Mathematics puzzle

[Iwantmoney]

I already know the answer which is cannot but I post here to see if there’s any mathematical genius who can have a solution.

Let’s say you have five of diamond at the end of each deck in a eight deck shoe. A 5Diamond at card number 52, another at 104, the third at 156. You get to cut the shoe. The rest are random cards. With a single cut is there ANY way or method to get two of the 5 Diamonds together?

[tr3b]

It's possible with some random chance, ex:

Full table (7 players), cut deck at 89 cards from front. 1 Burn card, last base is dealt 5Diamond on his second card (card 104), dealer's hole card is card 105. Players in first through sixth position draw a total of 50 more cards (unlikely, but possible with multiple splits and small cards). Last base draws another 5Diamond (card 156) on top of his previous.

[Iwantmoney]

Ah I was actually looking for with one cut and they set the shoe two five diamonds becomes adjacent cards aka 52nd card and 53rd card of the set shoe

It's possible with some random chance, ex:

Full table (7 players), cut deck at 89 cards from front. 1 Burn card, last base is dealt 5Diamond on his second card (card 104), dealer's hole card is card 105. Players in first through sixth position draw a total of 50 more cards (unlikely, but possible with multiple splits and small cards). Last base draws another 5Diamond (card 156) on top of his previous.

[CuriousOne]

I do not see this as a math problem but more of a card shuffling/process issue.

Just to clarify, when you say that you have a 5 of Diamonds at the end of each deck in an 8 deck shoe - you have a 5 of Diamonds at 52, 104, 156, 208, 260, 312, 364, and 416 positions of an 8 deck stack.

Assuming that the dealer always takes the front of the cut portion and appends it to the end of the 8 deck stack, as they do in every casino I have ever played, then it is not possible to cut the deck in a way to achieve consecutive 5 of diamonds. The reason being that the first card (top) of the uncut deck is appended to the last card of the 8 decks (the 416th position).

Even if the dealer takes the back portion of the cut stack and appends it to the front of the 8 deck stack there is no single cut that would align the two 5 of diamonds. The top card is the key to your goal.

You would need a 5 of diamonds at the top of the uncut stack and at the end of the uncut stack then you could cut anywhere and the dealer could move the front of the cut to the back or vice versa.

[LOVE]

With a single cut is there ANY way or method to get two of the 5 Diamonds together?

Mathematically no, with cheating yes. There's going to be a five of diamonds at the very bottom of the 8 deck shoe, to get away with connecting two five of diamonds, you could firmly hold the last 51 cards, excluding the exposed 5 of diamonds, with your pointer, middle, and ring finger on one small edge of the deck, your thumb on the opposite small edge, and applying constant pressure into the deck and downwards on the exposed five of diamonds with your pinky, your pinky remains pressed on the 5 with some downward pressure as you take out the other 51 cards, your pinky will connect the two last five of diamonds in the shoe, and no one would notice, like a magic trick being done when no one expects it, and nothing is revealed to anyone.

Can you consistently count 52 cards exactly, no, would it be possible to actually do this at a table one in 100 tries assuming the 5 was perfectly placed, yes.

[Iwantmoney]

Very nice idea but.. You are only allowed to touch the cutting blank card ain’t you? If you simply touch the deck they are going to call pit boss or reshuffle. Actually do not even have to touch the deck. You just cut too thin , two or one deck they will get pit boss, let alone touch the cards.

Mathematically no, with cheating yes. There's going to be a five of diamonds at the very bottom of the 8 deck shoe, to get away with connecting two five of diamonds, you could firmly hold the last 51 cards, excluding the exposed 5 of diamonds, with your pointer, middle, and ring finger on one small edge of the deck, your thumb on the opposite small edge, and applying constant pressure into the deck and downwards on the exposed five of diamonds with your pinky, your pinky remains pressed on the 5 with some downward pressure as you take out the other 51 cards, your pinky will connect the two last five of diamonds in the shoe, and no one would notice, like a magic trick being done when no one expects it, and nothing is revealed to anyone.

Can you consistently count 52 cards exactly, no, would it be possible to actually do this at a table one in 100 tries assuming the 5 was perfectly placed, yes.