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Wendy Testaburger and Sally Turner play a game with two unfair coins. A coin flip with Wendy's coin shows head with probability $\frac {1} {100}$. A coin flip with Sally's coin shows head with probability $p$. Wendy does the first coin flip. Wendy and Sally flip their coins alternately until a coin shows heads. The girl with the coin showing head wins the game. What value of $p$ makes this game a fair game?

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  • $\begingroup$ Could someone explain why them having the exact same coin doesn't make this a fair game? Does fair have some sort of second meaning in this context? $\endgroup$ – user1717828 Mar 16 '16 at 19:10
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    $\begingroup$ @user1717828 it's not fair when having identical coins because Wendy gets to start first always. It would only be fair with identical coins if they both get a chance to flip and then compare results but this is not the case. When Wendy flips her coin and gets heads she won the game without Sally even having an opportunity to flip. $\endgroup$ – Ivo Beckers Mar 16 '16 at 22:17
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    $\begingroup$ Since this is from a 2D animated universe, do the coins have a thickness? Can you even flip them? Once they rotate 90 degrees horizontally, they will disappear!? I'm flipping out!! $\endgroup$ – Doug.McFarlane Jun 27 '16 at 18:21
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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, the game returns to the start, which has no effect on fairness. So, to be fair, these two probabilities should be equal, which happens for $p=1/99$.

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  • $\begingroup$ I think this is the simplest answer. $\endgroup$ – Fimpellizieri Mar 15 '16 at 21:45
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The answer is

$p = \frac{1}{99}$

Proof

Let $p(W,k)$ be the probability of Wendy winning on her $k$th turn and $p(S,k)$ be the probability of Sally winning on her $k$th turn. Then,

$p(W,1) = \frac{1}{100}, p(W,2) = \frac{99}{100}(1-p)\frac{1}{100}$ and, by simple induction $p(W, k) = \left(\frac{99}{100}(1-p) \right)^{k-1} \left(\frac{1}{100} \right)$.

Also, by a simple induction it can be shown that
$p(S, k) = \left(\frac{99}{100}(1-p) \right)^{k-1}\left(\frac{99p}{100} \right)$.

We must then enforce

$\sum_{k=0}^{\infty} p(W,k) = \sum_{k=0}^{\infty} p(S,k)$
which implies $\frac{1}{100} = \frac{99p}{100}$ as required

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The game is fair for

p = $\frac1{99}$

Proof:

If Wendy's coin has odds $\frac1n$, a fair game is achieved when Sally's coin has odds $\frac1{n-1}$. It's easy to see that if Wendy's coin has $\frac12$ odds, Sally needs a coin that always comes up heads. If Wendy's coin comes up heads $\frac13$ of the time, and Sally's comes up heads $\frac12$ of the time, there's a $\frac13$ chance Wendy wins on turn 1, and a $\frac12 \times \frac23=\frac13$ chance Sally wins on turn 2 ($\frac12$ as her odds of getting heads, and $\frac23$ as the odds of the game lasting long enough for there to be a turn 2). Then there's a $\frac13 \times \frac13 = \frac19$ chance Wendy wins on turn 3, a $\frac12 \times \frac29=\frac19$ Sally wins on turn 4... for any pair of turns, Wendy and Sally will have equal odds. We can see this trend holds for any $\frac1n$, leading us to the above answer of $\frac1{99}$.

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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:

$$p(W)= w + w's'w + w's'w's'w + w's'w's'w's'w ... = w ( 1 + (w's')^2 + (w's')^3 ... )$$

Sally wins with probability:

$$p(S) = w's + w's'w's + w's'w's'w's ... = s ( w' + w'^2s'^1 + w'^3s'^2 + ....) = sw'(1 + (w's')^2 + (w's')^3 ... $$

Set them equal to each other:

$$sw'(1 + (w's')^2 + (w's')^3 ... = w ( 1 + (w's')^2 + (w's')^3$$ $$sw' = w$$ $$w= 0.01, w'=0.99$$

Therefore, $s = 0.01 / 0.99 = 1/99 = 0.0101010101 ...$.

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The key simplification is that:

since the game repeats from the beginning if no-one wins after they each have a turn, we only have to make sure that they chance of them each winning on their first toss is equal.

Therefore it must hold that:

1/100 = (1 - 1/100)p

So

p = 1/99

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    $\begingroup$ p cannot be 99 :) $\endgroup$ – Oray Mar 16 '16 at 12:22
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    $\begingroup$ @Oray 'p' is the denominator in Ergwun's formulation - the probability is 1/p. $\endgroup$ – Lawrence Mar 16 '16 at 13:07
  • $\begingroup$ @Oray Lawrence is right, I was using 1/p. I've changed it to match the use of p in the question. $\endgroup$ – Ergwun Mar 16 '16 at 22:52
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First instinct, but I may be wrong on this:

1/100. Then both of them have the same chance of winning.

[EDIT]

Second instinct :

1/99

Why my second instinct?

let's limit to 1 toss each.
Wendy's chance. $1/100$.
Sally's chance $99 * p/100 $. For this to be equal chances $p = 1/99$.

Note: will try my third instinct for 2 tosses each.

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  • $\begingroup$ I was thinking the same $\endgroup$ – Ivo Beckers Mar 15 '16 at 16:15
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    $\begingroup$ I think the problem is that since wendy goes first she has a higher chance of winning. e.g. if they were normal coins, she would have 0.5 + (0.5)^3 + (0.5)^5 .... etc chance of winning. So you have to increase sally's chance to compensate for this. $\endgroup$ – astralfenix Mar 15 '16 at 16:17
  • $\begingroup$ @astralfenix. That's what I'm trying to calculate right now :) $\endgroup$ – Marius Mar 15 '16 at 16:17
  • $\begingroup$ If you think this is fair, I'd like to play with both of you for money. Of course I would always go first, like Wendy. $\endgroup$ – Sleafar Mar 15 '16 at 16:19
  • $\begingroup$ @Sleafar. LE'S DO IT...I feel lucky. :) $\endgroup$ – Marius Mar 15 '16 at 16:36
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Let $f(n)$ be Wendy's chance of winning at her $n$th turn. $f(1)=1/100$ with $f(n+1) = f(n)*(1-p)*99/100$ in general. Her total chance of winning is the sum of the infinite series consisting of all $f(n)$ values. If we let it be $A$, we get $A*(1-p)*99/100 + 1/100 = A$. Since $A$ is supposed to equal 1/2, $p=1/99$.

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You can't make this a fair game because if you give 1/100 to one person and 1/99 to another, after like 200 chances the 1/99 person would be in advantage. If the chances are the same(1/100), the person starting first have the advantage.

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