Solve problem 493.

python/e493.py

@@ -0,0 +1,63 @@
+from functools import lru_cache
+
+
+BALLS_PER_COLOR = 10
+N_COLORS = 7
+TO_DRAW = 20
+
+
+@lru_cache(maxsize=10**9)
+def count(urn, to_draw):
+    if to_draw == 0:
+        distinct_colors = sum([1 if c != BALLS_PER_COLOR else 0 for c in urn])
+        return distinct_colors
+
+    remaining_balls = sum(urn)
+    expected = 0
+    for color_i, color_remaining in enumerate(urn):
+        if color_remaining == 0:
+            continue
+        new_urn = list(urn)
+        new_urn[color_i] -= 1
+        expected_for_color = count(tuple(sorted(new_urn)), to_draw - 1)
+        probability = color_remaining / remaining_balls
+        expected += (probability * expected_for_color)
+    return expected
+
+
+def binomial_coefficient(n, k):
+    if k < 0 or k > n:
+        return 0
+    if k == 0 or k == n:
+        return 1
+    k = min(k, n - k)
+    result = 1
+    for i in range(k):
+        result *= n - i
+        result //= i + 1
+    return result
+
+
+def math():
+    # Find probability that a given color is chose in TO_DRAW draws.
+    combs_without_color = binomial_coefficient((N_COLORS - 1) * BALLS_PER_COLOR, TO_DRAW)
+    combs_with_color = binomial_coefficient(N_COLORS * BALLS_PER_COLOR, TO_DRAW)
+    p_single_color = 1 - (combs_without_color / combs_with_color)
+    # Then, expected colors is that probability times the number of colors.
+    return p_single_color * N_COLORS
+
+
+def euler_493():
+    urn = [BALLS_PER_COLOR for _ in range(N_COLORS)]
+    c = count(tuple(urn), TO_DRAW)
+    c = round(c, 9)
+    m = round(math(), 9)
+    assert c == m
+    return c
+
+
+if __name__ == "__main__":
+    solution = euler_493()
+    print("e493.py: " + str(solution))
+    # assert(solution == 0)
+