I have question at probability that I need to use the "Inclusion–exclusion principle"...
Hand of bridge is 13 cards that picked up randomly. What is the probability that we will have a King and Ace from the same kind?
I have one question before: I try to find what is the probability that we have at a hand heart King and Ace?
Thank you!!
There are $\binom{52}{13}$ bridge hands. We assume they are equally likely. (Since real shuffling is not an reliable randomizer, this is not necessarily a realistic assumption.)
Now we count the "favourables," the hands that have a King and Ace of the same kind.
Consider first the hands that have both the $\heartsuit$ King and the $\heartsuit$ Ace. These are easy to count: The remaining $11$ cards can be chosen in $\binom{50}{11}$ ways.
Similarly, there are $\binom{50}{11}$ hands that contain both the $\spadesuit$ King and the $\spadesuit$ Ace. We get the same counts for $\diamondsuit$ and for $\clubsuit$.
So it is tempting to say that the number of favourables is $4\binom{50}{11}$.
However, when we add our counts of $\binom{50}{11}$ for $\spadesuit$ and $\binom{50}{11}$ for $\heartsuit$, we are double-counting the hands that have King and Ace of **both $\spadesuit$ and $\heartsuit$. There are $\binom{48}{9}$ of these. Similarly, in $4\binom{50}{11}$ we have double-counted the hands that that have King and Ace of possible combination of two suits. So we must subtract $6\binom{48}{9}$.
But we have subtracted too much! For we have subtracted once too many times the hands that, for example, contain King and Ace of all of $\spadesuit$, $\heartsuit$, and $\diamondsuit$. We leave it to you to compute how much we should add back to compensate.
And if you are continuing the above pattern, you will have added back too much, for you will have counted more than once the hands that have all the Kings and Aces. Consider carefully how much we must subtract.