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If I can name any card and any position don't you need 52 ^ 2 decks? Or at least that many divided by 2 for counting off the top or bottom.
by ALittleLight 5y ago
If I can name any card and any position don't you need 52 ^ 2 decks? Or at least that many divided by 2 for counting off the top or bottom.
- anandoza 5y agoIf you want to frame it that way, then each physical deck can handle 52 distinct queries. So you have 52^2 / 52 = 52 decks required.
- rk06 5y agoNo, because a single deck can answer for 52 queries. and 52 decks (offset 0-51) will be sufficient for any card and number combo
- ALittleLight 5y agoOh, I see. For what it's worth, here is someone else performing the trick and it doesn't look like that's what's happening. https://www.youtube.com/watch?v=mdXIVQ-asqU https://www.youtube.com/watch?v=mdXIVQ-asqU
- perilunar 5y agoIt still requires memorising the positions of 52^2 cards though.
- stouset 5y agoNo, because all decks have the same order (just rotated).