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January 15th, 2019, 09:38 AM  #11 
Senior Member Joined: Sep 2015 From: USA Posts: 2,408 Thanks: 1309 
In order to derive expectations we need to find the distribution of a win in $k$ tries. This leads to a partition of $(k1)$ 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. $P_r(m,n)=P[mn \text{ hiders eliminated this round}]$ You then end up with a probability chain $P_r(m_j,1)P_r(m_{j1},m_j) \dots P_r(m_1, N_h)$ 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: 638 Thanks: 85 
Thank you for whatever you can do. I don't think I ever learned probability chains.

January 17th, 2019, 10:43 AM  #13  
Senior Member Joined: Sep 2015 From: USA Posts: 2,408 Thanks: 1309  Quote:
$P_1(0) = \dfrac{3}{4}$ $P_1(1) = \dfrac{1}{4}$ $\text{for }n>1$ $P_n(k)=P[k \text{ eliminated} n \text{ start in round}] = \dfrac{1}{4^n4}\begin{cases} 3^n3 &k=0 \\ \dbinom{n}{k}3^{nk} &k=1,2,\dots,(n1) \end{cases}$ One issue I'm running into now is that $P_n(0) >0$ and so there is nonzero 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: 2,408 Thanks: 1309 
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 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 638 Thanks: 85  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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amount, corners, expected, game, rounds, takes 
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