Goblin Game is a Magic: the Gathering card. The full text of the spell is:

Each player hides at least one item, then all players reveal them simultaneously. Each player loses life equal to the number of items he or she revealed. The player who revealed the fewest items then loses half his or her life, rounded up. If two or more players are tied for fewest, each loses half his or her life, rounded up.

(Rulings on the card permit players to simply write down a number rather than actually hide physical objects.)

I am currently building a deck built upon the plan of casting Goblin Game, and thus will generally be "prepared" to play the game, while my opponents think on their feet. However, I'm not certain what the ideal choice is when playing the game.

The format of Magic game I play has all players starting at 40 life, and there are 3 or 4 players at the table.

Q1: What optimal strategy for picking my number can I use when playing the Goblin Game, assuming all participants start at 40 life?

The goal, obviously, is to be the player with the highest life total after the Game is complete.

The next part I call "Nobody wins the Goblin Game!" Or, you can consider it an iterated version of the above.

The secondary aim in the deck I am building is not only to play Goblin Game, but to copy it a number of times. 2 or 4 iterations of the Game will be easy for me to accomplish, but the number of iterations is theoretically unbounded, depending on what resources I have available to me. (Without "going infinite", the limit on iterations is 24.)

Q2: Given N iterations of the Goblin Game, assuming all participants are at 40 life before the first Game begins, what is the ideal strategy to selecting numbers for each Game?

There are five basic outcomes of the iterated Goblin Game from my perspective, in order of preference:

  1. I win the Game: all opponents reach 0 or less life, while I have a positive life total
  2. I end the Game with the most life: myself and at least one opponent have a positive life total, but mine is highest
  3. I end the Game, but don't have the most life: myself and at least one other opponent have a positive life total, but mine isn't the highest
  4. Everybody loses the Game: I reach 0 or less life on the same iteration as all of my remaining opponents
  5. I lose the Game: I reach 0 or less life while at least one opponent has a positive life total

Note that if I, as the person who controls the Game, reach 0 or less life, all remaining iterations of the Game are canceled, so any remaining opponents won't lose any more life. Thus, simply producing an infinite number of iterations of the Game is not sufficient to guarantee outcome #4. On the other hand, once all of my opponents reach 0 or less life, any remaining Games are also canceled, so I don't have to worry about hitting #4 when I've already achieved #1.

Finally, "Goblins Cheat at Games". My deck includes three cards which give me an advantage over my opponents:

Platinum Angel:

You can't lose the game and your opponents can't win the game.

Platinum Emperion:

Your life total can't change.

Lich's Mirror:

If you would lose the game, instead shuffle your hand, your graveyard, and all permanents you own into your library, then draw seven cards and your life total becomes 20.

With the Angel, my life total can go negative without me being eliminated. With the Emperion, my life total simply stays where it is, no matter what number I choose for the Goblin Game. With the Mirror, when and if I do lose due to the Goblin Game, my life is reset to 20 no matter how far negative it was at the resolution of the Goblin Game, thus giving me a few more iterations during which I can play. (Of course, since the Mirror is a permanent I own, it will get shuffled into my deck and won't save me a second time.)

Q3: Given the presence of one of the above advantages, how does my strategy for the iterated Goblin Game change?

Note that my opponents will all be fully aware that I have the advantages, which could naturally affect their decision-making process. Of course, in the case of the Angel and the Emperion, as far as I'm aware I should simply be considered as a non-participant in the Game; I could choose 100 (no opponent can beat that without losing), and then it's a fight between my opponents to see which will be lower than the other.

When I play this for real, I will of course be playing against people who may make illogical or incorrect choices – they're not mathematicians, and they will generally have to come up with an answer without an overabundance of time to think about it. However, for the purposes of the answer here, you may assume ideal logic (or even cooperation, since they're free to talk to one another) on the part of my opponents.

