Six boxes

Q

quasiplot

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Hello, everyone. I'm having quite a bit of trouble with this, as my intuitive feeling tells me that I don't have the tools to approach this problem correctly. Here it is:

You have six boxes, 1, 2, 3, 4, 5, and 6. Contained within them are: 1 sphere, 2 (identical) cubes, and 3 (identical) cones. That is, one sphere is in one, one cube in another, and so on. By throwing a die, you are to open the box determined by how the die lands; if it lands on 1, then you open box 1. The task is to determine the probability that there will be containing exactly 1 sphere, 1 cube, and 1 cone remaining - there being three throws of the die.

Here's the sticking point: I know that one could compute the probability from the combinations for this problem (there being 20 combinations, and 6 of those being relevant, .3 would be the probability), but it would appear that having to throw a die introduces an additional constraint or conditionality of some kind. In other words, I don't think 3/10 is the real answer to this problem, but is something else.

Can anyone help me? Am I on the right track? If so, what must I do to get the correct answer?
 
There is a bit of confusion. If you roll a 1 and open that box, what happens to that box? Do you put it back or not? If you put it back, it's simple enough. If you don't put it back, what do you do if you roll another 1? I'm going to assume that they are not returned to their boxes and a re-roll is not a favorable outcome.

Tree

First 3 Branches:

1) Sphere (1/6)
2) Cube (2/6)
3) Cone (3/6)

Next

1) Sphere (1/6)
---1) Nothing (1/6)
---2) Cube (2/6)
---3) Cone (3/6)
2) Cube (2/6)
---1) Sphere (1/6)
---2) Cube (1/6)
---3) Cone (3/6)
---4) Nothing (1/6)
3) Cone (3/6)
---1) Sphere (1/6)
---2) Cube (2/6)
---3) Cone (2/6)
---4) Nothing (1/6)

It is rather troublesome that a selection that was not available in the first round "Nothing" becomes available later.

You do round 3.
 
Yes, the boxes will still be there after they've been opened, but the items themselves will be gone.

Apparently, it looks as though that with each successive step, the boxes that contain nothing will increase by 1, until by the last step, half of the boxes will have nothing.

So, it becomes:

SP---CU---CO---NO
1/6 ; 1/3; 1/2 ; 0/6

1/6 ; 1/6 ; 1/2 ; 1/6

1/6 ; 1/6 ; 1/3 ; 1/3

1/6 ; 1/6 ; 1/6 ; 1/2

The issue is how do I compute an objective result? I don't think computing the combinations successively makes much sense...
 
Hello, quasiplot!

You have six boxes, 1, 2, 3, 4, 5, and 6.
Contained within them are: 1 sphere, 2 identical cubes, and 3 identical cones, one per box.
By throwing a die, you are to open the box determined by how the die lands; if it lands on 1, then you open box 1, etc.

You throw the die three times.
Determine the probability that you will get exactly 1 sphere, 1 cube, and 1 cone.


Here's the sticking point: I know that one could compute the probability from the combinations for this problem,
there being 20 combinations, and 6 of those being relevant. .0.3 would be the probability.
But it would appear that having to throw a die introduces an additional constraint or conditionality of some kind.
In other words, I don't think 3/10 is the real answer to this problem, but is something else.

Can anyone help me? .Am I on the right track?
If so, what must I do to get the correct answer?
Click to expand...

Do you know the correct answer? .If you do, why keep it a secret?

There are other questions which arose . . .


Suppose the contents of the boxes are as follows:

. . \(\displaystyle \begin{array}{cc} \text{Box} & \text{Content} \\ \hline 1 & \text{sphere} \\ 2 & \text{cube} \\ 3 & \text{cube} \\ 4 & \text{cone} \\ 5 & \text{cone} \\ 6 & \text{cone} \end{array}\)


You said that, on a roll of the die, the indicated object is removed.

Suppose you roll a "1". .You take the sphere.

Suppose you roll another"1". .What happens?

Does this count as your second roll?
So you have only one roll left?

Or do you ignore this roll and take another second roll?

 
GEESHHHHH....why not 1 black candy, 2 white candies and 3 green candies ? On a diet?
 
