# Dating problem probability

If you dismiss someone without a job offer, they’ll be snapped up by a rival company: You cannot go back to them later and offer them the job. Logic suggests that you shouldn’t offer the job to the first person you interview, because you have no idea what the general caliber of the candidates is.Nor do you want to wait until the 10 person, because if they’re the only one left you’re going to be forced to offer them the job regardless of how well suited to it they are.The Event Alex is looking for is a "double", where both dice have the same number.

To calculate the expected value from is definition, we have to compute the probability with which 'k' men would get their correct hat back. I know that this problem can be solved by using 'Linearity of Expectation' in a much simpler way, but i would like to know the way to compute this probablity.Therefore, any optimal strategy must consist of probabilities $p_k$ for

To calculate the expected value from is definition, we have to compute the probability with which 'k' men would get their correct hat back. I know that this problem can be solved by using 'Linearity of Expectation' in a much simpler way, but i would like to know the way to compute this probablity.

Therefore, any optimal strategy must consist of probabilities $p_k$ for $1\le k\le N$ such that when the $k$th date is better than all before, we marry (and stop) with probability $p_k$ and continue otherwise.

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.

||To calculate the expected value from is definition, we have to compute the probability with which 'k' men would get their correct hat back. I know that this problem can be solved by using 'Linearity of Expectation' in a much simpler way, but i would like to know the way to compute this probablity.Therefore, any optimal strategy must consist of probabilities $p_k$ for $1\le k\le N$ such that when the $k$th date is better than all before, we marry (and stop) with probability $p_k$ and continue otherwise.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.

\le k\le N$ such that when the $k$th date is better than all before, we marry (and stop) with probability $p_k$ and continue otherwise.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.