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You can make arbitrarily large sets of dice with this property. Start with Efron's dice: A: 4, 4, 4, 4, 0, 0 B: 3, 3, 3, 3, 3, 3 C: 6, 6, 2, 2, 2, 2 D: 5, 5, 5, 1, 1, 1 A beats B, B beats C, C beats D, and D beats A with probability $\frac{2}{3}>60\%$. Now add many copies of die B, each using a different value between 2 and 4. For example, one of ...

42

As xnor points out in his answer, this question is basically asking for the way to most evenly distribute $6^n$ results among $100$ bins, and gives a very brief description of the solution. I'll go into a bit more detail here. If you're not interested in the proofs, skip to section 2.3 Summary In order to construct a "best" algorithm for converting rolls ...

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The answer is indeed...             ...because the question is equivalent to...   Calculations:

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This surprisingly beguiling puzzle may also be solved with a surprisingly unsophisticated approach. Symmetry, by itself, predicts the average length of evens-only sequences ending with 6 to be... Start with T  many random throws: 2153664315121226553111444142566363625461525 . . 3644464461 Sift them into 4 groups that, due ...

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I believe this set of dice satisfies all your requirements:

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Get 10 different d6 dices and describe them on paper. Next to each description, associate a unique number from 0 to 9. Put all those dices in an opaque bag (you should have one to transport that near-infinite number of dices). Shuffle the bag. Pick one die blindly and retrieve its value on the paper you wrote. Multiply that value by 10. Shuffle the bag. ...

27

To explain, here is a more formal way to state Ian's answer: Let the probability of die A rolling $i$ be $a_i$ for $1\le i\le6$, and similarly for die B. If every sum has the same probability, then they must each have probability $\frac{1}{11}$. Because 2 is only possible from double 1s, and 12 is only possible from double 6s, $a_1b_1=a_6b_6=\frac{1}{11}$. ...

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As already mentioned, the smallest possible sum is 27, resulting from the numbers 1, 2, 4, 5, 7, 8 arranged like this: 1 4758 2 Optimality is a consequence of the observation that no three sides of the die may show three consecutive numbers (e.g. 1, 2, 3 or 3, 4, 5 are not possible). To see this, suppose that $n$, $n+1$ and $n+2$ were all on different ...

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@Deusovi's answer is totally correct, but I want to add here the general approach for solving such problems as well. No need to upvote, since I did not invent the technique, and you can see it described in this puzzle as well. The idea is to use generating functions. Basically, we try to check if there is a factorization of $x^1 + x^2 + x^3 + x^4 + x^5 + x^... 23 The answer is Proof Alternative proof 23 Answer: Explanation: 21 All addition is modulo 6 (e.g. 4 + 3 = 1, 3 + 3 = 6, 5 + 3 = 2, 6 + 1 = 1). 3 Dice Roll Resulting 2 Dice Roll Two same, one different: AAB AB 135 14 246 25 All same: AAA 36 Three in a row: A, A + 1, A + 2 AA All different, two of which differ by 3:A, B, A + 3where B = A + 1 or B = A + 2 AB Credits to Scott M for a suggestion which simplified this ... 20 Figured i'd add my comment as an answer: We roll 3 dice$(D1,D2,D3)$, which we seperate spatially or in some other way (we used to use 3 different colors). We reroll D1 and D2 if they land on 6, and we reroll D3 if we it lands on a 5 or a 6. We repeat these rerolls until we have 3 numbers,$D1 \& D2 <= 5$, and$D3 <= 4$. We then form our dice ... 19 There is a way that doesn't use the dice (3+ coin-flips): Flip the coin three times and record 1 or 0 for heads or tails. If all three numbers are zero start again. Otherwise read the 1s and 0s as a binary number. Its value will be from 1 to 7 inclusive with each number having an equal probability of occurring. Here's a summary of Mike's answer (2+ dice-... 19 19 Strategy: Expectated gain: Conclusion: 18 I'll use a different spoiler for each level of solving the puzzle, such that you can read only as far as you want to be spoiled, if you want to only look at the first few steps for hints. Note that there is some Mathematica code further down, which is not inside a spoiler tag. Making sense of the hints Putting the hints together Finding the tour These ... 16 The probability is exactly$\frac{1}2$in the original game. The crucial fact is that rolling a$5$is a loss and that rolling a$6$is a win - so we can consider the game over once one of those appears. However, at any given roll, the probability of rolling a$5$equals the probability of rolling a$6$. We can quickly extend this local property to learn ... 16 As @MikeEarnest says, unless you put an upper limit on the number of rolls, you can always do better with more rolls. Since$6^n$is never divisible by$100$, you can never be completely fair. With more rolls you can always get the unfairness lower. The fact that$6^4=1296$is very close to a multiple of$100$means you can get very close to fair with$...

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The post linked in rand al'thor answer optimized the worst case pick the visitor of the sphinx can make. But the question asked for the best average case. A better solution in that case is: A: 1,1,5,5,5,5 B: 2,2,2,2,6,6 C: 3,3,3,3,3,3 D: 4,4,4,4,4,4 B defeats A with a probability of $\frac{5}{9}$ C defeats B with a probability of $\frac{2}{3}$ D defeats ...

15

Let $X_n$ be the event that the dice takes $n$ rolls to get the first 6, given all the rolls are even. Let $A_n$ be the event that it takes $n$ rolls to get the first 6, and let $B$ be the event that all rolls up to the first 6 are even. $P(X_n)=P(A_n|B)=\dfrac{P(B|A_n)P(A_n)}{P(B)}$ (using Bayes' theorem) Now: $P(A_n)=\dfrac{1}{6}\cdot\left(\dfrac{5}{6}\... 14 The answer is First I show that a lower number is not possible: This establishes a lower bound. The upper bound is established by providing an actual solution. First die: Second die: Third die: To find the numbers, I first tried base-6 with the numbers on each die offset to get the correct maximums. But that approach did not work as the one die with ... 14 Easy: 14 I rotated the grid 90 degrees clockwise before solving the grid deductions, in order to reduce vertical space. Let's solve Statue Park first. Now to the Fillomino. Now that the two grids are solved, we can combine them in two ways: Turns out that the second interpretation is correct, and the column sums are which spells out... As @oAlt noticed, the ... 13 A partial solution You asked for the existence of a$(2n+1)$-sided polytope/die, so that all$2n+1$sides are congruent and so that the probability of landing on each of these sides is precisely$1/(2n+1)$. I do not see how to mathematically model the property that the probability of landing on each of the sides is precisely$1/(2n+1)$. This seems to need ... 13 Let$n$be an arbitrary positive integer. Start with a right cylinder with cross section a regular$n$-gon. To each of the bases, attach a pyramid (with regular$n$-gon base). The resulting polyhedron has$n$rectangular faces and$2n\$ triangular faces. If the cylinder and pyramids are tall enough, it will be impossible for the die to land on the triangular ...

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Firstly arrange the dice into a 3x3 matrix. If the sum of all dice is even, choose the column of first cup according to the total sum mod 3, and the row according to the sum of its own column. Let's say the first row/column means mod 3 = 0 and the second means mod 3 = 1. If the first cup is O (its own column denoted by o), choose the second cup between A, B ...

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