One-card poker

Originally posted by iamatiger
Yes, sorry folks. I did make some mistakes, working out algebra in the posting window is rather too error prone for me it seems.

Revised algebra indicates the following strategies are "optimal"
P1:
raise with king
pass with queen
raise on 1/3 of jack deals
after a pass with a queen and a raise by P2, see with prob 50%
after a pass with a jac
So the expected loss by P1 is (1/6)*(-1/3) = minus one eighteenth of a dollar per deal.

Edit - Scratch that. Great job iamtiger!

Edit2 - However, even though the expected value is -1/18, there's seems to be a mistake with those 50%. When P1 always raises with King is then he calls with a Q on 2/3 of the occasions. Also, P2 must call with Q on 1/3 of the time.

Here's the Mathematica code I used to solve the algebra (does anybody know how to set the domain of the probabilities between 0 and 1 easily?). I used what the technique I mentioned earlier of writing the value functions, maximizing and finding a fixed point. It was very rushed, so it's a crappy code but it works anyway.

To clarify:
a - probability of p1 raising with K
b - probability of p1 raising with Q (DOMINATED, b=0)
c - probability of p1 raising with J
d - probability of p1 calling with Q after 2 raises
e - probability of p2 calling with Q after a raise
f - probability of p2 raising with K after a pass (DOMINATING, f=1)
g - probability of p2 raising with Q after a pass (DOMINATED, g=0)
h - probability of p2 raising with J after a pass (trying to bluff)

Clear["Global`*"];
p1k = FullSimplify[ a*(1/2*(2*e + (1 - e)) + 1/2) + (1 - a)*(1/2*(2*g + (1 - g)) + 1/2*(2*h + (1 - h)))];
p1q = FullSimplify[(1 - b) (1/2 (h (2*d - (1 - d)) + (1 - h)) +1/2 (f (-2*d - (1 - d)) - (1 - f))) + b*(1/2 - 1)];
p1j = FullSimplify[-(1 - c) + c (-1 - e + (1 - e)/2)];
valuefunction1 = FullSimplify[(p1k + p1q + p1j)/3];
probraise = (a + b + c)/3;
p2kp = 1/2 (2 - b - c)/(3 - a - b - c);
p2qp = 1/2 (2 - a - c)/(3 - a - b - c);
p2jp = 1/2 (2 - a - b)/(3 - a - b - c);
p2kr = (b + c)/(2 (a + b + c));
p2qr = (a + c)/(2 (a + b + c));
p2jr = (a + b)/(2 (a + b + c));
p2raise =
FullSimplify[2*p2kr + p2qr (e ((2 c)/(a + c) - (2 a)/(a + c)) - (1 - e)) - p2jr];
p2pass = FullSimplify[p2kp ((1 - b)/(2 - b - c) (2 d + (1 - d)) + (1 - c)/(2 - b - c)) +
p2qp (g ((1 - c)/(2 - a - c) - 2 (1 - a)/(2 - a - c)) + (1 - g) ((1 - c)/(2 - a - c) - (1 - a)/(2 - a - c))) +
p2jp (h ((1 - b)/(2 - a - b) (1 - 3 d) - 2 (1 - a)/(2 - a - b)) - (1 - h))];
valuefunction2 = FullSimplify[probraise*p2raise + (1 - probraise)*p2pass];
foca = D[valuefunction1, a] == 0 ;
focb = b == 0;
focc = D[valuefunction1, c] == 0;
focd = D[valuefunction1, d] == 0;
foce = D[valuefunction2, e] == 0;
focf = f == 1;
focg = g == 0;
foch = D[valuefunction2, h] == 0;
eqs = {foca, focb, focc, focd, foce, focf, focg, foch,
a <= 1, a >= 0,
b <= 1, b >= 0,
c <= 1, c >= 0,
d <= 1, d >= 0,
e <= 1, e >= 0,
f <= 1, f >= 0,
g <= 1, g >= 0,
h <= 1, h >= 0};
vars = {a, b, c, d, e, f, g, h};
Reduce[eqs, vars]


Output:
0<=a<=1
b=0
c=a/3
d=(1+a)/3
e=1/3
f=1
g=0
h=1/3
valuefunction1 = -1/18
valuefunction2 = 1/18

Again, kudos to iamtiger for solving it first.

Originally posted by Palynka
Edit - Scratch that. Great job iamtiger!

