Fragen in Vorstellungsgesprächen für Trader in Region Yung Shue Wan, Hongkong | Glassdoor.de

# Fragen in Vorstellungsgesprächen für Trader in Yung Shue Wan, Hongkong

65

Fragen aus Vorstellungsgesprächen für Trader, von Bewerbern geteilt

## Top Vorstellungsgespräch-Fragen

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17. Okt. 2010

### Ein Bewerber für eine Stelle als Quantitative Trader bei Jane Street wurde gefragt...

17. Okt. 2010
 There are 25 horses. Each time you can race 5 horses together. Now you need to pick the top three horses among them. How many races do you need to conduct? 13 Antworten7 times.after the 5th race the top five will be established; so, after the 6th the top three could be established. Answer is 6.doesn't that assume that 1st in one race is faster than 2nd/3rd/4th/5th in all the other races? can we be sure of that? 1st round: 25 horses and 5 heats => drop bottom 2 in each race 2nd round: 15 horses and 3 heats => drop bottom 2 again 3rd round: 9 horses and 2 heats => drop bottom 2 from large heat and 1 from the other 4th round: 6 horses and 2 heats; first heat has 5 and drop bottom 2 then race again with the 6th horse included 12 heats in total but there's probably a more efficient wayMehr Antworten anzeigenI think it is 7. Kannp, the top five are not established after 5 because the second in one race might be better than the first in another race. Have five races of five each and keep the top 3 in each races. Then, take each of the winners and race them against eachother. The two bottom and the 4 who lost to them are discarded. The two who lost to third place is discarded. And, the one who got 3rd in the race with the horse who gets second in the 6th race is discarded. Now, there are six horses left. Race all but the horse who won twice and keep the top two, combined with the horse who sat out. Now you are done in 7 races.you're over thinking this...... run 5 heats with 5 horses in each one. Using a stopwatch, you take the 3 fastest........ but if you don't have a stop watch - anonymous is wrong... marty is right - anonymous you can't take the winner of each race, because you need the top 3. if a 2nd place from one race is faster than the other 4 winners, you don't have it right. 1st time thru 5 horses by 5 heats drop bottom 2 = 15 left 2nd time thru 5 horses by 3 heats drop bottom 2 = 9 left 3rd time thru one race with 5, other with 4. keep top 3 of each race = 6 left now run any 5 horses, keep top 3 and for last one, run remaining 4,drop last. 12 heats total71) 5 heats take the first 3 =15 horses left 2) run 3rd against each other, take the first, same thing for 2nd and for the 1st take the best 3 = 3 heats 5 left 3) final heat 9 totalHe gave a very good explanation. I'll decline to explain why prm is incredibly wrong.7 is correct in assumption that each round ends up with a clear 1-st, 2-nd and 3-rd places. If horses can come simultaneously - than we need 11 rounds (When we have 9 horses left - we set up a round for first 5 and set up a round for the last 4 and a horse that took 3-rd place in the previous round. This leaves us 5 horses to race in the final 11th round).I got stuck with his explanation but once I understood it, it seems very brilliant. Let me put it in this way: 5 groups, each of 5 horses -> race (heat) take the 1st from each group and race (semi-final) discard the 4th and 5th one and all other horses which lost to them in the heat race keep the 3rd one but discard all horses lost to it in the heat race keep the 2nd one but discard the 3,4,5-th horses who raced against it in the heat race keep the 1st one but discard the 4,5-th horses who raced against it in the heat race Now there are 6 horses, namely the 1st from semi-final and the two just following him in heat the 2nd from semi-final and the one just following him in heat the 3rd from semi-final the 1st from semi-final is champion. ignore it and race all other 5 horses in final so 7 in total definitely Russ considers more cases but I am not interested in making it too complicated7Hey guys, revicing the things here. Can you guys be more explicit on the 7 figure ? Even with the explainations it’s not clear for me. You say run the 5 winners of the 1st round and discard the last 2, but what if the 5 winners of the first round are actually the 5 firsts ?Forget it I read with my eyes closed, I was lookin for top 5 and it’s top 3

### Ein Bewerber für eine Stelle als Trader Intern bei Jane Street wurde gefragt...

16. Aug. 2013
 Toss a fair coin between two people. If it end with HHT, A wins the game. If it ends with HTT, B wins the game. Whats the probability of A wining the game. 9 Antworten2/3if order isn't important, than you have for possible outcomes. HHH, TTT, HTT, TTH. You can see immediately there's 1 chance out of 4 for A to win. If order is important, there are 8 possibilities. 1 chance out of 8 for A to win.1/4 srsly? http://www.glassdoor.com/Interview/Flip-a-coin-until-either-HHT-or-HTT-appears-Is-one-more-likely-to-appear-first-If-so-which-one-and-with-what-probabili-QTN_46824.htmMehr Antworten anzeigenOnce the second person flips their first H, they cannot ever go back to the state where they must roll an H to move on. So after P1 and P2 have both flipped their first H, P2 needs 2 flips to go his way, P1 needs 3, thus has half of the probabilty of A to win.Prob(HHT) * 0.5 = Prob(HTT), Prob(HHT) + Prob(HTT) = 1, ===> Prob(HHT)=2/3Pa=2/3, Pb=1/3Pa = 3/8Let p be the probability player 1 wins. Nothing matters until you roll a H, so let's fast forward to there. If another H is rolled, then player 1 wins automatically so that's (1/2). If a T is rolled, then if T rolls again you lose but if H rolls now then you reset to the past situation. So p = 1/2 + 1/4p so p = 2/3.Clearly A has the upper hand, because A retart from 2 if 3 fails, but B restarts from 1 if 3 fails

