1-Armed Bandits && Demorgan's Law

When simulating the spin of a three-wheel slot machine, how do we detect three of a kind?





The following condition recognizes three of a kind:

var s1 = 'lemon', s2 = 'lemon', s3 = 'lemon';
console.log((s1 == s2) && (s1 == s3) && (s2 == s3));

We can modify it to control a loop that spins the wheels:

       s1 = RandomOneOf(['lemon', 'cherry', 'bar', 'donut']);
       s2 = RandomOneOf(['lemon', 'cherry', 'bar', 'donut']);
       s3 = RandomOneOf(['lemon', 'cherry', 'bar', 'donut']);

       while (!((s1 == s2) && (s1 == s3) && (s2 == s3))) {
           //spin again...
       }

Applying Demorgan's Law we get:

       while (!(s1 == s2) || !(s1 == s3) || !(s2 == s3)) {
           //spin again...
       }

Equivalently:

       while ((s1 != s2) || (s1 != s3) || (s2 != s3)) {
           //spin again...
       }

To hit the jackpot, we must spin the wheels at least once. This leads to a final, simpler version using a do-while:

do {
       s1 = RandomOneOf(['lemon', 'cherry', 'bar', 'donut']);
       s2 = RandomOneOf(['lemon', 'cherry', 'bar', 'donut']);
       s3 = RandomOneOf(['lemon', 'cherry', 'bar', 'donut']);
} while ((s1 != s2) || (s1 != s3) || (s2 != s3));

The do...while loop will always be executed at least once, even if the condition is false, because the statements are executed before the condition is tested.

It is a post-test loop; the while loop and the for loop are pre-test loops.

=> Use a do-while when an action must be performed at least once.