[theorie des jeux] equations d'indifferences

[pariAAboy]

bonsoir


au passage "solving these 6 equations gives......" j'ai pas compris comment il arrivait à déduire la valeur de, par exemple, a = 2B/(2+B)²(1+B)

[mathelot]

Bonsoir,
pour que les jeux soient équilibrés, il faut et il suffit que les espérances de gain du joueur soient nulles.
ce qui donne six équations ( égalités). On égalise les espérances de gains gagnants et les espérances de pertes.

[pariAAboy]

re

tart with the given 6 equations:

(1) d = B / (2 + B)

(2) (2 + B)e + Bf = B

(3) d + e - f = 0

(4) (2 + B)a + Bc = B

(5) (2 + B)b - Bc - 2e = 0

(6) b + c - 2f = 0

We are trying to find expressions for a, b, c, d, e, and f just in terms of B. Notice first that equation (1) accomplishes this for d, so one down, 5 to go.

Next, substitute (1) into (3) to arrive at B / (2 + B) + e - f = 0

Multiply by (2 + B) to get B + (2 + B)e - (2+B)f = 0

This looks similar to (2), so add Bf to both sides to get B + (2+B)e + Bf - (2+B)f = Bf

Now, use (2) to replace a portion of the left hand side: B + B - (2+B)f = Bf

Moving the f term to the right side and factoring: 2B = f(2 + 2B)

Divide both sides by (2 + 2B) and cancel a factor of 2 to get f = B / (1 + B)

Now, sub into (3): B/(2+B) + e - B/(1+B) = 0

Rearrange: e = B/(1+B) - B/(2+B) = [(2 + B)B - (1+B)B]/[(1+B)(2+B)] = [B^2 + 2B - B^2 - B][(1+B)(2+B)] = B/[(1+B)(2+B)]

So far, we have successfully solved for d, e, and f. Now for a, b, and c.

Start with equation (6) and sub in the value for f: b + c - 2B/(1+B) = 0 [A]

Then take equation (5) and sub in the expression for e: (2+B)b - Bc - 2B/[(1+B)(2+B)] = 0 [B]

Multiple [A] by B to get Bb + Bc - 2B^2/(1+B) = 0

Add this to [B] (notice that the c terms cancel) to get (2 + 2B)b - 2B^2/(1+B) - 2B/[(1+B)(2+B)] = 0

Simplify a little bit: (2 + 2B)b = [2B^2(2+B) + 2B]/[(1+B)(2+B)] = [2B(B^2 + 2B +1]/[(1+B)(2+B)] = [2B(1+B)^2]/[(1+B)(2+B)] = [2B(1+B)]/(2+B)

Divide both sides by (2 + 2B): b = B/(2+B)

Now, plug this back into (6): B/(2+B) + c - 2B/(1+B) = 0

Rearrange: c = 2B/(1+B) - B/(2+B) = [4B + 2B^2 - B - B^2]/[(1+B)(2+B)] = [B(3+B)] / [(1+B)(2+B)]

Now, all that is left is to solve for a.

Use (4): (2+B)a + B[B(3+B)] / [(1+B)(2+B)] = B

So (2+B)a = B - [B^2(3+B)] / [(1+B)(2+B)]

Divide by (2+B): a = B/(2+B) - [B^2(3+B)] / [(1+B)(2+B)^2] = [B(1+B)(2+B) - B^2(3+B)]/[(1+B)(2+B)^2] = [B(B^2 + 3B + 2 - 3B - B^2)]/[(1+B)(2+B)^2] = 2B/[(1+B)(2+B)^2]

j'suis bloqué à l'étape "Multiple [A] by B to get Bb + Bc - 2B^2/(1+B) = 0"
j'ai pas du tout compris ce passage

[mathelot]

re,
que veut dire check-folds et check-calls ?
quelle est la nature du jeu ? cartes, poker ? roulette ? jeu abstrait ?

[pariAAboy]

re,
que veut dire check-folds et check-calls ?
quelle est la nature du jeu ? cartes, poker ? roulette ? jeu abstrait ?

poker (check/fold = laisser la parole et coucher contre une mise et check/call = laisser la parole et payer contre une mise et mise = grand B dans la démo et B > 0)
le pdf complet, j'suis à la page 4
https://www.tomsferguson.com/papers/poker2.pdf

[pariAAboy]

en fait je comprend plus les calculs à partir de tout ce qu'il a en dessous de "Multiple [A] by B to get Bb + Bc - 2B^2/(1+B) = 0"

meme avec ces explications je comprend absolument rien à la manip' :

Now, let's look at what we have:

[equation 1] b + c - 2B/(1+B) = 0

[equation 2] (2+B)b - Bc - 2B/[(1+B)(2+B)] = 0

One manipulation I can do is to add the left sides of these two equations and also add the right sides together. But before I do that, I'm going to plan ahead so that when I add them, one of the variables cancels out. Notices that we have a c term in [equation 1] and a -Bc term in [equation 2]. These wouldn't cancel yet, but if I multiplied [equation 1] by B, when I add them, the c terms would do away. Let's do that:

B*[equation 1] Bb + Bc - 2B^2 / (1+B) = 0
[equation 2] (2+B)b - Bc - 2B/[(1+B)(2+B)] = 0
Now notice that if I add the top equation to the bottom, the c terms cancel and I'm left with: Bb + (2+B)b - 2B^2/(1+B) - 2B/[(1+B)(2+B)] = 0