since you guys knocked the last thread out of the park, let's see you take this one on
>>29325713
>100%
He's a man
Oregano commentolini
Albert wins by default because he is a man and the Patriarchy is on his side
They stopped playing because the game got really intense and they stabbed each other to death.
addendum: albert is ftm
>>29325713
95% chance. There's a 5% chance he fails at the fourth round:
Round 1 - 40% chance to fail
Round 2 - 20% chance to fail
Round 3 - 10% chance to fail
Round 4 - 5% chance to fail
where's my prize, la?
Binomial tree. Well, almost. I'm bruteforcing it on a spreadsheet. Will have the answer in a few.
>>29325713
Too hard, next question.
Okay, I bruteforced it with a binomial tree. Maybe someone can come up with a smarter way. The trick is that you have to take out the "Albert has won 4 in a row" tree path.
Albert 3 wins after round 4: 33.06% .
0.2592, right? or am I being stupid?
>>29325910
wouldn't you also have to remove the path with B winning 4 in a row as well?
>>29325974
Oh... yes, you're right. You have to remove both paths. So in round 3, both "Albert has won 3" and "Bertha has won 3" have to be removed.
Seems like it becomes 0.36 then. Okay, now to figure out where it comes from exactly, and how to simplify that tree formula...
>>29325974
>>29326002
The only possibilities at the end of round 3 are:
A-B 2-1: three routes to come there, all with the same probability (60% * 60% * 40%)
B-A 1-2: three routes to come there, all with the same probability (40% * 40% * 60%)
Which does then add up to, at the end of round 3, A-B being 60%. Now the only thing that remains is to apply one last 60% to that path. Meaning, 36% is the answer.
60% chance of winning each round * 4 rounds = 240% chance Albert won
>>29325713
So basically, this is a bernoulli trial where we take it as a 60% chance that albert succeeds, and a 40 % chance he fails. the order is not specified and as such does not matter. As such, its (4C3)(.6)^3(.4). Answer is .3456 or 34.5%, approximately.
>>29326147
>the order is not specified and as such does not matter
Yes, it matters. You can't have Albert or Bertha with 3 wins at the end of round 3, because then the game would end and there would be no round 4.
>>29325713
60% + 60% + 60% = 180%
fucking easy
next
(1-.06) * 0.6^3 * (4 choose 3) + 0.6^4 = 0.4752
The rest of you are retarded.
>>29325713
albert can win with
>A A B A 8,64%
>A B A A 8,64%
>B A A A 8,64%
and can lose with
>B B A B 3,84%
>B A B B 3,84%
>A B B B 3,84%
lose % 11,52,win % 25,92
>>29325713
2000 hours in MsTex
>>29326208
hmm, forgot to address that, i concede. And in hindsight, i didn't even acknowledge Bertha winning. There may be a little more to this problem than i thought.
>>29326305
idiot, you forgot how to use conditional probability.
it's 9/13, or about 69%
probability that albert wins, given that the game takes 4 rounds =
(0.4*0.6^3)
-------------------------------
(0.4*0.6^3)+(0.6*0.4^3)
~= 69%
>>29326315
>>29325910
22 posts and only these guys remembered that Albert can win 4 times. (And only the first guy knew how to compute it efficiently.)
>>29326350
You need to take out AAAA and BBBB. They can't play a 4th round if one of them wins in round 3.
>>29326208
>>29326358
If one of them has won after round 3 what difference does it make whether or not you have a round 4? Just leave it in to make the math easier.
>>29326481
As long as AAAB and BBBA are in there it'll work out. Adding the probabilities of AAAB and AAAA will work out to the same as just cutting it off at AAA.
>>29326501
>the math is too hard for me so I'll just change the problem
You can't have one of them winning after round 3. Because the problem clearly states that they play 4 rounds in this situation.
>>29326519
I'm not changing the problem. AAAB has .4 times the probability of AAA, AAAA has .6 times. So AAAB + AAAA is the same as just using AAA, except the former lets you use a closed expression while the latter forces you to make these giant trees.
>>29326518
Wut? You can't just add and subtract at will like that. For your tree to be correct, it must include every possibility, and exclude every impossibility.
>>29326559
The question doesn't ask for a tree. It asks for a probability. Which you get by summing over the leaves of the tree.
Having one leaf AAA, or two AAAB and AAAA whose probabilities sum to the same thing, gives the same sum.
>>29325713
>>29326481
>>29326350
Wait, i'm retarded.