Edit2 - However, even though the expected value is -1/18, there's seems to be a mistake with those 50%. When P1 always raises with King is then he calls with a Q on 2/3 of the occasions. Also, P2 must call with Q on 1/3 of the time.

Hmm, sorry, I lazily didn't explan the 50% "call" reason. Its because, with the 1/3rd (exactly) chance of raising with a jack and the 100% chance of raising with the king it makes no difference at all whether the player with the queen folds or sees.

Say the player with the queen folds, he loses $1

Say he sees: he has a 3/4 chance of seeing a king and losing $2, but a 1/4 chance of seeing a jack and winning $2. So his expected loss is 2*3/4 - 2/4 = $1

In fact this means that the ratio of calls to passes of the player with the queen can be anything at all, it doesn't affect his profit.

This equality of seeing/passing is balanced on a knife-edge. If the player with the king/jack shifts his probabilities so that more than 1/4 of the raises are jacks, then the player with the queen should see every time for optimal profits, similarly if the player with the king/jack under-raises jacks, then the player with the queen should pass every time. Each of these deviations reduce the profit of the player with the king/jack as long as the queen player responds appropriately, but in play such a shift might take time to detect, and in the interim the profits of the king/jack player are increased.

I thought 50% seeing might be wise as it gives no obvious "way to go" for the king/jack player if he is thinking about shifting the probs.

So the 1/4 jack point is the only point that works for a 50% seeing strategy, but is also the point that maximises the profits of the player with the king and jack.

Nice work with the mathematica solution by the way. Sadly I don't have mathematica,and I'm not a student so I can't get it cheap 🙁

Also, I think a should definitely be 1 and c should definitely be 1/3. The equation for profit is readily differentiable by a or c and in either case has a constant positive result. This means, that for any a&c a larger a or c will acrue a larger profit (as long as the c=a/3 relationship is maintained) this takes us to A=1 C=1/3 which is the maximum size of both.

That's incorrect. The value function for P1 is differentiable for a but it's increasing only for non-optimal choices of P2's strategy. P2 strategy will involve choosing his values such that P1 cannot exploit a to his advantage. This is not uncommon when choosing mixed strategies in a sequential game.

By his choice of strategy, he can force P1 to have to optimally raise with J one third of the time that he raises with K (c = a/3) AND make it impossible for him to manipulate a such that he can gain an advantage.

It's not the most intuitive of explanations, but I got it by brute force of maximization. PBE6's link also mentions this. This is also the reason why it's not optimal to choose e=1/2. In that case, it's optimal for P2 iff P1 doesn't adjust his strategy, but it would no longer be optimal to P1 to choose c=a/3 given e=1/2.

Edit - What's your field iamatiger? Just curious, if you don't want to answer don't feel pressured to.

Originally posted by Palynka
That's incorrect. The value function for P1 is differentiable for a but it's increasing only for non-optimal choices of P2's strategy. P2 strategy will involve choosing his values such that P1 cannot exploit a to his advantage. This is not uncommon when choosing mixed strategies in a sequential game.

By his choice of strategy, he can force P1 to have to r field iamatiger? Just curious, if you don't want to answer don't feel pressured to.

I'm a systems/software engineer. Currently I work on radars.

When you say "that's incorrect" did you just mean the bit about the differentiation of the value function? Which value function did you get?

Originally posted by iamatiger
I'm a systems/software engineer. Currently I work on radars.

When you say "that's incorrect" did you just mean the bit about the differentiation of the value function? Which value function did you get?

Value function for 1 (first term is if he holds a K, second a Q and third a J):

1/3{a(1/2(2e + (1 - e)) + 1/2) + (1 - a)(1/2(2g + (1 - g)) + 1/2(2h + (1 - h)))}+
+1/3{b(1/2 - 1)+(1 - b) (1/2 (h (2d - (1 - d)) + (1 - h)) + 1/2 (f (-2d - (1 - d)) - (1 - f)))}+
+1/3{-(1 - c) + c (-1 - e + (1 - e)/2)}

If you differentiate this with respect to a to find the first-order condition and do some simple algebra you get:

FOCa: 1/6(e-g-h) =0

P2 at the optimum will set g=0, e=h=1/3. At this value, the value function is still differentiable for a but NOT with a constant positive result. With the choices you propose for P2, then the FOC wrt C is always negative and the FOC wrt A is always positive so it's optimal for P1 to decrease C to 0. Of course, then it's not optimal for P2 to keep his choices. P2 has to keep P1 at knife-edge so he can take advantage of playing second in this game.