3-Card Deck Poker

Suppose you have a poker-based game with the following rules:

1. The deck has three cards -- J,Q, and K.
2. Each of two player antes $1.
3. Each will get one card, face down.
4. Player 1 may fold or raise $1.
5. If player 1 raises, player 2 may call or fold.

What should be the strategy of player 1? Assume both players are perfect logicians. To be more specific, what should player 1 do if he gets a jack? If he raises, and player 2 has the queen, then player 2 might fold, thinking player 1 might have the king. Randomized strategies are allowed. Also, what is expected value of player 1?

player 1 should raise all hands, player 2 should call with Q or K.

Can't player 2 re-raise? Can't player 1 check? Can player 2 raise if player 1 just checks?


"Randomized strategies" suggests bluffing. Therefore the reply that player 1 should ALWAYS raise, is not right, because it makes him predictable. Or meaningless. I.E. With player 2 then calling with every king and queen, the game will always be a 50/50 proposition.



I have no data to back this up, but here's my strategy for player 1 and various rules.


If bluffing, checking and re-raising are not allowed:

Card	Action
Jack	Fold
Queen	Raise
King	Raise

This just feels more right than raising with anything.

maybe I'll play with the math...


Other rules:

Player 1 can only fold or raise. With the card shown, he should:

Card	Fold	Raise
Jack	66%	33%
Queen	33%	66%
King	0%	100%

If you always fold the jack, player 2 will only call with a king. By occasionally showing the jack bluff, you'll get player 2 to call with a queen more often. Why not raise 100% with the queen? I don't know.


Player 1 can fold, check or raise, but player 2 cannot re-raise:

Card	Fold	Check	Raise
Jack	25%	50%	25%
Queen	25%	25%	50%
King	0%	25%	75%

With more options, you need to mix it up more. Yeah, occasionally check the king.


Both players can fold, check, raise and re-raise, player 1 should:

Card	Fold	Check	Raise
Jack	50%	50%	0%
Queen	25%	25%	50%
King	0%	0%	100%

When your opponent has as many options as you do, you have to be both more conservative, and more aggresive.




FYI:

Although I didn't respond, I could have sworn I saw this question recently, but the three cards were Ace, Deuce, Trey. I believe it was in the Monty Hall thread. Then again, I could have seen it somewhere on the internet while the Monty Hall question was active...

I think I agree with rudeboyoi's strategy. Unfortunately, I just don't see it as being optimal for Player 1, as he is in a bad position.

I can't remember which poker write wrote that 3 different cards were needed to induce the concept of bluffing into poker, which is a very key element in the game. Unfortunately, the way this hypothetical game is setup doesn't allow for much bluffing, seeing as how Player 1 only has 2 options.

If regular poker rules were used, I would advise Player 1 to check to Player 2 when he's holding a Jack. Then, a raise after Player 2's bet (check-raise) could scare Player 2 if Player 2 is holding a Queen. Conversely, an immediate re-raise by Player 2 could imply Player 2 has the King, or that Player 2 has the Queen and is trying to call Player 1's bluff. Of course, Player 1's re-re-raise, could REALLY scare Player 2 if he's holding the Queen.

With the rules of this hypothetical game, there is no good solution for Player 1. rudeboyoi's strategy provides Player 1 with a 0-sum game, while Player 2 has the advantage with a slightly positive outcome. Thus, nobody would be advised to play the game, unless they could secure Player 2 position.

Quote: rudeboyoi

player 1 should raise all hands, player 2 should call with Q or K.

Why should player one raise with all hands? Yeah, I get the obvious stuff (P1 always raise Q or K), but why should player 1 always raise with a J?

If logic points to player 1 always raising with a J, then Player 2 should always call with a Q. But this then implies that Player 1 should never call with a J (since E[P1 raise with J | P2 always call Q] = -2 < E[fold] = -1).

No, I can't see P1=J would always raise. I think it has to be P1 raise x% of the time, (randomly determined).

Just to elaborate on my post some more, I've convinced myself that P1=J should neither always raise or always fold.

Since both players are perfect logicians:

1. P1=J always raises is the most logical decision. This implies that P2=Q always call. But this implies P1=J should always fold. Therefore, our hypothesis P1=J always raises is false.

2. P1=J always fold. This implies that P2=Q always folds. But this implies P1=J should always raise. Therefore, our hypothesis P1=J always folds is false.

If you don't understand the above, work out the EV's. You'll see the contradictions.


The perfect strategy is for P1=J to raise some amount x % of the time. But what is x?

i guess the key is to figure out what ratio of Ks and Qs to Jacks in your raising range is going to make it unprofitable for your opponent to call or fold with his Q regardless of what decision he makes.

