@[email protected] to xkcdEnglish • 4 days agoxkcd #3015: D&D Combinatoricsxkcd.comexternal-linkmessage-square43fedilinkarrow-up1399arrow-down17file-text
arrow-up1392arrow-down1external-linkxkcd #3015: D&D Combinatoricsxkcd.com@[email protected] to xkcdEnglish • 4 days agomessage-square43fedilinkfile-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)linkfedilinkEnglish9•4 days agoThe 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-squareRentlarlinkfedilinkEnglish3•4 days agoThat 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.