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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 ...


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 ...




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 ...




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 ...


(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 ...


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


Familiar indeed.


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 ...


This is because


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




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, ...


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 ...


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 ...


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


Another way to think about it


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 ...


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 ...


Strategy: How this works:


I believe this set of dice satisfies all your requirements:


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 ...


(a) Edit: This is incorrect, see comments


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. ...


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 ...


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


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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