Unfortunately, I don't see any reason why Player 2 should every fold the Queen.

If you are P2 holding the Queen, and you know P1 folds Jack 100% of the time, then you are guarantted to win 100% of the time. On the other hand, if P1 folds Jack 0% of the time, then you have a 50% shot of beating P1. And, if P1 folds some percentage of the time, then you have somewhere between 50% and 100% of a chance to beat P1. Since your odds are not going to get worse than 50%, you might as well go ahead and call, since 50% of the time you will win 2 units, and 50% of the time you will lose 2 units, and thus it's a 0-sum game for you.

Knowing that P2 should call all the time with Q or K, and fold with J, makes calculating the x% of time that P1 should raise with a J easy: P1 should raise with J an infinitely small percentage of the time, but not 0% of the time.

Creating a small spreadsheet, it's easy to see that the expected value of the game for P1 if he raises 100% of the time is -0.333. Conversely, if P1 raises 0% of the time, the expected value is -0.167. (This is lower because knowing that P1 will never raise with a J makes P2's decision with a Q that much easier.)

If P1 raises 50% of the time, his expected value becomes -0.167. Raising 25% of the time, his expected value becomes -0.083. Raising 10% of the time, -0.033. And raising 1% of the time -0.003.

Thus, P1 needs to raise an infinite small amount of time, without raising 0% of the time. This is to get P2 to call with the Q. Thus, I guess, P1 wants to fold the J as much as possible, but still give P2 the idea that he MIGHT raise with the J.

Wow, a lot to catch up on here.

Quote: rudeboyoi

player 1 should raise all hands, player 2 should call with Q or K.

I think there is a better strategy than that for player 1. With that strategy, if player 1 has a J, and player 2 has a Q, player 1 will lose 2 units. What if player 1 folds 1% of the time with a J, and player 2's strategy stays the same as you indicate. Won't player 1 save 1 unit 1% of the time in that situation, increasing his EV (expected value)? Or do you think player 2 would change his strategy to counter player 1's 99%/1% strategy?

Quote: DJTeddyBear

Can't player 2 re-raise? Can't player 1 check? Can player 2 raise if player 1 just checks?

No, no, and N/A. I may add those options later.

Quote: DJTeddyBear

Player 1 can only fold or raise. With the card shown, he should:

Card	Fold	Raise
Jack	66%	33%
Queen	33%	66%
King	0%	100%

Why would the player 1 sometimes fold with a Q?

Quote: konceptum

If P1 raises 50% of the time, his expected value becomes -0.167. Raising 25% of the time, his expected value becomes -0.083. Raising 10% of the time, -0.033. And raising 1% of the time -0.003.

I agree with the 50% EV figure, but not the 25% one. What do you show player 2 will do with a Q under the 25% case?

Quote: Wizard

Quote: DJTeddyBear

Player 1 can only fold or raise. With the card shown, he should:

Card	Fold	Raise
Jack	66%	33%
Queen	33%	66%
King	0%	100%

Why would the player 1 sometimes fold with a Q?

Logic says that if you hold the Q, you will only get called when facing a K. Who wants that? While you won't ever get called when facing a J, you need to get your K called as much as possible.

So you sometimes fold the Q, for the same reason that you would sometimes raise with a J.

To introduce unpredictability.


If you fold 1/3 of your hands (or less), player 2 will know that all, or at least most, of your folds are a J.

If you always fold J, and only fold 33% of your cards, player 2 will know you never play your J and always play your Q. Therefore if you raised, you're holding a K or a Q. If he has a Q, he knows he's against a K and will always fold.

By sometimes raising with a J, and getting called, you're forced to show those bluffs, which proves that if you fold 1 out of 3 hands, it's not necessarily a J being folded. That puts player 2 into an uncertain situation when facing a raise and holding a Q.


You won't get the same unpredictable 'read' if you always raise with Q and K, and sometimes raise with J.

I suppose I could have make those numbers look like:

Card	Fold	Raise
Jack	75%	25%
Queen	25%	75%
King	0%	100%

But I feel the 1:2 fold/raise ratio is the best in this 3 card scenario.

Quote: DJTeddyBear

So you sometimes fold the Q, for the same reason that you would sometimes raise with a J.

To introduce unpredictability.

If player 1 has a Q, player 2 either has a J or K. Either way, player 2's play is obvious. Player 2 will always fold with a J, and call with a K. So, if player 1 has a Q there is a 50/50 chance player 2 has a J or K. His expected value by raising is 0.5*1 + 0.5*-2 = -0.5. That is greater than the expected return of -1 by folding. Thus, I maintain, player 1 should always raise with a Q.

