Simulation Results
 

pvn,

I wrote a quick simulation for an additional comparison. I don't think my Mathematica code is really readable, but it beats writing a longer c++ program. Here is the code and the results:

<font class="small">Code:</font><hr /><pre>
(* deln computes the differences between elements of the
vector v whose indices differ by n *)
deln[v_List, n_Integer] := Drop[v, n] - Drop[v, -n];

vOverlaps = Table[
(* generate 400 start times in seconds past Jan 1 *)
vStartTimes = Table[ Random[Integer, {0, 365*24*60*60}], {400}] // Sort;

(* count the overlaps where the start times differ
by less than 3600*3 seconds *)
iOverlaps = Count[ Join @@ Table[
deln[vStartTimes, n], {n, 5}],
x_ /; Abs[x] &lt; 3600*3],
{10000}];

(* print the average number of overlaps *)
Print[ Plus @@ vOverlaps /10000. ];
</pre><hr />

I ran the code above twice and got the following results:

first run - ave 54.6250 overlaps/year over 10000 years.
second run - ave 54.6962 overlaps/year over 10000 years.

I think I can use the reasoning in your previous post to generate a number near 54.

As uuDevil pointed out, the expected number of failures in a three hour period is 400/(24*365/3)=0.136986.

We expect to have 400 failure events (averaging three hours each) in a year. During each one of those, the expected number of other machines that will start fail is 399* 3/365/24 = 0.136644. (Except after 9 pm December 31st.)

Now if we multiply 400*0.136644 = 54.6575 we get the expected number of failures. We have not counted any overlaps twice because for each failure start time we only counted overlaps with failures that began after the original start time.

Re: Simulation Results
 

[ QUOTE ]
Now if we multiply 400*0.136644 = 54.6575 we get the expected number of failures. We have not counted any overlaps twice because for each failure start time we only counted overlaps with failures that began after the original start time.

[/ QUOTE ]

Very nice. I agree with your line.

Now, if I want to expand this to see about 3 simultaneous failures, it should be pretty simple, right?

When two machines fail simultaneously, the average length of overlap should be 1.5 hours.

So we have 54.6575 events where two failures overlap. The expected number of failures during that overlap period should be 398 * 1.5/365/24 = 0.06815

54.6575*0.06815= 3.725 triple failures. This seems about right.

Extending from here should be obvious.

Re: Simulation Results
 

[ QUOTE ]
(Except after 9 pm December 31st.)

[/ QUOTE ]

BTW, I don't think this matters.

Re: Simulation Results
 

[ QUOTE ]



Now, if I want to expand this to see about 3 simultaneous failures, it should be pretty simple, right?

When two machines fail simultaneously, the average length of overlap should be 1.5 hours.

So we have 54.6575 events where two failures overlap. The expected number of failures during that overlap period should be 398 * 1.5/365/24 = 0.06815

54.6575*0.06815= 3.725 triple failures. This seems about right.
Extending from here should be obvious.

[/ QUOTE ]

Your reasoning looks good to me. I ran some more simulations and got an average near 3.55 triple failures per year, but I am not confident that the code was perfectly correct.

Re: reliability (no poker content)
 

[ QUOTE ]
i think uuDevil is on the right track... I simply used the Poisson formula, and got a bit different number, with the probability being ~.0081, resulting in about 23.9, so 24... basically the same though...


[/ QUOTE ]

look, will one of you two tell me why you're disregarding the poisson? it was designed for situations of this sort...

and btw, for a triple occurence, only approximately 1 will occur in a year...

Re: reliability (no poker content)
 

[ QUOTE ]
look, will one of you two tell me why you're disregarding the poisson? it was designed for situations of this sort...

and btw, for a triple occurence, only approximately 1 will occur in a year...

[/ QUOTE ]

The poisson distribution assumes discrete intervals, not continuous ones (sorry if my terminology is awkward) - i.e. poisson will give you the expected number of failures in three hour intervals such as 00:00 - 03:00, 03:00 - 06:00 etc. A failure at 2:30 and another at 3:30 would overlap, but in the poisson they would be in different three-hour periods, and wouldn't get counted.

I think a better way to phrase it might be that the poisson is concerned with events in a given period, whereas I'm concerned with the proximity of events.