In addition to combinatorial/hypergeometric approach in my comment:
By simulation, the probability of a number higher than 95, out of 23 draws without replacement from among numbers $1,2,3,\dots,94,96,97,98,99,$ is $0.663\pm 0.001.$

    set.seed(2020)
    mx =replicate(10^6, max(sample(c(1:94,96:99), 23)))
    mean(mx > 95)
    [1] 0.663096
    2*sd(mx > 95)/1000
    [1] 0.000945304

Unconditionally, the chances the highest number out of $1,2,3,\dots,99$
exceeds 95 is $0.677\pm0.001$ by simulation. [The probability is a little
larger because the guest in question also _might_ have gotten a number
above 95.]

    set.seed(928)
    mx = replicate(10^6, max(sample(1:99, 24)))
    mean(mx > 95)
    [1] 0.677377       # aprx P(max nr > 95)
    2*sd(mx > 95)/1000
    [1] 0.0009349601   # aprx 95% margin of simulation err

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_Note:_ A somewhat related combinatorial problem, also easily handled by simulation, is the (very low) probability of winning [frustration solitaire](https://en.wikipedia.org/wiki/Frustration_(solitaire)).

Turn over cards in a well-shuffled deck. Call out $1$ and turn over the first card, call out $2$ and turn over the second card, and so on to the $13$th card. Then start again calling out $1, 2, \dots, 13$ in sequence; repeat two more sequences of 13 to finish the deck. If, at any point, the denomination of your card matches the denomination you call out (a "hit"), you lose the game. Otherwise, you win.

    set.seed(1234)
    m = 10^6;  deck = rep(1:13, times=4)
    x = replicate( m, sum(deck == sample(deck)) ) 
    mean(x==0);  2*sd(x==0)/sqrt(m);  mean(x);  2*sd(x)/sqrt(m)
    [1] 0.015997       # sim P(Win)
    [1] 0.0002509272   # aprx margin of sim error for P(Win)
    [1] 4.004366       # sim E(Hits)
    [1] 0.003883274    # aprx margin of sim error for E(Hits)

The exact probability of winning $0.01623$ is found by an inclusion-exclusion procedure described at `https://arxive.org/pdf/math/0703900.pdf.` [Incorrectly] assuming independence, one might use $X \stackrel{aprx}{\sim}\mathsf{Pois}(52/13 = 4),$ so $P(Win) = P(X = 0) \approx e^{–4} = 0.018,$ but $X \stackrel{aprx}{\sim}\mathsf{Binom}(52, 1/13)$ gives a better approximation $P(X = 0) \approx (12/13)^{52} =  0.0156.$