Probability of a boring afternoon I am extremely bad in probabilities (as in so many other areas) and I fully understand that this question is off-topic and/or missing context (may I underline that this is not homework).
Could you tell me how to compute the probability that, in a bridge tournament consisting in $n$ deals, it can happen that I never get more than $p$ points in my hand ?
If needed, each player has $13$ cards in hand and the points are counted as ${Ace}=4$, $King=3$, $Queen=2$, $Jack=1$.
This happened to me last week with $n=28$ and $p=8$ and this has been a very boring tournament for me (not for my opponents !).
Thanks in advance.
 A: For a single round, the situation is not that uncommon at all. If you draw $13$ cards from a suit of $52$ cards. The number of ways to get $p$ points, $\mathcal{N}_p$ is given by the expression
$$\mathcal{N}_p =[s^p][t^{13}] \bigg((1+s^4t)^4 (1+s^3t)^4 (1+s^2t)^4(1+st)^4(1+t)^{36}\bigg)$$
where $[x^k]$ stands for extracting the coefficient for the term $x^k$ in a polynomial in $x$.
Instead of expanding it by hand, I throw above expression to an CAS and get
$$\mathcal{N}_8 = 56466608128$$
The corresponding probability is 
$$\frac{\mathcal{N}_8}{\binom{52}{13}} = \frac{56466608128}{635013559600} \approx 
0.08892189351605147$$
i.e. you have less than $1\%$ of chance to get a hand with $8$ points. However, this is not that uncommon because the possible range of $p$ is from $0$ to $37$.
For a fair comparison welther your hand is bad or not. One should compute the
probability you get a hand like or even worse than what you get. It turns out
the probability of getting a hand of at most $8$ points is
$$\frac{\sum_{p=0}^8 \mathcal{N}_p}{\binom{52}{13}} = \frac{237982921392}{635013559600} \approx 0.3747682514715234$$
i.e. you have about $37\%$ of chance to get what you have or even worse.
So if you get a hand with $8$ points or less, this is nothing uncommon at all.
However, if you entering a tournament with $n = 28$ rounds and never get a hand with more than $8$ points, assuming all the rounds are independent, the probability is (as pointed out by another answer)
$$0.3747682514715234^{28} \approx 1.162406926911035 \times 10^{-12}$$
which is ridiculous small.
A: As an example, let me illustrate how to compute for your luckless run. Denoting the 36 non honor cards as N, you can get $8$ points in various ways:
$2A, 11N: \dbinom42\dbinom{36}{11} = 3604831776$
$1A, 1K, 1J, 10N:\dbinom41\dbinom41\dbinom41\dbinom{36}{10} = 16267958784$
$1A, 2Q, 10N: \dbinom41\dbinom42\dbinom{36}{10} = 6100484544$
$1A, 1Q, 2J, 9N: \dbinom41\dbinom41\dbinom42\dbinom{36}{9} = 9037754880$
We continue like this for all such hands that give 8 points and add them up to get the numerator, X
Total possible hands, $\dbinom{52}{13} = 635013559600 = Y$, say
$P$(get a hand of $8$ points once) $= \dfrac{X}{Y}$
$P$(your luckless run of $28) = \left(\dfrac{X}{Y}\right)^{28}$ 
Better luck in future ! ☺
ps:
Of course, you would be knowing best how to set it up for maximum efficiency !
A: According to Wikipedia, the probability of getting no more than $8$ "high-card" points in a bridge hand is approximately $37.48$%.  There is a link there to a site on high card expectancy which has a table listing the number of hands for each possible point total, from $0$ to $37$.
The probability of not getting more than $8$ points in $28$ hands is approximately
$$(0.3748)^{28}\approx1.165\times10^{-12}$$
so it would seem you were unlucky indeed!
