I feel like I know what game this is mimicking but I cannot put my name on its title:

There are two players in the game (one is Even, the other is Odd). Even starts off with 30 blank cards, and Odd starts with 40 blank cards. Each turn, the players write down a number on there respective card and both lay it in the middle. They reveal them both at the same time and if the sum of the cards is even, then Even receives the two cards; while Odd would receive the two cards if they were odd. The player that ends up with all of the cards is the winner. Who will win the game? and how many turns are needed to produce a winner?

  • $\begingroup$ From my little understanding of probability, even should win the game. Since, if both even and odd cast odd numbers, it will go to even player and if both cast even numbers then also it will go to even player. The only time when odd player will receive cards is when there is even and odd numbers cast. $\endgroup$ – Ramesh Apr 27 '14 at 19:11
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    $\begingroup$ What happens with the cards that has already a number? Can be re-written? If not, what when there are no cards left? $\endgroup$ – ajotatxe Apr 27 '14 at 19:13
  • $\begingroup$ @Ramesh that is not a correct analysis, because they might not be picking between writing an even number or an odd number at random, and even if they were, even and odd is twice as likely as either even and even or odd and odd. $\endgroup$ – 6005 Apr 27 '14 at 19:28
  • $\begingroup$ @Goos, yeah you are right. I missed that part. My bad :) $\endgroup$ – Ramesh Apr 27 '14 at 19:42
  • $\begingroup$ What happens after 30 rounds have been played. Does Even start playing the cards won? In some particular order, or in the order won? $\endgroup$ – Ross Millikan Apr 27 '14 at 19:43

There is no answer to this problem, I think, unless you make the assumption that unmarked cards have to be played first before marked cards. If unmarked cards have to be played first, then odd can keep playing even numbers and the only way even can win is if even writes down an even number, but then eventually even will be left with only even cards whereas odd will have blank cards left, so at that point odd can keep writing down odd numbers and/or playing odd cards and win the rest of the hand. It's also easy to calculate the worst-case minimum number of moves for odd to win if they use this strategy. If players are allowed to play marked or unmarked cards, then consider the following. If even has an even or odd marked card in their hand, then they play that card with 50/50 probability, else they play the opposite card, possibly marking a card if necessary. Now if odd decides to always play even, and even has an even card, then even will win but even will still have unmarked cards so that they have the ability to play odd in the future. In essence, if one player decides to play even/odd with 50/50 probability using marked cards when they can, then I don't think you can predict the winner of the game, or how many moves it will take to win.


This answer is from the user that asked the question so it may still be wrong*

No finite solution to this only best possible outcome/worst possible outcome scenarios

Best = Odd wins after 30 rounds of winning consecutively

Worst = No winner after infinite rounds

Odds chances for winning any given round (42.857142%):

even + odd =odd

odd + 0 = odd

0 + odd = odd

Evens chances for winning any given round (57.142857%):

even + even = even

odd + odd = even

even + 0 = even

0 + even = even



Zero is completely equivalent to even, so there are actually only three possibilities and evens chances are 66.6%.

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    $\begingroup$ It's not clear to me that this answers the question. Are you making the same argument as Ramesh (in the first comment)? $\endgroup$ – Eric Stucky May 11 '14 at 23:07

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