In my effort to work on my probability and combinatorics skills before the semester starts, I have come across this problem:
What is the probability that a bridge hand will contain 13 cards of the same suit?
In my opinion, the answer should be 4/(52C13), i.e. 6.3*10^(-12), which is consistent with this Question and many others of the kind but apparently, the correct answer is: 5.25*10^(-13)
What am I doing wrong? I am probably confused because, well, frankly speaking, my understanding of the game of bridge is limited to everything I learned about it from probability problems.
Under the usual assumptions, the probability that the South hand has $13$ cards of the same suit is indeed $\frac{4}{\binom{52}{13}}$.
For there are $\binom{52}{13}$ equally likely ways to choose the South hand, and $4$ of them are "favourable."
The probability is about $6.3\times 10^{-12}$. (I have relied on software to compute $\binom{52}{13}$.)
We could interpret the problem differently, and ask for the probability that at least one of the four hands consists of cards of the same suit. But this probability is clearly bigger than the probability that South's cards are all of the same suit. The calculation for the probability is interesting, for multiplying our answer by $4$ overcounts the situations in which more than one hand is one-suited. So we need to use Inclusion/Exclusion.
I cannot think of a variant of the problem that produces a probability of size about $5.25\times 10^{-13}$.