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If you can do a little multiplication and adding in your head this is not too difficult. First get a rough estimate about where you need to be. Try 40 x 40 = 1600. Good but you can do better. Try 44 x 44 = (44x40) + (44x4) = 1760 + 176 = 1936. Now try 45 x 44. All this is saying is we need 45 sets of 44. So take 1936 and add 44 to it. The result is 1980. Now try 45 x 45. All this is saying is that we need 45 sets of 45. Since we can reverse numbers in multiplication we already have 44 sets of 45 which is 1980. Just add 45 to 1980 and the result is 2025. So we know the square root is somewhere between 44 and 45. The range from 1936 to 2025 is 89. 2000 - 1936 = 64. So what is 64/89? Round it up so we get a fraction like 70/100 which is .7. Therefore, 44.7 is a good estimate. The actual root is 44.72135954.... Hope that helps. Moins

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The square root of 2000 is a number that when multiplied by itself (once) is equal 2000 Moins

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Agreed. I double counted. 3.5*.4 + 7.5*.6 3.5 was given 7.5 =(5+6+7+8+9+10)/6 Moins

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"You decide on taking the dollar value on the card or $3.50 *before* seeing the number on the card." 2nd game: Obviously, the game price should be no less than 3.5, otherwise there is an opportunity for arbitrage. Now that the price is higher than 3.5, it makes no sense for a rational player to choose just take 3.5 dollar and exist the game. The player should always choose to bet. And the expectation for betting is (5+6+...+10)/10 = 4.5. Moins

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For reference to the 3-points-on-same-semicircle problem, see e.g., http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html Moins

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This problem is also equivalent to the probability that, if you have a line segment from 0 to 1 and you make 2 random cuts on that line segment, what is the probability that the three resulting pieces do NOT make a triangle? Moins

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None, you don't need to move any of the fruit to correctly label the barrels. You need to change the labels. Moins

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All three are mislabeled and you cannot look inside the barrels. If you take one out of the apples+oranges barrel, whatever fruit you pull out is the fruit of the barrel. And Since you know the other two are mislabeled, you would switch the labels. So you need to take out only one fruit Moins

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the infinite roll approaches the expectation of profit to be 86.3571 please run the following matlab code i wrote % this calculate the expecation of the profit when you throw a 100-faced % dice - given the value of the accepted tossed points - the first toss is % free - starting from the 2nd toss you pay 1£ to play again % input: Nmax -- the largest max number of tosses allowed % output: % N -- number of maximum allowed tosses % P -- expectation of the profit function [N, P] = one_hundred_faced_dice(Nmax) N = linspace(1, Nmax, Nmax); P = N - N; %---if we only throw it once, the expectation is 50.5 P(1) = 50.5; %---throw more times-- P(i-1) is known for i = 2:Nmax, %----if the first toss is lower than N(i-1), toss it again P(i) = floor(P(i-1))*P(i-1)/100; %----if the first toss is higher than N(i-1), accept it for j = (floor(P(i-1))+1):100, P(i) = P(i) + 0.01*j; end %----however from the 2nd toss, we need to pay to play it P(i) = P(i) - 1; end %----draw the curve plot(N, P); end Moins

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if you are going to roll an M-faced dice for Nmax times with a fee starting from the 2nd roll please run the following matlab code % this calculate the expecation of the profit when you throw a 100-faced % dice - given the value of the accepted tossed points - the first toss is % free - starting from the 2nd toss you pay 1£ to play again % input: Nmax -- the largest max number of tosses allowed % M -- no. of faces % output: % N -- number of maximum allowed tosses % P -- expectation of the profit function [N, P] = M_faced_dice(Nmax,M,fee) N = linspace(1, Nmax, Nmax); P = N - N; %---if we only throw it once, the expectation is 50.5 P(1) = (1+M)/2; P(1) = P(1) -1; %---throw more times-- P(i-1) is known for i = 2:Nmax, %----if the first toss is lower than N(i-1), toss it again P(i) = floor(P(i-1))*P(i-1)/M; %----if the first toss is higher than N(i-1), accept it for j = (floor(P(i-1))+1):M, P(i) = P(i) + j/M; end %----however from the 2nd toss, we need to pay to play it P(i) = P(i) - fee; end %----draw the curve plot(N, P,'o-'); end Moins

