Thezzaruz wrote:detlef wrote:
I'm sorry, I missed the day in class when "very, very slim" was specifically defined. I fail to see what sort of argument you're trying to make if the best you can do is question whether a proposition that fails over 90% of the time is not "very, very slim".
I agree, discussing semantics won't do us any good. But as you still haven't realized that the way you tried to calculate the original question is wrong I'll give it one last try.
Take a 100v100 battle and ask "what is the odds of being left with at least 25% of the armies?". We'll start by doing it your way (splitting it into several small battles, lets say 5 this time) and that gives us a 45.15% chance of ending with at least 5 armies. That gives us 0.4515 x/y 5 = 0.0187, i.e almost a 2% chance. And the odds when doing a straight calculation is (as we know from earlier in this thread) 25.97%.
So we have:
(5 left in 20v20) * 5 = 2%
25 left in 100v100 = 26%
That huge difference means that your way of doing it isn't a good or even a reasonable estimate but rather incredibly wrong both intuitively and mathematically.
Well then, I'll concede here. Though I don't think Pi's assertion that losing 120 is "average" is correct either. According to another odds calculator, attacking 3v2, the attacker loses .85 armies for every 1 army the defender loses. Thus, "average" losses on such attack would be 505 or 95 armies left over. So, by that, MeDeFe ended up with more than 50% more guys than he "should " have. So, I tried to think about what could be wrong with this and the first thing that came to mind is the linear nature of the equation vs the fact that the attacker has an ever increasing armies advantage over the defender. However, regardless of how many more guys he has, he's still rolling 3v2, so assuming that the .85 number is true, he's never going to out pace that ratio until he's rolling 3v1 which won't happen until the very last roll or couple. Ironically, the longer he's given the advantage of the increased odds of 3v1, the worse his winnings will be.
I thought I'd try another stab at approximating the chances. I took a smaller battle 150 v 150 and broke that up into 6 pieces like I tried with the first and looked at how many left over after a 25v25 would have the same odds of having 25 after 100v100. At 25v25, there's a 27% of having 9 left over. So, I ran a 150v150 to see what the odds of having 54 left over on that attack. It came to 3%. Which is obviously several orders of magnitude more likely than the .02% that I originally thought.
There seem to be two factors at play here. One being that there's a rather unavoidable trend for the attacker to amass extra armies vs the defender. However, there's also the fact that the longer you roll dice, the more likely you are to trend back to neutral odds. So, any proposition requiring favorable odds is less and less likely to work out over the long haul.
Doing something that has a 1 in 4 chance of working out once is no big deal. However, defying those same odds continually over a long period of time is.
The first factor obliterates my theory, but the second seems to do the same for Pi's
Of course, as has been rather exposed by Thezzaruz, my math is incredibly rusty.
At first glance that didn't seem so extraordinary as most people tend to forget how much the attacker is favored in large battles. I dropped it into the odds calculator but it can't handle battles that large so I broke it down into 100 v 100 battles looking at what the odds would be of ending up with an extra 25 guys per batch of 100. Well, long story short, that's some nice freaking dice! Though I'm sure there's some reason why working it out this way isn't 100% accurate, it would have to be way off. After all, I came up with something like .02% chance of that happening! 
)