>They stop playing after four rounds
Its a conditional problem again.
>>29326557
Not him but the problem is that it states that they play exactly four rounds and after AAA happens, the game just ends so the condition of having played 4 rounds is false. Therefore it is just 60%^3 * 40% * 3 = 25.92%
>>29326666
forgot my super original math pic
>>29326670
>>29326317
just as i thought original friend
>>29326670
The problem clearly states that they play to 4 rounds, but I'm telling you it does not affect the final probability whatsoever whether or not they quit.
Do you think that if Albert wins 3 rounds then playing another round will magically undo that?
>>29326677
Your math is way off. There is no way it can be as high as 69%.
yeah, backing original statement of bernoulli trial, but its a win on 4th game, guaranteeing 2 wins and 1 loss prior, meaning its (3C2)(0.6)^2(0.4)*(0.6) = .2592
>>29326832
You need to scale it back up. You're still including the win-in-3-games possibilities. That would only be correct if the game didn't stop after 3 wins.
>>29325713
P(win after 3 rounds) + P(win after 4 rounds) = 0.6x0.6x0.6 + 0.6x0.6x0.4x0.6 = 0.216 + 0.086 = 0.302
>>29326918
never mind, 4 rounds played, can't win after only 3 rounds
>>29326918
1) your 2nd term must be multiplied by 3
2) it's wrong anyway because they play 4 founds, eliminating the 3 win in 3 games possibility
>>29326779
It is because OP specified that it ends after 4 rounds.
All paths that aren't green or red aren't even a possibility anymore, because they would end too soon (3 turns) or too late (5 turns).
If you would ask
>What is the chance that Bertha won?
it would be
P(Bertha won | end in 4 turns) =
P(Bertha won in 4 turns)/P(ends in 4 turns) =
green / (red+green) =
(0.6*0.4^3*3)/(0.6^3*0.4*3 + 0.6*0.4^3*3) =
30.77% =
100% - 69.23%
>>29326909
did i not account for that by establishing it as a 3C2? basically looking at it from the point between the 3rd and 4th game, where the 4th was taken as an absolute certainty for the purpose of the problem.
> inb4 the "they stop at 4 rounds regardless of the games' outcome" people start arguing with the "they stop at 4 rounds because somebody won" people, only using numbers
>>29326918
B A A A = .4x.6x.6.6 = .0864
A B A A = .6x.4x.6x.6 = .0864
A A B A = .6x.6.x.4x.6 = .0864
---------------- +
0.259
>>29326978
They stop playing after 4 rounds. There may not be a winner. 2-2 is still a possibility.
>>29326918
>>29326943
correction
>>29327004
Everyone confused whether a 4th round is played after AAA or BBB
The phrasing should be changed "the winner is the one who wins 3 rounds"
>>29327059
Is it not naturally assumed that it is only played until there is a victor?
Threads like this remind me why I can't stop teaching math.
>>29327059
if they played over 5 games then you could have 2 winners, so it is assumed that it has ended by the end of the 5th game at the latest
>>29325713
0% because Bertha cried patriarchy and our gynocentric society automatically awarded her the winnings while destroying Albert's life and career under accusations of muhsoggyknee
>>29327088
No, because it's written that they stop after 4 rounds, which could mean (a) they play exactly 4 rounds, (b) they play until a winner or until 4 rounds, (c) they stop at 4 rounds BECAUSE there was a winner.
Options (a) and (b) give the same number but (c) does not.
It's 64.8% you retards.
>>29327132
So, the question is stupid.
I got it faggots
.6x.6.x.6 = 21% chance he wins
>>29326317
Forgot the draw ops,but doesn-t matter
albert can win with
>A A B A 8,64%
>A B A A 8,64%
>B A A A 8,64%
and can lose with
>B B A B 3,84%
>B A B B 3,84%
>A B B B 3,84%
a draw when
>A A B B 5,76%
>A B B A 5,76%
>A B A B 5,76%
>B B A A 5,76%
>B A A B 5,76%
>B A B A 5,76%
DRAW 34,6%,WIN 25,92,LOSE 11,52%
>>29327020
You're right.
I thought that they would play until someone wins.It's pretty poorly worded desu.Like all 4chan math bait threads.
>>29325713
There are 3 different outcomes resulting in albert's victory
A : He wins games 1, 2 and 4
B: He wins games 1, 3 and 4
C: He wins games 2,3,4
Each of these outcomes's P = (6/10)^3 * (4/10)
Multiply P by 3 and there you have it
Final result is 0.2592 chance that albert has won
>>29326463
>albert can win 4 times
but we had 4 rounds.
