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77

No questions are required!

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What you're missing here is the chance of playing at all, given that the game ends when someone finds the prize. (or, chance of finding a prize goes to 0, which is the same thing) Person 1 has a 100/100 chance of playing, and a 1/100 chance of winning. Person 2 has a 99/100 chance of playing, and a 1/99 chance of winning. Person 3 has a 98/100 chance of ...

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It doesn't matter which option you choose, because Your probability of survival if you're one of n players left is as follows: Informal proof It was established in the question that if there are only 2 players left, they must annihilate each other. So if there are 3 left, your only hope of survival is to annihilate both your opponents, which can only ...

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Explanation:

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because:

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Satan should stick to fiddling. You will win, and here is a simple proof. Consider the game $n$ turns at a time. After each cycle of $n$ turns, all the coins are in their original position (though not necessarily flipped the same way). Replace $H$ with $0$ and $T$ with $1$. In each cycle, you flip all $1$'s to $0$'s, until Satan flips a $0$ to a $1$. Once ...

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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 ...

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(a) I claim that the expected typing length are the same for both monkeys. I guess something in my argument will be incorrect, as jafe's answer has 9 approvals, but finding that incorrectness would be helpful to me. I prove that the probabilty that a monkey ends his typing after exactly $n$ letters is the same for both monkeys, which then implies that the ...

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Leaving aside the dubious assumption that Monty is entirely on the up-and-up...

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Familiar indeed.

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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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This is because

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

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The answers of rand al'thor and Callidus are great; I just want to give a different argument for the result. Claim: After each round, the number of surviving players is even. Proof: Let $f_i$ be the flip of player $i$, with tails $1$ and heads $0$ (it's arbitrary which is which). Let $s_i$ be $1$ if player $i$ survives the round and $0$ otherwise. Then, ...

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Actually,

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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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Let $S_n$ be the state where we have $n$ candies out on the table. We want to find the expected cost in eaten candies to advance from state $S_n$ to $S_{n+1}$. (This may, by chance, involve us having to move back by eating candies before moving forward again.) Let this expected value be $\Delta_n$. $\Delta_0 = 0$ since from $S_0$ (0 candies out) you will ...

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They could toss it twice HT = first player wins; stop TH = second player wins; stop HH or TT = ignore these two tosses; repeat the procedure

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Another way to think about it

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Strategy: How this works:

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If the king sits in his own seat, then each guest will sit in their own seat and Ophelia will always sit in her own seat. This occurs with probability $1/2015$. If the king sits in Ophelia's seat, then each guest will sit in their own seat and Ophelia will end up sitting in the king's seat. This also occurs with probability $1/2015$. If the king sits in ...

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The probability is $1/2$. We have a permutation that maps each box to the box whose key it contains. Once we open a box, we can open the box it maps to. So, we can open all the boxes exactly if there is no all-steel cycle. Label the boxes $1$ through $100$. We denote the permutation in cycle format like $(31)(542)(6)$. To make this canonical, write each ...

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This is a perfect opportunity to use the theory of Markov Chains. The states are the number of candies currently on the table (either 0, 1, 2, 3, 4, 5, or 6 candies). If all 6 candies are present, then the game is over (this is called an "absorbing state"). Otherwise, if there are $n$ candies on the table, then the probability of matching one of them and ...

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I believe this set of dice satisfies all your requirements:

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(a) Edit: This is incorrect, see comments

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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. ...

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OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that. The simplest somewhat-plausible way to handle #1 ...

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Step 1. Step 2. Step 3. This works as follows:

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The average number of sick ants will be We start with the classic ant-on-a-stick observation that Now, make an observation about how ants get infected:

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