Winning with a worse deck

For the deck consisting of 20 cards, there are $\binom{20}{5}=15504$ possible hands of 5 cards. The number of combinations for each type of hand is as follows:

Straight flush: $\binom{1}{1}\binom{4}{1}=4$

Four of a kind: $\binom{5}{1}\binom{4}{1}\binom{4}{1}=80$

Full house: $\binom{5}{1}\binom{4}{3}\binom{4}{1}\binom{4}{2}=480$

Straight (excluding straight flush): ${\binom{1}{1}\binom{4}{1}}^5-\binom{1}{1}\binom{4}{1}=1020$

Three of a kind: $\binom{5}{1}\binom{4}{3}\binom{4}{2}{\binom{4}{1}}^2=1920$

Two pair: $\binom{5}{2}{\binom{4}{2}}^2\binom{3}{1}\binom{4}{1}=4320$

Two of a kind: $\binom{5}{1}\binom{4}{2}\binom{4}{3}{\binom{4}{1}}^3=7680$

You can calculate the corresponding probabilities by dividing by $15504$ .

For the deck consisting of 32 cards, there are $\binom{32}{5}=201376$ possible hands of 5 cards. The number of combinations for each type of hand is as follows:

Straight flush (including royal flush): $\binom{4}{1}\binom{4}{1}=16$

Four of a kind: $\binom{8}{1}\binom{7}{1}\binom{4}{1}=224$

Full house: $\binom{8}{1}\binom{4}{3}\binom{7}{1}\binom{4}{2}=1344$

Flush (excluding straight flush): $\binom{4}{1}\binom{8}{5}-\binom{4}{1}\binom{4}{1}=208$

Straight (excluding straight flush): $\binom{4}{1}{\binom{4}{1}}^5-\binom{4}{1}\binom{4}{1}=4080$

Three of a kind: $\binom{8}{1}\binom{4}{3}\binom{7}{2}{\binom{4}{1}}^2=10752$

Two pair: $\binom{8}{2}{\binom{4}{2}}^2\binom{6}{1}\binom{4}{1}=24192$

Two of a kind: $\binom{8}{1}\binom{4}{2}\binom{7}{3}{\binom{4}{1}}^3=107520$

No pair/high card: $\left[\binom{8}{5}-4\right]\left[{\binom{4}{1}}^5-4\right]=53040$

You can calculate the corresponding probabilities by dividing by $201376$ .

Your friend will always have higher cards, so if you have the same kind of hand, your friend will win. The probability to win with each type of hand is:

$P($ you have that type of hand $)\times P($ your friend has a worse hand $)$

Summing these probabilities over each type of hand yields $P\approx0.56625$ . Therefore $\lfloor1000P\rfloor=\boxed{566}$ .

Here is a python solution to this problem - https://github.com/moyi-d/poker/blob/master/poker.ipynb

The class "deck" calculates the number of possible combinations for each possible hand given a particular deck. You can create the deck by specifying the different ranks and suits that are in the deck, as well as, optionally, the highest and lowest cards if necessary. (i.e. specify 14 as Ace). You can calculate for a full deck, or the cut decks in the problem. See the notebook for more details. Running the notebook would give the answer to this problem.