You can't predict the expected value of Player 1 because you don't know what Player 2 will do!

Say that Player 1 plays "perfect strategy" which is to always fold on J's and always raise on Qs and Kings.

The Expected Value with the J is -1.
The Expected Value with the Q is -0.5 (Player 2 will always call or fold based on his card, and you lose 2 units when he has the K but win 1 when he has the J).
The Expected Value with the K is 1. Since Player 2 will quickly learn your strategy and has the Q, he knows that you are raising with a K (since he has the Queen) and therefore he will never call a raise and he is holding a Queen. Therefore, the EV for Player one is -.5/3 = -1/6.

Can Player 1 improve his strategy?

Let's say that you play an "always raise" strategy which is to raise everything. Player two will not know what to do with his Q should he have one.

The Expected Value with the J is 2 units x (-1/2 [Player 2 has K]-1/2 x (% Player 2 Raises).
The Expected Value with the Q is -0.5 as above.
The Expected Value with a K is 2 units x (1/2 [Player 2 has J] + 1/2 x (% Player 2 Raises).

When you put it all together, the EV is -0.5/3 = -1/6 as the Percent that Player 2 raises with a queen doesn't matter as he will win 1/2 the time and lose 1/2 the time. Player 2's optimal strategy when holding the Queen doesn't matter.

So the EV with a no bluffing strategy is -1/6 and the EV with a raise everything strategy is -1/6, the same.

So the hint though is to raise a certain % of hands by randomizing when you raise the Jack. You would always raise the Q and K (you can't improve the strategy with a Queen as Player 2 will either raise (he has the king) or fold (he has the Jack).

When Player 1 raises the Jack an unknown percentage of the time, Player two will fold on hands he shouldn't fold on on the 50% of the time that he is holding the Queen while you are holding the Jack. He will also call on Queens when playing the King. By calling a % of Jacks, you will get Player 2 to call some Queens until he can figure out how often you call with Jacks. He can then elect to fold on a percentage of queens based on Player's one strategy.

Quote: Wizard

If player 1 has a Q, player 2 either has a J or K. Either way, player 2's play is obvious. Player 2 will always fold with a J, and call with a K. So, if player 1 has a Q there is a 50/50 chance player 2 has a J or K. His expected value by raising is 0.5*1 + 0.5*-2 = -0.5. That is greater than the expected return of -1 by folding. Thus, I maintain, player 1 should always raise with a Q.

This is exactly right. The only 2 plays that are not obvious in this problem is what should P1=J do and P2=Q do. And in fact, these two plays are equivalent, if you can figure out one you will know the other.

And we also know that neither always fold or always raise/call is the solution (you can see my earlier post for why). All that's left is to figure out what % of the time should P1=J raise.

Quote: boymimbo

You can't predict the expected value of Player 1 because you don't know what Player 2 will do!

Since both are logicians, you can predict that player 2 will act in his own best interests, according to what player 1's strategy is.

Quote: boymimbo

Let's say that you play an "always raise" strategy which is to raise everything. Player two will not know what to do with his Q should he have one.

I disagree. Player 2 will always call with a Q if player 1 raises everything. In that situation, there would be a 50/50 chance player 1 has a J or K. Player 2's EV by calling would be .5*2 + .5*-2 = 0. That is higher than the -1 by folding, so player 2 would always raise.

Ok, so this is what I've got so far. Does this make sense:

Assume P1=J will raise with probability p, and P2=Q will call with probability q.

E[P1=J] = p * E[P1=J raise] + (1-p) * E[P1=J fold]
E[P1=J] = p * E[P1=J raise] + (1-p) * (-1)
E[P1=J] = p * E[P1=J raise] + p - 1

Examining E[P1=J raise]:

E[P1=J raise] = 50 % * E[P1=J raise | P2=Q] + 50 % * E[P1=J raise | P2=K]
E[P1=J raise] = 50 % * E[P1=J raise | P2=Q] + 50 % * (-2)
E[P1=J raise] = 50 % * E[P1=J raise | P2=Q] - 1
E[P1=J raise] = 50 % * [q(-2) + (1-q)(+1)] - 1
E[P1=J raise] = -1.5q - 0.5

Therefore,
E[P1=J] = -1.5pq + 0.5p - 1

P1=J will chose p that maximizes the above with respect to q. But this does not fully answer the question yet because we need to quantify q. So q will base his decision on the following:

E[P2=Q] = P(P1J and raised) * E[P2=Q | P1=J] + P(P1K and raised) * E[P2=Q | P1=K]
E[P2=Q] = 50% * p * E[P2=Q | P1=J] + 50% * E[P2=Q | P1=K]
E[P2=Q] = 50% * p * [q*2 + (1-q) * (-1)] + 50% * [q * (-2) + (1-q) * (-1)]
E[P2=Q] = 50% * p * [q*2 + (1-q) * (-1)] + 50% * [q * (-2) + (1-q) * (-1)]
E[P2=Q] = 1.5qp - 0.5p - 0.5q - 0.5

And P2=Q will chose q that maximizes the above with respect to p.

