Post by @ifixcoinops@retro.social

2026-09-28 23:57 UTC

I bet there's a formula that can tell me roughly how many pennies it'd cost before you got the same combination again, but it's the sort of thing where because I've always been shite at maths I'd just write a computer program to brute force play the situation out a few thousand times, record all the results, lop off the top and bottom and then average them out and go Probably Roughly About That

Replies (2)

@SnoopJ@hachyderm.io 2026-09-29 00:29

@ifixcoinops@retro.social assuming an infinite supply of jellybeans and uniform probability of any given flavor (and that the machine spits out the same number every time because fuck modeling that), the chance of drawing the same combination of m flavors in a single pull is 1/49^m (i.e. 49^m combinations, only one of which is *your* combination) So the probability of *not* getting the same combination is 1-that, and across k trials, that's (1-1/49^m)^k for the probability that you haven't drawn, which is just kinda an easier thing to work with here. EDIT: this calculation is missing the nCr factor and my estimate is excessively pessimistic, see: https://infosec.exchange/@whbboyd/117351781880656742 for m=5, after k=3 million pulls, you have about a 1% chance of having pulled the same combination. after k≈640 million pulls, you have about a 90% chance, and after k≈1.3 billion pulls you have about a 99% chance. so, let's be safe and say around £65 million. for bonus marks, the total number of flavor combinations is (49!)/(49-m)!, which for m=5 is 228,826,080 combinations. Assuming it took you the full 1.3 billion pulls, you would see repeated flavors after a bit less than 1/10th of the way into the process.

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@ifixcoinops@retro.social 2026-09-28 23:59

Primary school maths teachers of Fedi, feel free to use my jellybean machine to confuse some small children

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