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carapace: 2013-01-29 06:09:08 am
As I´ve been curious for a long time why it´s so rare to get atleast 4-1-1 at Palace 1 crystal, I decided to break down the probability from a statistical viewpoint.
These calculations assume you do not lose exp to tinsuits and that you have 2 chances on a 200 pbag and 3 chances on a 50 pbag.

Case 1: You get 2x200 pbags. If the probability of getting a pbag is 1/2, the probability of getting 2x200 pbags is 1/2 * 1/2 = 1/4 or 25%.

Case 2: You get 1x50 pbag and 1x200 pbag. If the probability of getting a 50 pbag is 1/8, it makes the probability of getting atleast 1x50 pbag 1-(7/8)^3 = 33%. But since you still need 1x200 pbag (exactly 1x200 pbag has a probability of 1/2 to occur) you must multiply the chances of getting 1x50 pbag (which was 33%) with 1/2 (50%). 0.33 * 0.5 = 16.5%.

Ergo: The chances of getting 4-1-1 is therefor 16,5%+25% = ~41,5%.

I made two savestate in the first cave. One savestate where a dagger was in the air, about to hit an enemy with a drop. The enemy got killed at the exact same frame and pixel every time; the drop was the same every time. I got a pbag 50 times in a row reloading this savestate. The second savestate I made did not have a dagger in the air, but just when I loaded the savestate I manually fired a dagger. The drops from the second savestate were completely random.

Conclusion: The random number generator is most likely memory based, and will thus pick a number from the last line of RAM. So a dagger in the air, replayed a million times, will always yield the same result, because the random number generator will always pick the same number from the memory. This also means that it is extremely difficult to manipulate drops.
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PresJPolk: 2013-01-29 05:33:44 am
PresJPolk: 2013-01-29 05:33:41 am
PresJPolk: 2013-01-29 05:33:41 am
PresJPolk: 2013-01-29 05:32:52 am
PresJPolk: 2013-01-29 05:32:31 am
This math is inaccurate somewhere. Let me walk through it. :)

There are 4 possible combinations of Big 6 drops.  One of them guarantees Attack 4, two of them give you a shot at Attack 4, and one of them precludes attack 4.

When you get one 200 Pbag, it's actually easier to calculate the probability of *not* getting a random 50, and inverting it: 1-(7/8)^3 is your shot at getting a 50.

So, your chance of getting 4-1-1 is 1/4 + (1/2)(1-(7/8)^3 ) = ~41.5%.

Oh, I see what you did! You took a 3/4 chance of getting 1 200.  It's actually a 3/4 chance of getting AT LEAST 1 200 pbag. But you already used up the 2-200 combination in the 4-1-2 case, so you're double counting that.
Yea, it was a long time since I did statistics and I was actually hoping somebody would find an error. I was thinking to myself that it felt weird a to count that scenario twice, but I wasn´t sure. Thanks for refreshing my memory! I am going to correct the math in my first post.
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Svenne: 2013-01-29 12:06:52 pm
One of the more reasons why I actually hope to get some more bots in my way. So I get a bigger chance on the 50 bag. I don't like killing one of those skulls for 50 points in a run: In races I rather just take the 3-1-1 then hit that guy 100 times! XD
58.5% chance of it being no fun.