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November 10th, 2009, 02:30 AM
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| | Interview Questions - Help 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 | 
November 10th, 2009, 04:08 AM
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| | Quote:
Originally Posted by Garfield 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 | It would be better if you say what your answers are (along with the working that lead to them) and then we can tell you if they're right or not.
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November 10th, 2009, 04:21 AM
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| | 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! | 
November 10th, 2009, 04:32 AM
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| | Quote:
Originally Posted by Garfield 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%] ??
[snip] | Mixing percentages and numbers is a bad idea. If the event is certain, use 1 NOT 100%. 1/(X+1)/[X/(X+1)*0.5^10 + 1/(X+1)] is correct and can be simplified.
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti) To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
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November 10th, 2009, 04:39 AM
|  | Flow Master | | Join Date: Dec 2007 Location: Zeitgeist
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| | Quote:
Originally Posted by Garfield 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?"
[snip] | 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 ....
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti) To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
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November 10th, 2009, 06:01 AM
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| | 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? | | The following users thank BERRY for this useful post: | |  | 
November 10th, 2009, 11:57 AM
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| | Thank you Berry. Its an investment banking job. | 
November 10th, 2009, 05:17 PM
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| | Quote:
Originally Posted by Garfield 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?" | 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? | | The following users thank theodds for this useful post: | |  | 
November 10th, 2009, 06:04 PM
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| | I had kept this thread open so that the OP had the opportunity of showing his/her attempt at a solution, NOT so that someone else could post a solution. It should have been very clear that posting a solution was not appropriate.
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti) To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
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