m = random(1, n);
if(m < k)
return m;
else -1;
k/n的概率返回k内的数字，k内的数字m被选中的概率是多少?
是k/n呢还是k/n * 1/k？
count = 0
count2 = 0
summ = 10000000
for i in range(summ):
k = 6
m = random.choice(range(10))
if(m < k):
count = count + 1
if m == 1:
count2 = count2 + 1
print count * 1.0 / summ 0.5998247 ~~ 6/10
print count2 * 1.0 / count 0.166646688608 ~~ 1/6
###结论 k内的数字m被选中的概率是k/n * 1/k = 1/n