I want to model and eventually solve for equilibrium the Magic the Gathering card "Wheel of Misfortune" Link to card
The card can be summed as such: players reveal a number simultaneously, players with the highest discard their hand and draw seven cards and lose life in proportion to the number, players who bet the lowest don't discard their hand and don't draw cards. In the event of a tie, the players lose life in the amount of the number and don't discard and draw.
For ease of modeling, we will only be considering the situation in which 2 players who both have 20 life, want to draw seven cards. I would like to eventually add the clause that a single card drawn is equal to 1 life, but that is besides the point. The objective of the game is to pay the least amount of life required in order to ensure you draw seven cards.
My question is, how would one go about modelling this situation? what would the players payoff equations look like? and how might one go about solving for the equilibrium.
The part about how ties resolve isn't mentioned on the linked page - on the card, or on the rulings. (I'm ignoring the other differences which don't matter when there's only 2 players.)
Assuming you are wrong about ties: both players should choose 0. They draw 7 cards, and lose 0 life.
Assuming you are right about ties: Let's say each player chooses a number, n, with a probability p_n. Here is where things get tricky - if either player can play a card before this resolves that affects the life totals, that will change things. Otherwise, provided taking 20 damage will kill a player - before drawing new cards, with no last minute saves or attacks from stuff on the field - then if both players choose p_20 to be 1, then both die. (Likewise, if one or both players chooses 0, then they won't take damage or draw cards.) This limits the range of number to 0<n<20. Whether or not the number must be an integer is kind of important.
So here is where analysis might break with reality:
In practice, you are guessing about whether the other player will have something, and you are guessing what number they will choose. You probably aren't going to 'randomize'.
Continuing the analysis:
Your clause 'a single card drawn is equal to 1 life' would limit things a lot more - possibly to 0<n<7, though it doesn't entirely characterize a solution because it's not clear what you mean. Are the players indifferent between A) getting 0 cards (keeping their current hand) and losing 0 life and B) getting 7 cards but losing 7 life? Assuming yes, then the players seem to be maximizing U:
U(losing x life, getting new cards) = 7-x
From this we can guess if both players choose 7, then for both of them U is -7 (because they lose 7 life, and they don't draw new cards). U could also be defined in a relative way, as the difference between Us, i.e., if I choose 5 and you choose 9, then I like that because my U is 0, but yours is -2.
If I only care about my U function then:
The payoff functions are just 7-x, where is the amount of life you lose. What is an equilibrium? Case 1: both players choose 0. U = 0 for both players: no cards drawn but no life lost. Both players choose 1, U = -1 for both players, and so on. Things get interesting at both players choose 7 though: U = -7 for both players. But, if one player knew they were both about to play that way, switching to choosing 6, or anything less than 7, would result in both players get U = 0, because the player who switches loses no life but draws no cards. The player who stays draws 7, but 7-7=0, so U=0 for that player as well.
As long as only integers (non-negative) may be chosen, then an 'equilibrium' is a probability distribution for each player where no player gains (or stays the same?) between any change they could make.
Weirdly, it's not obvious we can rule out numbers above 7. If you knew the other player was going to play 7, and you chose 8, then U=-1, and that's still better than -7 (choosing 7), though not as good as choosing a number below 7 (U = 0 for you). So maybe the probability distribution could go as high as 19 (assuming both players aren't fine with both dying, or something).
So your question is: what sets of two (possibly symmetric, possibly not) distributions over p_0 to p_19, for both players is an equilibrium?
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If we ignore mixed strategies, then...
Whether the other player chooses 0 or 1, you are better off choosing 1, if you don't want the other player to have an advantage. (This payout function might be: U for you - U for the other player.) This logic doesn't get you all the way to 20 however. If you were both going to play 20, you'd be better off choosing 19. If you were both going to play 19, you'd be better off switching to 18 (the other player get U=7-19=-12, you get U=0). If both players were both going to play 6 or 7 (the same exact number for both), then your payout would increase if you switched, but the relative payoff would be the same.
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For mixed strategies, let's start small. Both players choose 0 and 1 with some probability.
For player 1 that's a and b respectively. a + b = 1, b = 1-a
For player 2 that's c and d respectively. c + d = 1, d = 1-c
The payouts are:
. a b
c 0 6
d 6 -1
The 0 and below, mean both players get that. The positive numbers mean the player who chose the higher number won and got that, the other player just got zero.
So for player one the payouts are: $ 0*(a*c) + 6*(b*c) + 0*(a*d) -1*(b*d) = 6bc-bd = b(6c-d) $ If we assume a=c, and b=d (which seems likely if we solve this), then the payout for player one is b(6a-b). If we set a = 0, then this becomes -1. If we set a = 1 this becomes 0. In between..a=1-b, so 6ab-b^2 = 6(1-b)b-b^2 = 6b-7b^2. The first derivative with respect to b is 6-14b. If we set that equal to 0, then we have:
$6-14b = 0$
$6 = 14b$
$b = 6/14 = 3/7$
And the payout is $3/7*(6*4/7-3/7) = 3/7*6/7 = 18/49$
You can probably calculate this for more possibilities (choosing the numbers 0 to 19), but I'm not going to calculate 20+ partial derivatives at the moment. I'm not sure if the way the pure strategies turned out implies the mixed strategies will converge to there (choosing 6 or 7). If it turns out players can choose any number, integer or not, then you're probably going to have to start from scratch with continuous distributions, although that might still converge to somewhere around 6 to 7.