Weighted War is a game of bidding, where:

  • Both players have cards valued from $1$ to $11$ in their hands
  • There is a third pile of cards from $1$ to $11$ face down on the table and shuffled, with one random card being removed from it at the beginning of the game
  • Each beginning of a turn one random card from the table pile is turned face up, and the players offer one of their own cards face down.

  • When both players decided on their bid card, their cards are flipped and the higher value takes the table card. The bid cards are put aside, and new turn begins.

  • If the bid cards are equal, they are put aside and they start a next turn by flipping the next table card. This turn they bid for both cards. If the equal value repeats, they continue to add table cards to bid pile until someone wins it. (If both players run out of bidding cards and the bid pile hasn't been won yet, it goes to no one and stays aside.)

  • When all cards are won, players count their points by adding the values of the table cards they won. The winner is one with more points.

I'm wondering what would be the optimal strategy that maximizes your chances of winning and minimizes your chances of losing?

I could find barely anything on this game online.

One trivial things is that playing $1$ doesn't make sense since there is one card less at the table than in your hand.

Also, playing a random card won't be any good for you since cases like bidding $11$ for a low valued card will rarely have any good effects for you.

The video also mentions that they found that it's the best to play the same valued card as the table card but I couldn't find any proof of that. A Counter to that would be playing one card higher, and counter to that would be occasionally sacrificing some low cards to gain advantage in end game.

Anyway, I'm also interested in how much luck has an effect here.


I attempted to develop the optimal strategy for when there are only $2$, $3$ or $4$ cards in hopes of helping me to find a strategy for the $11$ card game. Assuming both players use the optimal strategy:

For $2$ card case, there is no point in playing $1$ so both play $2$ and end up in a draw.

For $3$ card case, a draw is forced $\frac{2}{3}$ times, and $\frac{1}{3}$ times happens when the first table card is $2$, then both players have equal chance of either winning,losing or ending up in a draw again. That depends on the table card that was removed at the beginning of the game.

For $4$ card case, a draw is forced $\frac{3}{4}$ times, and $\frac{1}{4}$ times happens when the first table card is $4$, draw should also be forced since $4$ is the best choice to play for both players, but if a $4$ is countered with a $2$ then both players have again equal chance of either winning,losing or ending up in a draw if they continue to play perfectly. ($3$ always beats $2$ and $4$ always beats $3$, if the rest of choices are best possible from both players.)

That means the safest option in this $\frac{1}{4}$ case is $4$ and also the best option against a random play; but that always results in a draw if both players play perfectly. Thus, if both perfect players play a set of games until the match is resolved, and observe that they both keep drawing with a $4$, one might attempt to break the draw chain by playing $2$ and give both players an equal chance to resolve the match without needing to worry that the opponent might suddenly play $3$, making that the optimal strategy? But if other can predict your $2$ and counter with a $3$ he beats your method.

That's why I'm not sure about the strategy for $4$ case if players are going for a win rather than accepting the draw as the best way of minimizing their losing chances, so I decided to post a separate question.

All in all, I have solved these $2,3,4$ cases by observing each possible game state to determine what would be the optimal play. If the pattern holds, the original game of $11$ cards should have a optimal strategy which if used by both players, always results in a draw and/or equal chances for both players to win, lose or end up in a draw.

But I still don't know how to create a general optimal strategy other than evaluating all possible states by brute force. I wonder If this can be solved by a optimal set of rules other than a brute force approach? (Either yes or no, I would need a proof of the solution)

  • 3
    $\begingroup$ Nice game. Why don't you start with the modified version where there are only cards numbered 1 to 2? $\endgroup$ Commented Jul 16, 2016 at 19:46

1 Answer 1


Some work has been done on this game, under the name Goofspiel or Game of Pure Strategy.

As of writing, that wikipedia article references Rhoads and Bartholdi 2012 (doi:10.3390/g3040150), which claims to have found the optimal strategy for maximizing score.

Unfortunately, their solution is numerical, and it's hard to pick out much intuitive advice for a human player.


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