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Oct 5, 2009
 Here is an example of a brainteaser during the interview: You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How will the coins end up being divided, assuming all the pirates are rational and want to end up alive?15 Answers54321If the pirates are RATIONAL- then they would all agree with taking just 20 coins a peice and splitting it evenly. That is the short and sweet answer. Unfortunately these nerd interviewers may prefer a pointlessly long and drawn out answer to demonstrate your irrelevant mathematical reasoning skills/ability In which case you may answer with the following Give one coin each to the 2 lowest ranked pirates. and split the remaining between the top 3( Maybe 32 for the top pirate and 33 for pirate 2 and 3 ). The bottom 2 will certainly vote against you- but now atleast you are increasing your odds of the other 2 pirates agreeing with the plan. ( which are the 2 minimum votes needed for pirate one to survive) Obvious explanation If there are 5 pirates, one comes up with the plan and the other 4 vote. If fewer then half agree with the top pirate- he gets killed. That means if less than 2 agree he would get killed. He need at the minimum 2 pirates to agree with him to live.the answer isnt that obviouse.....you give the 3rd and 1st one coin and keep 98....start at the beginning. if theres 2 pirates, 100 for 2, 0 for 1 (one doesnt like this) If there is three 1 will be happy with just a single coin, b/c he does not want it to go down to 2. If there is 4 pirates, 2 will be happy with a single coin, b/c he does not want it to get down to 3 pirates where he will receive 0. So he gets 1 and 4 gets 99. At 5 it changes a bit. Here 1 and 3 will be happy with single coins b/c if it goes down to 4 they will receive 0 coins. So 5 takes 98, and 1 and 3 take 1 one eachShow More ResponsesIf there are 4 pirates and you give 1 coin to 2 and 99 to yourself....surely 1 and 3 will vote against you and it will go down to 3 pirates??0 to the lowest 2. 1/3*100 to the three others. half would disagree but the top pirate would live.highest rank keeps 98, two get 1 each, two receive nothing. now it's just up to the first two to be happy with 1 coin each, but then, if they were not, the other two would be more happy with 1 instead of 0 coins. one could argue that if all 4 are rational, they would demand 25 each or be unhappy alltogether.lol i hope that you are kidding....otherwise it's better to find a manual-farmer job.... btw sean is rightsean is not right. The top pirate DOES NOT vote. If there are two pirates and the top decides to take the 100 coins for himself, the other one will vote against and the top pirate will be killed98, 0, 1, 0 ,1This is just a classic game theory question and you have to work backwards through it starting with the base case to understand the pirates motivation. If you have two pirates, p1 and p2, and if p1 is the final voter then he will always vote negatively unless p2 gives him all $100 since he can kill him anyways and take the whole loot. This means p2 is in a compromised position and does not want the game to go down to 2 pirates and will take any value greater than 0 from any other pirate, or will vote yes if he receives at least$1. When p3 is introduced, he knows p2 will need at least $1 to vote for the plan therefore he keeps$99 and gives away the last dollar to p2. This means p3 is in a dominant position and will vote down any plan that grants him less than $99. When p4 is introduced then he needs two of the three voters to vote for his plan. Granted p1 will decline unless he receives all of it and p3 will decline unless he receives at least$99 of it then he will give p3 exactly that and p2 $1 otherwise he is killed. p4 is in a compromised position so he will accept any offer where he receives something greater than 0. When p5 is introduced he knows p4 and p2 are screwed and the maximum they can earn if it bypasses him is$1. Therefore granting them each that money will guarantee their vote leaving