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:
- 18.2k
26
votes
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
- 9,860
26
votes
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, ...
- 26.2k
20
votes
Accepted
Simulating a biased coin with an unbiased one
Yes, you can do it like this:
Why does this work?
- 145k
13
votes
Accepted
A coin flip's long tail
$p$ is a random variable, chosen with uniform probability over the interval [0,1]. The probability of two heads is $\int_0^1 p^2=\frac{1}{3}$, and by symmetry so is the probability of two tails.
- 7,319
12
votes
12
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11
votes
Accepted
Coin Flipping Puzzle
In this paper about Levine's hat puzzle there is a better strategy with a winning probability of $0.7$.
Let $a_i$ be the coin toss outcomes that are told to $A$, and $b_i$ the ones that are given to $...
- 50.3k
10
votes
Frequentists be damned! Design an evil coin to prove a point
Frequentists won't be damned. This is a trick question.
In looking at making a coin, you can't bias it by weighting. There is an entire class on such myths and methods of bias at Berkley (at least ...
- 1,263
8
votes
Unfair coins at South Park Elementary
The answer is 0.0101010101...
Proof
Say that $w$ is the chance of Wendy’s coin coming up heads; $s$ is the chance for Sally’s coin. Wendy will win with probability:
$$p(W)= w + w's'w + w's'w's'w + ...
- 2,849
8
votes
Accepted
The asymmetric coin game
Suppose instead that
But in this new situation,
For the game to be fair, the amount he wins should be
- 33.6k
7
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7
votes
Accepted
Seven cowboys play a coin tossing game
It's pretty easy to calculate the exact result using a computer. I haven't found an easy way to calculate it by hand (yet), but if someone wants to check his result before posting:
Update: Optimized ...
- 18k
7
votes
Accepted
7
votes
Frequentists be damned! Design an evil coin to prove a point
I have an idea.
You take a normal coin. It is evenly weighted on each side (if you cut it exactly in half along the side the two halves would weigh the same).
Now
So
However
So
Diagram of idea (...
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7
votes
Simulating an unbiased coin with a biased one
A straightforward answer (actually, a generalisation of loopywalt's answer):
Example:
- 11.3k
7
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6
votes
Accepted
No two heads in a row
The number of possible outcomes for $100$ coin tosses is simply $2^{100}$, which will be our denominator.
For the numerator we must count the number of outcomes without $2$ successive heads. Let us ...
- 18k
6
votes
Unfair coins at South Park Elementary v2
Yes, we can have a game in which
So,
And
Thus,
Assuming that the results from two coins tosses are independent.
In simple words,
Thus, there's no need of biased coins.
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6
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6
votes
Seven cowboys play a coin tossing game
Making the events independent, we get a probabilty that is
$p_{i,j} = \mathbb{P}(\text{player }i\text{ wins }j\text{ games}) = {6 \choose j}\cdot 2^{-6}$.
(note: if six players wins 3 games, then ...
- 5,260
6
votes
Accepted
A jar of mixed coins
The probability is given by $P(fair|heads^m) = P(heads^m|fair) \cdot P(fair) / P(heads^m)$, where
$P(heads^m) = P(heads^m|fair)\cdot P(fair) + P(heads^m|nonfair)\cdot P(nonfair) = 1/2^m \cdot K/N + 1^...
- 5,260
5
votes
A coin flip's long tail
This seems to be mathematically impossible. The probability of two heads is $1/3$ and the probability of two tails is $1/3$.
Since they're equal, this gives us
$$
p^2 = (1-p)^2
$$
$$
p = 1-p
$$
$$
p ...
- 391
5
votes
More coin flipping
None: this is Penney's game, and the second picker always has an advantage because the win condition is nontransitive: that is, there are several instances of "Option A beats option B, which beats ...
- 145k
5
votes
Frequentists be damned! Design an evil coin to prove a point
Without any materials changing or falling apart—
I see a lot of solutions based on decaying materials (whether radioactively or otherwise, e.g. glue). I want to post one idea that doesn't involve any ...
- 304
5
votes
Frequentists be damned! Design an evil coin to prove a point
You can avoid the need for degrading materials:
As a bonus:
- 151
4
votes
Unfair coins at South Park Elementary
The key simplification is that:
Therefore it must hold that:
So
- 2,730
4
votes
Drawing uniformly using a single biased coin
It is
This is not my own work, I got this answer from here.
Let's say we flip the coin $F$ times. For each $k$, we will take the $\binom{F}k$ coin flip sequences which have exactly $k$ heads and ...
4
votes
The asymmetric coin game
The probability of Alice getting $i$ heads after $n$ flips is:
The probability of Bob getting $i$ heads after $2n$ flips is:
Then the probability of Alice and Bob getting the same number of heads ...
- 1,867
4
votes
Simulating an unbiased coin with a biased one
Combining the ideas from loopywalt and trolley813's answers, start by:
After that:
If this does not make a decision,
Generalising this to n
There is then also the possibility of further shortening ...
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