It was not my intention to keep the "true" answer a secret. The answer I have is .3, but I am doubtful of that being the true answer.

However, this closer inspection of the empty-box seems to point to .3 as being the correct answer, even though the problem isn't on its face the same sort of combinatorial problem of just opening the boxes in a predetermined order (not involving a die). I think this because, as you mention soroban, rolling for a box that has already been opened can be ignored in the analysis since it has no effect (no change in the condition of the problem results) on the state of the problem.

I'm just trying to bounce my observations off anyone who has familiarity or more confidence in their computation of the answer.

As everyone knows, this problem would be like a simple combinatoric computation. Given 6 items, and the selection of 3 of those items at any given time, 20 combinations are possible, but only 6 count (1*2*3), so there is a 6 by 20 chance of the three different items being left over.

Here is the relevant formula:

COMBINATION = N! / ( S! (N - S)! )
N is the total number of items.
S is the number of items selected at a time.

Given this information, is there a flaw with my answer? Am I right to doubt its validity?
 
Denis said:
GEESHHHHH....why not 1 black candy, 2 white candies and 3 green candies ? On a diet?
Click to expand...

Well, we have 1 spherical candy, 2 cubical candies, and 3 conical candies! :p
 
Hello, quasiplot!

If the correct answer is 3/10. I think I know how we can get it.


\(\displaystyle \text{We can get a sphere, a cube and a cone in }3! = 6\text{ possible orders.}\)

\(\displaystyle \text{The probability of getting sphere-cube-cone (in that order) is: }\:\frac{1}{6}\cdot\frac{2}{5}\cdot\frac{3}{4} \:=\:\frac{1}{20}\)

\(\displaystyle \text{Therefore: }\:p(\text{sphere-cube-cone, any order}) \:=\:6\cdot\frac{1}{20} \:=\:\frac{3}{10}\)

 
soroban, yes, I understand that that is another way to get .3. However, I don't actually know what the real answer is. Do you suppose there is a way to determine that it is the real answer? That's what I'm asking, because I don't want to apply algorithms blindly. I want to make sure I'm not misunderstanding the problem.

Thanks!
 
Here's a new observation: there are 14 unique arrangements of three items among the boxes, and there are 6 different patterns of items for each pattern, so there are 6*14 = 84 different ways the final answer can be expressed. Taking n^r, I divide 84/216 to yield .38889.

Is there anything wrong with this approach? Is it better or more correct?

It's hard for me to say!
 
Denis,

I am surprised that you did not write a simulation program (one of those loopy things) and confirmed or debunked some of the answers.
 
quasiplot said:
Hello, everyone. I'm having quite a bit of trouble with this, as my intuitive feeling tells me that I don't have the tools to approach this problem correctly. Here it is:

You have six boxes, 1, 2, 3, 4, 5, and 6. Contained within them are: 1 sphere, 2 (identical) cubes, and 3 (identical) cones. That is, one sphere is in one, one cube in another, and so on. By throwing a die, you are to open the box determined by how the die lands; if it lands on 1, then you open box 1. The task is to determine the probability that there will be containing exactly 1 sphere, 1 cube, and 1 cone remaining - there being three throws of the die.

Here's the sticking point: I know that one could compute the probability from the combinations for this problem (there being 20 combinations, and 6 of those being relevant, .3 would be the probability), but it would appear that having to throw a die introduces an additional constraint or conditionality of some kind. In other words, I don't think 3/10 is the real answer to this problem, but is something else.

Can anyone help me? Am I on the right track? If so, what must I do to get the correct answer?
Click to expand...

I don't think the throw of the die introduce any other event or choices. Say if the die was a dodecahedron - then you would throw a wrench in the mix. As it is - the die is just tool to choose numbers from 1 to 6 - randomly.
 
Subhotosh Khan said:
Denis,

I am surprised that you did not write a simulation program (one of those loopy things) and confirmed or debunked some of the answers.
Click to expand...
can't...still standing in the corner :cry:
 
So, now what? Am I to sit here and agonize over my answers?

