He should've pre bought a lot of i
Basically a guy named mistafuccyou on twitter did a livestream doing roulette to promote a coin he made of himself, saying that if he dies to make sure it rockets.
He pulls the trigger three times and then hits the live, killing himself with a .22 revolver
(Kiwi link has image of the gun firing and censored video but not any gore but some might find it disturbing the moment before a man's death)
https://kiwifarms.st/threads/pump-fun-memecoins.206260/page-10#post-20656526
the stream lasted an hour and over the course of it he died, with reports of
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1:6 chance to lose, but then he pulled the trigger 3 times
Absolute state of cryptobros.
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At least he wasn't dumb enough to play with a magazine
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ARE reported my comment saying that mags hold more ammo. As if that statistically decreases his odds of getting shot.
Holy frick, please don't play Russian Roulette with a glock.
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Well yes magazines hold more ammo, but you don't fill it up, that would be silly. Heck if you're really smart, you can leave the first couple spots empty.
Guaranteed win.
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!r-slurs
PSA
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Hmm those odds are off since you are sampling without replacement.
Intuitively, for K trigger pulls it's just the odds that one of the first K cylinders contains the cartridge, so (number of trigger pulls) / (capacity of the cylinder). Which in this case is 3/6.
Another way to express it is like this. He dies if any of the first K cylinders have a cartridge, so his chance of dying after a given trigger pull is:
1 - (5/6 * 4/5 * 3/4 β¦ * K-1/K)
Rather than 1 - (5/6)^K
Which again is 1/2 in this case, not ~.43 or whatever it is.
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Good analysis. How about the chance to pull the ace of spades in 26 tries?
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Without replacing the cards?
To calculate it for
ktries, you take the product of not drawing the ace of spadesktimes in a row and subtract it from 1. On each attempt, the probability of drawing the ace of spades is:1 / remaining cards in the deckSo given 52 cards in a deck, the odds that you'll draw the ace of spades in
ktries is:For
iin0 .. k-1:P = 1 - product((52 - i - 1) / (52 - i))(The same intuition applies though: this simplifies to "what are the odds that the ace of spades is in the first
kcards?", which is justk / 52)Jump in the discussion.
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