My Math Forum Expected Amount of Rounds the Game 4 Corners Takes

 Probability and Statistics Basic Probability and Statistics Math Forum

 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 $(k-1)$ 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: 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:
 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: 2,408 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

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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top