# Tag Info

Accepted

### 3 doors, three guards, one stone

No questions are required!
• 3,615
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) ...
• 7,779
Accepted

### 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 ...
• 33.7k
Accepted

because:
• 2,847
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 ...
• 566
Accepted

### How many tries to roll a 6?

The answer is indeed...             ...because the question is equivalent to...   Calculations:
• 21.9k
Accepted

### Monty Hall Revisited: Winning Both Goats!

Leaving aside the dubious assumption that Monty is entirely on the up-and-up...
• 2,772

### 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 ...
• 21.9k
Accepted

• 976
Accepted

### A Bridge Problem

The answer is To see this,
• 21.8k

Actually,
• 34.2k
Accepted

Alice Bob
• 120k
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:
• 21.3k
Accepted

### How to simulate one die with three dice?

I believe this set of dice satisfies all your requirements:
• 147k

### 2 Monkeys on a computer

(a) Edit: This is incorrect, see comments
• 77.5k

### Coin Game with infinite paradox

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 ...
• 120k
Accepted

### Three horse race

A slightly different approach
• 136k

### Lottery strategy

The average payout for each ticket is So Thus
• 5,132
Accepted

### Unfair coins at South Park Elementary v2

As Timmy has a "head" on his shoulders with probability of 1, he always wins at his turn. Then
• 9,900
Accepted

### Unfair coins at South Park Elementary

Wendy wins on the first flip with probability $1/100$. Otherwise, the game keeps going and Sally has probability $p$ to win on the next flip, which has overall probability $99/100 \times p$. If not, ...
• 27k

### 2 Monkeys on a computer

Monkey problem To settle down which monkey is faster on average, I'll use Markov chains and Mathematica. Define a state $i$, for $i = 0..6$, as that the monkey 1 has currently written $i$ correct ...
• 361
Accepted

### A Short Dice Puzzle

The answer is Proof Alternative proof
• 136k

### Three horse race

It is The following scenarios are compatible with the information given: Scenario 1: Scenario 2: Scenario 3:
• 21.3k

### How to simulate one die with three dice?

@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 ...
• 6,964
Accepted

• 7,282

### What is the probability that your life will have lasted for 100 years once you die?

The frequentist answer to this question is This is because
• 488
Accepted

### Make 2 dice out of 3 dice

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 +...
• 32.4k

### 100 pieces 1 opportunity, choose wisely!

Other people have already given correct answers, but I wanted to suggest a different way of thinking about the question that involves less calculation:
• 120k
Accepted

### frog on a number line

It's easy to see by transforming the problem into a symmetric one - instead of a 1/3 vs. 2/3 jump to B or D, make a third branch so it's a uniform 1/3 chance of going to B, D, or D' (which in turn has ...
• 4,804