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40 votes
Accepted

Winning chance in coins game with fixing

GoblinGuide's user avatar
34 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:
loopy walt's user avatar
  • 21.3k
23 votes

You are at a crossroads with five roads leading away from it

I can do it in To show we can't do it in less
Ross Millikan's user avatar
22 votes

You are at a crossroads with five roads leading away from it

Bonus question What if there is a single coin? There is a simple way to prove that to obtain 5 equally probable choices is... And here is why With a computer I can improve this result.
Florian F's user avatar
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20 votes
Accepted

Simulating a biased coin with an unbiased one

Yes, you can do it like this: Why does this work?
Deusovi's user avatar
  • 147k
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 $...
Jaap Scherphuis's user avatar
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 ...
Keeta - reinstate Monica's user avatar
9 votes
Accepted

You are at a crossroads with five roads leading away from it

A solution is:
Kombajn zbożowy's user avatar
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
f'''s user avatar
  • 33.7k
7 votes

Simulating an unbiased coin with a biased one

A straightforward answer (actually, a generalisation of loopywalt's answer): Example:
trolley813's user avatar
  • 11.3k
7 votes

Simulating a biased coin with an unbiased one

This works because
hdsdv's user avatar
  • 5,190
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 (...
Beastly Gerbil's user avatar
7 votes
Accepted

More coin flipping

Fabich's user avatar
  • 7,165
6 votes

You are at a crossroads with five roads leading away from it

AnimeNate's user avatar
6 votes

Winning chance in coins game with fixing

This is not nearly as elegant as GoblinGuide's simple answer: https://puzzling.stackexchange.com/a/126192/1777 The probability of winning is always: Strategy: For any given N coins, there is: This ...
LeppyR64's user avatar
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5 votes

Frequentists be damned! Design an evil coin to prove a point

You can avoid the need for degrading materials: As a bonus:
cobbal's user avatar
  • 151
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 ...
Andrew Cheong's user avatar
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 ...
Deusovi's user avatar
  • 147k
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 ...
Tony Ruth's user avatar
  • 1,867
4 votes
Accepted

Choose the wine

Toss the coin twice... ...yielding four equally likely outcomes. HH: wine 1 HT: wine 2 TH: wine 3 TT: failed, so toss twice again Each trial succeeds with probability $3/4$, so expect $4/3$ trials (...
RobPratt's user avatar
  • 14.2k
4 votes

Simulating a biased coin with an unbiased one

If $p$ is $\frac{n}{m}$ for some $n,m$: Then just Example with $p=\frac{1}{7}$: Some optimization (thanks to @steve in the comments). Thanks to the lexicographical order, sometimes we can stop the ...
melfnt's user avatar
  • 5,132
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 ...
Steve's user avatar
  • 3,885
3 votes

Simulating an unbiased coin with a biased one

[An earlier version of this answer had the wrong formulae - these are now corrected] Assuming that $p$ is known, we can proceed as follows: This procedure is optimal where $p$ is known, and in ...
Steve's user avatar
  • 3,885
3 votes

Simulating an unbiased coin with a biased one

Similar to the previous and accepted answer:
Retudin's user avatar
  • 9,278
3 votes

Frequentists be damned! Design an evil coin to prove a point

Following other thoughts here, a coin cut in half and "altered". Unfortunately due to the way coin flipping works, all the solutions involving unbalanced sides would only affect things when the coin ...
Rycochet's user avatar
  • 241
3 votes

Frequentists be damned! Design an evil coin to prove a point

Well, no matter how hard I look at it, this coin needs something that changes, at least slightly, after enough tosses. So, it has to be electronics or degradable material. At least in some way. Sure, ...
Zizy Archer's user avatar
  • 2,140
3 votes

You are at a crossroads with five roads leading away from it

This is not an original answer. It only adds how to derive @Florian F's 5-toss answer analytically. Let $B = P - \frac 1 2$ be the coin's bias. We'll reference the 5 groups of outcomes by the numbers ...
Albert.Lang's user avatar
  • 6,195
2 votes

Frequentists be damned! Design an evil coin to prove a point

So, The nice thing with this you don't even need to toss, but you can and won't see any anomalies for a while.
shabunc's user avatar
  • 121
2 votes

Find the unseen numbers of two tokens after then have been tossed, given sums

I wouldn't look at this as a probability question. What I'd do is (with small correction pointed out by @anodyne): So to elaborate further, here's the basic template that you would need to fill out: ...
Dr Xorile's user avatar
  • 23.7k
2 votes
Accepted

Guess the coin toss

TakingNotes's user avatar
  • 3,449

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