Really minor nit: You could drop one of your nows.
Tangentially, my sons told me that when they start card tricks with an unopened pack of cards and strongly highlight that fact like it proves the cards have not been tampered with, the trick there is that card decks all start in the same order. So an unopened deck will come with a particular card order and they can know the order of the cards and can use that fact to help them with their card trick. It is part of how they fool the audience.
>> ... because now it would now take 31.5 dominos.
> Really minor nit: You could drop one of your nows.
Good catch - change in the process of being uploaded.
WRT cards in an unopened deck, there are two different orderings that are used. Sometimes there are "kissing kings" and sometimes not. If you always use the same make of cards, probably the order is always the same.
For dealing out the cards, it seems that the only thing that would stop you from drawing a straight via one card from each pile would be if any one pile had one card in all 4 suits, along with one of its adjacent cards in the straight; e.g. 4 of hearts, clubs, diamonds and spades, and any 5 or 3. (If you pluck off all the same-suited cards first, then you'll end up with two adjacent cards for the final draw in the same pile). Since the piles are constrained to have no more than 4 cards, this situation can't happen.
It would be helpful to explain why the answer is not convincing. After all, you challenged in your article to find a proof. Replying with "Nope, that's not it" is not really helpful.
In this case, if we consider the situation where we take a card from a pile, this leaves 3 cards left in the pile. We can't take another card from this pile this round, but there are only 3 cards left which guarantees that there is still one card available in another pile. This much is correct, but as you say it is not enough to prove the issue.
Once we take a card from one pile, three other cards are not available to draw this round. When we take another card, three more cards are not available. How do we know that there were not 2 of one card in the first pile and 2 of another card in the second pile (or some other annoying combination)? Indeed, this situation will exist. In order to prove the original assertion, you have to prove that there will always be another solution that will allow us to avoid doing that. That's much harder.
Readit News
This would be clearer if the word "closed" were inserted before "tour".
Cheers.
Really minor nit: You could drop one of your nows.
Tangentially, my sons told me that when they start card tricks with an unopened pack of cards and strongly highlight that fact like it proves the cards have not been tampered with, the trick there is that card decks all start in the same order. So an unopened deck will come with a particular card order and they can know the order of the cards and can use that fact to help them with their card trick. It is part of how they fool the audience.
> Really minor nit: You could drop one of your nows.
Good catch - change in the process of being uploaded.
WRT cards in an unopened deck, there are two different orderings that are used. Sometimes there are "kissing kings" and sometimes not. If you always use the same make of cards, probably the order is always the same.
Deleted Comment
What you say is true, but on its own it's not enough to be convincing.
In this case, if we consider the situation where we take a card from a pile, this leaves 3 cards left in the pile. We can't take another card from this pile this round, but there are only 3 cards left which guarantees that there is still one card available in another pile. This much is correct, but as you say it is not enough to prove the issue.
Once we take a card from one pile, three other cards are not available to draw this round. When we take another card, three more cards are not available. How do we know that there were not 2 of one card in the first pile and 2 of another card in the second pile (or some other annoying combination)? Indeed, this situation will exist. In order to prove the original assertion, you have to prove that there will always be another solution that will allow us to avoid doing that. That's much harder.