the remaining $98 for himself and half the votes are positive thus he is not killed and gets to keep$98. So the distribution for p5, p4, p3, p2, p1 should be 98, 1, 0, 1, 0.If we assume that once the top pirate is killed the next one takes his place and decides how the money is allocated we can solve it like this: Five Pirates: || 34 || 33 | 33 | 0 | 0 | - pirates 4 and 3 make more than the 20 they should earn if the money is distributed equally and they should vote to pass while 2 and one vote against. Split 50/50 it should pass. BUT - could pirates 4 and 3 make more money if they voted against and then split it up among fewer people? Four Pirates: || 34 || 33 | 33 | 0 | - Again, everyone gets 33 and we assume pirates 3 and 2 vote in favor, outweighing pirate1. BUT - pirate 3 sees an opportunity Three Pirates: || 50 || 50 | 0 | - with only one vote to sway in order to get 50%, pirate 3 only has to give pirate 2 half of the total. So at best, pirate 3 will vote NO unless he gets his 50 up front. Pirate 4 will vote NO unless he gets his 34 up front. In order to survive, pirate 5 should pay accordingly to those pirates and take home a meager 16. Though 16<20, the other greedy pirates will always vote NO unless they get the amount they know they can receive. If Pirate 5 wants to stay alive that's what he'll do. TL;DR Pirate 5 - 16 Pirate 4 - 34 Pirate 3 - 50 Pirate 2 - 0 Pirate 1 - 0Wait, that's wrong I just realized. The buck stops with pirate 3. He can get 50 and he will, so pay him 50 first. In the Four pirate scenario, Pirate 4 will also have to pay out 50 to pirate 3 or he'll be killed, so the most he can allot to himself is 17, which is all Pirate 5 should give him. New Tally: Pirate 5 - 33 Pirate 4 - 17 Pirate 3 - 50 Pirate 2 - 0 Pirate 1 - 0 Pirate 2 is always going to be looking for that $50 payday in the Three Pirate scenario, so it's important to neutralize his vote immediately by having 4 and 3 agree with 5 in the first place.If the top pirate gets killed, the rest of the pirates gets to divide the 100 coins which is 25 coins per person. but the higher ranking pirates 3,4 will probably get more so the best case scenario for pirates 1,2 is 25 coins per person. If the top pirates give 1 and 2 26 coins per person, they will vote for him and he gets to keep the rest of the 48 coinsShow More ResponsesWe have the following fundamental assumption: Any pirate will not vote down as long as his payout is greater or equal to that of what his payout would be with 1 fewer pirate. As such, we product the following chart, working up from 1 pirate: (Payout,Expected of 1 less pirate) , we require that (Payout >= Expected of 1 less pirate) to get their vote # pirates | Rank: 1 2 3 4 5 1 (100,) 2 (100,100)(0,0) 3 (0,100) (1,0) (99,0) 4 (1,0) (2,1) (0,99) (97,0) 5 (1,1) (0,2) (1,0) (0,97) (98,0)20 each ### Assistant Trader at Jane Street was asked... Apr 7, 2011  what is the expected number of flips of a coin to simulate a 6 sided die.12 AnswersE(x) = .75*3 + .25*E(X+1). From there it simplifies to E(X) = 10/3E(x)=3+0.25E(x), so E(x)=4.Hi Can you please explain this in detail ? I dont seem to follow how did u guys do this ? ThanksShow More Responses11/3. E(x) = .75*3 + .25*E(X+2).The probability of a six sided die is 1/6, while the probability of a coin flip is 1/2, thus, in order to simulate a six sided die the coin must have the same probability, so 1/2 has to become 1/6. From there, you can write the equation1/2^x=1/6, or to make it simpler, 2^x=6. Without using guess and check, you can rewrite the equation to log base 2 6=x (or log2(6)=x), and from there solve: log(6)/log(2), which comes out to 2.58496, or 2.585 (2^2.585=6.000). So, to answer your question, the coin would have to flip 2.585 in order to simulate the flip of a coin. x=2.5853 flips. The coin gives 1 out of 2 possibilities per flip. 3 flips will give you 6 possible combinations like the 6 sided die (1 heads, 2 tails, 2 heads, 1 tail, etc.).It is easier to look at it as almost 3 flips set up as a binary number. It would be perfect if you were representing an 8 sided die, but since you want only 6 possible results and 2 flips is too little and 3 flips is too big you have to adjust your fomula as such: 0=tails 1=heads 000 = Null this means you scratch these results and flip again three more times. 