Every day man puts 10 1 £ coins and 8 2 £ coins in his pocket. Each day, in the morning he losses 8 coins in the afternoon he losses 6 coins. Find its morning and afternoon loss correlation coefficient.
I have found a similar solution on the internet: Expected value, dispersion and correlation coefficient of losses during the day, however, the solution is a brute-force eyesore calculations. Implementing same solution would be a painful experience since in my case the task has a higher input. Thus, there will be much more combinations.
My current solution: In the morning, if the losses are:
$8£ = C(10,8) = 45$
$9£ = C(10,7)\times C(8,1) = 960$
$10£ = C(10,6)\times C(8,2) = 5880$
$11£ = C(10,5)\times C(8,3) = 14112$
$12£ = C(10,4)\times C(8,4) = 14700$
$13£ = C(10,3)\times C(8,5) = 6720$
$14£ = C(10,2)\times C(8,6) = 1260$
$15£ = C(10,1)\times C(8,7) = 80$
$16£ = C(8,8) = 1$
Overall, there $C(18,8) = 43758$ different combinations of him losing 8 coins in the morning, thus:
$$E(X) = 8 \times \frac{45}{C(18,8)} + 9 \times \frac{960}{C(18,8)} + \dots 16 \times \frac{1}{C(18,8)} = 11.5556$$
However, I am having issues calculating losses on the evening for the reason, that to calculate each possible 6 coin loss, morning losses should be accounted for, then there will be over $6*16$ calculations. My solutions stops there, any support would be appreciated.