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    Just been invited for a 15min over the phone maths interview:

    "We would like to invite you to an initial fifteen minute telephone interview. This is a maths interview using mental arithmetic based on probabilities and possibilities. The use of a calculator or pen and paper is NOT permitted. "

    Has anyone had any experiences of this / any advice to give to me / have any websites that you particularly recommend for practise?

    Thanks in advance
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    :lolwut:

    How would they know if you're using a calculator or not? Presumably, the questions test the approach to probabilities questions more than the answers, but still...

    Sorry, bit of a useless post for you.
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    Starts off with some metal maths, lie 29*31 , there is always a trick - eg (30-1)*(30+1)

    Some dice rolling expectation questions eg what is the expected value of a dice roll, what about two dice rolls where you take the highest etc.

    You have a pirate chest of gold with a number of coins in it between 0 and 500, you bid for the chest, if your bid exceeds the value of the chest you win it and the guy selling it throws in an extra 1/2 the number of coins in the chest for you. How much should you bid for the chest. etc etc etc

    The final rounds get very hard, let us know how you get on.
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    (Original post by andiroo)
    Just been invited for a 15min over the phone maths interview:

    "We would like to invite you to an initial fifteen minute telephone interview. This is a maths interview using mental arithmetic based on probabilities and possibilities. The use of a calculator or pen and paper is NOT permitted. "

    Has anyone had any experiences of this / any advice to give to me / have any websites that you particularly recommend for practise?

    Thanks in advance
    Example first round questions start off with simple 'add/divide/multiply these two numbers' and then go on to probability style questions.

    Eg if I roll 3 fair dice and add the 3 numbers up, what is my expected result? They then go into more complicated questions based on this, although I don't really want to say more on that. You can find out more by going up to them in careers fairs/company presentations - they give a load of example questions there.
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    First round is fairly simple - They'll start with mental arithmetic questions, then move on to stuff about probability.

    Second Round and onwards - No more mental arithmetic, just probability based questions. (e.g. I have a best of 7 series, where each team is equally matched. What is the probability that we go to the decider?)

    As TheTallOne said, it does get much more complicated than the example above, but don't think you have to get every question right. I got through 2nd round answering about half of the questions correctly, and stumbling through the other half.

    Also after each question, be prepared for "How confident are you in your previous answer?"
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    (Original post by paulak1985)
    Starts off with some metal maths, lie 29*31 , there is always a trick - eg (30-1)*(30+1)
    Cheers for this advice an amost identical question came up! Interestingly he asked me how I did it. I got 100% of the mental maths correct.

    I didn't get past the first probability question, it was a simple enough question but had me stumped at the time, I just drew a blank and didn't have any method as such to discuss with them. I could hear him sigh down the phone as I was thinking about it, he eventually held my hand and guided to me to the correct answer after which I got "I'm going to end the interview here, this was a basic question and you took too long to get it wrong, they only get harder from here"

    Was a fair outcome really, atleast I'm more prepared/experienced for my next interview
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    I can remember one of the questions...

    You're in a coin tossing game, best out of 9 (or whatever), so first to 5 wins! What is the probability that all 9 tosses are required to determine a winner.

    If you're fine for Tibra/Optiver, you'll destroy JSC
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    (Original post by The IC Guy)
    I can remember one of the questions...

    You're in a coin tossing game, best out of 9 (or whatever), so first to 5 wins! What is the probability that all 9 tosses are required to determine a winner.

    If you're fine for Tibra/Optiver, you'll destroy JSC
    Optiver numerical testing is a completely different ball game to JSC isn't it though? It's just basic arithmetic, which I've found people that have studied in China seem to be extremely good at (possibly because their education system is rote learning or so I've heard). Where as JSC seems to involve more thinking, which I assume Optiver makes up for in the final round.
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    (Original post by andiroo)
    Cheers for this advice an amost identical question came up! Interestingly he asked me how I did it. I got 100% of the mental maths correct.

    I didn't get past the first probability question, it was a simple enough question but had me stumped at the time, I just drew a blank and didn't have any method as such to discuss with them. I could hear him sigh down the phone as I was thinking about it, he eventually held my hand and guided to me to the correct answer after which I got "I'm going to end the interview here, this was a basic question and you took too long to get it wrong, they only get harder from here"

    Was a fair outcome really, atleast I'm more prepared/experienced for my next interview
    That's rough; I hope something ellse works out!
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    (Original post by Swayum)
    Optiver numerical testing is a completely different ball game to JSC isn't it though? It's just basic arithmetic, which I've found people that have studied in China seem to be extremely good at (possibly because their education system is rote learning or so I've heard). Where as JSC seems to involve more thinking, which I assume Optiver makes up for in the final round.
    As someone mentioned above, with JSC there is always a trick. The old (X-Y)*(X+Y) will definitely come up. In this sense it is much harder to prepare for than rival firms, but the questions in general are much easier. I'm trying to rack my brain for other stats questions they asked
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    (Original post by paulak1985)
    Some dice rolling expectation questions eg what is the expected value of a dice roll, what about two dice rolls where you take the highest etc.

