The expectation of the minimum of two IID binomial distribution Bi(m,1/M) only depends on m/M

Question

I want to prove following proposition:

$X,Y \sim Bi(m,1/M)$, $X$ and $Y$ are independent. Prove that $f(m,M)= \mathbb{E} \min(X,Y) $ only depends on $\dfrac{m}{M}$.

I have done the following derivation:

$f(m,M)= \mathbb{E} \min(X,Y)=\sum_{n=0}^m P(\min(X,Y)>n)$

$=\sum_{n=0}^m P(X>n)P(Y>n)$

$=\sum_{n=0}^m (1-\sum_{k=0}^{n} \binom{m}{k}(\frac1M)^k(\frac{M-1}{M})^{m-k})^2$

But I'm stuck here. I don't know how to prove $f(m,M)=g(m/M)$ in this complicated expression.

I wrote a python program to verify it, and the result showed the proposition is true. And it seems that $\mathbb{E} \max(X,Y)$ also only depends on $m/M$.

import numpy as np
import matplotlib.pyplot as plt

m = 100
M = 5
T = 100
N = 40000
ans = np.zeros(T)
for k in range(1,T):
    ans[k] = 0
    for i in range(N):
        a = np.random.binomial(k * m,1/(k * M))
        b = np.random.binomial(k * m,1/(k * M))
        c = min(a,b)
        ans[k] += c
    ans[k] /= N
plt.plot(ans)

Any help would be appreciated!