3 cards betting game

Question

Bob and John are playing a betting game of multiple rounds. Each round, they both draw a single card from a deck consisting of only three cards: 1 Queen, 1 King, and 1 Ace. Whoever reveals to have the higher card wins the round.

The round begins with a $\$1$ bet from each player. After both players look at their own card, Bob has the option to raise an additional $\$1$. If Bob raises, John has the option to match by putting in $\$1$ too (and seeing who has the higher card), or fold and lose the bet without revealing any cards.

Assume that John can play the best possible strategy to counter what Bob does, what is the highest $ amount that Bob could be expected to win from this game per round?

here is my rough attempt

Bob	John	Expected value for Bob
A	K	$+3$
A	Q	$+3$
K	A	$-2$ or $-1$
K	Q	$+3$ or $+2$
Q	A	$-1$
Q	K	$-1$

I tried to see all the possible values like this, but I am not sure how to model the conditional probability, like when Bob gets the king, he can either raises 1 extra dollar or not. If he raises (not) and John gets an Ace, then he loses -2 (-1), if he raises (not) and John gets a Queen, then he gets +3 (+2)

How do I model this? Like the amount of I win depends on the whether John gets a queen or an ace? Otherwise, how would you do this?

Answer 1

We have three cases to consider: Bob gets a Queen, Bob gets a King or Bob gets an Ace.

Case 1: Assume Bob gets a Queen, then Bob knows that John has an Ace or a King and Bob will definitely lose the bet. In this case to minimize his loss Bob should not raise and lose 1 dollar. Since Bob has 1/3 probability to get the Queen, the expected pay-off of this case is $E_Q=-1/3$.

Case 2: Assume Bob gets a King, then Bob knows that John has an Ace or a Queen. Now we need to consider 2 subcases: John has an Ace and John has a Queen.

Assume John has an Ace then John will know that he will win the bet so he will always call if Bob raises. So if Bob raises he loses 2 dollar and if Bob does not raise he only loses 1 dollar. Because John has 1/2 probability to get the Ace, the expected pay-off in this case is -0.5 dollar if Bob does not raise and -1 dollar if he does raise.

Now Assume John has a Queen then John will know that he will lose the bet so he will never call if Bob raises. So no matter what Bob does Bob will gain 1 dollar. Again John has a probability of 1/2 to have the Queen as such Bob will have an expected pay-off of 0.5 dollar.

Bob has 1/3 chance to get the King and let's call the probability that he raises when he has a King $p_R$. The total expected pay-off of this case is $E_K=\frac{1}{3}(-p_r-0.5(1-p_r)+0.5)=-\frac{1}{6}p_r$.

Case 3: Assume Bob gets an Ace, then Bob knows for certain he will win the bet and he should raise. On the other hand we have again 2 subcases.

Assume John has a Queen then John knows he is going to lose and will not call. In this case Bob will gain 1 dollar. Since John has 1/2 probability to gain the Queen, the expected pay-off for Bob will be 0.5 dollar.

Assume John has a King then John does not know for certain if he is going to lose or win. If he calls Bob will win 2 dollars, if he does not call Bob will only win 1 dollar. If we call the probability with which John calls if he has a king $p_c$, then the expected pay-off for Bob in becomes $0.5(1-p_c+2p_c) = 0.5(1+p_c)$.

Since Bob has 1/3 chance to get an Ace the expected pay-off in this case $E_A =\frac{1}{3}(0.5 + 0.5 + 0.5p_c)=\frac{1}{6}(2+p_c)$.

Hence the total expected pay-off of Bob over all the cases will be \begin{equation} E = E_Q + E_K + E_A = -\frac{1}{3} - \frac{1}{6}p_r + \frac{1}{6}(2+p_c) =\frac{1}{6}(p_c-p_r) \end{equation}

Since Bob will try to maximize $E$ by adjusting $p_r$ and John will try to minimize $E$ by adjusting $p_c$, the optimal strategy for both will be given by $p_c = p_r = 0$. Bob should never raise and furthermore will on average not make any gains by playing this game.

Edit: I forgot to to take bluffing into account, which is discussed in Piita's answer.