If no one wants to confirm or invalidate my answers (30% or 38.9%), then please tell me what programs are out there that I could use to look into this problem. I want it solved for good.
 
Okay, I've determined there are exactly 55 unique arrangements of items:

x, y, y, z, z, z

{x,y,y,z,z,z} {x,y,z,y,z,z} {x,y,z,z,y,z} {x,y,z,z,z,y} {x,z,y,y,z,z} {x,z,y,z,y,z} {x,z,y,z,z,y} {x,z,z,y,y,z} {x,z,z,y,z,y} {x,z,z,z,y,y} {y,x,y,z,z,z} {y,x,z,y,z,z} {y,x,z,z,y,z} {y,x,z,z,z,y} {y,y,x,z,z,z} {y,y,z,x,z,z} {y,y,z,z,x,z} {y,y,z,z,z,x} {y,z,x,y,z,z} {y,z,x,z,y,z} {y,z,x,z,z,y} {y,z,y,x,z,z} {y,z,y,z,x,z} {y,z,z,x,y,z} {y,z,z,x,z,y} {y,z,z,y,x,z} {y,z,z,y,z,x} {y,z,z,z,x,y} {y,z,z,z,y,x} {z,x,y,z,y,z} {z,x,y,z,z,y} {z,x,z,y,y,z} {z,x,z,y,z,y} {z,x,z,z,y,y} {z,y,x,y,z,z} {z,y,x,z,y,z} {z,y,x,z,z,y} {z,y,y,x,z,z} {z,y,y,z,x,z} {z,y,y,z,z,x} {z,y,z,x,y,z} {z,y,z,x,z,y} {z,y,z,y,x,z} {z,y,z,y,z,x} {z,y,z,z,x,y} {z,y,z,z,y,x} {z,z,x,y,y,z} {z,z,x,z,y,y} {z,z,y,x,z,y} {z,z,y,y,x,z} {z,z,y,z,x,y} {z,z,y,z,y,x} {z,z,z,x,y,y} {z,z,z,y,x,y} {z,z,z,y,y,x}

If I multiply this by 6, then I get all of the possible combinations that could result. Does that sound right?
 
Actually I just realized that list hasn't been vetted for cyclic permutations. I'll have to cut those out before the list is truly representative.
 
quasiplot said:
Okay, I've determined there are exactly 55 unique arrangements of items:
x, y, y, z, z, z
{x,y,y,z,z,z} {x,y,z,y,z,z} {x,y,z,z,y,z} {x,y,z,z,z,y} {x,z,y,y,z,z} {x,z,y,z,y,z} {x,z,y,z,z,y} {x,z,z,y,y,z} {x,z,z,y,z,y} {x,z,z,z,y,y} {y,x,y,z,z,z} {y,x,z,y,z,z} {y,x,z,z,y,z} {y,x,z,z,z,y} {y,y,x,z,z,z} {y,y,z,x,z,z} {y,y,z,z,x,z} {y,y,z,z,z,x} {y,z,x,y,z,z} {y,z,x,z,y,z} {y,z,x,z,z,y} {y,z,y,x,z,z} {y,z,y,z,x,z} * {y,z,z,x,y,z}
Click to expand...
Methinks there's 60.
As example, this one was missed: * y,z,y,z,z,x :...too lazy to find the others !
 
Yes, there's actually 60. But one has to eliminate the cyclic permutations to obtain the truly "distinct" ones.
 
quasiplot said:
The task is to determine the probability that there will be containing exactly 1 sphere, 1 cube, and 1 cone remaining - there being three throws of the die.
Click to expand...
Whadda heck does that mean? :shock:
The REMAINING (UNOPENED) boxes will contain these?

Doesn't matter anyway; still .30

The use of a dice here simply serves to "confuse".
"Pick 3 boxes at random" is same thing...
 
Denis said:
The REMAINING (UNOPENED) boxes will contain these?

Doesn't matter anyway; still .30

The use of a dice here simply serves to "confuse".
"Pick 3 boxes at random" is same thing...
Click to expand...

Yes, that's exactly what was meant, or that is what the problem is.

I guess 30% is the right answer and my second-guessing was wrong. Hrm.
 
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