For the game Caribbean Stud Poker, compute the expected value of the call option for the following cases.
I try to solve part (a). There are $\mathsf{A}\mathsf{K}\mathsf{J}$ there. If $\mathsf{Q}$ comes it will complete the game.
Getting $\mathsf{Q}\spadesuit$ has probability $\frac{1}{47}$; getting other cards has probability $\frac{46}{47}$.
Hence the expected value is $1 \frac{1}{47} + 0 \frac{46}{47} = \frac{1}{47}$.
But I'm not sure about this.
The rules for Caribbean stud poker are that the player places an ante to be dealt a hand. The player is then dealt five cards and the dealer displays one of the house's cards. The player has the option to either fold, forfeiting his ante, or call and raise his bet. The dealer then reveals the rest of his facedown cards and must get a hand at least as good as an AceKing high or the dealer "doesn't qualify" and had you called your net winnings are zero (i.e. you get back your bet). Otherwise compare the dealers hand to your hand and pay out accordingly. (Hands are 5-cards each and handrankings are the same as in normal poker).
For part (a), any time that the dealer qualifies you will lose (since you have AKJ-high and dealer will have at least an AKQ-high or better). Let us try to calculate the probability that the dealer doesn't qualify.
That will happen whenever the dealer does not have an AceKing high, or a pair or better. This can be broken into three cases:
(1) All numbers different, Ace, no king, (no flush), (no straight) [flush and straight are impossible in these cases]
(2) All numbers different, King, no ace, no flush, no straight
(3) All numbers different, No ace, no king, no flush, no straight
And each of these cases will be broken down further into sub-cases based on whether any of the numbers match one (or more) of the numbers in your hand.
(1a) Ace, no king, no jack, no 8, no 3: $3\cdot \binom{7}{3}\cdot 4^3$
(1b) exactly one of {jack, 8, 3} along with Ace, no king: $3\cdot 3\cdot 3\cdot \binom{7}{2}\cdot 4^2$
(1c) exactly two of {jack, 8, 3} along with Ace, no king: $3\cdot 3^3\cdot 6\cdot 4$
(1d) all three of {jack, 8, 3} appear along with Ace, no king: $3^4$
For a combined total of $17817$ possible dealer's hands of the first case
(2) same as (1) except we need to remove the straights and flushes, applying inclusion-exclusion: $17817 - 3^2\cdot 4^2 - \binom{11}{4} + 1 = 17344$
(3a) no ace, king, jack, 8, or 3:(straights are impossible without a jack) $\binom{7}{4} \cdot 4^4 - \binom{7}{4}$
(3b) no ace or king and exactly one of {jack, 8, 3}:(straights are impossible without both a jack and 8) $3\cdot 3\cdot \binom{7}{3}\cdot 4^3 - \binom{7}{3}$
(3c) no ace or king and exactly two of {jack, 8, 3}: $3\cdot 3^2\cdot \binom{7}{2}\cdot 4^2 - 3^2\cdot 4^2 - 3\cdot \binom{7}{2} + 1$
(3d) no ace or king and all three of {jack, 8, 3}: $3^3\cdot 7\cdot 4 - 7$
For a total of $38665$ hands of type 3
For a combined total of 73826 dealer hands which do not qualify if the face up dealer card is a Queen of Spades. There are $\binom{46}{4}=163185$ possible dealer hands given what cards are currently face-up. Thus, the probability that the dealer doesn't qualify is approximately $0.4524$. The probability (as mentioned) for part (a) that you win is zero (as any qualifying hand will be better than yours), and the probability that the dealer wins is then approximately $0.5476$.
Depending on the card table you play at, if dealer doesn't qualify you get back your money for a net winnings of 0, if dealer wins you lose your money for a net loss of 1, and if you win you get double your money for a net gain of 1.
As such, your expected value for the call option for part (a) is $(-1)\cdot 0.5476 + (0)\cdot 0.4524 + (1)\cdot 0.0 = -0.5476$
The other cases can be calculated similarly, but you will need to also figure out in which situations you will win (dealer must qualify but must have a worse hand than you, so in part (b) would be an AKJ82 or AK with next highest card a 10 or lower) and also figure out the situations in which you tie the dealer. As mentioned, this problem is incredibly tedious to do by hand, and would be much better handled by a computer.