reliability (no poker content)

[irchans]

[ 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.

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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.