Four Coins Against Five

I toss four fair coins and you toss five. You win if you get strictly more heads than I do. What is the probability that you win?

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#A tails mirror

Let $H$ be your heads and $G$ be mine, and set $D = H - G$ , so you win exactly when $D \ge 1$ . Now turn every coin over to its opposite face. Heads and tails are equally likely, so the flipped table of outcomes is exactly as probable as the original. Your head count becomes $5 - H$ and mine becomes $4 - G$ , so the difference transforms as

D = H - G \;\longmapsto\; (5 - H) - (4 - G) = 1 - D. \tag{1}

Because the flip preserves probabilities, $D$ and $1 - D$ have the same distribution.

Flipping every coin sends the difference D to one minus D, mirroring the bars about the dashed line at one half. The winning scores, a difference of one or more, sit entirely on one side of that line and carry exactly half the total weight.

#Reading off the answer

The reflection $D \mapsto 1 - D$ centres the distribution on $\tfrac12$ , so

\PP(D \ge 1) = \PP(1 - D \ge 1) = \PP(D \le 0). \tag{2}

Those two events are complementary, because the integer $D$ is either at least $1$ or at most $0$ . Complementary probabilities that are also equal must each be one half,

\PP(\text{you win}) = \PP(D \ge 1) = \tfrac12. \tag{3}

#Why the extra coin matters

The reflection centre sits at $\tfrac12$ precisely because you hold one more coin than I do. That centre falls strictly between the losing scores $D \le 0$ and the winning scores $D \ge 1$ , splitting the mass into two equal halves. With equal coin counts the centre would sit at $0$ , a genuine tie outcome, and the clean halving would break.