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There is a square of side $n$ units. Join the diagonals. Now the square is divided into 4 regions of equal area. Each of them is coloured differently. Given 2 points that can lie within any of the four regions, what is the probability that two points are in the same coloured region?

The problem:

I and my friend get two different solutions using two different lines of reasoning. It has to do with the order of the points being placed:

Solution 1 (mine): The order in which the points are placed does not matter. The number of ways 2 identical objects can be placed in 4 distinct regions is 10 (stars and bars). Out of those, there are 4 cases which satisfy the condition. Hence the probability is $2/5$.

Solution 2 (my friend's): The order of the points being placed does matter. That means there are 16 ways to place the points (4 orientations × 4 regions). Out of those, 4 cases satisfy the condition. Hence the probability is $1/4$.

Now of course, one line of reasoning must be wrong here. Which one is wrong? An explanation is appreciated.

Parcly Taxel
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user406287
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  • Presumably each point has a 1/4 chance to be in any given region, independently of the other point? Hence the probability for both points to be in region 1 should be 1/16. The problem about your approach is that it does not give that. You would calculate the probability of both points being in region 1 as 4/10. – almagest Feb 20 '18 at 13:03
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    "The number of ways 2 identical objects can be placed in 4 distinct regions is 10" are all those ways equally likely? – Matthew Towers Feb 20 '18 at 13:04
  • Your method assumes that the probabilities for different possible pairings are uniformly distributed over these 10 cases. This is not what you are given - you are given that the probabilities for single points is uniformly distributed. In fact, these probabilities are not uniformly distributed, which is what gives you your mistake – John Doe Feb 20 '18 at 13:05
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    Ah, I think I understand now. Say the four regions are $a,b,c,d$, then the points are twice as likely to not be in the same region because for example $ab,ba$ (2 possibilities) but only $aa$ (1 possibility). Is that right? EDIT: Yes it is. Thank you everyone. – user406287 Feb 20 '18 at 13:08

1 Answers1

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The first solution is wrong and the second is correct. Whichever region the first point lies in, there is a $\frac14$ chance the second point will be in that same region.

The first solution assumes that the probabilities of having the two points in one given region and those of having them in two separate, given regions are the same, when in fact the second is twice the first.

Parcly Taxel
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