001 = 1 010 = 2 011 = 3 100 =4 101 = 5 110 = 6 111 = is another Null result Of course the Nulls can be made to be anything as are the numbers it is all relative. This give you a 1 in 6 chance everytime.Wsc is on the right track. You have to flip 3 coins at least, but the probability of a null result is 1/4 every time, so you would need to flip 3 more coins. This is a geometric sequence 3+3*1/4+3*(1/4)^2 etc, so the answer is 3/(1-1/4) = 4 expected coin flips. 4The correct answer is 11/3, as written by "Someone" on July 9, 2011. Both William's and P's answers are close, but not quite accurate, because they assumed that it takes 3 flips to discard a null result. Instead, it only takes two flips to discard a null result (HHT or HHH) since we don't have to flip the third coin if the first two are already heads (HH). That's why the answer is slightly less than 4. Indeed, note that we have a 3/4 chance of being able to get one of the 6 good results (HTH, HTT, THH, THT, TTH, TTT) that yields a random number from 1 to 6, and a 1/4 chance of having to reflip the coins (HHH or HHT). The key is that for those two results to be discarded, we only needed to flip 2 coins to determine that! There's no need to flip a 3rd coin if the first two were already HH (it'd be a waste of a throw!). Hence, 3/4 of the time, the number of flips required is three and 1/4 of the time, the expected number of flips required is 2+E[X]: E[X] = 3/4*3 + 1/4*(2+E[X]). Alternatively, we can flip the coin twice, and then note that 3/4 of the time (for HT,TH,or TT), we only flip one more coin, and 1/4 of the time (for HH), we have to re-flip, so E[X] = 2 + (3/4*1+1/4*E[X]) Either way, we have E[X] = 11/4 + 1/4*E[X], which yields the desired result, E(x) = 11/3.4.5It's 11/3. Say you want HHH and TTT to be your rolls to start over so that the last roll is 50-50 to be H or T and you'll only have to roll twice more since there will be no bias. Roll your first roll, it's H or T. After that there is a probability of 1/4 that you'll have to 2 times again. So there a 3/4 chance you won't have to roll again. So using the equation 1/p to find EN we get 1+2(4/3)=1+8/3=11/3You can't flip a die because you only have a coin. DUH ### Assistant Trader Summer Internship at Jane Street was asked... Feb 26, 2010  Flip a coin until either HHT or HTT appears. Is one more likely to appear first? If so, which one and with what probability?13 AnswersLet A be the event that HTT comes before HHT. P{A} = P{A|H}P{H} + P{A|T}P{T} = .5P{A|H} + .5P{A|T} P{A|T} = P{A} therefore, P{A|H} = P{A|T} P{A|H} = P{A|HH}P{H} + P{A|HT}P{T} = (0)(.5) + P{A|HT}(.5) Therefore, 2P{A|H} = P{A|HT} P{A|HT} = P{A|HTT}P{T} + P{A|HTH}P{H} = (1)(.5) + P{A|H}(.5) 2P{A|H} = .5 + P{A|H}(.5) P{A|H} = 1/3 and P{A|H} = P{A}, therefore, P{A} = 1/3 So, HHT is more likely to appear first and it appears first 2/3 of the time.P{A|H} = P{A|HH}P{H} + P{A|HT}P{T} = (0)(.5) + P{A|HT}(.5) Need help - - why is P{A|HH} = 0 ?P(A|HH) = 0 because after a sequence of consecutive heads, you can no longer achieve HTT. The moment you get a tail, you will have the sequence HHT. This the reason HHT is more likely to occur first than HTT.Show More ResponsesP(A|HH) = 0 because after a sequence of consecutive heads, you can no longer achieve HTT. The moment you get a tail, you will have the sequence HHT. This the reason HHT is more likely to occur first than HTT.HHT is more likely to appear first than HTT. The probability of HHT appearing first is 2/3 and thus the probability of HTT appearing first is 1/3. Indeed, both sequences need H first. Once H appeared, probability of HHT is 1/2 (b/c all you need is one H), and probability of HTT is 1/4 (b/c you need TT). Thus HHT is twice is likely to appear first. So, if the probability that HTT appears first is x, then the