# Tag Info

## Hot answers tagged probability

Accepted

### 3 doors, three guards, one stone

No questions are required!
• 3,290
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### 100 pieces 1 opportunity, choose wisely!

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) ...
• 7,524
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### Coin Flipping Game with the Devil

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 ...
• 7,688
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### Left coin, right coin, last coin?

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 ...
• 114k
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Explanation:
• 33.3k
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### Deceptive dice game

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 ...
• 33.3k
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because:
• 2,819
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### Aproximating 100 by 6

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 ...
• 16.2k
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### 2 Monkeys on a computer

(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 ...
• 556
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Familiar indeed.
• 1,058
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### How many tries to roll a 6?

The answer is indeed...             ...because the question is equivalent to...   Calculations:
• 21.5k
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### Monty Hall Revisited: Winning Both Goats!

Leaving aside the dubious assumption that Monty is entirely on the up-and-up...
• 2,497

### Left coin, right coin, last coin?

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 ...
• 23.8k

### How many tries to roll a 6?

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 ...
• 21.5k
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This is because
• 7,300
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### A lonely pawn on the chessboard

Strategy: How this works:
• 5,846
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### Two men for one gold coin

They could If the result is
• 44.8k
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### Eccentric Millionaire Probability Paradox

Another way to think about it
• 16.9k

Actually,
• 33.7k
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Alice Bob
• 112k
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### The "M&M Sugar Rush" game

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 ...
• 1,225
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### One Hundred Lockboxes of Wood and Steel

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 ...
• 23.8k
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### How to simulate one die with three dice?

I believe this set of dice satisfies all your requirements:
• 140k
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### Simulating an unbiased coin with a biased one

One possibility: This works because: EDIT: Inspired by @trolley813's answer here is a way to recycle the rejected entropy:
• 13.6k

### The "M&M Sugar Rush" game

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, ...
• 933
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### Do better than chance

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