If albert won rounds 1,2,3 it means that we wouldn't have a 4th round.
>>29327244
Wrong. You need to scale it up. This would only be correct if every single combination of wins in 4 games were possible. Which it's not.
>>29326677
see
>>29327020
There are six 2-2s with the probability of 0.4^2 * 0.6^2 each.
See your super originial math pic changed in a super original way.
>>29327292
Yay, finally someone else got the same answer I got way early in the thread. Most likely the correct one.
>>29325713
i think i got this
>A A A A not gonna happen 0%
>A A A B not gonna happen 0%
>A A B A (B wins third round 100%) 21,6%
>A B A A 8,64%
>B A A A 8,64%
WIN % 38,88%
>>29325713
Let's look at all of the possible combinations Albert could have used to win:
LWWW
WLWW
WWLW
Now, let's look at the probabilities associated with every combination:
LWWW: 0.40*0.60*0.60*0.60 = 0.0864
WLWW: 0.0864
WWLW: 0.0864
Now lets add the associated probabilities up:
0.0864+0.0864+0.0864 = 0.2592
Actually this is incorrect. This is just the probability that he would win after four rounds, not given that they went for four round what is the chance Albert has won. This is a conditional probability question!
So, let's first calculate the probability that of all combinations that lead to them playing four games.
LWWW
WLWW
WWLW = 0.2592
WLLL
LWLL
LLWL = 0.1152
Total = 0.2592 + 0.1152 = 0.3744
We already have the probability that Albert would win after four games (0.2592). Therefore given that the games went on for four rounds, the chance that Albert won is expressed as his chances of winning anyway, over the probably that the game would end after four rounds:
0.2592 /0.3744 = 0.69230769230769230769230769230769
or...
0.682
That's Albert chances of winning.
Hopefully I didn't make a mistake.
I would tackle this with binomial probability. (I fucked up in the last thread but i think binomials are appropriate in this thread).
Probability of albert winning on any given round (S): 0.6
prob of bertha winning(F): 0.4
Rounds played: 4
so the 2 cases in which he wins is when he wins three or four rounds.
using the notation:
nCr*F^(n-r)*S^r,
we have n as rounds played (4), r as rounds won by albert, and F and S the probability of albert failing and winning on any given round.
i.e.
4C3*0.4*0.6^3 + 4C4*0.6^4
=0.4752
so he has a 47.52% chance of winning when four rounds are played.
>>29328496
please note that i assumed from the question that they would keep playing even if albert wins three in a row. it says they play four rounds, which is stated with no exceptions in the question.
>>29325713
i'm fucking done
>A A A A not gonna happen
>A A A B not gonna happen
>A A B A (B wins third round 100%) 21,6%
>A B A A 8,64%
>B A A A 8,64%
WIN 38,88%
>B B B B not gonna happen
>B B B A not gonna happen
>B B A B (A wins third round 100%) 6,4%
>B A B B 3,84%
>A B B B 3,84%
LOSE 14,08%
>A A B B (B wins third round 100%) 14,4%
>A B A B 5,76%
>A B B A 5,76%
>B B A A (A wins third round 100%) 9,6%
>B A B A 5,76%
>B A A B 5,76%
DRAW 47,04%
38,88+14,08+47,04=100% checkmate
>>29328636
incorrect
originalo comentalo
>>29325713
8.64%? REEEEE It's not low in content. Math is pure content
>>29328691
bullshit,it is perfect
Well the probability that the other guy won is .6*.4*.4*.4 = .0384
1-.0384 = .9616
so....96.16% then, right?
>>29328526
It doesn't matter if they keep playing or not mate. Continuing to play after you win does not affect your chances of winning. P(AAA*) =P(AAAB)+P(AAAA). Congrats on getting the right answer though.
Cba to do it but it is bernoulli
If the game stops after a winner is decided:
You have a conditional probability problem.
P(Four games) = 1 - P(Not four games) = 1 - (0.6^3+0.4^3) = 0.72
P(Albert wins AND Four games) = 0.6^3*0.4*3 = 0.2592
P(Albert wins | Four games) = 0.2592/0.72 = 0.36
If they play four games regardless of whether a winner was determined by the third game or not:
Simply add all the probabilities of Albert winning together.
P(Albert winning) = 0.6^3*0.4*3 + 0.6^3 = 0.4752