probability that HHT appears first is 2x. Since these are disjoint and together exhaust the whole probability space, x+2x=1. Therefore x=1/3.You guys seem to be mixing order being relevant and order being irrelevant. If order is relevant (meaning HHT is not the same as HTH) then this has a 1/8 of occuring in the first 3 tosses. Also HTT has a 1/8 chance of occurring in the first 3 tosses, making them equally likely. Now, if order is not relevant. (so HHT and THH are the same), then this has a (3 choose 2) * (1/8) probability of happening in the first 3 tosses. The same goes for HTT (which would be the same as THT etc and others) so this has a (3 choose 2) * 1/8 probability of happening in the first 3 tosses as well. Either way they come out to being equally likely, please comment on my mistake if I am doing something wrong.Ending up with HHT more likely (with probabilty 2/3).HHT is more likely (2/3) probability. People with wrong answers: Did you not Monte Carlo this? It takes 5 minutes to write a program, and you can then easily see that 2/3 is correct empirically.I don't get it. Shouldn't P{A|HH} = P{A} in the same sense that P{A|HTH} = P{A} from both HH and HTH we have get the first H from HTT and so it should be P{A|HH} = P{A|HTH} = P{A} Am i wrong?sorry, i meant: I don't get it. Shouldn't P{A|HH} = P{A|H} in the same sense that P{A|HTH} = P{A|H} from both HH and HTH we have get the first H from HTT and so it should be P{A|HH} = P{A|HTH} = P{A|H} Am i wrong?Above link is the best solution I have seen for this problem http://dicedcoins.wordpress.com/2012/07/19/flip-hhh-before-htt/Apologies, Below*Here's my answer. Let x = probability of winning after no heads (or a tail). y=probability after just one heads. z=probability after two heads. w=probability after HT. Thus x=(1/2)x+(1/2)y, y=(1/2)z+(1/2)w, z=1/2 + (1/2)z, w=(1/2)y. Therefore, z=1, y=2/3, w=1/3, x=2/3. We wanted x at the beginning, so it is 2/3 that HHT comes up first. ### Assistant Trader at Jane Street was asked... Apr 7, 2011  Simulate a 6 sided die with a coin.13 Answerssplit into two sets {1,2,3,4} {5,6} with one assigned to heads and the others tails. If in the {5,6} set flip again. If in the {1,2,3,4} set assign {1,2} to heads and {3,4} to tails. then split again.Toss three coins, we get 8 outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT, for the first two, we toss again until neither HHT or HHT appears, for the remaining 6, we assign 1~6 individually.I propose another solution. Use 4 coins. Flip 2 coins together and separately another 2 coins together. Sample space for a 2 coin flip is: HH, TT, HT, TH (assume that HT=TH) For the two 2-coin flips assign value of dice, for example (column 1: 2 coin flip, column 2 other 2-coin flip) HH HH => 1 HH HT/TH => 2 HH TT => 3 TT TT => 4 TT HT=> 5 HT HT=> 6 with TH=HTShow More ResponsesWith 1 coin, flip three times and HHH=1,HHT=2, HTH=3,HTT=4, THH=5, THT=6. If the first twice is TT, then abandon the first T and start with the second T.The candidate and William have problems. I think Ey's solution works. For the candidate's solution, you will get way more 5s and 6s, since fully 1/2 of the results will be either a 5 or a 6. Similarly, for William's solution, if you toss out results starting with HH, then you end up with ANY result that starts with a tail, and only half the results that start with a H. That means you will end up with more 3s, 4s, 5s, and 6s than you should. I think by simply changing what you throw out you could fix William's solution. If you threw out all HHH and all TTT you would be closer, but a solution where you don't toss out any data is still better.Nope. I'm a moron. Can't edit earilier post, but Ey, sorry, I gave you too much credit. Can't count HT and TH as the same thing because you get too many of them. No idea why I didn't see that 10 minutes ago. You can't map 4 coin results to three numbers fairly. You will end up with too many 2s and 5s and way too many 6s with your solution. Now we have no valid answers. I think someone proved it can't be done