Hi everyone,
I have an interview for a job coming up and they have asked me to bring in answers to a couple of conditional probability questions.
Would any of you guys be able to tell me your answer to the following questions so that I can double check my answer
1. You and your opponent have the goal of forcing the other guy to be the first to say '60'. You must announce numbers in ascending order. You can add to whatever number is said in increments of 1-10. You can choose to go first or second, and the person who goes first can begin with any number between 1 and 10. What is the perfect strategy for winning?"
2. one 2-headed coin with x fair coins in a jar, coin is picked at random and flipped 10 times, all heads: what is the probability that you flipped the biased coin).
Thanks for your help
Hi Mr Fantastic,
For the first Q I am using the conditional probability formula
P(B given A) = P(A & B occuring)/P(A occurring)
So if I let
A be the probability of getting 10 heads and
B be the probability of picking the biased coin
then I can use the formula to calculate the probability that I flipped the biased coin given 10 heads occurred.
P(A&B) is 1/(X+1)*100%
P(A)= X/(X+1)*0.5^10 + 1/(X+1)*100%
So my answer should be 1/(X+1)/[X/(X+1)*0.5^10 + 1/(X+1)*100%] ??
I'm stumped on the 2nd Q!
I assume there are very good reasons why your prospective employer wants you (and not me or someone else) to solve this question, so if you have no idea then, I'm sorry too say, my post is not going to change that ....
However, it's possible that if you look at the optimal strategy for the game of Nim, you might be able to shed some light on the question all by yourself ....
Define the following:
P(H) = {10 heads in a row}
P(B) = {selecting a biased coin from the jar}
P(F) = 1 - P(B) = {selecting a fair coin from the jar}
Find the conditional probability that given the biased coin, you get heads 10 times = P(H|B)
Find the conditional probability that given a fair coin, you get heads 10 times = P(H|F)
You're looking for P(B|H) so this is just basic Bayes' Theorem
As for the other problem, think of it this way. If you want to arrive at 60 and both of you can only add numbers between 1-10, you want to arrive at 49 before he does since you automatically win if you're at 49. He can only add somewhere between 1-10 which means he'll have a number between 50-59 meaning you'll win in the next step. Now, take it 1 step further. If you "win" at 49, you'll also "win" if you arrive at 38 first by the exact same logic. Now keep applying the same logic. You "win" at 27, 16, 5. So in order to guarantee you win, you select to go first by picking 5, and make sure you choose whatever number that would add to 16, 27, 38, 49, and lastly 60.
Damn, I should get this job.
Mind me asking what type of job this is?
I'm almost positive I have the right answer for this, but it seems a little...uhh...morally questionable to give you the answer. I really want to know if my answer is right though, so could you post your answer (or if someone else did it, PM the first move) so I can know if I did it right?
EDIT: Nevermind, I did it the same way Berry did. Except he got the rules a little mixed up Seriously though....what's up with giving him the answer? If his prospective employer found out he got the answer this way, there's no way they would hire him. Don't we already have enough investment bankers with questionable morals?