Let’s look at a relatively simple game along these lines:
Person A either cheats an outcome or rolls a d4. Then person B either accuses, or doesn’t. If person B accuses, the game ends immediately, with person B winning (losing) if their accusation was correct (incorrect). Otherwise, repeat a second time. At the end, assuming person B accused neither time, person A wins if the total sum is at least 6. (Note that person A has a lower bound of a winrate of 3/8ths simply by never cheating.)
Let’s look at the second round first.
First subcase: the first roll (legitimate or uncaught) was 1. Trivial win for person B. Second subcase: the first roll was 2. Subgame tree is as follows:
The resulting equlibria are:
Person A always rolls.
Person B allows 1-3, and either always calls 4, or calls 4 1⁄4 of the time.
Either way, expected value is −1/2 for Person A, which makes sense given person A plays randomly (1/4 die rolls win for person A).
Third subcase: the first roll was 3. Simplified subgame tree is as follows:
There are 5 (five) equlibria for this one:
Person A always plays randomly.
Person B always allows 1 and 2, and either always calls both 3 and 4, or always calls one of 3 or 4 and allows the other 50% of the time, or 50⁄50 allows 3/calls 4 or allows 4/calls 3, or 50⁄50 allows both / calls both.
Overall expected value is 0, which makes sense given person A plays randomly (2/4 die rolls win for person A).
Fourth subcase: the first roll was 4. I’m not going to enumerate the equlibria here, as there are 40 of them (!). Suffice to say, the result is, yet again, person A always playing randomly, with person B allowing 1 and calling 2-4 always or probabilistically in various combinations, with an expected value of +1/2.
And then the first round:
Overall equlibria are:
Person A plays randomly 3⁄4 of the time, cheats 3 3/16th of the time, and cheats 4 1/16th of the time.
Person B always allows 1 and 2, and does one of two mixes of calling/allowing 3 and 4. (0 | 5⁄32 | 7⁄16 | 13⁄32, or 5⁄32 | 0 | 9⁄32 | 9⁄16 of call/call | call/allow | allow/call | allow/allow).
Either way, expected value for person A is −5/32.
Tl;DR:
This (over)simplified game agrees with your intuition. There are mixed strategies on both sides, and cases where you ‘may as well’ always call, and cases where you want to cheat to a value below the max value.
(Most of this was done with http://app.test.logos.bg/ - it’s quite a handy tool for small games, although note that it doesn’t compute equilibria for single giant games. You need to break them down into smaller pieces, or fiddle with the browser debugger to remove the hard-coded 22 node limit.)
Let’s look at a relatively simple game along these lines:
Person A either cheats an outcome or rolls a d4. Then person B either accuses, or doesn’t. If person B accuses, the game ends immediately, with person B winning (losing) if their accusation was correct (incorrect). Otherwise, repeat a second time. At the end, assuming person B accused neither time, person A wins if the total sum is at least 6. (Note that person A has a lower bound of a winrate of 3/8ths simply by never cheating.)
Let’s look at the second round first.
First subcase: the first roll (legitimate or uncaught) was 1. Trivial win for person B.
Second subcase: the first roll was 2. Subgame tree is as follows:
The resulting equlibria are:
Person A always rolls.
Person B allows 1-3, and either always calls 4, or calls 4 1⁄4 of the time.
Either way, expected value is −1/2 for Person A, which makes sense given person A plays randomly (1/4 die rolls win for person A).
Third subcase: the first roll was 3. Simplified subgame tree is as follows:
There are 5 (five) equlibria for this one:
Person A always plays randomly.
Person B always allows 1 and 2, and either always calls both 3 and 4, or always calls one of 3 or 4 and allows the other 50% of the time, or 50⁄50 allows 3/calls 4 or allows 4/calls 3, or 50⁄50 allows both / calls both.
Overall expected value is 0, which makes sense given person A plays randomly (2/4 die rolls win for person A).
Fourth subcase: the first roll was 4. I’m not going to enumerate the equlibria here, as there are 40 of them (!). Suffice to say, the result is, yet again, person A always playing randomly, with person B allowing 1 and calling 2-4 always or probabilistically in various combinations, with an expected value of +1/2.
And then the first round:
Overall equlibria are:
Person A plays randomly 3⁄4 of the time, cheats 3 3/16th of the time, and cheats 4 1/16th of the time.
Person B always allows 1 and 2, and does one of two mixes of calling/allowing 3 and 4. (0 | 5⁄32 | 7⁄16 | 13⁄32, or 5⁄32 | 0 | 9⁄32 | 9⁄16 of call/call | call/allow | allow/call | allow/allow).
Either way, expected value for person A is −5/32.
Tl;DR:
This (over)simplified game agrees with your intuition. There are mixed strategies on both sides, and cases where you ‘may as well’ always call, and cases where you want to cheat to a value below the max value.
(Most of this was done with http://app.test.logos.bg/ - it’s quite a handy tool for small games, although note that it doesn’t compute equilibria for single giant games. You need to break them down into smaller pieces, or fiddle with the browser debugger to remove the hard-coded 22 node limit.)