Probability of a boring afternoon

Question

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.

Side note: You are probably the highest-ranked user on this website who is "extremely bad in probabilities (as in so many other areas)". — barak manos, Oct 26 '15 at 10:24
Side note #2: It would be nice if you listed the rules of Bridge that are relevant to this question. — barak manos, Oct 26 '15 at 10:25
@barakmanos. This is the privilege of age (if any !). I know now that I know less than $\epsilon$. — Claude Leibovici, Oct 26 '15 at 10:26
@barakmanos. Each of the four players is given $13$ cards (among $52$) and the strength of your hand is primarly defined by the number of points which is $4\times N_A+3\times N_K+2\times N_Q+1\times N_J$. That's all. — Claude Leibovici, Oct 26 '15 at 10:30
+1 even though I strongly disagree this is off-topic. BTW, what is $n$? — achille hui, Oct 26 '15 at 14:22

achille hui · Answer 1 · 2015-10-26T15:23:46.207

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.

@BarryCipra okay, I'm missing the $n = 28$ part in question. — achille hui, Oct 26 '15 at 15:24
@user281591 It is a standard technique of constructing generating function for combinatorial problems. Suppose you have a set of $n$ elements $X = { x_1, x_2, \ldots, x_n }$ and for each element $x_i$, you have several types of weights $a_i, b_i, \ldots, c_i$ associated to it. If you want to count the number of subsets of $X$ such that the total weight for $a_i$ is $a$, for $b_i$ is $b$, $\ldots$ and for $c_i$ is $c$. You can expand the polynomial $$\prod_{i=1}^n \left(1 + u^{a_i} v^{b_i} \cdots w^{c_i}\right)$$ and the coefficient for the term $u^a v^b \cdots w^c$ is the number you want. — achille hui, Oct 27 '15 at 12:40

true blue anil · Answer 2 · 2015-10-26T14:25:32.540

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 !

score 1 · Answer 3 · answered Oct 26 '15 at 14:46

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!

Probability of a boring afternoon

3 Answers3