### Ein Bewerber für eine Stelle als Trader bei Jane Street wurde gefragt...

8. Juni 2011
 We play a game. Let’s guess the sum of two dices. Who make a guess that is closer to the outcome is the winner. What’s your strategy?5 AntwortenI choose 7 because if you pick say 6, the closest digit with lower value than 7 (meaning greater range of winning numbers), I get from [7,12], but you only get from [2,6]. And if I choose 7 and you choose say 8, the closest digit with higher value than 8, I get from [2,7], but you only get from [8,12]. Therefore, if I pick 7 I always have a greater chance of being closer to the correct number.the EV of rolling a dice is 3.5, so if you roll a dice twice the number with a greater chance is 7I have you pick first and bet one step closer to 7 . eg you say 4 I say 5, unless you say 7 then I say 7.Mehr Antworten anzeigenBen's answer was what I came up. But Pierreou's one is an even better strategy.Pierreou's answer assumes that your opponent does not have the concept of expectation. It is so easy to win if your opponent is ignorant......So I suggest you take the initiative to bet a guess on 7 then the probability of winning is >=6/11 no matter what your opponent's guess is

8. Juni 2011

### Ein Bewerber für eine Stelle als Quantitative Trader bei Jane Street wurde gefragt...

17. Okt. 2010
 It is noon. The hour hand and minute hand overlap. When is the next time these two hands overlap again? 4 Antworten13:05:300/11 You need to be very accurate into fractions of seconds1:05:5/1113:05:5/11 is correct. there is a simple way to think of it. during 12 hours, 11 times of overlapping. so each duration of overlapping is 12/11=1 and 1/11 hour. So the first time of overlapping after noon is 13:05:5/11Mehr Antworten anzeigenHere is the longer way. The minute hand of a clock travels at a rate of dM/dt=(360 degrees)/(60 minutes)=6 degrees/minute. Note that every hour of a clock spans (360/12 degrees)=30 degrees. The hour hand of a clock travels at a rate of dH/dt=(30 degrees)/(60 minutes)=(1/2) degrees/minute. Now, in your head, imagine the minute and hour hand at 12:00. The minute hand will travel faster than the hour hand and wrap around the clock once. When the minute hand is at 0 minutes or 0 degrees again, the hour hand is at 5 minutes or 1 o'clock or 30 degrees. Now we determine the point at which the minute hand catches-up to the hour hand. To do this, we set the two "degree" functions equal to each other. For 0<=t<=60, M(t)=6t and H(t)=30+(1/2)t. Setting M(t)=H(t), we get that t=60/11 minutes= 5 + 5/11 minutes. Now, we find out how many seconds, approximately, are in 5/11 minutes. There are 60 seconds in a minute, and hence 5/11 minutes= (60*5)/11 seconds= 300/11 seconds= 27 + 3/11 seconds. Therefore, the time at which the minute and hour hand meet is 1:05:27.

### Ein Bewerber für eine Stelle als Assistant Trader bei Jane Street wurde gefragt...

22. Okt. 2012
 If you have four coins and I have four. We both throw the four and if your four sides equal to mine, I will give you 2 dollar and otherwise you give me 1. Will you do it?4 AntwortenThe answer is no, as the expected value of the game is (1/256 + 1/16 + 9/64 + 1/16 + 1/256) * 2 - (1 - (1/256 + 1/16 + 9/64 + 1/16 + 1/256)) = -1/40 0.0625 0.0625 0.9375 -0.8125 -0.05078125 1 0.25 0.25 0.75 -0.25 -0.0625 2 0.375 0.375 0.625 0.125 0.046875 3 0.25 0.25 0.75 -0.25 -0.0625 4 0.0625 0.0625 0.9375 -0.8125 -0.05078125 Roll P(Roll) P(WIN|Roll) P(LOSE|Roll) Expected Gain Given Roll Expected Gain Expected Return: -0.1796875In this game of you four tosses, there are five possible combination {HHHH, HHHT, HHTT, HTTT, TTTT}. The probability of my four tosses matching any particular combination is 1 out of 5 (20%). So you shouldn't engage in any bet unless there's atleast a 5-for-1 payout. In the question there's only a 2-for-1 payout so you shouldn't engage in the bet.Mehr Antworten anzeigenoutput probability 0000 (1) --> 1/16 0001 (4) --> 1/4 0011 (6) --> 3/8 0111 (4) --> 1/4 1111 (1) --> 1/16 \sum \sqr{prob} = 2 / 16 / 16 + 2 / 16 + 9 / 64 = 1/128+1/8+9/64 = 0.27 (+2 dollars) < 1/3 so No.

### Ein Bewerber für eine Stelle als Quantitative Trader bei Jane Street wurde gefragt...

17. Feb. 2011
 hh ht ht tt, one h, probability of the other is h?3 Antworten1/2, not to use bayes rule?? Dont understand the question...Assuming your question is 'flipped two coins, one is a head what is the probability the other is a head' then it is 1/2 by independence. If the question is 'flipped a coin twice, one of the outcomes was a head what is the probability the other outcome was a head' then the required probability is P(HH | H) = 1/3

### Ein Bewerber für eine Stelle als Quantitative Trader bei Jane Street wurde gefragt...

17. Feb. 2011
 4 coins, at least 2 T?2 Antworten(16-1-4)/1611/16; or 1-1/4-1/16;

### Ein Bewerber für eine Stelle als Trader bei Jane Street wurde gefragt...

27. Sept. 2013
 15% of 3673 Antworten55.05 It was not difficult, but I was very nervous that it took me some time to figure it out.10% + 5% (half of 10%) 36.7 + 18.35 = 55.05360*0.15+7*0.15=54+1.05=55.05
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