80 votes
Accepted

3 doors, three guards, one stone

No questions are required!
  • 3,290
63 votes
Accepted

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) ...
59 votes
Accepted

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
58 votes
Accepted

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 ...
56 votes
Accepted

Crippled King Crossing a Canyon

Explanation:
  • 33.3k
53 votes
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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
52 votes
Accepted

Say 100 and win

because:
  • 2,819
44 votes
Accepted

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
44 votes
Accepted

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
43 votes
Accepted

A Strangely Familiar Probability Problem

Familiar indeed.
  • 1,058
42 votes
Accepted

How many tries to roll a 6?

The answer is indeed...             ...because the question is equivalent to...   Calculations:
  • 21.5k
42 votes
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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
41 votes

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
40 votes

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
38 votes
Accepted

My grandfather's socks

This is because
  • 7,300
38 votes
Accepted

A lonely pawn on the chessboard

Strategy: How this works:
37 votes
Accepted

Two men for one gold coin

They could If the result is
  • 44.8k
37 votes
Accepted

Eccentric Millionaire Probability Paradox

Another way to think about it
  • 16.9k
37 votes

100 pieces 1 opportunity, choose wisely!

Actually,
  • 33.7k
35 votes
Accepted

The "Loop of rope" dilemma

Alice Bob
34 votes
Accepted

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
32 votes
Accepted

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
32 votes
Accepted

How to simulate one die with three dice?

I believe this set of dice satisfies all your requirements:
  • 140k
32 votes
Accepted

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
31 votes

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, ...
30 votes
Accepted

Do better than chance

Step 1. Step 2. Step 3. This works as follows:
  • 12.7k
30 votes

2 Monkeys on a computer

(a) Edit: This is incorrect, see comments
  • 68.3k
28 votes

Aproximating 100 by 6

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-...

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