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# Interview questions in London, England

Bloomberg L.P. Interviews in London

www.bloomberg.com /  HQ: New York, NY

751 Interviews in London (of 4,281)

3.1 Average

PwC Interviews in London

www.pwc.com /  HQ: New York, NY

427 Interviews in London (of 6,874)

3.1 Average

Goldman Sachs Interviews in London

www.goldmansachs.com /  HQ: New York, NY

412 Interviews in London (of 4,064)

3.2 Average

## Interview Questions in London

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Apr 27, 2011

Dec 29, 2012
 We consider numbers from 1 to 1 million. How many digits 2 are there??10 Answers600 000940951654321: The total number of digits 1-999,999 is 5888889 Total 1-9 = 9 10-99 = 2*9*10 and so on so number of 2s is 5888889/9Show More Responses654321: The total number of digits 1-999,999 is 5888889 Total 1-9 = 9 10-99 = 2*9*10 and so on so number of 2s is 5888889/9Each time you fix a digit 2 (in units, tens...), then you change the other 5 digits. So you'll have 100 000 number. Having that for the six placements, the result would be 600 000.600,000 6C6 with 6 digits, 9(6C5) with 5, (9^2)(6C4) with 4...etc. Add them up 600,000.11111110^6-9^6observe the recurrence relation. 1-9 there is 1 from 1-99 there is 10 1-9 intervals plus 10 digits from 2x. and then it continuous in exactly the same fashion add one 0 you get 10 times as many plus 10^(n-1) from the nth digit you add. (((((10+10)*10+100)*10+1,000)*10+10,000)*10+100,000)=600,000Consider from 0000000 to 9999999, the number of 0,1,2,3,4,5,6,7,8,9 present must be the same since they are all symmetric. prove: the position can be filled with 0 can be filled with 1, so the number of 0=the number of 1; the number of 1=the number of 2 for the same reason. ... so the number of 0-9 are all the same. There are 6*1million positions in total, thus 6 million positions, divided by 10, is 600,000.

Jan 2, 2011
 You roll two die: What is the probability of rolling a 10 and an 11 before rolling a 7?8 Answers17/132How did you get that?Individually 10 comes up 3/11 (4+6, 5+5, 6+4), 11 comes up 2/11, 7 comes up 6/11. Consider 10 first: Then only interested in 7,11: Probability 11 is 2/8, 7 is 6/8. Consider 11 first: Then only interested in 7,10: Probability 10 is 3/9, 7 is 6/9. Consider 7 first: Busted. So add 10 then 11, and 11 then 10: (3/11)*(2/8)+(2/11)*(3/99)=17/132.Show More Responses5/11 let's say d is the probability that 10 or 11 comes up before 7. probability of rolling 10: (6+4,4+6,5+5) = 3/36 probability of rolling 11: (6+5,5+6): 2/36 probability of rolling 7: (6+1,1+6,5+2,2+5,4+3,3+4): 6/36 d = you roll 10 or you roll 11 or (you roll something other than 10,11,7 and roll 10 or 11 before 7) d = (3/36) + (2/36) + (25/36)*d (11/36)*d = (5/36) d= 5/11Anonbyfar is correct, 5/11 is the right answer but for the wrong questiondoesnt the answer by 'Anonbyfar' (17/132) give you the probability of rolling a 10 and 11 within two rolls? Shouldnt this be then related to the probability of rolling a 7 (6/11). I.e 17/72?P(10 or 11 before 7) = P(10 or 11 | 10 or 11 or 7) = [P(10) + P(11)]/[P(10)+P(11)+P(7)] P(10) = 3/36 4,6 5,5 6,4 P(11)= 2/36 5,6 6,5 P(7) = 6/36 1,6 2,5, 3,4 4,3 5,2 6,1 Answer: 5/11roll 10 or 11: 5 cases roll 7: 6 cases roll others: 25 in total: 36 cases So Pr = 5/36 + 25/36 *Pr Pr = 5/11

Oct 6, 2013
 If I write down all of the numbers from 1 to 1,000,000 on a page, how many times do I write down the digit 2?9 Answers(base 10 )log(x)/10*xI got 468599600000 (verified through brute force)Show More Responsesimagine the numbers written as 000001 000002 000003.....999998 999999, then we 1 million numbers containing 6 digits, each digit of 10 digits we have, is used as much as any other number. Therefore, you write down the digit 2 10% * 6 million - 600,000the minus must be an equal sign on the end, so 600,000 is the right number[1 x 10 x 10 x 10 x 10 x 10] + [10 x 1 x 10 x 10 x 10 x 10] - 1 + (bcos 222222 covered above) [10 x 10 x 1 x 10 x 10 x 10] - 2 + (bcos 222222 & 22222 covered) [10 x 10 x 10 x 1 x 10 x 10] - 3 [10 x 10 x 10 x 10 x 1 x 10] - 4 [10 x 10 x 10 x 10 x 10 x 1] - 5 = 600000 - 15 = 599985Continuation from above from 1 upto 1 with n 'zeros' (e.g. 10 has 1 zero, 100 has 2 zero, etc), answer = n(10^[n-1]) - 0.5n(n-1) So upto 1000000 would give 6 x 10 ^ 5 - 0.5 x 6 x 5 = 599985The idea is to exploit the symmetry among 0,1,2...9. Instead of 1->10^6, one should think of 000000 -> 999999, where 0-9 occurs with the same probablity . Altogether 000000 -> 999999 has 10^6 numbers, 6*10^6 digits; divided by 10 this is 600,000.The idea is recursion, let x_n be number of times that digit 2 appear in 0 to the largest n digit number. For example, x_1 = number of times that 2 appears in 0-9. x_2 is up until 99, etc. you work out to see x_1 = 1. x_2 = 10*x_1 + 10 etc. finally, I think you will get 600000

