Getting the highest amount of positives on one set with a 50% random chance

Question

If you had 2 sets of numbers, that both start at 0 (each set can only go up from 0, but never lower than 0), and you run a 50 percent coin flip computer program and if it is heads, you add +1 to a chosen set, and tails -1 from a chosen set, and you have to choose which set of numbers to add/subtract. What would be the best way to alternate between the two sets so you can get the highest number in the first set?

Answer 1

STRATEGY:

When you flip HEADS → Add +1 to Set A

When you flip TAILS → Subtract -1 from Set B

RESULT: Set A will reach approximately 500 (the number of heads you flip) Set B will bottom out at 0

WHY THIS WORKS:

The key insight is the asymmetry created by the zero boundary (can't go below 0).

With ~500 heads and ~500 tails from 1000 flips:

1. Set A gains: +500 (all the heads)
2. Set B tries to lose: -500 (all the tails)
3. Set B hits 0 and stops there
4. Once B is at 0, remaining tails disappear (can't go negative)
5. Set A keeps all 500 gains intact

COMPARISON:

If you ALWAYS choose Set A instead:

1. 500 heads add to A
2. 500 tails subtract from A
3. A ends up near 0 (wins and losses cancel)
4. Result: A ≈ 24

Buffer strategy: A ≈ 500 Always choose A: A ≈ 24

The buffer strategy is about 20 times better.

THE MATHEMATICAL PRINCIPLE:

Without boundary: 500 wins minus 500 losses = 0 With boundary at 0: Losses get trapped at the floor once you reach it. By putting losses in B and gains in A, you trap the losses in B and preserve the gains in A.