# B Mines in a crossroad

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1. Oct 25, 2017

### Kumar8434

I thought of this problem:

The roads AB and CD have block E in common. There are 6 blocks in road AB and 13 blocks in road CD.

Someone has planted a mine in some of the blocks. He gives us this information:

1. There's only one mine in road AB in one of its blocks.
2. There's only one mine in road CD in one of its blocks.

Clearly, if there's a mine in block E then it satisfies both of the above conditions, hence there are no more mines in the roads. And, if there's no mine in E then there must be two mines, one in each road.

What is the probability of that there's a mine in E? If we consider only road AB, then the probability that E has the mine is $\frac{1}{6}$. And, if we consider road CD, then it's $\frac{1}{13}$. How do we combine these two probabilities? I don't think it should be the product of these probabilities, because the product is smaller than either of the probabilities, and intuition tells us that the mine is more likely to be in E than the rest of the blocks in road CD and less likely to be in E than the rest of the blocks in road AB. I think it should be the arithmetic mean of the probabilities. But how?

2. Oct 25, 2017

### Staff: Mentor

It is impossible to tell without knowing the rules how the mine(s) were placed.

I picked a number at random. What is the probability that I picked 3?
If I don't tell you how I picked that number (e.g. from 1 to 5? From 1 to 100?) you can't assign a reasonable probability value to it.

3. Oct 25, 2017

### Kumar8434

In your example, the probability that you picked 3 is 0.
I've inserted a picture and have written two rules about how the mines were placed:

1. There's only one mine in road AB in one of its blocks.
2. There's only one mine in road CD in one of its blocks.

And we know the number of blocks in each road.

Is this information not sufficient to calculate probabilities?

4. Oct 25, 2017

### Staff: Mentor

These are not rules about how mines were placed, these are rules telling us something about the positions later.

I picked a number at random. The number is not 4. What is the probability that I picked 3?

Consider the following scenarios ("random" means with a uniform distribution here):
"Pick a random block of street AB, put a mine there. If it is not at E, put an additional mine at a random other block of road CD" -> probability 1/6
"Pick a random block of street CD, put a mine there. If it is not at E, put an additional mine at a random other block of road AB" -> probability 1/13
"Put a mine under block E" -> probability 1
"Put a mine at the blocks closest to A and C" -> probability 0
All four scenarios are consistent with the given information, and it is easy to make up even more like them. How likely are these different scenarios? We have no way to tell.

5. Oct 25, 2017

### Kumar8434

I think the person who put the mines could have followed this rule:
Number the blocks on road A from 1 to 6, up to down.
Number the blocks on road B from 1 to 13, left to right.
Block E is numbered 2 on road A and 7 on road B.

Now, roll two die simultaneously. Die P has 6 faces. Die Q has 13 faces. If Die P shows 2 or Die Q shows 7 or both, then put a mine on block E and leave the other blocks alone. If both the dice show some other numbers, then put two mines on the corresponding blocks.

6. Oct 25, 2017

### Kumar8434

Maybe the law of total probability can be used.

A= put a mine in block E.
$E_1$ = put a mine in a random block on road AB

$E_2$ = put a mine in a random block on road CD

The person who put the mines could have made any decision between $E_1$ and $E_2$. So, $P(E_1)=P(E_2)=\frac{1}{2}$.

So, $P(A)= P(E_1)P(A|E_1)+P(E_2)P(A|E_2)$

$P(A)=\frac{1}{2}\frac{1}{6}+\frac{1}{2}\frac{1}{13}$

$=\frac{1}{2}(\frac{1}{6}+\frac{1}{13})$

which is the arithmetic mean of the two probabilities.
But I think this has flaws.

7. Oct 25, 2017

### SSequence

I suppose the following could also be another "reasonable" way to place the mines:
(1) Count the total number of blocks and then place first the mine in one of the blocks (with uniform probability). Since the number of blocks is 18, after the first mine is placed, each block has 1/18 probability of having a mine.

(2a) If the first mine is in block E, don't place any second mine.
(2b) If the first mine is on the road AB (but not on the block E), then place the second mine with uniform probability on any of the remaining blocks. In this case 12 blocks will be left as candidate, so we get 1/12 for the second mine on any of these blocks.
(2c) If the first mine is on the road CD (but not on the block E), then place the second mine with uniform probability on any of the remaining blocks. In this case 5 blocks will be left as candidate, so we get 1/5 for the second mine on any of these blocks.

==========

It would be interesting to see whether this method leads to identical or different probabilities compared to other "reasonable" methods for placing the mines.

While it seems that the obvious way to make the question concrete might be by fully specifying the process of placing the mines in advance, I think one might also try to see whether different reasonable methods/processes of placing the mines lead to same answers or not.

I think the above two methods give a special preference to one street over the other. It does seem conceivable that more "symmetric"/uniform sort of processes might lead to same answers (but this could also be false too I think).

Last edited: Oct 25, 2017
8. Oct 25, 2017

### Staff: Mentor

Could have - sure. But we don't know.
Well, it is your question, if you decide that, sure, but in the way the first post is written we don't know.

9. Oct 25, 2017

### Kumar8434

We calculate probabilities based on what we know and not on what more we could know.

Suppose we are asked to choose any number between 1 to 100. One correct number wins. So, for all we know, we have a $\frac{1}{100}$ chance of winning. But if we knew that the person, who designed the game, rolled a 6 faced die to decide the winning number between 11, 32, 56, 13, 78, and 100, then we'd know that each of these numbers have $\frac{1}{6}$ chance of winning and the rest of the numbers have 0 chance of winning.

10. Oct 25, 2017

### Kumar8434

I think there's not enough information in this problem to calculate probabilities. Thank you all.