Quantitative Trader was asked...2 December 2010

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I loved this question and want to renew this debate. What do you guys think about my two approaches to solve it: 1) If we can only play this game once AND our goal is to maximize profit (as the question states). I agree with above that expected value of a coin is 50. Given that we bid 51 to win auction and pocket 24. Problem is we only win if coin is (0:50) which gives us new expected value of 25, and so we lose. We can deduct this way all the way to zero bid. 2) Nothing beats little Monte Carlo experiment. I created a matrix of 100X1000000. Where 100 is the number of possible bids given certain price. 1M is the number of random uniformly distributed prices 0-100. Calculated expected gain at each bid level 0 to 100. I wish I could post a MATLAB graph here. It looks as downward facing 1/2 of parabola with max value of 0 and min of -25. Results: best gain of 0 achieved at 0 bid, worst average gain of -25 is at 100 bid. Comments appreciated! Less

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i had the longest argument with a friend on this. you cannot get a positive expected value no matter what you bet. if you bid $50, then you can discount the EVs if the value of the coin is 51-100 since that'll be 0 (you don't win the auction). if you bid $50 and the coin's worth $50, you sell for $75make $25. but if the coin's worth $0 you lose $50. keep comparing the extremities and you will see in almost all cases you will be losing more than you make...that's the best i can explain it. i had to use a spreadsheet to prove this to my friend. in order to get an EV of 0, you'd need to change the multiplier to 2. which makes sense. if X is your bid, your profit is (X/2) *1.5 - X. Less

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The price is NOT necessarily uniformly distributed between 0 and 100, therefore 0 might not be the right answer Less

Quantitative Trader was asked...17 October 2010

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2/3

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Because you have 1/3 chance to get double head coin and you will surely get head, 1/3 chance to get single head coin and then 1/2 chance to get head. So the probability of choosing double head coin and get head is 1/3, while choosing single head coin and get head is 1/6. Then, given you get head after tossing, then chance that you chose double head coin is (1/3)/(1/3+1/6) = 2/3 Less

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2 heads on double headed coin, 1 head on the other, P(head is coming from double headed) = 2/3 Less

Quantitative Trader was asked...2 January 2010

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it is "Moser's circle problem" 1+nC2+nC4

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This is a famous example in mathematics; it's often used as a warning against naive generalization. Here are the answers for the first six natural numbers: (# points) : (# regions) 1 : 1 2 : 2 3 : 4 4 : 8 5 : 16 6 : 31 Yes, 31. You can see, e.g., Conway and Guy's "The Book of Numbers" for an account of this. Less

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mingda is correct

Quantitative Trader was asked...17 October 2010

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I 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. Less

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He gave a very good explanation. I'll decline to explain why prm is incredibly wrong. Less

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7 times.

Quantitative Trader was asked...10 December 2016

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1/2. Whether even or odd heads is ultimately determined by the nth flip, for any n. Probability even to odd or vice versa is n. Less

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( (2^99)+1) / (2^100) Even numbers include 0 heads to 100 heads. There exist 50% even complements and 50% odd complements from 1 heads to 100 heads.. So half our sample space. Now we do our zero case which is no heads and there is only 1 outcome. Less

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1/2. Let p_(e,n) be the probability that we have an even number of heads given n tosses of a coin. We have the recursion relation p_(e,n) = (prob to get heads on toss N AND we have odd number of heads after n-1 tosses) + (prob to get tails on toss N AND we have even number of heads after n-1 tosses) = 1/2 p(o, n-1) + 1/2 p(e,n-1). But p(o, n-1) = 1 - p(e, n-1), so p_(e,n) = 1/2 (1-p(e, n-1)) + 1/2 p_(e, n-1) = 1/2. Less

Quantitative Trader was asked...1 October 2015

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Let x be the expected value of the game. Then, we can see that the following recursion holds: x=0.5(30)+.25(-30)+.25(x) It follows that x=10, so I'd pay no more than $10.00 to go first. Less

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no more than 7.5?

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The question is not very clear : does the game continue until someone gets heads ? Less

Quantitative Trader was asked...2 February 2016

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The highest information gain is when you split all possible sequences into two equal sized groups, so the simple question "is the first flip heads?" yields the best chance. Asking if there are more heads than tails is not correct since it does not split the possibilities into two equal groups (one group contains the 50/50 split sequences) Less

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I was also asked this question..... Totally lost..

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Any meaningful question will split the sample space into two non-empty parts. If the sample space has size X ( = 2^100 ), let's say the two parts have sizes A and X - A. Then the probability of guessing correctly is (1/A)*(A/X) + (1/(X-A))*((X-A)/X) = 2/X = 1/2^99. So asking about the value of the fist toss is as good as anything. Less

Quantitative Trader was asked...2 January 2010

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At least two are tails, no HHHH neither THHH, HTHH, HHTH, HHHT. 2^4-5=11 possible scenarios. Three tails, HTTT, THTT, TTHT, TTTH = 4 cases P=4/11 Less

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its 4/11... it'd only be one half if they told you the first two coins were tails. Use bayes rule or actually write out all the possibilities Less

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can you show me the math behind it? i got P(B|A) x P(A) / P(B) = (1x 1/4) / (5/8)=2/5... Less

Quantitative Trader was asked...17 April 2014

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To turn this into a probability, note the conditioning: Rolled 1, 9 left: 4 ways (out of 36) 8: 5 ways 7: 6 ways 6: 5 ways 5: 4 ways 4: 3 ways Now each of the first rolls is 1/6 probability, hence you have total probability: P(10) = P(9 from 2 dice|1 from first) * P(1 from first) + ... Which is 1/6 * 27/36 = 1/8 Less

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The combinations is not 6*3 but 6*3+3*3=27, which will give the probability 1/8

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You will have the following possibilities with 3! permutations: (1, 3, 6) ; (1, 4, 5) ; (2, 3, 5). Besides, there are other 3 vectors resulting into 10 but with only 3 permutations each: (2, 2, 6) ; (2, 4, 4) ; (3, 3, 4). That way, we have 6⋅3+3⋅3 = 27 possible combinations. This divided by 6^3, finally gives us the Probability{Sum(a,b,c) = 10}= 1/8. # Less

Quantitative Trader was asked...9 July 2015

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I think some of you guys have too many steps for mental math. Basically 12 inches diameter is 6 inches radius which is 1.5x the 8 inch diameter bowling ball (4 inch radius). The volume formula is cubed so 1.5 * 1.5 = 2.25 * 1.5 = 3.375 (it might be easier in your head if you do 15*15*15). So the 12 inch bowling ball is 3.375x heaver and they made it easy by making the original ball 10. 3.375 * 10 = 33.75. Less

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It's (12/8)^3*10=135/4

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I get 33.75 lbs as well: (12/2)^3 = 216. (8/2)^3 = 64. Answer is the weight of the small ball, multiplied by the ratio of the volumes of the two balls: (10 lbs) * (216 / 64) = 33.75 lbs. NOTE: Because we're only using a ratio of volumes, the (4/3)*pi (constant in the ratio of a sphere calculation) can be ignored because it will cancel out since it's on both the top and bottom of the volume ratio. Less