The Marble Jar Gamble

Place 1 red marble in jar A and all remaining 99 marbles (49 red, 50 blue) in jar B. This yields an overall winning probability of approximately 74.75% — the mathematically proven maximum for this setup.

This is an optimization puzzle disguised as a probability question. The goal is to arrange 50 red and 50 blue marbles across two jars to maximize the expected probability of drawing red, given uniform random jar selection. Step 1: Let jar A hold r1 red and b1 blue marbles, and jar B hold r2 red and b2 blue. Constraints: r1+r2=50, b1+b2=50, each jar non-empty. Step 2: Overall P(red) = 1/2 x r1/(r1+b1) + 1/2 x r2/(r2+b2). We want to maximize this expression. Step 3: To maximize jar A's contribution, push its red ratio as high as possible — ideally 100%. That means b1 = 0 and r1 = 1 (minimum marbles, all red). Step 4: Jar A: 1 red, 0 blue → P(red | jar A) = 1/1 = 1. Jar B: 49 red, 50 blue → P(red | jar B) = 49/99. Step 5: Overall probability = 1/2 x 1 + 1/2 x 49/99 = 0.5 + 0.2475 = 0.7475, roughly 74.75%. Step 6: Why not put more reds in jar A? Adding a second red marble makes jar A have 2/2 = 1 still (if no blue), but jar B loses a red: 48/99 < 49/99. The net effect decreases the total. The single-red isolation is strictly optimal. Key Insight: The trick is decoupling jar A's success rate from the main pool. A jar with exactly 1 marble (and that marble being red) acts as a guaranteed-win wildcard. The other jar absorbs the cost but retains a near-50% red ratio, resulting in a combined probability close to 75%.