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dont believe above answer is correct: as correctly mentioned, this option would only be used if you are 1-0 up. Of the next 2 coin flips for (assuming i want heads), after the 1-0 scenario is HH, HT, TH, TT. Of which, all but 1 will give me $2. So $2 x 0.75 = $1.5 in. compared to $-2 x 0.25 = -$0.5 averaging out the above gives you +$1 expected value. Going further (i.e. what is this option worth to you before the game starts), you start off 0-0, you wouldnt use it. you go 1 down, you wouldnt use it, you go 1 up you do use it. It will only be used in 50% of scenarios, therefore you pay no more than $0.5 for it. Moins

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option is worth 0.25 at time 0. at time 1 (after one flip) the option is worth zero or 0.5 (as mentioned above) depending on whether it will be exercised. draw out a tree and this can be clearly seen Moins

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SternH I think got the right answer but due to some arithmetic error... x = (1/100)*100 + (1/100)*99 + ... + (1/100)*x + (x/100)*(x-1) is not quite correct. Two issues I see: 1) Need to consider floor/ceil issues. 2) there are 100-x+1 terms, not 100-x terms, with the 1/100 factor. One way you could adjust for this is make (x/100) into (x-1)/100 Moins

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E[X] = 50.5 thus you stop if you get X > 50.5. Thus, P(X) = 0.5 to get E[X] = 75.5 and P(X) = 0.5 when we re-roll first time. Now, second roll has E[X] = 49.5 as all outcomes are reduced by value 1 needed to pay for a re-roll. Thus, P(X) = 0.5*0.5 for E[X] = 74.5 when we accept the second roll. And P(X) = 0.5*0.5 when we re-roll for 3rd time. In this way, E[X] = Sum( (0.5^x )*(75.5-x) ) where x goes from 1 to 75 Answer = 74.5 + 2^(-76) Moins

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Using a Monte Carlo Method, the answer is roughly 7.7. I'm not too sure how you would do this analytically though :( Moins

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do you replace the card if it isn't a new suit?

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sorry for all the typos, the solution should be understandable in spite of that.

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This can be solved by considering the number of sequences of heads/tails that do NOT have consecutive heads. If N coins are tossed, you can have 0 heads in 1 way. You can have 1 head in N ways--choose 1 from any of the N positions in the sequence. You can have 2 heads as long as they are not consecutive. Imagine you have N-2 tails and you need to decide where to place the 2 heads. You can insert them before any of the tails, or after the last tail--so you need to choose 2 out of a possible N-1 positions. Similarly, for any H less than or equal to N/2, you can have H heads by selecting H positions from a possible (N-H)+1 positions. So the number of sequences which do NOT have two consecutive heads can be found by the sum: 1 + nCr(N,1) + nCr(N-1,2) + nCr(N-2,3) + ... Evaluating this starting with N = 2 gives the values 3, 5, 8, 13, 21, 34, ... These are Fibonacci numbers. The number of sequences of length N without 2 consecutive heads is given by F_{N+2}, where F_1 = 1, F_2 = 1, and F_N = F_{N-1} + F_{N-2}. It follows that the probability for obtaining two consecutive heads in N flips of a fair coin is given by 1 - ( F_{N+2}/ 2^N). Note: j-dw has a correct solution. The solution given by Charles ignores the fact that many sequences will have BOTH two consecutive tails AND consecutive heads. He treats these as non-overlapping sets. Delusion ignores all cases such as HTTHTTHTT in which there is more than one T separating the heads. Moins