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January 15th, 2019, 10:38 AM   #11
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In order to derive expectations we need to find the distribution of a win in $k$ tries.

This leads to a partition of $(k-1)$ and a probability chain corresponding to that partition.

Central to this is $P[\text{n hiders left at end of round}|\text{m hiders at start of round}]$

i.e. $P_r(m,n)=P[m-n \text{ hiders eliminated this round}]$

You then end up with a probability chain

$P_r(m_j,1)P_r(m_{j-1},m_j) \dots P_r(m_1, N_h)$

these probabilities aren't too hard to solve for.

More on this as I have time.
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January 16th, 2019, 01:16 PM   #12
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Thank you for whatever you can do. I don't think I ever learned probability chains.
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January 17th, 2019, 11:43 AM   #13
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Quote:
Originally Posted by romsek View Post
In order to derive expectations we need to find the distribution of a win in $k$ tries.

This leads to a partition of $(k-1)$ and a probability chain corresponding to that partition.

Central to this is $P[\text{n hiders left at end of round}|\text{m hiders at start of round}]$

i.e. $P_r(m,n)=P[m-n \text{ hiders eliminated this round}]$

You then end up with a probability chain

$P_r(m_j,1)P_r(m_{j-1},m_j) \dots P_r(m_1, N_h)$

these probabilities aren't too hard to solve for.
bit of an update

$P_1(0) = \dfrac{3}{4}$
$P_1(1) = \dfrac{1}{4}$


$\text{for }n>1$
$P_n(k)=P[k \text{ eliminated}| n \text{ start in round}] =

\dfrac{1}{4^n-4}\begin{cases}
3^n-3 &k=0 \\
\dbinom{n}{k}3^{n-k} &k=1,2,\dots,(n-1)
\end{cases}$

One issue I'm running into now is that $P_n(0) >0$ and so there is non-zero probability that the game will never end. This can be handled but I've got to think about how.

For example suppose you start with 5 hiders. A typical sequence of how many left and a win might be

5 4 2 2 1 1 1 W

but there are also sequences such as

5 4 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 etc. for quite a while and these have nonzero probability of occurring.

More as I figure things out.

Last edited by romsek; January 17th, 2019 at 12:39 PM.
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January 17th, 2019, 01:27 PM   #14
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After playing with this for a bit I can say that for 2 people initially hiding

$E[\text{#rounds until win}] = 6$
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January 22nd, 2019, 05:10 AM   #15
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Quote:
Originally Posted by romsek View Post
After playing with this for a bit I can say that for 2 people initially hiding

$E[\text{#rounds until win}] = 6$
I disagree. The 2 people will pick different corners, and the probability of a winner in each round is 2/4 = 1/2. Therefore it simplifies to the expected number of flips needed to get 1 head, which is 2.
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