4 ms·

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

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.

ColinWright 9y ago
That's very much an argument along the lines of: "I can't imagine anything going wrong, so it must be OK." What you say is true, but on its own it's not enough to be convincing.
- barrkel 9y ago
  I could flesh it out, but I think there's enough in my comment to get all the way for anyone interested.
  - ColinWright 9y ago
    Nope. I've tried, and I can't make your argument work. Just because you think that's a plausible way for it to go wrong, and you claim that it can't go wrong like that, I see nothing in what you've said to convince me that it can't go wrong in other ways. You might be right that it can't go wrong like that, but why can't it go wrong in other ways?
    - barrkel 9y ago
      It isn't my job to convince you of anything, FWIW.
      - deleted 9y ago
        [deleted]
      - ColinWright 9y ago
        Absolutely true, but I'm just trying to point out that there are more things that can go wrong than you seem to be covering. You said: > barrkel: I could flesh it out, but I think there's enough in my comment to get all the way for anyone interested. I'm letting you know that this turns out not to be the case, for I am interested, and I've read your comment in detail, and for me, there isn't enough there to get all the way. But you're right that it's not your job to convince me of anything, I'm just trying to let you know that, in my opinion, for what it's worth, you haven't solved the challenge.
- mikekchar 9y ago
  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.
  - deleted 9y ago
    [deleted]
  - ColinWright 9y ago
    This isn't the right place to have this discussion because it's going to end up long, threaded, context ambiguous, and generally frustrating for all concerned. However, this is the only place we have, so I'll try to address your points here. But let me begin by saying that it's true that a straight can always be drawn, and showing that someone's false argument of a true statement is indeed false is hard, because you can't give counter-examples. Even so, let's look at what's been said: > barrkel: 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. > ColinWright: That's very much an argument along the lines of: "I can't imagine anything going wrong, so it must be OK." What you say is true, but on its own it's not enough to be convincing. So here barrkel is describing the only way it can go wrong. Why is this the only way it can go wrong? What if I choose a card from each pile bar one, and then find that the remaining pile doesn't have any of the final card I'm looking for? Why can't that happen? Well, it can happen, and I need to fix my previous choices. How do I know I can fix my previous choices? Why can't it be the case that my first 12 choices always leave me in this predicament? [X] In short, the situation described is only one of the gazillion things that might go wrong, and there is no discussion of why they can't happen. That's why it's not convincing. Then, to go on with your comment: > if we consider the situation where we take a card from a pile, this leaves 3 cards left in the pile. Yes. > 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. I don't understand at all what you're trying to say here. What is available in the other piles? (Based on text later in your comment I suspect this doesn't matter.) Added in edit: Ah - I think you're saying that for any card we want to pick next it must be available somewhere, because there are four of them, and only three can be in the now-forbidden pile. OK. > 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. Yes. > When we take another card, three more cards are not available. Yes. > 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. Yes, exactly. So you have outlined one way that it can all go wrong that isn't covered by the original argument put forth by barrkel. I hope you agree that my comment [X] above also provides enough explanation of why barrkel's argument is unconvincing.

ColinWright 9y ago

That's very much an argument along the lines of: "I can't imagine anything going wrong, so it must be OK." What you say is true, but on its own it's not enough to be convincing.

barrkel 9y ago
I could flesh it out, but I think there's enough in my comment to get all the way for anyone interested.
- ColinWright 9y ago
  Nope. I've tried, and I can't make your argument work. Just because you think that's a plausible way for it to go wrong, and you claim that it can't go wrong like that, I see nothing in what you've said to convince me that it can't go wrong in other ways. You might be right that it can't go wrong like that, but why can't it go wrong in other ways?
  - barrkel 9y ago
    It isn't my job to convince you of anything, FWIW.
    - deleted 9y ago
      [deleted]
    - ColinWright 9y ago
      Absolutely true, but I'm just trying to point out that there are more things that can go wrong than you seem to be covering. You said: > barrkel: I could flesh it out, but I think there's enough in my comment to get all the way for anyone interested. I'm letting you know that this turns out not to be the case, for I am interested, and I've read your comment in detail, and for me, there isn't enough there to get all the way. But you're right that it's not your job to convince me of anything, I'm just trying to let you know that, in my opinion, for what it's worth, you haven't solved the challenge.
mikekchar 9y ago
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.
- deleted 9y ago
  [deleted]
- ColinWright 9y ago
  This isn't the right place to have this discussion because it's going to end up long, threaded, context ambiguous, and generally frustrating for all concerned. However, this is the only place we have, so I'll try to address your points here. But let me begin by saying that it's true that a straight can always be drawn, and showing that someone's false argument of a true statement is indeed false is hard, because you can't give counter-examples. Even so, let's look at what's been said: > barrkel: 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. > ColinWright: That's very much an argument along the lines of: "I can't imagine anything going wrong, so it must be OK." What you say is true, but on its own it's not enough to be convincing. So here barrkel is describing the only way it can go wrong. Why is this the only way it can go wrong? What if I choose a card from each pile bar one, and then find that the remaining pile doesn't have any of the final card I'm looking for? Why can't that happen? Well, it can happen, and I need to fix my previous choices. How do I know I can fix my previous choices? Why can't it be the case that my first 12 choices always leave me in this predicament? [X] In short, the situation described is only one of the gazillion things that might go wrong, and there is no discussion of why they can't happen. That's why it's not convincing. Then, to go on with your comment: > if we consider the situation where we take a card from a pile, this leaves 3 cards left in the pile. Yes. > 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. I don't understand at all what you're trying to say here. What is available in the other piles? (Based on text later in your comment I suspect this doesn't matter.) Added in edit: Ah - I think you're saying that for any card we want to pick next it must be available somewhere, because there are four of them, and only three can be in the now-forbidden pile. OK. > 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. Yes. > When we take another card, three more cards are not available. Yes. > 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. Yes, exactly. So you have outlined one way that it can all go wrong that isn't covered by the original argument put forth by barrkel. I hope you agree that my comment [X] above also provides enough explanation of why barrkel's argument is unconvincing.