20200401, 12:15  #1 
"Mike"
Aug 2002
1111111100101_{2} Posts 
April 2020

20200401, 18:43  #2 
Apr 2020
5_{10} Posts 
is there someone get 29.16521896% after day 10?
i'm still getting 24.25371788% ._. 
20200401, 22:12  #3 
Jan 2017
2×3^{2}×5 Posts 

20200401, 22:19  #4  
Nov 2003
2^{2}×5×373 Posts 
Quote:
This problem seems very underposed! Some questions: (1) What are the initial conditions? If all are healthy at t=0, then they remain healthy. (if one is adjacent to 2 rather than 3 people the chance that at t+1 a neighbor will get sick is lower). (1a) How MANY are sick at t= 0??? Just one? Do we get to select which one is sick at t = 0? (1b) Can a sick person infect more than one adjacent person? (2) If someone is sick at time t do the people immediately adjacent become sick with equal probability? (assuming the healthy ones have no other sick neighbors) (3) Is it possible that once infected, someone will heal? If so, how long does it take? [I assume not, but it needs to be stated] (4) If sick at t, how is the probability determined that someone adjacent becomes sick at t+1? Is it constant throughout the 10 days? Is it a necessity that someone adjacent does become sick? i.e. if at time 0 someone is sick and does not infect a neighbor must another adjacent neighbor become sick? (4) If someone is sick at time t do the people adjacent become sick with equal probability? (5) Are the probabilities independent? If someone is healthy and two adjacent people are sick at t, does one add the probabilities (p1+p2) that the person becomes sick at t+1 or is the probability 1 ( (1p1)(1p2))? I assume the latter. I can see an argument for either if it is possible to get infected by two people at the same time.... Or am I just plain missing the obvious? Last fiddled with by R.D. Silverman on 20200401 at 22:25 

20200401, 22:33  #5  
Nov 2003
16444_{8} Posts 
Quote:
Is it possible to infect more than one person in a time period?? 

20200401, 23:44  #6 
Jan 2017
2·3^{2}·5 Posts 
I think the initial conditions are the only particularly unclear part. I assumed that the first person (corresponding to the first row of the adjacency matrix) is always the only person infected at start.
For the rest, I think the following is the most obvious natural way to calculate how the set of infected people changes from one day to the next, and does match the probability given for the example: Given a set of infected people S, find all edges in the graph that connect to an infected person. Color each such edge red with 10% probability. The next set of infected people is people who are either already in S or have a red edge. 
20200402, 01:51  #7 
Apr 2020
5_{10} Posts 
If I want to calculate probability of all infected at day t, should I just product all edges infection probability?
If there are only 3 edges and infection probabilities are like (a:0.1, b:0.1, c:0.1) all edges infected probability @ day t = (0.1)^3 ? or should I also concern all edges infected probability @ day 0 to t1 ? its rly confusing 
20200402, 01:52  #8  
Nov 2003
2^{2}·5·373 Posts 
Quote:
An interesting question: Given a graph X. Delete an edge. Prove or disprove that the resulting graph always has a lower probability. This seems intuitively true. The resulting graph has two people with no direct contact, thus lowering the probability for at least those two. This immediately suggests that the maximal probability graph should be complete. (every node connected to every other). Now all one need do is check the complete graphs to size 8. 

20200402, 01:59  #9  
Nov 2003
2^{2}×5×373 Posts 
Quote:
process/martingale. I wish I could remember the material..... Its been 45 years since I last looked at/took a course....Can you say "ossification of the intellect"? Last fiddled with by R.D. Silverman on 20200402 at 02:00 Reason: typo 

20200402, 02:07  #10  
Nov 2003
1D24_{16} Posts 
Quote:
herein. Another interesting question. Among all graphs, which is maximal? I think I know the answer. Last fiddled with by R.D. Silverman on 20200402 at 02:10 

20200402, 02:10  #11  
Jan 2017
2×3^{2}×5 Posts 
Quote:
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