Probability of winning a number draw between 1 and 99

Question

Say you are attending a social gathering. (pre-/post-pandemic, perhaps?)

There are 24 guests at the gathering. The host announces a raffle where each guest must draw a number from a hat, each with a single number between 1 and 99. There are no duplicate numbers.

One guest draws the number 95. They are quite confident in winning the raffle. However, another guest draws 98, which ends up being the winning draw. The guest who drew 95 is frustrated, stating they had a 95% chance to win and they still lost.

Is this guest's reasoning mathematically sound (drawing 95 in this scenario equates to a ~95% chance of winning)? What statistical/probabilistic models would best fit this scenario?

Answer 1

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 to find the (very low) probability of winning 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 $0.01623$ of winning 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 $P(X = 0) \approx (12/13)^{52} = 0.0156.$