elsewhere. Interesting problem though.Toss 1: Heads=even, start with 2, tails=odd, start with 1 Toss 2: Heads=0, tails=2, add to total Toss 3: Heads=0, tails=2, add to totalIt cannot be done. The odds of any set of coin flips is always a power of 2, and there is no power of 2 for which 6 is a factor, because 6 factors to 2 * 3 and both are prime numbers.Eric, you're right, I thought my solution worked as well. My idea was that we do 2 independent 2-coin flips. Each 2 coin flip has the following possible outcome: HH, TT, HT(or TH, same thing). When combining the outcomes of the 2 independent 2 coin flips this is what we get: (1) HH (2) HH (1) HH (2) HT (1) HH (2) TT (1) TT (2) HH (1) TT (2) HT (1) TT (2) TT (1) HT (2) HT (1) HT (2) HH (1) HT (2) TT we get 9 possible outcomes, even if we set for example outcomes (1) HH (2) TT to be equivalent to (1) TT (2) HH, and similarly with others to get 6 outcomes, it won't work...HHH, TTT, start the process again. Assign the rest values of dice. HHT = 1 HTH = 2 HTT = 3 THH = 4 THT = 5 TTH = 6Geez, they should really allow for nested responses for better discussions. I agree that with the approach first stated by William (except he had a typo, "HHT or HHT appears" should read "HHH or HHT appears" he was just referring to first two (or any two of the eight combination that one chooses not to assign values to). The way that "Do three flips" describes is clearer. Other approaches are problematic in that the chances of getting 1-6 are disproportionate leading to an unfair dice (for reasons that others stated). "Binary Thinkers" approach is also incorrect as for similar reasons. there 8 possible accounts (3 flips 2^3 = 8). Although it's a unique way to account for all the numbers 1 through 6... 3's are and 4's are precisely twice as likely to occur (2/8 or 1/4 chance) vs the other numbers (1/8 chance).You may ask for a probability of THHHHH if we get a consequent combination with 5 heads and one tail. Here are the 6 possible combinations: THHHHH HTHHHH HHTHHH HHHTHH HHHHTH HHHHHT so the probability of each is 1/6Not a precise solution, but good enough for many purposes: Flip the coin a large number of times, interpret the result as number in base 2, and take it modulo 6. Some of the six numbers will be slightly disadvantaged, but you can make that difference arbitrarily small by increasing the number of coin flips. ### Assistant Trader at Jane Street was asked... Oct 4, 2010  You have a box filled with cash. Cash value is uniformly randomly distributed from 1 to 1000. You are trying to win the box in an auction: you win the box if you bid at least the value of the cash in the box; you win nothing if you bid less (but you lose nothing). If you win the box, you can resell it for 150% of its value. How much should you bid to maximize the expected value of your profit (resale of box minus bid)?13 AnswersMy bad, I meant "0 to 1000" not "1 to 1000". Not that it affects the answer (hint hint).the expectation is always negative?bid 0. The expectation is negative.Show More Responsescan someone explain why it is the case mathematically?? I thought your expected payoff is (0+1000)/2 x 1.5=750.. so if you bid 500. you will get expected 250 profit...Anon, If you bid 500 and it is worth 1, you only get paid$2 so you lose 498. \$0 is correct. Generalize this for 1,2,5,10 maybe and find the EV of all those, or just 1,2,10, and then you will be able to "guess" that it isn't getting more and more negative and you should bid 0you should bet 1, that is the minimum in the box, so you will win 0 or .5if you bet one you'll never win. if you bet 500, you're missing out if value(box) >500 and <750 since those are +EV too. believe the answer is 749.The answer is bid 0 dollars. Soln: If you bid X dollar, you get the box if if it contains Y dollars, where Y < X. Or you nothing, which doesn't hurt anyway. Now think of possible values of Y in PAIRS, (0, X), (1, X-1), (2, X-2)... in each pair (a, b), expected payoff is P(Y being a or b) * E(payoff given Y = a or b). but the second term E(payoff given Y = a