Mom look, I’m on the Newsletter! 
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So piggybacking of my husbands answer (love you) (DrewskiLuv)
He was on the right track but slightly off.
So as he stated, McDonald’s probability would remain 1/9, or 11% since bfi told them they wouldn’t say anything about their chances,
And the remaining three were found to still be in the odds because they were not eliminated by BFI’s information. BUT all three of them share similar odds because of this.
So splitting the remaining odds between them (88.88%) leaves the probability at:
McDonald’s: 11.11% (1/9)
BFI: 29.63% (2.66/9)
ROSE:29.63% (2.66/9)
Alex 29.63% (2.66/9)
My answer 
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Scenario 1: (8 / 9 chance of occurring)
Lets assume that McDonalds was NOT picked. BFI told McDonalds that he wouldn’t say anything about his chances. so when BFI sums up 5 people that were not picked, McDonalds is guaranteed to not be in the list. 4 candidates remain and McDonalds is not it (because we assumed that). So the other candidates each have a 1/3 chance.
Scenario 2: (1 / 9 chance of occurring)
Lets assume that McDonalds WAS picked. 4 candidates remain, 3 of which have a 0 % chance of being, McDonalds has a 100% of being picked
Now lets combine these two scenarios:
McDonalds: 8/9 * 0 + 1/9 * 1 = 1 / 9
BFI: 8/9 * 1/3 + 1/9 * 0 = 8 / 27
Rose: 8/9 * 1/3 + 1/9 * 0 = 8 / 27
Alex: 8/9 * 1/3 + 1/9 * 0 = 8 / 27
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Alright Alright. So.
The question says, “What is the probability for each of them that they were selected at random by BFI.” I’m inferring that BFI will not chose himself as the wording kind of plays it as he won’t.
So I may just be an idiot but, wouldn’t
Rose, McDonalds, and Alex all have a 1/3 chance of getting chosen, aka 33.33% chance?
Most likely a dumb and wrong answer but here:
For basically the same reasons as some of the people above I don’t think Beefy included himself in the competition, and since Alex posted this one I don’t think BFI would choose him to post the next one as well so he would probably be excluded too. That’d leave Rose and McDonalds with 4.5/9 or 50% each but gambling on the assumption Rose was excluded too since she posted the last newsletter that would leave everyone else with 0% and McDonalds with 100%.
This sounded smarter in my head
edit
Nvm I think I got the actual answer now but it’s too late ;(
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blue, final answer
(JOKE ANSWE DONT COUNT THIS PLEASE)
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Mcdonalds still has a 1/9 chance of winning as it has not been confirmed if he is part of the people who are in the final group. as for the other three, they each have a 8/27 chance of winning
this adds to a total of 27/27
bruh i thought i was clever 2 other people have the same answer
this is the longest newsletter i have ever read on this website.
Since there were only 3 minutes between your response and @Hikket I would like to split the reward. Your answer is correct but the reasoning isn’t really water-proof because it relies on the information that McDonald’s chances do not change, which can be checked simply but is not trivial (even though the fictional BFI claimed this).
Hikket took a tiny bit longer but gave a satisfactory response. I will split the prize 50/50 between you, both of you let me know which of the prizes you are interested in.
As for anyone interested, I will add a formal solution to the problem. You will see that it is pretty much just a formalized version of @Hikket’s calculations, but it requires some more advanced concepts.
Formal Solution
We will use Bayes’ theorem which can be derived easily from the concept of conditional probabilities:
P(A|B) = P(B|A) P(A) / P(B).
Assume the following events:
A: McDonalds wins
A1: BFI01 wins
A2: Rose wins
A3: Alex wins
A4: Someone else wins
B: BFI, Rose, McDonalds, Alex are the remaining options given the information in the question
Note also that: P(B) = P(B|A) + P(B|not A) = P(B|A) + P(B|A1) + P(B|A2) + P(B|A3) + P(B|A4) as well as: P(A) = P(A1) = P(A2) = P(A3) = 1/9.
Now start with some calculations:
If we know that McDonalds wins, the chances of B are simply:
P(B|A) = 5! (1/8 * 1/7 * 1/6 * 1/5 * 1/4)
because we pick 5 losers randomly out of the 8 remaining options.
Very obviously we can see that P(B|A4) = 0 by analyzing the structure of the problem.
