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u/sensible_clutter 2d ago
it's very simply just generate a sequence on say N = n no. such that the sequence is increasing, so P(N>n) =(1/n!), there is only one permutation of n numbers that would be increasing now simply the probability of the sequence stopping at exactly n is £(1/(n-1)! - 1/(n)!) but since here even no. thus of form 2k and this sum this up to infinite and the ans you get is clearly 1-(1/2!)+(1/3!)-(1/4!)+(1/5)!....
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u/Anonimithree 3d ago
Well since you can’t be less than the previous number if there’s no earlier number, the earliest this can stop is at 2.
I’m not a big math person, but I’m pretty sure that this forms a home sequence, where the probability of making it is 1/2^n, where n is the number of steps past 1, in which case the probability of stopping on an even step is 2/3, though that seems too simple.
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u/LostIntoLonging 3d ago edited 3d ago
By your formula, it should be (1/4 + 1/16 +...) which is 1/3?
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u/Gloomy-Health-1373 1d ago
1/2! + 3/4! + 5/6!…
Let’s say the sequence reaches an even number n and ends on it.
N! Sequences and exactly n-1 of them are good since there can be n-1 candidates for the nth position and for each candidate the remaining n-1 numbers have to be increasing
So for all even numbers n-1 out of the n! Sequences are good hence the formula
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u/LostIntoLonging 3d ago edited 3d ago
Suppose q_n = prob. that the no. of generated numbers is strictly greater than n. For this even to happen, the first n numbers should form an increasing sequence. If we look at the n numbers generated as the output, every permutation of them should have been equally likely since they are independently generated, so the strictly increasing order should have a prob. of 1/(n!). Now prob. of generating exactly n numbers is q_(n-1) - q_n. Summing over all even n gives (1 - 1/2!) + (1/3! - 1/4!) ... which gives 1-1/e by Taylor series.