Calculate probabilities of the number of uniformely distributed items ending up in a bucket

def p(items_in_bucket, cur_items, m):
 
    if items_in_bucket not in m:
        m[items_in_bucket] = {}
 
    if cur_items in m[items_in_bucket]:
        return m[items_in_bucket][cur_items]
 
    if items_in_bucket == 0:
        # Out of all cur_items, none ended up in the bucket
        m[items_in_bucket][cur_items] = pow(1-1/buckets,cur_items)
    elif items_in_bucket == cur_items:
        # Out of all cur_items, all ended up in the bucket
        m[items_in_bucket][cur_items] = pow(1/buckets, cur_items)
    else:
        # P(already have enough items in this bucket)*P(item will hit another bucket) +
        # P(missing only one item in this bucket)*P(item will hit this bucket)
        m[items_in_bucket][cur_items] = \
            p(items_in_bucket, cur_items-1, m)*(1-1/buckets) + \
            p(items_in_bucket-1, cur_items-1, m)*1/buckets
 
    return m[items_in_bucket][cur_items]
 
items = 996
bucket_total_colors = {
    100:"\033[92m", # green
    1000:"\033[91m", # red
    10000:"\033[94m"} # blue
 
for buckets in [100,1000,10000]:
    print(bucket_total_colors[buckets])
    print(f"{buckets} buckets, {items} items (Poisson λ={items/buckets:0.4})\n")
    memo = {}
    for i in range(20):
        prob = p(i,items,memo)
        print(f"{i}-item buckets: {prob*buckets:.3f}, P={prob:.7f}, {prob*100:.5f}%")
print("\033[0m")