Now for A1, A2, A3 we each get:
P(B|A3) = P(B|A2) = P(B|A1) = 5! (1/7 * 1/6 * 1/5 * 1/4 * 1/3)
because in this case we pick 5 losers randomly from 7 remaining options (the winner and McDonalds are excluded).
Putting this all together we now get:
P(A|B) = P(B|A) P(A) / P(B) = [5! (1/8 * 1/7 * 1/6 * 1/5 * 1/4) * 1/9] / [5! (1/8 * 1/7 * 1/6 * 1/5 * 1/4) + 3 * 5! (1/7 * 1/6 * 1/5 * 1/4 * 1/3)] = [5! (1/7 * 1/6 * 1/5 * 1/4)] / [5! (1/7 * 1/6 * 1/5 * 1/4)] * [1/8] / [1/8 + 3 * 1/3] = [1/8] / [9/8] = 1/8 * 8/9 = 1/9.
For A1 (and equivalently for A2, A3):
P(A1|B) = P(B|A1) P(A1) / P(B) = [5! (1/7 * 1/6 * 1/5 * 1/4 * 1/3) * 1/9] / [5! (1/8 * 1/7 * 1/6 * 1/5 * 1/4) + 3 * 5! (1/7 * 1/6 * 1/5 * 1/4 * 1/3)] = [5! (1/7 * 1/6 * 1/5 * 1/4)] / [5! (1/7 * 1/6 * 1/5 * 1/4)] * [1/3] / [1/8 + 3 * 1/3] = [1/3] / [9/8] = 1/3 * 8/9 = 8/27.
Note that 5! marks the permutations of picking the 5 losers because they are no picked in a fixed order. If this confuses you, you can just assume that the order in which they were picked is fixed. In that case you can just leave out this computation and the result does not change.
If the formal solution is too complicated for you but you are still interested in the solution, Hikket’s response gives a good intuition:
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I didn’t quite read other answers too much, didn’t want to get distracted by senseless math equations that add and then subtract the same value.
Information
Go up ↑
9 Staff members.
The draw is done before the Host leaks information.
5 Staff members are ruled out after the draw.
SierAlex posted the newsletter.
At the start, everyone had an equal chance of 11.11%, McDonalds asked BFI01 for leaks on his chances.
BFI01 didn’t want to leak Information about McDonalds to McDonalds, but he leaked 5 other people who definitely didn’t make it.
The chances however didn’t change, they are still 11.11%, but with a hole of 55.55%.
My Answer: 11.11%
Or 1/9.
Analysis of other people's math equation:
You can only put whole numbers in fractional numbers.
1/3.5 is wrong, and everyone should know this because the actual fraction is 2/7.
Since people now know it is wrong, a joke answer would be 1/π.
Speculations
I don’t know the relationships between the staff members just like everyone else who is allowed to participate in this competition, so BFI might not have used random.org to choose someone at “random.”
It’s also possible that BFI wanted to lure McDonalds on a red herring and exclude everyone who didn’t win from the list of people who didn’t win that he leaked so that Alex would win with a 0% probability.
If they don’t have a solution yet, and choose the most convincing one, I might have a chance of 1/7 to win as cheater_ ↑ stated that all above his comments are wrong.
But that of course could also be a red herring, as he stated that they seem to be on the right track.
PS: It took me an hour to write this and another hour to click on Reply.
PPS: I saw cheater_ write a reply and tried to halt my reply for another few minutes to see if he’s gonna drop the first hint. I guess not. q-q
At least my post uses HTML features. (I could have made it use only HTML and no markdown since I analyzed details and the separating lines)
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slightly annoyed i had the answer correct for the maths problem before anyone else did but i went with the incorrect solution instead
smh my head (AEST time btw so i was a bit ahead of the others)
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i want 30 swamp crates cheater ;D
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lol pce L
Congrats @Spocknado and @Hikket !! I really liked this math problem and it brought me on an idea: what if you make some sort of scavenger hunt for the next newsletter. You would have to solve riddles and the solutions bring you to unlisted topics with new clues and the problems become increasingly harder etc. It would be a lot of work to set up but I think it could be a lot of fun.
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yes pls
The Munchy Minecraft Forums ARG. I would love to take part in designing that :3
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For the math: bfi lied cheater wrote the newsletter
Omg staff member of the month?? 
Ty 
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