What is the probability of having a pentagon in 6 points

If the probability that $5$ random points in the plane whose horizontal coordinate and vertical coordinate are uniformly distributed on the interval $\left(0,1\right)$ occur to be the vertices of a convex pentagon is $\frac{49}{144}$, what is the probability that a subset of $6$ random points in the plane whose horizontal coordinate and vertical coordinate are uniformly distributed on the interval $\left(0,1\right)$ occurs to be the vertices of a convex pentagon? Thanks a lot.

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By the way, I mean the exact value of the probability in the question. Thanks. – kejma Jan 18 '13 at 5:27
Where does the $49/144$ come from? Maybe the place where that's proved would be a good place to start on the 6-point problem. – Gerry Myerson Jan 18 '13 at 6:21
It would have made sense to provide a link to this related question of yours, both initially and in particular in response to @Gerry's question. – joriki Jan 18 '13 at 7:26
I think this is considerably harder than the corresponding question with $5$ and $4$ points that you posed in the comments under my answer to the other question, because if the convex hull is a quadrilateral the remaining two points may or may not be part of a convex pentagon, and if the convex hull is a pentagon, the remaining point may be part of $0$, $1$ or $2$ convex pentagons; I think these cases will be hard to deal with. In case someone does come up with an answer, you can check it against the value $0.7565$ estimated by this code. – joriki Jan 18 '13 at 8:11
Also posted to MO, mathoverflow.net/questions/119300/… – Gerry Myerson Jan 19 '13 at 3:49

Your original set has cardinality 6, your target set has cardinality 5, set member order does not matter, so you have $C^6_5=6$ possible subsets. Each of these has a $\frac{49}{144}$ probability of forming a convex pentagon. What is the probability that at least one of these forms a convex pentagon? It is 1 minus the probability that none of them form a convex pentagon:
$1-(1-\frac{49}{144})^6\approx1-65.97\%^6=1-8.2445\%=91.76\%$
To see that this argument does not work, consider the same situation when we ask what is the probability that out of $5$ random points we can pick $4$ which are in convex position. Your argument would give $1-(1-p_4)^5<1$, where $p_4$ is the probability that $4$ random points are in convex position (it is strictly smaller then $1$). But this is false: Every set of five points in general position contains the vertices of a convex quadrilateral. (see en.wikipedia.org/wiki/Happy_ending_problem) – Gilles Bonnet May 9 '14 at 19:58
Even if we where talking about whether the $6$ random points are independent among themselves, I don't think assuming that has anything to do with Occam's razor. – user133281 Jun 21 '14 at 12:55