[rtpg]

this is super interesting! The deeper explanation[0] really gets to the essence of what's happening here.

An interesting corrolary: the deeper down the deck you go, the less number of paths are actually available. For example, if you have 10 cards on the first row, then there are at most 10 paths to the end, but those can merge. So if you have many cards, paths will merge and you might just end up with one path.

So most cards towards the end of the deck are totally unreachable! For example the next to last card is most likely not reachable through this walking algorithm in most distributions.

I'm not sure of the implications but it sure feels important for things like randomly generated worlds or whatnot.

[0]: http://faculty.uml.edu/rmontenegro/research/kruskal_count/in...

[irishsultan]

> For example, if you have 10 cards on the first row, then there are at most 10 paths to the end, but those can merge

In fact they must converge, suppose that the first card on the first row is less than 10, then it will merge with another path on the first move (because it will land on the first row), so it must be 10 to have no merging paths. The next card will merge with another path on the first row if it's less than 9, but if it's 9 it will merge with the first path, so it too must be 10. Same reasoning goes for the third, fourth and fifth card, however there are only 4 cards with value 10, so at least two of the paths on the first row merge within one move.

[rtpg]

the fact that they must merge is really dependent on the distribution of cards within the deck, though. For example, imagine if your deck was filled with 9s. You would get 2 paths merging but not much else.

Of course with the standard 52-pack you'll end up with a high degree of merging. I just think it's important to say that the necessity of merging is really dependent on the instruction set, so to speak.

[dalore]

I think the fact the deck doesn't have many repeats of 9s. It has lots of repeats of 5s which causes merges and the other numbers will eventually merge. You have 4 each of 6,7,8,9. If you repeat 9 4 times there will be no more 9's and it will merge.

[Sean1708]

> In fact they must converge

Are you sure? Is this board not a counter-example if A starts on the right-most column and B starts on the second from the right:

  x x x x x x x 8 9
  x x x x x x 8 x 9
  x x x x x 8 x x 9
  x x x x 7 x x x 9
  x x 7 x x x x x A
  B x x x x x x

[falsedan]

Yes, the claim is 'at least two of the paths taken from the 10 cards on the first row must converge', not 'there are no paths taken from the 10 cards on the first row which do not converge'.

[irishsultan]

I may have not made it clear, but like falsedan said, I'm not claiming that all paths must converge, only that at least two paths must converge.

edit: That being said, I made a mistake by following the claim from above that there are 10 possible paths, in fact there are only 9 cards in a row.

[sp332]

TFA says the two-deck version is successful about 95% of the time. If you click through, it says the one-deck version works 80% of the time.

[SiempreViernes]

I don't really see an explanation for the crucial fact that walks merge, which together with the fact that walks can't diverge means that the 1/10 chance of picking the same card at the start invariably increases towards the end.

[rtpg]

>. Their walker is equally likely to be on any type of card from the deck (it was shuffled), so there is at least a 1/13 chance of stepping to the spot your walk visited. Since you both use the same rule for taking steps then they will subsequently continue along the same route as your walker and so end at the same card