0 to 1 game: three players

The famous 0 to 1 game in a pre-flop no-raise version for three (well, actually two) players:

Each player draws a random number between 0 and 1.
First player A pays "small blind" $1.
Second player B pays "big blind" $2.
Third player C may call for $2 or fold.
First player A my call for $1 or fold.
(B gets no action).
Highest number in the game wins the pot.

With only two players A and B, obviously A should call at 0.25 or higher. EV for both positions is zero.

Whats the best strategy for player A and C, and whats the EV for each position ?

I've already got some "exact" numbers, different approaches wouldwould be more interesting to discuss.

I'll take a stab at this, only because no one else has answered yet, although it's very likely I have no idea what I'm talking about. :)

As you mentioned, if Player C doesn't call, and if it's just Player A and Player B, A should call with .25 or better.

If Player C DOES call, A only has to put an additional $1 in the pot, which now has $5 in it. If so, now he only needs .1667 or better to call. (I think.)


As far as Player C is concerned, he doesn't know yet if he will be facing Player A and Player B, or just Player B.

If he knew ahead of time A was going to fold, and thus he would just be up against B, he would be risking $2 to win $3, and thus would need .4 or better to call. (I think.)

But of he knew ahead of time that Player A WAS going to call, he would be risking $2 to win $4, and thus would only need .3333 or better to call.

Thus, the average of the two might be the right play for him. If so, he might need .3666 to call. (.4 + .3333) / 2

Either that or since he can assume that when he does call, A will be also be calling about 83% of the time. So maybe that's figured into what he needs, which should drop his requirements a tad.



I suspect there is more to this puzzle/problem than this, but at least this gets the ball rolling and gives others a chance to jump in.

This a very worthy project, but I just don't have the time right now. Hopefully, I can spend some time on this in November.

Quote: EdCollins

As you mentioned, if Player C doesn't call, and if it's just Player A and Player B, A should call with .25 or better.

If Player C DOES call, A only has to put an additional $1 in the pot, which now has $5 in it. If so, now he only needs .1667 or better to call. (I think.)

Maybe a small hint: The probabiltiy of being ahead with a hand of x against n unknown opponents is x^n.

Quote: MangoJ

Maybe a small hint: The probabiltiy of being ahead with a hand of x against n unknown opponents is x^n.

Ah.

So if Player A is getting 5 to 1 on his money, (when he calls the additional $1 to win the pot of $5, when Player C calls) then he needs to win approx .166 of the time to properly make that call. So he would need about .408 or better, since .408^2 = .166.

Yes? Right track?

Yeah thats the right track.

Now, C won't call any hand, so A will still be behind with .408. But let's say A will fold 0 to .4 to a call of C. When should C call ?

Quote: MangoJ

When should C call ?

Yea, that's the part I still haven't figured out yet. :)

My reasoning goes like this.

If C new that A wouldn't call, C would be risking $2 to win a pot pf $3, so he would only have to win 40% of the time to break even. So with one opponent, B, he would need a .4 or better to call.

And if C new that A WAS going to call, C would be risking $2 to win a pot of $4, so he would only have to win about 33% of the time to break even. But with now two opponents against him, he needs a higher total. It seems that .5745^2 = .3335, so needing .5745 or greater is probably in the ballpark.

But C doesn't have ESP, so I still don't know if C should take the average of this, ((.4 + .5745) / 2) or determine his calling number in another fashion.

If A and C play optimally, this has GOT to be a negative EV game for B, I'm pretty sure about that.

Haha, no ESP yeah :)

My personal approach was different to that problem, but your ideas are inspirating.

An average seems like a reasonable guess, so let's see:
If C calls, A will fold at least for 40%, since A will need .4 (actually more) to call 2 opponents. So 40% of the threshold (.4 for $2 to win $3) we found.

Let's concentrate on the remaining 60%. So we assume A has .4 or higher and will call, giving us $2 to win $4.
We need p=1/3. But p is no longer x^2, since A already has .4. In fact C will still beat B with probability x. But to beat A we will have probability (x-.4)/.6.
(just a guess, should be linear, and 0 for x=.4 and 1 for x=1).
So we would have p=1/3 = x * (x-.4)/.6, or x>0.690.

I*m not sure if simply taking the (weighted) average is a good guess, but then it would be 40% * 0.4 + 60% * 0.690 = 0.574

A more sophisticated calculation would probably be
0 = 40% * (x * $5 - (1-x)*$2) + 60% * (x * (x-.4)/.6 * $6 - (1 - x * (x-.4)/.6) * $2)
yielding x = .525 for C calling...