  • $\begingroup$ I guess I'm unclear on why you wouldn't just write the biggest number you can think of? In order for there to be a reasonable notion of strategy, there needs to be some tradeoff for picking a large number. You mention 100 as an upper bound, but I don't see why that is true, given what you've told us. $\endgroup$ Commented Jul 24, 2014 at 6:59
  • $\begingroup$ You lose life equal to the number you pick, then after that the person who picked the lowest loses half their life. If you pick a number equal or greater than your life total, you'd lose immediately. (I mention 100 as an arbitrary large number. The players begin at 40 life. Nobody can match 100 without losing immediately... but they can't match 40, either.) $\endgroup$
    – Brian S
    Commented Jul 24, 2014 at 7:00
  • $\begingroup$ From what you're saying, the equilibrium is to play 0 all the time. You're sure to be the smallest hence to lose at most 1. Everybody else who plays strictly more will lose more than you. $\endgroup$
    – Matt B.
    Commented Jul 24, 2014 at 9:02
  • $\begingroup$ @Matt: Disregarding goblin powers, if you're the smallest number, you lose half your life. After six iterations, that costs you the game. $\endgroup$ Commented Jul 24, 2014 at 11:24
  • $\begingroup$ I am fairly sure this game doesn't have a solution that helps you. It reminds me of an experiment by Selten I participated in: a repeated game in which in each round each participant writes down a number from 1-20, and whoever is closest to half the numbers' average wins that round. The averages. maybe not surprisingly, slowly approached zero; and in the last round one sociology student proudly told me he figured out that using 20 would maximize his chances! (Having observed winning numbers of about 7, 5, 3 before). Even assuming you'd find a NE, your friends are unlikely to play by it. $\endgroup$ Commented Jul 24, 2014 at 15:28

1 Answer 1


Let's call $P_1,...,P_n$ the players, $\alpha_1,...,\alpha_n$ the total life points they have, $b_1,...,b_n$ the amount of life points they're ready to bet.

Without loss of generality, let's suppose $\alpha_1 \leq...\leq\alpha_n$.

EDIT: this is based on a payoff which would be $-\frac{\alpha_i}{2}$ if $b_i$ is the minimum, $-b_i$ otherwise, which is not what is requested.

The pay-off here seems to be $-\frac{-\alpha_i-b_i}{2}$ if it is the minimum or $-b_i$ otherwise.

Q1/2: It feels to me that the Nash equilibrium is to bet a number of points $b_i$ between $\frac{\alpha_{i}}{2}$ and $\frac{\alpha_{i-1}}{2}$ with $\alpha_0=1$ (imagining that you can have decimal lives for simplicity).

It is obvious that $P_1$ cannot improve his situation from that strategy. If he bets less, he will be losing the same because his bet will be strictly lower, if he bets more, well, he is worse off.

Knowing this, $P_2$ will want to bet as little as possible but ensuring he's staying ahead of $P_1$ so betting:

  • at least $\frac{\alpha_{2-1}}{2}+1$ ensures he is not the smallest (or if he is, his loss will be only 50% vs $P1$ strictly more than 50%).
  • at most $\frac{\alpha_2}{2}$, otherwise he is taking a hit for no reason

By induction, everyone will play along those lines, and the final bet is $\min(\frac{\alpha_{i-1}}{2}+1,\frac{\alpha_i}{2})$.

If everybody starts at the same level, then everybody declares 20 in your example, and everybody's life is divided by 2 at each stage. Any smart ass who plays more will lose more, and who plays less will not change the game.

That's a classical example of a Nash equilibrium where the equilibrium is to bring everyone down with you.

You can also note that if you get an edge, people can't team up against you, you're perfectly immune to their actions, there is no cooperation possible.


  • The angel doesn't change anything as your pay-off is unchanged from above.

  • The Emperion: conratulations, you won the game no matter what, because everyone else will be losing at least 1 at each turn.

  • The mirror: doesn't give you an edge either in the game, but will push you above everyone else at the last round, making you win the tournament.

  • $\begingroup$ If I understand the math correctly, this suggests to pick min(a,b), where a is 1 greater than 1/2 an opponent's life total, where that opponent is someone with equal or less life than me (the closer to my life total the better), and b is 1/2 my life total. Do I understand that correctly? (Also, if everyone picks 1/2 each stage, they'll be losing 3/4 of their life each iteration, not 1/2, since ties for lowest choice also lose 1/2 life. 40->20->10; 10->5->2; 2->1->0) $\endgroup$
    – Brian S
    Commented Jul 24, 2014 at 14:53
  • $\begingroup$ My bad, it does change the solution quite dramatically. I would need to give it more thoughts, as the equilibrium I described above doesn't exist. $\endgroup$
    – Matt B.
    Commented Jul 24, 2014 at 15:17
  • $\begingroup$ Uhh with platinum angel you can just keep picking 40 because your life total can drop arbitrary low. $\endgroup$
    – Joshua
    Commented Nov 20, 2015 at 23:46

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