Expectation of Minimum Value of 4 Distinct Integers from the Set $ \left\{1, 2, 3, \ldots, 47, 48 \right\}$. Suppose I choose $4$ distinct integers from the set $\{1,2,3,...,47,48\}$. What would be the expectation of the minimum of the four numbers (say $X$).
My attempt: Using tail sum formula we get:
$P(X\geq1)+...+P(X\geq45) = \frac{48C4}{48C4}+\frac{47C4}{48C4}+...+\frac{4C4}{48C4}.$
Where can I go from here?
P.S. is there a way to approach this question from a more intuitive sense?
 A: Defining $ X $ as the minimum value of the chosen 4 numbers.
By definifion:
$$ \mathbb{E} \left[ X \right] = \sum_{i = 1}^{45} \mathbb{P} \left( X = i \right) i $$
The probability $ \mathbb{P} \left( X = i \right) $ is choosing the number $ i $ and counting the number of combinations of choosing 3 other values higher than $ i $ divided by the number of combinations to have 4 numbers out of 48:
$$ \mathbb{P} \left( X = i \right) = \frac{ \binom{48 - i}{3} }{ \binom{48}{4} } $$
This yields:
$$ \mathbb{E} \left[ X \right] = \sum_{i = 1}^{45} \mathbb{P} \left( X = i \right) i = \sum_{i = 1}^{45} \frac{ \binom{48 - i}{3} }{ \binom{48}{4} } i = 9.8 $$
The above is easy to evaluate using MATLAB:
% Monte Carlo Simulation:
numTrials = 5e6;

vX = zeros([numTrials, 1]);

for ii = 1:numTrials
    vX(ii) = min(randperm(48, 4));
end

mean(vX)

% Analytic Simulation

vX = [1:48].';
vP = zeros([48, 1]); %<! Probability of X
denVal = nchoosek(48, 4); %<! Number of combinatirons of 4 values out of 48

for ii = 1:45
    numVal = nchoosek(48 - ii, 3);
    vP(ii) = numVal / denVal;
end


sum(vP .* vX)

A: Using the familiar combinatorial identity
$$\binom{n+1}{k+1}=\sum_{i=0}^n\binom ik$$
with $k=4$ and $n=48,$ the expectation is
$$E(X)=\sum_{i=1}^{48}P(X\ge i)=\sum_{i=0}^{47}P(X\gt i)=\sum_{i=0}^{47}\frac{\binom{48-i}4}{\binom{48}4}=\frac{\sum_{i=1}^{48}\binom i4}{\binom{48}4}=\frac{\binom{49}5}{\binom{48}4}=\frac{49}5=9.8$$
