# Card Game Probability Problem

i have a card game which involves a dealer and player who both have two seperate decks of 16 cards. The decks have the four suited Tens, Jacks, Queens and Kings (values are the same as normal poker style games, king beats queen ect).

The dealer deals himself 2 cards face down then deals the player 3 cards, the player discards 1 card of his choice and cards are revealed, the highest value hand wins and the dealer wins in the event of a draw. Being able to discard a card is a huge advantage for the player but i cannot work out how big of an advantage it is (the math).

My question is what is the probability of the player winning and dealer winning?

If i could get some help with this that would be amazing. Sorry for the long post :)

• Are pairs better than not pairs? Are there flushes and straights and do they beat pairs? There are $10$ or $16$ possible hands for each person, so you need to find the (independent) probabilities for these for each person, and then combine them to answer your question. Commented Dec 6, 2017 at 16:44

Assuming standard poker rules if there are only two cards, the better hand is the one with the higher pair and in absence of pairs, the better hand is the one with the highest unpaired card. It is assumed that the player is smart enough to throw away the right card to maximize the hand value.

Consider the following exclusive possibilities, which shows the type of hand the dealer and player gets, the probability of that combination, and the fraction of those types of hands in which the player wins. \begin{array} {|l|l|c|c|} \hline dealer & player & p & player\ win\ prob\\ \hline single & single & {4\over5} {16\over35} & {1\over4}{1\over6} + {3\over4}{1\over2}\\ \hline single & pair & {4\over5} {19\over35} & 1\\ \hline pair & single & {1\over5} {16\over35} & 0\\ \hline pair & pair & {1\over5} {19\over35} & {3\over8}\\ \hline \end{array}

These values are derived as follows:

Probability that the dealer gets pair: After playing first card, there are three cards remaining in the deck that will pair up. $p={3\over15}={1\over5}$

Probability that the player does not get a pair: After playing first card, there are 12 of 15 that do not match. For the third card there are 8 of 14 that do not match either of the first two cards. $p={12\over15}{8\over14}={16\over35}$

If both dealer and player have pairs, there are 16 combinations of dealer vs player pair values. For 6 of these combinations, the player wins. $p={6\over16}={3\over8}$

If both dealer and player do not have pairs, work out best single card probabilities. Overall the probability for the player to have Q-high (lowest possible hand) is 6 permutations of 4 Q of 16 cards, 4 J of 15 cards, 4 10 of 14 cards. $p=6{4\over16}{4\over15}{4\over14}={4\over35}$.

The second lowest hand for the player is K-high, and all the other possibilities have pairs. $p={16\over35}-{4\over35}={12\over35}$

Given that there are no pairs, the probability for a player Q-high and K-high is therefore $p_Q=1/4$ and $p_K=3/4$ respectively.

Similarly for the dealer, given no pairs, the probability for a J-high, Q-high, K-high is $p_J=1/6$, $p_Q=1/3$ and $p_K=1/2$.

The single-single player win probability is found by the product of the player-Q-high probability with the dealer-J-high probability plus the product of the player-K-high probability with the sum of the probabilities for dealer to have J-high or Q-high. $p={1\over4}{1\over6} + {3\over4}({1\over6}+{1\over3})={1\over4}{1\over6} + {3\over4}{1\over2}$

To work out the overall winning probability for all possible hand combinations, sum the products of $p$ and $player\ win\ prob$ in the table. $$P= {4\over5} {16\over35} \left({1\over4}{1\over6} + {3\over4}{1\over2}\right) + {4\over5} {19\over35} + {1\over5} {19\over35} {3\over8}={527\over840}=0.627$$ The player has a substantial advantage.

• This is exactly what is was looking for, thanks heaps! Commented Dec 7, 2017 at 10:20