I think the answer is p=q=1/3 (I got this by differentiating both equations and setting to 0...) So I would get Player 1 with a J should raise 1/3 of the time, and Player 2 should call with a Q 1/3 of the time. Does this make sense?

Quote: cardshark

I think the answer is p=q=1/3 (I got this by differentiating both equations and setting to 0...) So I would get Player 1 with a J should raise 1/3 of the time, and Player 2 should call with a Q 1/3 of the time. Does this make sense?

Assuming you are correct for p, I would suggest trying other values for q, and see how the overall EV of the game changes.

OOPS...double post

Quote: Wizard

Assuming you are correct for p, I would suggest trying other values for q, and see how the overall EV of the game changes.

Hmmm...ok, I set up a spread sheet, set p=1/3. When I set q=1/3 I get EV for Player 1 of -1/9. (And as a check, I get an EV for player 2 of +1/9).

Now, when I play around with q and fix p=1/3, the EV for player 1 doesn't change!! At first this wasn't intuitive, but now when I look at it, it makes perfect sense, because there is symmetry in the equations.

Similarly, when I play around with p and fix q at 1/3, EV for player 2 doesn't change.

So, that's it. That's the answer - optimal play for player 1 is to call with J 1/3 of the time (and always call with Q and K) which gives him an EV of -1/9. It doesn't matter how player 2 plays, his EV will always be -1/9.

Similarly, Player 2 should call with Q 1/3 of the time (and always call with K and fold with J). This gives him an EV of +1/9. No matter how Player 1 plays, Player 2 will always have an EV of +1/9 playing this way.

Very cool problem!

Quote: cardshark

Hmmm...ok, I set up a spread sheet, set p=1/3. When I set q=1/3 I get EV for Player 1 of -1/6.

Now, when I play around with q, the EV for player 1 doesn't change!! At first this wasn't intuitive, but now when I look at it, it makes perfect sense, because there is symmetry in the equations.

So, in fact, Player 1's EV is not influenced by Player 2's strategy when he chooses optimal strategy.

Yes, the key is to force player 2 to an indifference point. However, are you sure about that EV for the whole game?

Quote: Wizard

Yes, the key is to force player 2 to an indifference point. However, are you sure about that EV for the whole game?

Sorry Wizard, I typed incorrectly - its -1/9, I edited my post to fix this.

Thank you so much for sharing this problem!

Quote: cardshark

Sorry Wizard, I typed incorrectly - its -1/9, I edited my post to fix this.

Thank you so much for sharing this problem!

Ding, ding, ding, you got it!

I'm glad you liked the problem. Perhaps I'm pressing my luck, but what about a 5-card deck game (10-A)?

Quote: Wizard

Ding, ding, ding, you got it!

I'm glad you liked the problem. Perhaps I'm pressing my luck, but what about a 5-card deck game (10-A)?

Alright, I like a challenge. I will look at a 5-card game. Truthfully, I spent the last hour of work today on this problem (instead of finishing an important actuarial valuation!) Probably will do the same tomorrow with the 5-card deck version...

Now, it wouldn't by any chance be EV=+/- 1/25??? Just a shot in the dark.

I liked reading this problem. Even though I have no ability to figure out an answer, it's still a lot of fun to think about!

Quote: cardshark

Alright, I like a challenge. I will look at a 5-card game. Truthfully, I spent the last hour of work today on this problem (instead of finishing an important actuarial valuation!) Probably will do the same tomorrow with the 5-card deck version...

Now, it wouldn't by any chance be EV=+/- 1/25??? Just a shot in the dark.

You can justify the time by all the time you spend thinking about work when you're off the clock. I have the answer on my other computer, but I'm pretty sure it isn't -1/25.

Here's where I'm at so far:

Player 1 should always raise with Q, K, A.
Player 2 should never call with T, and always call with K, A.

After some mixed Nash calcs, I get that Player 2 should never call with J, and call with Q 2/3 of the time. You get this by considering the expectation of player 1 raising on T or J, and making him indifferent between raising and folding.

I haven't worked out the slightly more complicated calcs for what player 1 should do with a T or J yet... maybe soon