Talk:Pokéwalker: Difference between revisions

Talk:Pokéwalker (view source)

Revision as of 06:28, 15 April 2010

1,768 bytes added , 15 April 2010

→‎Dowsing Probabilities

Exawatt

6

edits

@@ Line 330: / Line 330: @@
 Others I spoke to--and even myself--originally thought that the leftmost or rightmost grass tufts had better chances (when taking second pick into account), but I did the probabilistic math and it turns out that no matter which tuft you pick, your second pick choice will always give you the same opportunity. If that's not making sense; basically there's no difference in your overall chance of finding an item between ''any'' of the grass tufts. All assuming completely random item placement, of course. I can post the detailed math if anyone's interested. I know it's not really important, but it seems to be a point of confusion among people I've spoken to. Might be worth clearing it up in the argticle.. - [[User:Exawatt|Exawatt]] 19:41, 14 April 2010 (UTC)
 :Having worked it out myself, the probability of finding an item for the end most tufts are actually 1/2 compared to 2/3 for the 4 middle tufts. Therefore it is better to take the middle tufts. Sorry this is so unclear. Will post a more detailed argument later.--[[User:Beligaronia|Beligaronia]] ([[User talk:Beligaronia|talk]]) 21:34, 14 April 2010 (UTC)
+::I've already done it, it was just on another computer:
+::Assuming completely random placement of the item, there is no difference in probability based on the first chosen (i.e. a side vs a middle).
+:::Initial pick is easy:
+::::P(pick1) = 1/6
+:::If first pick fails--these are all P(*|¬pick1):
+::::If choosing the side:
+::::P(near|side) = 1/5, P(pick2|nearΛside) = 1,
+::::P(far|side) = 4/5, P(pick2|farΛside) = 1/4,
+::::P(pick2|side) = P(pick2|nearΛside) * P(near|side) + P(pick2|farΛside) * P(far|side)
+:::::= (1)(1/5) + (1/4)(4/5) = (1/5) + (1/5) = 2/5
+::::If choosing a middle square:
+::::P(near|middle) = 2/5, P(pick2|nearΛmiddle) = 1/2,
+::::P(far|middle) = 3/5, P(pick2|farΛmiddle) = 1/3,
+::::P(pick2|middle) = P(pick2|nearΛmiddle) * P(near|middle) + P(pick2|farΛmiddle) * P(far|middle)
+:::::= (2/5)(1/2) + (3/5)(1/3) = (1/5) + (1/5) = 2/5
+::::.'. P(pick2|side) = P(pick2|middle).
+::::Therefore, probability of finding the item after picking a side square is equal to
+::::the probability of finding the item after picking a middle square.
+::I've tried to explain it in English but I end up complicating it. Basically the additional knowledge of where the item is doesn't help, since it still has a 1/5 placement. If you get 1/5 and 4/5 of 1/1 and 1/4 chances, it's the same as 2/5 and 3/5 of 1/2 and 1/3. The probabilities sort of "cancel" each other... But like I said I'm terrible at explaining this in English. If there's a serious question about whether the math is right, I can double check it with my probability professor from last semester. If it comes to it I'll see if I can work with him to determine this and the expected return of the entire game. But if nobody cares I won't bother. - [[User:Exawatt|Exawatt]] 06:28, 15 April 2010 (UTC)