# Third of Many: Rubies

"abdul the jeweler has in his home three chests of drawers; each chest contains two drawers. in one of the chests, each drawer contains a ruby. in another of the chests, each drawer contains an emerald, and in the third chest, one drawer contains a ruby and the other drawer contains an emerald. suppose you pick one of the three chests at random and open one of the drawers and find a ruby. what is the probability that the other drawer in the same chest will also contain a ruby?"

this one's hard.

journeyfulloflaughter 26-30, F 26 Responses Dec 31, 2007

First, I apologize for the weird alignment in the previous post; I hope it's still understandable.<br />
<br />
I can think of only one reason people are getting confused by this riddle:<br />
Readers aren't realizing they can only pull drawers FROM THE SAME CHEST. This results in a fixed possibility experiment instead of one that's truly random, like pulling marbles out of a bag. This is the same reason the experiment journeyfulloflaughter suggests is incorrect: the card experiment described will produce a 2 of 3 result but it is a different experiment given that the set up is completely random. To create an experiment to test the riddle and give an acurate probability one must take into account that after opening the first drawer only one of two outcomes is possible. There are three gems/cards left but again counting the 2nd ruby/red card is incorrect because of how the gems were placed into the drawers (fixed position) in the riddle originally. The original set up of the gems in the riddle combined with the condition that you may only open drawers from one chest creates the 50% probability. Journeyfulloflaughter, I don't mean to be condescending or snide but the explanation you posted is flawed. I don't know where you got that explanation but it is. For the record: I wouldn't have posted all this if I didn't know what I was talking about or felt I might be wrong. If anybody still has any questions as to how I got 50% please comment with your question and I'll be happy to answer it.

Chest 1 Chest 2 Chest 3<br />
R R E E R E<br />
Chest 2 can be eliminated from the calculations because it contains no rubies, leaving us with the following set of parameters:<br />
Chest 1 Chest 3<br />
R R R E<br />
Because you can only open the drawers of ONE chest: it must be one of these two possibilities (if you can't see why then you need to reread the riddle). Given that you have already pulled a ruby out of one of the drawers you are left with the following scenarios:<br />
Chest 1 OR Chest 3<br />
R E<br />
Two possible outcomes each with an equal chance of happening, thus a 50% probability the other drawer in the chest will contain a ruby. (One more post after this.)

I'm not trying to be stubborn (I read the long explanation) but the answer should actually be 50%. While yes between the two chests that do contain rubies there is a total of four gems (3 rubies 1 emerald) and you have already opened one drawer and found a ruby (leaving 3 gems), the fact is that you have only two possible outcomes for the other drawer: ruby or emerald. The answer is NOT 1 in 3 because because placement of the gems is not entirely random (i.e. having 3 gems in a bag). Counting the 2nd ruby is an incorrect calculation because you are working within a fixed pattern and set of conditions, meaning it doesn't count towards the probability. Diagram in next post.

The chance of finding another ruby in the 2nd drawer of the same chest is 1 in 3 tries.

it's supposedly 2 out of 3. read previous comments re: extended explanation.

That would be a 1 in 3 chance.

seriously. math hurts my brain. i like word based riddles better.

oh yea, know i understand.....<br />
<br />
oh wait, what is this black and red stuff oozing out of my ears????

<br />
"if you are still unconvinced, i suggest you try the following experiment: take 3 red cards from a deck (to represent the rubies) and 1 black card (to represent the emerald); shuffle the 4 cards so you don't know where the black one is. then, put the 4 cards face down on the table in two piles. suppose you now pick one of the 4 cards, turn it over, and find it to be red. you really believe that the chance of the other card in that pile being red is one half?<br />
<br />
if you are STILL not convinced, i suggest that you try the experiment 60 times, and see whether the number of times the other card is red isn't close to 40!"<br />
<br />
thus concludes the long answer. smullyan, the author, was a prof of mathemathical logic, so i'm trusting that he knows what he's talking about. i can't provide further explanation. i'll have to defer to myspip ;)

long answer part 2: <br />
<br />
"another way to look at it is this: the emerald is with equal probability in any of the four drawers. if you pick a drawer and find a ruby, then the emerald is with equal likelihood in any of the other three drawers, and so WHICHEVER second drawer you open--whether in the same chest or in the other one--the chances of finding an emerald are only 1 out of 3, and so the chances for a ruby are 2 out of 3.<br />
<br />
(to be cont'd again)

okay, here's the long answer (and it's long):<br />
"the correct answer is that the probability is 2 out of 3, though i will have trouble convincing some of you. <br />
<br />
let's look at it this way: the chest with the two emeralds is really out of the picture, so we can forget about it completely. we thus consider the R R chest (both rubies) and the R E chest (ruby, emerald). let R1 be the ruby in the top drawer of the R R chest, and R2 the ruby in the bottom drawer of the R R chest. Let R3 be the ruby in the R E chest. now, if you pick one of the four drawers at random and find a ruby, then the chances that it is R1, R2, or R3 are equal, so the chances are 2 to 1 that it is EITHER R1 or R2. thus, the probability is 2 out of 3 that the other drawer in the same chest is a ruby.<br />
<br />
(to be cont'd, afraid i'll be cut off)

2 in five chances

common sense vs statisical analysis = one hurting brain.<br />
<br />
If you picked a chest with a ruby, then you can not possibly have the chest with two emeralds. So, logically (as Sir Bedevere would put it), to figure "what is the probability that the other drawer in the same chest will also contain a ruby?" actually has only 2 choices of chests: one with '2 rubies' in it or one with 'one of both' in it.<br />
<br />

not correct, lm, sorry!

one in four

wow. again, lots and lots of math in those answers! but you are correct. the answer is that the odds are 2 out of 3. i had to read the detailed explanation a couple of times before i could be convinced it wasn't 50%. <br />
<br />
let me know if anybody wants me to post the long answer explanation!

should be 33.33/(33.33 + 16.67)=66.66% of course...<br />
Mathematical explanation:<br />
<br />
(1/3x1 + 1/3x0)/(1/6+1/6+1/6)<br />
(Chance to have chosen drawer A x Chance for to get second ruby in Drawer A + Chance to have chosen drawer C x chance to get second ruby in Drawer C )/(All chances to get to a ruby)

100 isn't right either, sorry!

100%<br />
Good one.

for my own reference, this is puzzle 43.

not quite right either...

Oh, I stink at these, but will say 2 in 5 (40%). :-)

on the right track, shoreguy, but still not quite there. it's a hard one.

33%. You ignore the first drawer and read it as "What's the chance of getting the two rubies."

no.

50%