... but this assumes A will call at .4 - which is clearly stupid if C calls at .525. So a next level calculation would need a new threshold for A calling at least at .525 (instead of 0.). Rinse and repeat, let's hope it will converge.... does it ?

Okay, here is my solution. I must admit, it's a bit "mathematical" - it gives a unique (and most probably correct) solution. However the math is very shy with regard to proper reasoning and interpretation of the result. But that is typical for optimization problems, it gives you a solution, but not an explanation :)

Let us first define the players strategies, before solving it. a1 is the call threshold for A if C calls, a2 the threshold if if C folds. c1 is the threshold of C.

Following are three steps:
First step is calculating the EV(A) and EV(B) for players A and C, provided we know all player strategies.
This is basically pen&paper work: we integrate over all 3 hands, i.e. over a volume of [0..1]^3. There are basically 2*2 segments of known size of interest, corresponding to the 2 times 2 call/fold decisions of A and C. For each of these segments the remaining players, the call-range and potsize is known: it is straightforward what the EV for each player for each segment is. The sum off all segments constitute to the EVs of all players, all depending on all players strategies.

Second step is: each player simultaneous tries to maximize his EV using his own strategy set, i.e. player A maximizes EV(A) with respect to a1 and a2 (given c1) - and player C maximizes EV(C) with respect to c1. The idea is, once such a simultaneous optimal strategy set is found, any single player deviating from it will automatically pay in his EV for the deviation. The game is simple enough that no player would always call or always fold in any situation. I.e. the overall solution is always strictly located within the cube, never on the sides. This makes things quite easy.

Zeroing the derivatives dEV(A)/da1 = 0 (and likewise) gives 3 equations which have to be simultaneously fulfilled for the optimal strategy.

Third step: Once all equations are defined, simply "search" for the optimal strategy set. In this simple game, EV is a piecewise polynomal (of order 3), hence all
derivatives are also piecewise polynomals of 2nd order, and there are efficient algorithms available.

From the 3 equations, I found all 3 solutions:
#1: a1 = .120754, a2 = .25, c1 = 3.31251
#2: a1 = -.80585, a2 = .25, c1 = -.49637
#3: a1 = .685096, a2 = .25, c1 = .58386

Only set #3 gives compatible game parameters, as all threshold decisions need to be between 0 and 1. But again, these are numbers, not explanations.

BTW: is it luck that there is only a single compatible solution ? what if there is none, or more than one ?

Here is my solution. Basic arithmetic and algebra. Maybe not less mathematical than the other solution, but maybe easier to see what's going on in the game and how the decision points are arrived at.

Let a be the threshold for A to call assuming C has called. (I'm ignoring the case where C folds, since that is just original 2 player case, i.e. A should call with 0.25 or higher). Let c be the threshold for C to call (he doesn't yet know whether A will call). These thresholds will be the break-even points, i.e. where the EV is zero.

some pre-lims: if your score is x, the probability of beating a random player is x. The probability of beating a player when you know his score is at least y is (x-y)/(1-y), assuming x > y. Obviously, your chance of winning is zero if x < y. Also, probability of beating multiple players is the product of the probability of beating each.

Player A:
Get break-even value of a, as a function of c from:
6*(a-c)/(1-c)*a = 1 (i.e. total pot of 6 * prob(beating C) * prob(beating B) = cost of 1 to make the call)
-- also as a sanity check, a>c in the final solution, as A shouldn't call with less than c if that is C's minimum score to call

Player C:
A bit more complicated as there are 2 cases.
Case 1, a > c : C can only win if A folds, for a total pot of 5. Probability of A folding is a. So:
5*a*c = 2

Case 2, a < c: C can also beat A if A calls for a total pot of 6. Probability of A calling is 1-a
5*a*c + 6*(1-a)*(c-a)/(1-a)*c = 2
simplify: 6c^2 - ac = 2
-- note: because of what I said above, we could probably ignore this as making no sense. But I'll go through the math anyways.

Two equations in two unknowns. Simple algebra from substituting for a for of the two cases for player C, solving for c gives:
Case 1
c^3 - 3.4c^2 + 0.96 = 0
solve for c = 0.58386, a = 0.685096 (as MangoJ got)
-- valid since a > c as required

Case 2
180c^4 + c^3 - 133c^2 + 24 =0
get c = 0.145461, a = 0.457065
This solution is not valid since a is not < c, as required for Case 2. (also makes no common sense as c<1/3. C at best gets 2-1 on his money. Against only a single player, c would still have to be at least 1/3.)

Quote: MangoJ

However the math is very shy with regard to proper reasoning and interpretation of the result. But that is typical for optimization problems, it gives you a solution, but not an explanation :)

There IS usually a last step to a mathematical reasoning: trying to extract an 'explanation', like you say.