    You have a pirate chest of gold with a number of coins in it between 0 and 500, you bid for the chest, if your bid exceeds the value of the chest you win it and the guy selling it throws in an extra 1/2 the number of coins in the chest for you. How much should you bid for the chest. etc etc etc
    just slightly fascinated by all this really..
    the chest one is interesting...is there actually a proper answer to that? like if its worth value at the lower end dont you stand to lose a lot of money so it's hardly worth the risk?
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    (Original post by Lorchii)
    just slightly fascinated by all this really..
    the chest one is interesting...is there actually a proper answer to that? like if its worth value at the lower end dont you stand to lose a lot of money so it's hardly worth the risk?
    everything has a value (for a given utility function). That is fundamental.

    Pretty sure for the given 'multiplication factor' (3/2) the answer to the chest problem is zero though...
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    (Original post by Chewwy)
    everything has a value (for a given utility function). That is fundamental.

    Pretty sure for the given 'multiplication factor' (3/2) the answer to the chest problem is zero though...
    well if between 0 and 500 is inclusive, its the only sure fire bid to enure no loss...but then 0 isn't really a bid though is it? its not like some actually starts an auction at 0 and can win with that value. If it's a trick question then I'm heartily disappointed
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    a) can somebody explain why 0 is the sure fire bid, apart from the reason that by bidding 0 you will never lose anything? I am also curious about the derivation of this answer, is it a process of iterating through bids 0-500, and finding the expected value for each bid? if so, what assumptions are you making for the distributions of the outcomes from the chest has 0 coins to 500 coins?
    b) @Chewwy mentioned utility functions and 'multiplication factor'. Do these two terms belong to a specific branch in Maths that i can look up?
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    (Original post by TheBigNerd)
    a) can somebody explain why 0 is the sure fire bid, apart from the reason that by bidding 0 you will never lose anything? I am also curious about the derivation of this answer, is it a process of iterating through bids 0-500, and finding the expected value for each bid? if so, what assumptions are you making for the distributions of the outcomes from the chest has 0 coins to 500 coins?
    b) @Chewwy mentioned utility functions and 'multiplication factor'. Do these two terms belong to a specific branch in Maths that i can look up?
    The multiplication factor (3/2) is just a number which says that if you've won, you've actually won 3/2 times the value of the coins (amount + 1/2 amount).

    Utility functions just tell you how good you'd feel about taking part in a lottery.

    E.g. U(Lottery) = (probability state 1 happening x utility function value of final wealth in state 1) + (prob state 2 x utility value of final wealth in state 2) ...etc.

    I can't see how utility functions fit in with this question though, I would have thought they'd need to tell you which function to use (e.g. CARA, CRRA). There are also 500 possible states which would take hours to calculate?

    Chewwy, could you expand on your answer?
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    (Original post by TheBigNerd)
    a) can somebody explain why 0 is the sure fire bid, apart from the reason that by bidding 0 you will never lose anything? I am also curious about the derivation of this answer, is it a process of iterating through bids 0-500, and finding the expected value for each bid? if so, what assumptions are you making for the distributions of the outcomes from the chest has 0 coins to 500 coins?
    b) @Chewwy mentioned utility functions and 'multiplication factor'. Do these two terms belong to a specific branch in Maths that i can look up?
    the way i see it is if you bid 0, and thats the value of the box then you dont lose anything. If you bid 50 for a box worht 5, then you lose 42.5. The odds for the box containing any of the values from 0-500 are equally likely, so the opportunity to win big, is off put by the risk at losing alot!

    It just seems like a stupid question, but I'm no expert on probability and I could missing something! I just wanted to know out of curiousity...
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    You just need to work out the expected value of the game:

    Expected value is the integral between 0 and B (where B is the bid) of this:

    (1.5V - B)*(1/500)dV

    With V being the actual value of the chest and the 1/500 coming from the assumption that the probability for the value is uniformly distributed between 0 and 500.

    From differentiating this you find the only way to break even would be a bid of 0, since there is no maximum turning point in the expected value.
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    (Original post by Iron89)
    You just need to work out the expected value of the game:

    Expected value is the integral between 0 and B (where B is the bid) of this:

    (1.5V - B)*(1/500)dV

    With V being the actual value of the chest and the 1/500 coming from the assumption that the probability for the value is uniformly distributed between 0 and 500.

    From differentiating this you find the only way to break even would be a bid of 0, since there is no maximum turning point in the expected value.
    V isn't continuous... it's a discrete variable. Using sums for the expectation of a discrete variable, a quick calculation in my head says the expectation should have a -B^2 in it somewhere, so you'd expect a maximum (could be at 0 though, I haven't bothered working it out).

    *Edit*

    Of course I haven't got my edit in on time, but I remembered afterwards that the problem isn't as simple as you've made it out, as Chewwy correctly points out below.
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    Seem to be the only one on here with a JS offer. Ignore this^. No specific maths knowledge required, just the ability to actually *think*!! If you're good at problem solving, you'll be fine- just make sure you're in the right state of mind and be confident- there really isn't any way you can prepare tbh. Although be prepared to be challenged in the final round even when you may not be wrong, i.e. 'but what if...' and other such quick fire questions.
    Hope that helps.
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    (Original post by Swayum)
    V isn't continuous... it's a discrete variable. Using sums for the expectation of a discrete variable, a quick calculation in my head says the expectation should have a -B^2 in it somewhere, so you'd expect a maximum (could be at 0 though, I haven't bothered working it out).
    Yea basically the answer is zero, I was just trying (albeit wrongly) to provide some basis of a mathematical proof for people who wanted to see the reasoning behind it.
 
 
 
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