Anonymous at Sun, 3 Nov 2024 19:48:26 UTC No. 16461291

>>16461287
you would have to use mathematics

Anonymous at Sun, 3 Nov 2024 19:56:38 UTC No. 16461300

>>16461291
Yes and which ones would I use?
52 cards in the deck. 28 laid out on the table. 24 in the draw pile. I can only draw 3 at a time. I'm not sure where to go from here.

Anonymous at Sun, 3 Nov 2024 19:58:36 UTC No. 16461307

>>16461300
>I'm not sure where to go from here.
use those numbers to do some probability mathematics

Anonymous at Sun, 3 Nov 2024 19:59:49 UTC No. 16461309

>>16461287
Monte Carlo, then work backwards.

Anonymous at Sun, 3 Nov 2024 20:48:45 UTC No. 16461376

>>16461287
I'm pretty sure the number of possible games is just permutations of 52 cards, so 52! (big number)
Then just use the birthday problem and replace 365 with 52! https://en.wikipedia.org/wiki/Birthday_problem

Anonymous at Mon, 4 Nov 2024 03:53:51 UTC No. 16461763

>>16461376
It might be less because there's some symmetry. Swapping clubs<->spades, hearts<->diamonds or red<->black leads to the same gameplay. A (cyclic) permutation of the 24 card draw deck might be allowed too.

Anonymous at Mon, 4 Nov 2024 04:32:54 UTC No. 16461793

>>16461287
smoke weed every day

Anonymous at Mon, 4 Nov 2024 05:35:37 UTC No. 16461831

>>16461287
>The overall probability of playing two identical games would be
(
1
52
!
)
×
P
(
samemovepath
)
(
52!
1

)×P(samemovepath), where
P
(
samemovepath
)
P(samemovepath) accounts for the exact sequence of moves being replicated.