
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
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 $(k1)$ 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: 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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399  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: 2,549 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 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 661 Thanks: 87  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  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Would These Two Games Have The Same Expected Amount of Spins?  EvanJ  Probability and Statistics  0  April 9th, 2018 04:47 PM 
One car travels 140 miles in the same amount of time it takes a second car traveling  GIjoefan1976  Algebra  6  December 19th, 2016 10:58 AM 
Expected value  above a certain amount  Normal Distribution  Peno77  Probability and Statistics  4  February 19th, 2015 04:45 AM 
Expected Value from betting game  salzrah  Algebra  0  June 12th, 2012 07:42 PM 
the amount of money it takes the be served  fc_groningen  Advanced Statistics  0  February 4th, 2011 07:54 AM 