Talk:Charging

From NetHackWiki
Jump to: navigation, search

When we say "blessed charging" do we mean only the item doing the charging - say a scroll of charging - is blessed or does the BUC of the item to be charged factor in any case other than the lanterns? --FJH 14:16, June 1, 2010 (UTC)

Only the BUC status of the charging scroll (or PYEC) matters. Tjr 14:31, June 1, 2010 (UTC)

"Unlike oil, however, regular charging has no ill effect on a magic lamp."

What "ill effect" does oil have on a magic lamp, exactly? Does it turn it into a normal lamp? --Kahran042 15:52, August 18, 2010 (UTC)

yesTjr 17:20, August 18, 2010 (UTC)
Are you sure? I just tried it in wizard mode, and it doesn't seem to do anything special. -Ion frigate 19:13, August 18, 2010 (UTC)

Cumulative estimate

"And thus the expected average, per wand, is 4.62 charges before the wand explodes."

This was a somewhat tentative estimate; while the other was fairly easy to calculate (just the odds of it never exploding), this one was driving me nuts because of NetHack clipping it. Pretty sure the overall odds would be whatever averages to 50%*. It's just basic algebra to reduce this:

0.5 = \frac{x^3}{343}

0.5 \times 343 = x^3

171.5 = x^3

\sqrt[3]{171.5} \approx 5.555903682

However, since NetHack clips it (there is no way to charge something 4.2 times, for example) it muddies things. I just interpolated between the two known points. Something about that figure (the LaTeX above) just seems too high. I thought to be bold and just add the interpolated figure—if it's not correct, it's probably also not that far off.

* Through iteration, I'd be here until the end of time trying to get the average otherwise; nothing prevents it from failing at 0.29% forty billion times in a row, spare RNG's period.

Feagradze 05:47, June 26, 2010 (UTC)


The point at the 50% chance would actually give you the median, rather than the mean. I just worked out the correct mean to be about 5.2334. The way to calculate it is:

(chance of success at 0)*(1+(chance of success at 1)*(1+(chance of success at 2)*(...)))

or

(chance of success all the way to 1)+(chance of success all the way to 2)+(chance of success all the way to 3)+...

I have put it in. 220.255.2.119 16:15, 20 May 2012 (UTC)