Two-player game about betting on the sum of two dice

Question

There are two players and each one has a fair die with sides 1-6. The two players each roll their dice, and each player can only see the number rolled on their own die. They each come up with a bid and communicate it to the casino. Neither person can see the other’s bid. The casino chooses the player who proposed the higher bid, and that player pays their bid to the casino and receives the sum of the values of the two die. The other player neither pays nor receives anything. What is the optimal bidding strategy?

How could I reason about figuring out an optimal strategy for this game? I am not too familiar with game theory so it seems intractable. Does an optimal strategy even exist? Any hints/insight would be greatly appreciated.

Answer 1

OP is asking for "hints/insight" , hence I am Posting this.

The Game looks artificial & not very well though out.

Here are the main "Objections" I see :

(0) What if Both Bids are Equal ?

(1) Player1 has DieValue=3 & Player2 has DieValue=5. What makes them truthful ? Why not claim 5 & 6 ?
Players may throw the Die multiple times until the favourable number is thrown.
Who will check that ?

(2) Assume they are truthful. P1 makes Bid $B1=30$ & P2 makes Bid $B2=60$.
Casino truthfully chooses P2 to collect $60$ & Pay back $3+5=8$ ????
How will the Players ever Break Even ?

(3) Hence Bids are not going to be arbitrarily high , staying within DieValues.
So let then actual Bids be $B1=3$ & $B2=5$.
Casino truthfully chooses P2 to collect $5$ to pay back $3+5=8$ ????
How will Casino ever Break Even ?

(4) Casino may cheat & claim the lower as the higher.
That will save the larger Pay-Out.
Who will check that ?

(5) Players may collude to cheat the Casino.
Always alternate the Bids 1 & 2 , always alternate the DieValues 5 & 6.
Pay $1+2=3$ , collect $5+6=11$. Get rich quick.
Casino will be bankrupt quickly.
What actions Casino will take ?

(6) Echoing user lulu : Will Player 1 want to beat Player 2 ? Will Player 1 want to make maximum Pay-Out ? Will Player1+Palyer2 together want to extract maximum money from Casino ?
Each of that will have a Different Strategy !
The Primary Objective is not well Defined here.

There are other Issues , I will not list them all.

Solution thoughts :

Assuming each Player wants maximum Pay-Out.
No matter what Strategy Player1 comes up with , Player 2 can try the Same Strategy.

Best Strategy is like this :
Always Bid 1 & 2 , alternating turns.
Casino always get $1+2=3$
Casino will always have to Pay-Out DieValue Sum , which be $7$ , in the long run.
Player winning the turn will earn $7-3=4$.
Playing $2n$ times , each Player would have won $n$ times , earning $4n$ in the long run.

That Strategy is good & Both will make a lot of money via alternating turns.
Casino will go bankrupt eventually.

Answer 2

Let $X$ be the value of first die and $Y$ be the value of second die.

Lets say first player bid $\delta_1(x)$ and similarly second player bid $\delta_2(y)$. Assume that this strategy does not vary with time and assume that both players work out strategy independently and rationally. Let $1(.)$ be indicator function.

Gain for player 1: $$ = (x+y - \delta_1(x)) \times 1(\delta_1(x) > \delta_2(y))$$

Gain for player 2: $$ = (x+y - \delta_2(y)) \times 1(\delta_1(x) < \delta_2(y))$$

On an average, gain for player 1 after getting $X = x$ on die 1: $$ = \sum_{i = 1}^6 1/6 \times (x+i - \delta_1(x)) \times 1(\delta_1(x) > \delta_2(i))$$

On an average, gain for player 2 after getting $Y = y$ on die 2: $$ = \sum_{i = 1}^6 1/6 \times (i+y - \delta_2(y)) \times 1(\delta_1(i) < \delta_2(y))$$

Hence optimal strategy for player 1: $$\max_{\delta_1(1),....,\delta_1(6)} \sum_{i = 1}^6 1/6 \times (x+i - \delta_1(x)) \times 1(\delta_1(x) > \delta_2(i))$$

But this is for a particular value of $x$. In general assuming we want to optimize the gain on an average over all values of $x$, we get: $$\max_{\delta_1(1),....,\delta_1(6)} \sum_{i = 1}^6 \sum_{j = 1}^6 1/36 \times (j+i - \delta_1(j)) \times 1(\delta_1(j) > \delta_2(i))$$

Lets Assume $\delta_1 = \delta_2$,

Hence optimal strategy $\delta$ for both the players is obtained by maximizing the average gain of player $1$ or player $2$: $$ = \max_{\delta(1),....,\delta(6)} \sum_{i = 1}^6 \sum_{j = 1}^6 1/36 \times (j+i - \delta(j)) \times 1(\delta(j) > \delta(i))$$ where $1(.)$ is an indicator function. Note that, we assumed that the strategy $\delta$ does not vary with time and that both players work out strategy independently and rationally.

$$ = \max_{\delta(1),....,\delta(6)} \sum_{i = 1}^6 \sum_{j = 1}^6 1/36 \times (j+i - \delta(j)) \times 1(\delta(j) > \delta(i))$$

Let $s_j = |\{i: \delta(j) > \delta(i) \}|$

Let $S_j = \{i: \delta(j) > \delta(i) \}$

$$ = 1/36 \times \sum_{j = 1}^6 \sum_{i \in S_j } (j+i - \delta(j)) $$

$$ = 1/36 \times \sum_{j = 1}^6 ( s_j j + Sum(S_j)- s_j \delta(j))$$

$$ = 1/36 \times (s_1 + Sum(S_1)-s_1 \times \delta(1)$$ $$+ 2s_2 + Sum(S_2) - s_2 \times \delta(2) $$ $$+ 3 s_3 + Sum(S_3) - s_3 \times \delta(3)$$ $$ + 4 s_4 + Sum(S_4) - s_4 \times \delta(4) $$ $$ + 5 s_5 + Sum(S_5) - s_5 \times \delta(5)$$ $$ + 6 s_6 + Sum(S_6) - s_6 \times \delta(6))$$

You can get gains easily in this setting. For example by choosing $\delta(6) = 1$ and $\delta(j) = 0$ for $j \leq 5$, we get a gain of: $$ = 1/36 \times (25 + 20) = 1.25$$