For example:
1) As B has no decision to make, there is no reason for A to bluff, or to use a mixed strategy (= a probabilistic choice of action instead of the deterministic one: "if I have x, then I always act Z.")
The same is true for C. So the strategy space is limited to finding "calling thresholds" a and c . (a2 being 0.25 is trivial, I assume. I renamed a1 as a.)

2) When she sees C calling, A knows he's got more than c. She would be silly to call with less than c, so we know that a > c .

3) At the threshold, a player should be indifferent between calling or folding. Expected value must be equal.
Consider C: at c,

• he gets 0 if folding;
• if calling and A has less than a (i.e. folds), he battles it out with B: win 3 or lose 2. Expected gain is 3c - 2(1-c) = (5c - 2);
• if calling and A has more than a (i.e. calls), he loses 2.

a (5c - 2) - 2(1-a) = 0 yields a c = 2/5.

4) Now consider A. At a, knowing that C has called,

• she loses 1 if folding;
• if calling and C has less than a (i.e. between c and a ), she battles it out with B: win 4 or lose 2. Expected gain is 4a - 2(1-a) = (6a - 2);
• if calling and C has more than a (i.e. between a and 1 ), she loses 2.

(a - c ) (6a - 2) + (1-a)(- 2) = (1 - c )(-1) .
Using a c = 2/5, one can rewrite as a^3 - (17/30)a + 2/30 = 0.

Solving gives the three solutions :

Quote: MangoJ

#1: a1 = .120754, a2 = .25, c1 = 3.31251
#2: a1 = -.80585, a2 = .25, c1 = -.49637
#3: a1 = .685096, a2 = .25, c1 = .58386

But a c = 2/5 means a > 2/5, so the third one is the solution.

Quote: MangoJ

BTW: is it luck that there is only a single compatible solution ? what if there is none, or more than one ?

Nash proved that there is always a solution, albeit sometimes mixed. But there could be several. It is the specifics of this game that lead to a single solution.

Yes, I agree with all of your conclusions. But these could be drawn before any actual calculation is made. But what if your calculation of your point (3) is more complex, that you cannot solve for a or c ?

Quote: kubikulann

or to use a mixed strategy (= a probabilistic choice of action instead of the deterministic one: "if I have x, then I always act Z.")

Well, that's rather a consequence of the continuous x. For a probabilistic set of choices, all choices would need to have same EV (otherwise you would always pick the one with the higher EV). If for some interval you would have two probabilistic choices A and B, then for any x1 and x2 in this interval EV(A|x1) = EV(B|x1) and EV(A|x2) = EV(B|x2). Now put x1 to some value x in this interval, and x2=x+dx (for an infinitisimal amount higher or lower), then you find EV(A|x) - EV(B|x) = EV(A|x+dx) - EV(B|x+dx), consequently EV(A) = EV(B) over the *whole* interval. That would be a rather "boring" game.

Likewise, if you have a probabilistic choice between A and B only for a single x (i.e. no interval), you could simply make decision A for x, and B for x+dx, i.e. shift your decision threshold by an infinitisimal amount. Infinitisimal different thresholds only change EV infinitisimal. Or in other words: the strategy decision for a single x cannot yield a difference in EV for the whole strategy, as this single x will be drawn with probability 0.

Hence one can solve these problem without the concept of probabilistic choices. Which in essence makes the 0...1 game such a nice model :)

"Boring"? Well, personally, I view the simple threshold rule as more boring than teasing your adversary with your random choice.

Consider the first 0-1 game proposed by the Wizard.
Several answers were proposed, according to what player Y had to do. In some, he just had a threshold of 0.4 whenever X bet. In others, he developed a probabilistic scheme ( aka a mixed strategy). It COULD be a constant proba on a range, or it could be a very fanciful profile.
All in all, the condition was that the global probability of calling be 0.6, and that can be achieved in a number of different ways, mixed or deterministic.

So, even with continuous x, mixed strategies arise.


If the calculations offer no solution in a and c, then you must resort to mixed strategies. Nash (1951) says there is always a solution.

Quote: MangoJ

Whats the best strategy for player A and C, and whats the EV for each position ?

Expected values are:
EVA = - .145940
EVB = - .416926
EVC =+ .562866

(unless I made an error again...)

I get that the expected values should be:
EVA = -2a^3 + 3a^2*c - ac + a - 0.875c = -0.04677
EVB = a^3 - 3a^2*c + 2.5ac^2 + ac - 0.5a - 1.125c = -0.51609
EVC = a^3 - 2.5ac^2 - 0.5a + 2c = +0.562866

These also agree with some simulations I ran to check.