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You and your friend have come up with a new two-player card game. There are 15 distinct cards numbered 1 to 15. The first player orders the cards in some order of his choice and places them face down on the table, without showing them to the second player. Now the second player turns over the top most card and places it into one of 5 piles. He repeats the process by turning over the next card and placing it into another pile. Note that he cannot place a card into the pile that the previous card was placed in. The game finishes once all cards have been turned over and placed into piles. If every pile has the same sum of numbers then the second player wins; otherwise the first player wins. Should you go first or second and why?

Bonus: would your decision to go first or second change if the cards had to be placed into 4 piles?

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  • $\begingroup$ Note 120/5 is 24 :) $\endgroup$
    – Dmitry Kamenetsky
    Commented Nov 8, 2021 at 8:22
  • $\begingroup$ Yeah, as usual I noticed my silly mistake seconds after I pressed the submit button. $\endgroup$
    – Jaap Scherphuis
    Commented Nov 8, 2021 at 9:01
  • $\begingroup$ Don't worry it happens to the best of us. $\endgroup$
    – Dmitry Kamenetsky
    Commented Nov 8, 2021 at 9:58
  • $\begingroup$ Just to check, does each pile have to have 3 cards or may some piles have 2 and others 4? $\endgroup$
    – hexomino
    Commented Nov 8, 2021 at 12:48
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    $\begingroup$ It's extremely confusing to use "they" on a single person. Especially in this case. It's very unclear which player is supposed to do which steps, or whether "they" (the two of them) are supposed to make moves in turn. This kind of usage of "they" is destroying the grammar of the English language. $\endgroup$
    – WhatsUp
    Commented Nov 9, 2021 at 1:35

1 Answer 1

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Base question

Guaranteed win

Player 1 e.g. wins with: all odds in any order, then 10 14 4 12 2 8 6

Since each stack must total 24 = even, after playing the odds 1 stack must still be empty (since there are less than 10 odds).

An empty stack cannot be filled with only evens, since 24 cannot be made with a legal combination of evens (if ordered as above)

I determined the sequence of evens as follows: 1/ Write down all (8) possible combos 2/ The only pair (14,10) must be adjacent 3/ 14 can have only one other neighbor so 8,2 or 4,6 must be adjacent (or both); same for 12,2 and 8,6 - a guess: using 2,8,6 seems a good choice since it helps for both requirements. 4/ Trying to add the now most problematic 12 to 14,10 and 2,8,6 yielded my solution after some fiddling. - note that there are many? more possibilities, e.g. 4,12,2,8,6,14,10

Addition: Bonus question

The odds are in player 1s favor

I don't see a guaranteed win, but again player one can make sure the evens do not make the required total of 30: 4 14 10 12 8 6. (note: 2 is not needed to prevent 30, it can be 1st or 7th) If the odds are put behind those evens in random order, player 1 has 47/70 chance of winning according to my calculations.

Note that 2 odds must be assigned to each stack, as soon as the first is assigned, the other is fixed, and may be the next coming up, giving player 1 the win.

summary of calculation that sums up to 47/70:
Looking at the pairing of the first 4 odds coming up:
pairs chance chance of success for player 1
1234 - 8/35 - 1/4 (fifth odd is from pair 4)
1212 - 4/35 - 2/3 (only 3434 is a win for player 2)
1231 - 4/35 - 1/2 (pair 4 must be adjacent, 44XX X44X and XX44 are success)
1232 - 4/35 - 1/2 (same)
1213 - 4/35 - 7/12 (also 3424 is successful)
other- 11/35 - 1 (already an adjacent pair within the first 4 odds)

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  • $\begingroup$ Why must one stack be empty? Why can't the player lay the first five cards down into separate stacks? $\endgroup$
    – Jason Goemaat
    Commented Nov 8, 2021 at 19:23
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    $\begingroup$ @JasonGoemaat If you put the first five odd card in separate piles, you only have 3 more odd cards to follow, so some of the piles will remain odd and will never sum to 24. The 8 odd cards have to pair up to get even totals in each pile. $\endgroup$
    – Jaap Scherphuis
    Commented Nov 8, 2021 at 19:37
  • $\begingroup$ Ah, so "Nyy fgnpxf zhfg gbgny gb 24. N fgnpx pnaabg pbagnva 1 be 3 bqq ahzoref be vg pbhyqa'g gbgny gb 24, na rira ahzore. Fb gurer zhfg or 4 fgnpxf jvgu 2 bqq ahzoref va gurz, naq bar fgnpx pbagnvavat bayl rira ahzoref. Orpnhfr rnpu cbffvoyr cbfvgvir-bayl gevcyr pbagnvaf gjb pbafrphgvir pneqf va lbhe nafjre, gurl ner nyy vzcbffvoyr gb perngr." $\endgroup$
    – Jason Goemaat
    Commented Nov 8, 2021 at 20:03
  • $\begingroup$ @JasonGoemaat Yes, but you're not limited to triples so there are a few more cases to check. $\endgroup$
    – Jaap Scherphuis
    Commented Nov 8, 2021 at 20:22
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    $\begingroup$ How did you realise that the order of the even cards should be 10 14 4 12 2 8 ? Just wanting to understand the thought process. $\endgroup$
    – Hemant Agarwal
    Commented Nov 9, 2021 at 3:34

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