### Assista at Jane Street was asked...

May 14, 2012
 2nd Round: You have a deck of cards, 26 red, 26 black. These are turned over, and at any point you may stop and exclaim "The next card is red.". If the next card is red you win £10. What's the optimal strategy? Prove this is the optimal strategy.6 AnswersThe simplest strategy is not to wait until any card is open and say the next one is red, than you have 26/52 chance of success. All other strategies require conditional probabilities, so when multiplying by the probability of that condition the overall probability is not going to be greater than 1/2.Regardless of your strategy, your probability of winning is 1/2Nope the optimal strategy is to count the amount of black cards that come up until you hit 26 and obviously the next one is redShow More ResponsesAdrian is wrong, because the strategy only work 50% of the time, when the last card is not black. So you still get 1/2.The idea here is A LEAP OF FAITH: that is we have found the optimal strategy when there is X red and Y black left in the deck. let Strategy(X,Y) be that strategy: it is either "Wait" or "Say it now". Let P(X,Y) be the probability that we are right by following this strategy. now Strategy(X, 0) is definitely "say it now" and P(X, 0) = 1. what about Strategy(1, 1)? half the time it will show black and we win for sure; half the time it will show red and we are doomed. Both action, "wait" and "say it now" have the same expected chance of winning, which is 1/2. so P(1,1) = 1/2. What about Strategy(2, 1)? if it is "wait", 2/3 time we go to (1, 1) and 1/3 time we go (2, 0). the chance of winning if we wait is 2/3*1/2 + 1/3*1 = 2/3. if it "say it now", the chance of winning is 2/3 as well. so P(2, 1) = 2/3. As you can see, we basically have the following relationships: P(X, 0) =1 for any X. P(X, Y) = max( X/(X+Y), X/(X+Y) * P(X-1,Y) + Y/(X+Y) * P(X, Y-1)) and you can probably guess that P(X, Y) = X/(X+Y) so P(26, 26) = 1/2. That means calling out on first card or wait both gives you optimal strategy.always 1/2

### Analyst at J.P. Morgan was asked...

Apr 3, 2011
 puzzle 9 balls, identical in appearance. 8 are of same weight. 1 is heavier. Identify this heavier ball by using a scale twice.5 Answersput 9 balls in groups of 3.Split them into two sets of 4 and 1 individual ball. Place the two sets on each side of the balance. If the sets balance out, then the individual ball is the heavier one. If the sets don't balance, then take the heavier set and split it up into two sets of two and repeat.I believe that an explanation to the solution that Interview Candidate is suggesting can be found here: http://brainteaserbible.com/snooker-balls-weighing-scalesShow More ResponsesSplit the balls into groups of three, compare two of the groups. If the sets balance then the heavy ball is in the third group, otherwise it is in the group left out. Whichever it is, take two balls from that group to weigh. If they balance, you know it is the third ball, if they don't, you have the heavier ball.http://brainteaserbible.com/snooker-balls-weighing-scales

Jul 21, 2010
 If there was a drawer in front of me containing 4 socks, either black or white, and I know that I have a 50% probability of pulling out two white socks in a row, then what is the probability of pulling out two black socks?5 AnswersThere is zero probability of pulling two black socks.Assuming there are 2 whites and 2 blacks then getting 2 socks of the same colour is the sameAbove meant to say: Has the same probabilityShow More ResponsesYou can't assume there are two white socks and two black socks. This problem is an easy enumeration problem actually. From the problem, you have at least 2 white socks, otherwise it would be impossible to take out 2 in a row. So you have either 2 white socks, 3 or 4. Just compute the three prob of getting two white socks in a row for all case and see which is 50%. P(two w in a row| 2 white socks) = (2/4) * (1/3) = 1/6 => not two socks. P(two w in a row| 3 white socks) = (3/4) * (2/3) = 1/2 => this is our answer. Obviously, if there were 4 socks, we would have a 100% chance of getting two whites in a row. So as calculated, there must be 3 white socks.The probability is 0 as there is only 1 black sock. Solution: Assume there are k white socks and 4-k black ones in the drawer. => 0.5 = P(2 white socks in a row) = k/4 * (k-1)/3 => k^2 - k - 6 = 0 => k = 3 is the number of white socks. qed.

Jul 23, 2010