or b) = 1/2 (3/2 a - X) + 1/2 (3/2 b - X) = 3/4 (a + b) - X = (-1/4) X, which is always negative. If we sum over all possible pairs, we should get a negative number. Ofocz you need to handle the edge case, where X is small and the pairing does not really work.I think that since we are dealing with expected profit, we should ignore the case when n (our bid) is less than X (cash in the box) as nothing happens in this case. So we create a new distribution where X is uniformly distributed on [0, n]. Thus, E(Profit) = E(1.5X - n) = 1.5E(X) - n = 1.5*n*(n+1)/(2*(n+1)) - n = 0.75n - n = -0.25n <= 0. So the expected profit is maximised (=0) if n=0.Sorry, guys. I seem to have made a mistake in my solution above. Let X = cash value, n = our bid price. Then Profit = 1.5X - n. We want to find E(1.5X - n) = 1.5*E(X) - E(n). Now in order to find E(X), we notice that that X is uniformly distributed. So every outcome has a probability 1/1001. Also we only need to sum up over the values from 0 to n. Hence, E(X) = n(n+1)/(2*1001). For E(n), we see that we will pay the amount n with probability (n+1)/1001 and 0 otherwise. So E(n) = n(n+1)/1001. Therefore, E(Profit) = 1.5*n(n+1)/(2*1001) - n(n+1)/1001 = -0.25*n(n+1)/1001<=0. In order to maximize profit, we should bid zero.The answer is 0 if you ask me and here is the reason. let V be the value of the box and let x be your bid then you profit P is P=(1.5V-x)1_{V<=x}. we take expected value of this and use conditional expectation to get E(P)=E(1.5v-x|v<=x)p(vThe answer is 2 with EV 1/3996. Let x, t, w be your pick, the cost, and your winnings. If x 1000 we should just pick 1000. Otherwise 1 x we have E(w) = 1/999 int (1.5t - x)dt from 1 to x, which equals 1/999(3/4 x^2 - x^2 - 3/4 + x)=1/3996(-x^2+4x-3) = 1/3996(1-(x - 2)^2) <= 1/3996 with equality iff x = 2.Answer is 2. The key is that the amount of cash X is uniform on (1,100) rather than (0,100). If we bid y money, then are expected value of profit P is E[P | bid y]=(\int_1^{y}(1.5-y)dx)/99=-(y^2-4y+3)/(4*99). Here the limits of integration are because P will be 0 if X is smaller than y, and we will get profit 1.5x-y if X=x and x<=y. Solving the maximum of this function gives y=1, with an expected profit of 1/(4*99)=0.0025. If the value was in fact from (0,100), then our expected value for any y would just be -y^2/(4*99) so we wouldn't want to play that.

Feb 24, 2011

Mar 2, 2010

Feb 8, 2011
 10 lightbulbs in a row, on or off, no two adjacent lightbulbs can be on. how many combinations can we have? you may use a piece of paper8 Answersthe answer is the answer for 9 lightbulbs + the answer for 8 lightbulbsexpress as a recurrence f[n] = f[n-1] + f[n-2]. so it's a fibonnaci sequence. [think why?] base cases: f = 2, f = 3, ... f = 144er...why? is it a fibbonaci sequenceShow More Responsesok terrible grammar. why is it a fibbonaci sequence?start small. 1 bulb = it can be on or off. so that's 2. 2 bulbs = bulb1 can be on, bulb2 can be on, or they're both off. that's 3. 3 bulbs = bulb1 on, bulb2 on, bulb3 on, bulb1 + bulb3, or all off. that's 5. 4 bulbs = trust me that it's 8... 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.Awesome solution.At most 5 of them are on. In that case, 4 off-lightbulb are in between every two, so we are left with 1 off-lightbulb, which could be placed 6 places; Similarly, if there are 4 lights on, 3 off will be put in between and we are left with 3 off bulbs. They can be put at 5 places (namely, left to the first on bulb, between the 1st and the 2nd on bulb, etc), and this is reduced to the question: X1+...+X_5=3, how many non-negative solutions? The anwser is C_{3+5-1}^3. Then we induct on the number of bulbs that are on, from 5 to 0, and sum up all the possiblities, the result is: C_6^1+C_7^3+C_8^5+C_8^7+C_10^9+1 (1 is the case that all bulbs are off) =144willie's answer is correct. Basically, suppose for n bulbs, there are A(n) ways. consider the last bulb: if that one is on, then the number of combinations is A(n-2); if that one is off, then it's A(n-1), so A(n) = A(n-2) + A(n-1).