We go back in the time, to the old Greeks. In that time an unknown commander, lived. This man was a real fighter. Loved wars, feared little, and had only once lost a battle. That was in Egypt, never earlier he had this way seen an armed force. That bothers him. He keept dreaming about if for months, he want to put an end on it.
He would invade Egypt once again but this time he will win. Just to be sure he went to the Oracle... ...
As a reward he took a leaden cube, which he had plunged in gold so that it seemed it was made of gold.
The oracle immediately noticed the counterfeit but wasn't angry at all.
He told the man: Cut that cube in 27 equal smaller ones. Put all the cubes in a red bag and blindly take one out. Throw that cube away as if it's a die.
As the upper side is made of gold, you will win against the Egyptians.
However, as the upper side is made of lead, you will suffer the most terrible perdition you can imagine.

Question: What is the chance that the upper side of the die throwed by the commander, is made of gold?

 

1 in 3 chance the face will be gold. Or, 2 to 1 in favour of lead.

Rational,
The large cube has 9 smaller gold faces on each of its six faces. Therefore, there are 54 gold faces.
The big cube slices into 27 small cubes with 6 faces each.
Total faces 162.
Odds are 54 in 162, which is 1 in 3.

Mike.

 

I got 1/3 too.

My calculations:

P["he picks a cube with 1 gold face"] = 6/27
P["he picks a cube with 2 gold faces"] = 12/27
P["he picks a cube with 3 gold faces"] = 8/27

P[gold] = 1/6*6/27 + 2/6*12/27 + 3/6*8/27 = 1/3
P[lead] = 1 - P[gold] = 2/3

 

Although previous result are the same I have some doubts about it because:

With Pommie's reasoning don't you omit the fact that the commander has to throw the cube and need a golden face upward?
I mean you're right saying that there are 162 faces in the bag, 54 of them are gold but some of them "stick" together in one cube.
Isn't it so that with your reasoning there are "faces" in the bag, gold ones end lead ones, and not cubes with some golden faces?

With eng1 reasoning I miss the 3.7% chance the commander has to pick the cube with no golden face at all. What happens with those 3.7%? I think you can't just forget about them in your P[gold] addition just because, when picking that cube, he has 0% of gold upward.

 

I took the cube with no gold into account, in fact in my P[gold] calculations the denominator is 27 (not 26) and the chance of getting a gold face from that cube is 0.
As a proof, let's calculate P[lead] explicitly:

P[lead]= 1/27 + 5/6*6/27 + 4/6*12/27 + 3/6*8/27 = 1/27+5/27+8/27+4/27 = 18/27 = 2/3

(*) 1/27= 6/6*1/27

 

Following that line of reasoning, the following is true:
1) Eight of the cubes have 3 gold faces
2) Twelve of the cubes have 2 gold faces
3) 6 of the cubes have 1 gold face
4) One of the cubes has no gold faces

You could calculate the odds for each case and sum them:
[(3/6)*8 + (2/6)*12 + (1/6)*6 + (0/6)*1]/27 = 1/3

Same answer. The other way (54/162) is much simpler, and to me, more intuitive.

EDIT: I guess this is basically the same method eng1 used to explicitly calculate P[lead].

__________________
Ron (aka Rube)

 

I also don't think one has to consider the case of multiple gold faces on a cube instead of just the total number of gold faces vs black faces.

Although a cube has 6 faces, the problem effectively ignores five of them, regardless if they are gold or black.

Its like picking 6 sealed envelopes and then open just one. The chance of success is the same as picking just one envelope.

__________________
L.Chung

 

If you do not consider the distribution of the gold faces, the answer is always 54/162 = 1/3.
Suppose that eight cubes have 6 gold faces. The total number is still 54.
What's the probability in that case? I think it is 8/27.

 

Ok, enough with the math already, I want to know, did the Commander kick some Egyptian Butt or not

Lefty

__________________
Measurement changes behavior

 

Yes in 1 universe and no in the other two.

Mike.