Answer 2

The payoff matrix is determined by the strategies that Bob and John elect to play along. Here, I'll solve for optimal strategies for both players, so that you can derive the payoff matrix yourself.

But first, let me adress your question how to go about this. Generally, the method for AKQ games it to set up the complete ex-showdown payoff matrix from Bob's point of view, and find optimal strategies for the game by eliminating dominated strategies s.a. 'John folding with an ace'. Repeat from John's point of view and eliminate strategies that are dominated for Bob.

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In this example, you will be left with a set of easy-to-grasp strategies (always bet or call A, never bet K, never call Q) and a 2x2 matrix, which will give rise to four strategic options:

• b: Bob bluffs a queen,
• 1-b: Bob checks a queen,
• c: John calls with a king,
• 1-c: John folds a king.

From here, we will use some math to establish concrete values for their mutual optimality. This is done cross-wise. We express Bobs decision to find John's optimum, and vice versa.

Let us have a look at Bobs decision. It will be influenced by how often John would fold, so we express this strategic subdecision with variables $c$ and $(1-c)$ from above; they represent John's frequencies, or call-fold-ratio via their quotient.

1. Bob, bluff a Q or not?
Bob must decide wheter to bluff with a queen, which loses a bet to aces, loses a bet to kings that call, but wins the pot against kings that fold. Consider $c$ John's calling frequency, we get Bob's ex-showdown EV:

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Edit: Inserting a description of how to derive the expected value EV for "Bob bets a Q"-lines:

Bluff Q, lose to A:
EV is $(\frac{1}{2})\times(-1)$ since John has an A with $p_{A|Q}=(\frac{1}{2})$ and $(-1)$ reflects the outcome of losing the bluff bet against John with an Ace.

Bluff Q, lose to King that calls:
EV is $(\frac{1}{2})\times(c)\times(-1)$, where $c$ now reflects the strategic option of calling a bet with a King (intending to catch Bob bluffing).

Bluff Q, win to King that folds:
EV is $(\frac{1}{2})\times(1-c)\times(2)$, same idea, now $(c-1)$ reflects folding the King, (2) is pot-size, and John again holds a King half the time, so $(\frac{1}{2})$.
End edit.

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Summing up the three weighted outcomes of Bob's bluff gives
$EV_{bluff}=[(\frac{1}{2})\times(-1)] +[(\frac{1}{2})\times(c)\times(-1)] +[(\frac{1}{2})\times(1-c)\times(2)]$

Now that we know how to express the EV of a bluff with a Queen, we can contrast the line with its alternative, which was to not bluff. The EV of not bluffing is simply $0$, regardless of what John holds, so
$EV_{check}=0$

Setting these equal to get an indifference between the strategies, we obtain John's optimal calling frequency

$c=\frac{1}{3}$.

2. John, call a K or not?
Similarly, John has to decide wheter to call with a king. John holding a King faces a problem. He cannot always fold to a bet (could be Bob bluffs), nor can he always call (Bob only bets his Aces). He has to "schroedinger" himself into the ratio c-to-(1-c) from above, and every change of this ratio in turn influences Bobs optimal bluffing frequency.
We set John to be indifferent to calling with a king, again with the aim of finding the concrete value of Bob's bluffing frequency $b$. Consider $b$ Bob's bluffing frequency with a queen, we get John's ex-showdown EV

$EV_{call}=(\frac{1}{2})(-1)+(\frac{1}{2})(b)(1)$ $EV_{fold}=(\frac{1}{2})(b)(-2)=-b$

Setting these equal to get an indifference between the strategies, we obtain Bob's optimal bluffing frequency

$b=\frac{1}{3}$.

So, mutually optimised strategies in play, Bob bets all his aces, none of his kings, and he bluffs a third of his queens; and John should fold all his queens, call with a third of his kings and of course all of his aces.

PS: The general forms of strategically indifferent frequencies b and c in AKQ games, where P is the pot size, are $b=\frac{1}{(P+1)}$ and $c=\frac{(P-1)}{(P+1)}$.

PPS: AKQ games and strategies are well described in: Bill Chen, Jerrod Ankenman, The mathematics of Poker, 2006.