schnurrito@discuss.tchncs.de to xkcd@lemmy.worldEnglish · 4 个月前xkcd #3015: D&D Combinatoricsxkcd.comexternal-linkmessage-square41fedilinkarrow-up1170arrow-down14file-text
arrow-up1166arrow-down1external-linkxkcd #3015: D&D Combinatoricsxkcd.comschnurrito@discuss.tchncs.de to xkcd@lemmy.worldEnglish · 4 个月前message-square41fedilinkfile-text
Look, you can’t complain about this after giving us so many scenarios involving N locked chests and M unlabeled keys. https://explainxkcd.com/3015/
minus-squareDragon Rider (drag)@lemmy.nzlinkfedilinkEnglisharrow-up9·4 个月前The math is easier if you look for the probability that the arrows are safe. If you do it that way, it’s simply 1/2*4/9, or 2/9.
minus-squareRentlar@lemmy.calinkfedilinkEnglisharrow-up3·4 个月前That is true. Cueball had asked the question in the way I described, but the check was done in the inverse which is easier to do the math on.
The math is easier if you look for the probability that the arrows are safe. If you do it that way, it’s simply 1/2*4/9, or 2/9.
That is true. Cueball had asked the question in the way I described, but the check was done in the inverse which is easier to do the math on.