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That wouldn't be the result. The result is that the final card is swapped only once, and can never be swapped with itself, meaning it is always somewhere else i
by just2n 13y ago
That wouldn't be the result. The result is that the final card is swapped only once, and can never be swapped with itself, meaning it is always somewhere else in the deck. If the final card is the ace of spades, then the ace of spades can never be the last card in the shuffled deck.
In most casino card games, shuffles happen before this knowledge becomes particularly useful, which is why I don't think it's necessarily the major cause for concern here. Being able to easily guess the seed is, however, a pretty massive issue.
- IanCal 13y ago> Being able to easily guess the seed is, however, a pretty massive issue. Particularly since they initialise the deck each time (it'd be a massive issue anyway, but this just makes it worse). Starting with 2^32 states, and knowing their algorithm, you have ~4 billion possible shuffles. However, each card you see drastically reduces the search space. My reasoning is that each card you know reduces the possible number by about how many cards it could have been (so no knowledge is 2^32, one card is 2^32 / 52 ...) Knowing just one card brings you to one of 83 million possible decks. One more card and you're down to just 1.6 million. Once the flop is down, you know 5 cards and that means there are only about 15 possible decks. I think this reasoning is right, but it does seem rather dramatic. I'll have to write a simulation, I've been looking for a good blog post to work through & to work on my visualisations.