas no critical hit the first and second time cannot occur due to the stated condition 'At least one of the hits is a crit.'

So the answer can't be just 25%

This is fundamentally true.
The probability of a first hit being a crit is 1/2. As exemplified:

Crit/Normal
Crit/Crit
Normal/Normal
Normal/Crit

As we are talking crits here, lets imagine the calculations are being made by a Dungeon Master rolling dice. Everytime the result is a Normal/Normal, the DM makes a reroll.
Further on, the probability of rolling a crit on the first turn is 50%. If, however, you roll a Normal on the first throw, there is a 50% chance, over a previous 50% chance, totalling a 25% chance, that your roll will be invalidated, since it will result on a Normal/Normal, wich results on a reroll.
Therefore, the possibility of getting a Crit on the first roll is 2/4 or 50%:
Crit/Normal 1/4
Crit/Crit 1/4
Normal/Crit 1/4
Normal/Normal 1/4
a Normal/Normal outcome, however, results on a reroll, wich means the probability of a getting a Crit on the first VALID roll is 2/3, or 66%:

Crit/Normal 1/3
Normal/Crit 1/3
Crit/Crit 1/3
Normal/Normal 0

You can figure out the rest.

tl;dr: the probability of a double crit outcome is 25% or 1/4. The probability of a double crit VALID outcome is 33% or 1/3.

first you pick whether your first attack or your second attack is the guaranteed crit, so 2 ways. they are disjoint cases so you add them to get the sum total of possible outcomes. there are two ways to pick the remaining attack, so we get 2*2 = 4 paths, of which 2 of these are valid, so 1/2

>a 50% crit chance
Lrn2probabilly fgt pls

According to your experiment, the second crit is twice as likely as the first one. Good job.

Yes

>I'm not sure how you get 33%
Crit -nocrit
nocrit-crit
crit-crit

33% chance

There are 3 possible outcomes each with equal chance, only one is valid.
So 33%.

Critical hit : Normal hit
Critical hit : Critical hit
Normal hit : Normal hit
Critical hit : No hit

Are these variables?

>You hit an enemy twice. At least one of the hits is a crit. Assuming an n% crit chance, what is the probability both hits are crits?

The probability is n%. One crit is guaranteed (at least one is a crit). That leaves one more hit that will or will not be a crit. What's the probability that this one is a crit? It's n%.

About 69.2%

Interesting way of doing it. I believe you're correct.

What I did was:

Probability that Albert wins in exactly 4 turns: (3 choose 2)(0.6)^3(0.4) = 0.2592
Probability that Bertha wins in exactly 4 turns: (3 choose 2)(0.4)^3(0.6) = 0.1152

Probability that Albert wins in exactly 4 turns divided by the probability that the game ends in four turns: 0.2592/0.3744 = 0.6923

25%, dumbass

fucking stop posting this image it wasn't funny when it happened and isn't funny now

fucking stop posting in Veeky Forums in general you piece of shit

No idea why this thread is still alive, but anyone posting an answer to the original "question" needs to first read
then commit sudoku for getting trolled so easily.

50% retards he said the one crit's so don't worry that's guarenteed hit so just worry about other one is 50%

No this is right the possiblity is :
YN
YY
NY
p(NY)=0.5
p(YY)=0.25
P(YN)=0.25

at least*

1 0
0 1
1 1

1/3 * 100%

Consider each hit is a bernoulli trial with probability of success (crit) 1/2, the sum of two bernoulli trials is binomial with parameters (n,p) = (2,1/2).
So P(X = k) = (n!)/(k!(n-k)!) * p^k * (1-p)^(n-k)
= 2!/(k!(2-k)!) (1/2)^2
=1/(2(k!(2-k)!)
the question is thus, out of two trials (shots) given there is at least one success (crit), what is the probability of two success' (crits)
or
P(X=2|X>=1)
pretty simple, as P(X>=1)=1-P(X=0)
we have the definition of conditional probability giving
P(X=2 and X>=1)/P(X>=1)
=P(X=2)/(1-P(X=0)), since the event {X>=1}and{X=2} = {X=2}

=(1/4)/(1-1/4)
=1/3
as required.

wow this is spot on

really clever

At least one. Depends on if you know which one of them or if you don't know which one.

If you know that it was at least the first one that hit for sure then there's only

YN
YY

left, alas 50%.

thanks guys

I tutor grad level measure theoretic probability if you were wondering why it was so perfect

1/4
25%

There's the exact same argument on /v/ with "two coin tosses, one is heads".

Guy is saying

>The formula is more or less:

>P(HH|at least one heads)=P(at least one heads|HH)*P(HH)/P(at least one heads)

>| stands for "given"

>P that is at least one heads given TT is, of course, 1.

>P(TT) is 1/4.

>P of at least one heads is:

>1-P(TT)=1-1/4=3/4

>so

>P(HH|at least one heads)=1/4*4/3=1/3

>Source: I have a fucking exam on this shit tomorrow.

but I don't understand why you he's allowed to use P(TT)=1/4 given that at least one toss is head.

Imo

>You can't use values from prior the restriction to apply them in your formula. Let me put it this way then-

>You're obtaining the values by giving yourself two independant variables (T1 & T2, Where T1 and T2 ~B(1/2), and X~B(p) reads "X follows the independant variable of Bernoulli of parameter p")

>You then say " P( Y1=1, Y2=1) = 1/4 (true, they're independant)
>P(Y1+Y2>0)=3/4 (also true)

>But then you insert that in a formula which uses two different variables, let's call them Y1 and Y2, where Y1~B(1/2) but Y2 ISN'T independant from Y1 because of the restriction.

>You conclude by saying

>P(Y1 = 1 and Y2 =1) = P(T1 + T2 > 0 | T1 + T2 = 2) * P( T1 = 1 and T2 = 1) / P (Y1 + Y2 > 0) which is false.

The fact that he gets the correct result is just a coincidence of this being a problem with few possible answers- so is his reasoning just or flawed?