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| Thread: Question about "Death Stare" |  This thread is pages long: 1 2 · «PREV |
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Russ
Promising
Supreme Hero
blah, blah, blah
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posted April 07, 2006 05:06 PM |
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Edited by Russ on 7 Apr 2006
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I don't know much about stats. So Mr Russtats. If a single MG has 10% chance. What are the odds of getting 35 attacks in a row with 1 MG without a single death stare?
You were pretty close. The odds are: 0.9^35=2.5%
About splitting. Again the sample size was small, but I found going 1 over a multiple of 10 was the best bet. Like 11, 21, 31, etc.
Yep 
21 attacks with 11 MGs:
0 kills 19%
The odds of that are 0.9^11=31%, which seems quite different from your test results. Perharps there is something else to it. Maybe it has to do with HOMM3's random generator. HOMM3 has a predefined random pattern. For example, if you do same moves on a certain map square, the probabilities will always be the same. This may have affected your results. I think the best way to beat that would be to get expert fire, blind and 10000 mana to your hero and keep blinding and attacking the stack with your MGs. 50-100 attacks would make a decent sample.
1 Kill 33%
I am a bit rusty, but I think the odds of that are (11 choose 1)*0.9^10*0.1 = 11*0.9^10*0.1 = 38%, which looks pretty close to what you got. See my notes above.
2 Kills 48%
Ok, now this is the most "diffucult" one 
1-0.31-0.38=31%
42 attacks with 19 MGs
0 kill 0%
1 Kill 34%
2 Kill 71%
Amm... Binabik? 34%+71%=105%. Run the tests again when you are not under the influence
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supersonic
Famous Hero
being digested. E=mc^2, s=vt
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posted April 07, 2006 07:31 PM |
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Ok, I said it once and I am saying it again - don't overestimate mighty moos, because no matter how many of them you have, there's always a slight chance of no kills at all! It is simply because 0.9^x > 0 . X in this equation is the number of bulls and can be any natural number.
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I am having a new style 
Big, fat, naughty. Potential girlfriend - pm me. 
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