# Dating problem probability

When a coin is tossed, there are two possible outcomes: Each time Alex throws the 2 dice is an Experiment.

It is an Experiment because the result is uncertain.

How you estimate the size of your possible dating population is entirely up to your statistical skills and the level of your self-confidence, as is how you then collect your sample.

A “voluntary response sample” is generally regarded as socially acceptable, whereas a “stratified random sample” can land you in jail. Problems like deciding who to settle down with have the mildly disturbing math name of “optimal stopping problems.” The original optimal stopping problem was known as the secretary problem, and here it is as originally framed.

Many events can't be predicted with total certainty.

The best we can say is how likely they are to happen, using the idea of probability.

There are $n$ hats and each person picks a hat uniformly at random hence each gets their right hat back with probability $\frac1n$.

Let $q_k$ be the probability of picking the best provided we picked none of the first $k$.The famous 'Hat Check Problem' goes like this, 'n' men enter the restaurant and put their hats at the reception.Each man gets a random hat back when going back after having dinner.Note that the combinatorial species $\mathcal$ of permutations with fixed points marked is $$\mathcal = \mathfrak \left(\mathcal\mathcal \mathfrak_2(\mathcal) \mathfrak_3(\mathcal) \mathfrak_4(\mathcal) \ldots\right).$$ Hence the generating function of these marked permutations is $$G(z, u) = \exp\left(uz \frac \frac \frac \ldots\right)$$ which is $$\exp\left(uz-z \frac \frac \frac \frac \ldots\right) \= \exp(uz-z) \exp\log\frac = \frac \exp(uz-z).$$ We can recover the total number of permutations on $n$ with $k$ fixed points from this generating function, it is given by $$n! [z^n] \frac \exp(-z) \frac \= \frac [z^] \frac \exp(-z) = \frac \sum_^\frac.$$ It follows that the OGF of the expected number of fixed points is $$\left.\frac G(z,u)\right|_ = \left.\frac \exp(uz-z)\times z\right|_ = \frac.$$ Now since $$[z^n]\frac =1,$$ we expect there to be on average one person who recovers his hat. $which is necessary for the average is already present in the GF. \sum_^ \frac \end Since there are$\binom$ways to choose$S$, the total number of permutations in$\mathfrak_$with$k$fixed points equals$\$ \binom \cdot (n-k)!