 My Math Forum Expected Amount of Rounds the Game 4 Corners Takes
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 January 15th, 2019, 09:38 AM #11 Senior Member   Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 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. January 16th, 2019, 12:16 PM #12 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 661 Thanks: 87 Thank you for whatever you can do. I don't think I ever learned probability chains. January 17th, 2019, 10:43 AM   #13
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Quote:
 Originally Posted by romsek 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 11:39 AM. January 17th, 2019, 12:27 PM #14 Senior Member   Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 After playing with this for a bit I can say that for 2 people initially hiding $E[\text{#rounds until win}] = 6$ January 22nd, 2019, 04:10 AM   #15
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Quote:
 Originally Posted by romsek 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. Tags amount, corners, expected, game, rounds, takes Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post EvanJ Probability and Statistics 0 April 9th, 2018 04:47 PM GIjoefan1976 Algebra 6 December 19th, 2016 10:58 AM Peno77 Probability and Statistics 4 February 19th, 2015 04:45 AM salzrah Algebra 0 June 12th, 2012 07:42 PM fc_groningen Advanced Statistics 0 February 4th, 2011 07:54 AM

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