Hilbert's Casino (pt. 2)
Once realizing the flaw in his plan , David decided to contact the manufacturer of the slot machine and ordered an infinite amount of slot machines (one for each room); the company agreed. Before Hilbert seals the deal, he asks you to work out this deal, and make sure it works, and that at least some of the machines don't break.
Instead of no one winning, what is the probability that not everyone wins at the slot machines? In other words, if represents the event of everyone winning, then find .
The answer is 1.0.
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1 solution
Suppose there are n people playing at the slot machine. The probability of any one of them winning is n 1 . Hence, the probability of all of them winning comes out to be P ( E ) = ( n 1 ) n = n n 1 . Since there are an infinite number of guests, the probability that not everyone wins the game comes out to be
P ( E ) = n → ∞ lim 1 − P ( E ) = n → ∞ lim 1 − n n 1 = 1 .
The fun problem was the one posted before this one. The simplicity of the result in this problem comes out from the fact that there is only one arrangement no matter how large the number is, but the number of dearrangements varies with the number of articles to be chosen and is, in fact, a fun problem to solve, especially at infinity.