How can one simply calculate the probability of pulling one out of (52!) balls out of a bag TWICE in a row with replacement?
The probability of pulling any ball from a bag of n is 1/n. Doing it twice is 1/n *1/n = 1/n^2.
Since you’re talking about 52!, I assume your question is actually “what is the probability of shuffling a deck twice and getting the same shuffle?”. 1/(52!)^2 is approximately 1.5*10^-136.
For context there are about 10^68 atoms in the milky way, which is already an ungodly large number. imagine that every atom in the milky way contains an entire second milky way. If every atom in ALL those milky ways were assigned 2 random deck shuffles, only 33 of all those atom-atoms would be assigned the same deck twice (the math actually comes pretty close, it really surprised me).
The calculation to check that out end up being ( 10^68 )^2 / 52!^2 = 33
Yes, you saw through me. lol Thanks for the math lesson. Probability was never my strong suit.
A good way to think about probability is to accept that what you’re calculating isn’t really special.
The universe and maths don’t care that you shuffled the same deck twice. The probability I calculated is true for any 2 decks, not just the ones that are identical. Because we’re human we care about the edge case where they are identical but any one shuffle is equally as unlikely to turn up as all the others. You shuffled a deck once, and got one of 52! Results, and did it again, and got one of 52! Results. The chance of those 2 decks turning up is equally as mind-boggingly small as the identical case, but it’s not interesting to dopamine-driven brains.
Matters exactly what you’re looking for. Are you looking to pull exactly that one marked ball twice in two trials? Are you looking for the odds of pulling a ball that you just pulled on the prior trial?
