# One-card poker

PBE6
Posers and Puzzles 10 Mar '10 18:51
1. PBE6
Bananarama
10 Mar '10 18:51
Two players are playing a game of one-card poker, which uses a three-card deck (Jack, Queen, King). Each player antes \$1 for a total starting pot of \$2, and then each player is dealt one card. The betting works as follows:

1. Player 1 goes first, and may either bet \$1 or check.
2. If Player 1 bets, then Player 2 may either call the bet or fold. If Player 1 checks, then Player 2 may either bet \$1 or check.
3. If Player 1 checks and Player 2 bets, then Player 1 may either call the bet or fold.

After all bets have been made, the winner collects the pot.

Questions

1. What is the optimal betting strategy for Player 1?
2. What is the optimal betting strategy for Player 2?
3. What is the expected outcome for each player using their respective optimal betting strategies?
2. forkedknight
Defend the Universe
10 Mar '10 19:421 edit
Originally posted by PBE6
Two players are playing a game of one-card poker, which uses a three-card deck (Jack, Queen, King). Each player antes \$1 for a total starting pot of \$2, and then each player is dealt one card. The betting works as follows:

1. Player 1 goes first, and may either bet \$1 or check.
2. If Player 1 bets, then Player 2 may either call the bet or fold. If Player ...[text shortened]... What is the expected outcome for each player using their respective optimal betting strategies?
So if I'm understanding correctly, the final value of the pot is always either \$2 or \$4

As far as optimal strategy goes, I think the only thing that can be said for sure is that you should never call if you have a Jack, and never fold if you have a King; the rest depends on the other player.

*edit* Oh, and for player two, never check if you have a King.
3. PBE6
Bananarama
10 Mar '10 19:561 edit
Originally posted by forkedknight
So if I'm understanding correctly, the final value of the pot is always either \$2 or \$4

As far as optimal strategy goes, I think the only thing that can be said for sure is that you should never call if you have a Jack, and never fold if you have a King; the rest depends on the other player.

*edit* Oh, and for player two, never check if you have a King.
The final value of the pot will be \$2 if both players check, \$3 if Player 1 bets and Player 2 folds (or Player 1 checks, Player 2 bets and then Player 1 folds), or \$4 if both players bet.

I think the two rules you just stated ("never call with a Jack" and "never fold with a King" ) make good sense. Of course, the question is what to do in the other situations? ðŸ˜‰ The fact that the other player is trying to confound you is precisely what makes the problem more difficult.
4. 10 Mar '10 22:24
We can describe player 1's strategy with the bidding table:
Hold King, raise with probability A
Hold Queen, raise with probability B
Hold Jack, raise with probability C
To fully describe player 1's strategy we need to specify that he calls with probability D if he holds the queen and has previously passed. It is clear what he must do if he passes, is raised, and holds the king (see) or jack (fold).

Similarly player 2 strategy is:
Hear raise, hold king, see
Hear raise, hold queen, see with probability E
Hear raise, hold jack, fold
Hear pass, hold king, raise with probability F
Hear pass, hold queen, raise with pobability G
Hear pass, hold queen, raise with probability H

So the first question to consider, is, is there any combination of A, B, C and D which wins or breaks even against all combinations of E, F G and H?
5. wolfgang59
Mr. Wolf
10 Mar '10 23:01
Originally posted by iamatiger
We can describe player 1's strategy with the bidding table:
Hold King, raise with probability A
Hold Queen, raise with probability B
Hold Jack, raise with probability C
To fully describe player 1's strategy we need to specify that he calls with probability D if he holds the queen and has previously passed. It is clear what he must do if he passes, is r ...[text shortened]... mbination of A, B, C and D which wins or breaks even against all combinations of E, F G and H?
if player A has a Jack he should not always fold after a raise. (The other player may have a J and tyring to steal pot)

The problem is quite complicated!
6. randolph
the walrus
11 Mar '10 05:38
I think that there is only one Jack in the deck, though?
7. 11 Mar '10 10:59
Can I just check? Is this three cards in the deck, or three types of card? In other words, are there multiple Jacks, Queens and Kings?
8. Palynka
Upward Spiral
11 Mar '10 11:102 edits
Originally posted by iamatiger
We can describe player 1's strategy with the bidding table:
Hold King, raise with probability A
Hold Queen, raise with probability B
Hold Jack, raise with probability C
To fully describe player 1's strategy we need to specify that he calls with probability D if he holds the queen and has previously passed. It is clear what he must do if he passes, is r ...[text shortened]... mbination of A, B, C and D which wins or breaks even against all combinations of E, F G and H?
Yep, this will have to come downed to mixed strategies. We need to write a value function for both given the others strategy and then maximize both and try to find a fixed point.

Let S1 and S2 be the strategies (A,B,C,D) and (E,F,G,H) respectively:

E(S1) = 1/3*E(S1|K) + 1.3*E(S1|K)+ 1/3*E(S1|K) (note |S2 was omitted for notational brevity)

E(S1|K) = A*{1/2*[E*2+(1-E)*1] + 1/2*1*1}+(1-A)*{1/2*[G*2+(1-G)*1]+1/2*[H*2+(1-H)*1]}

and so on... if you have the patience to continue! ðŸ˜€

(I'm assuming only 3 cards total in the deck)
9. forkedknight
Defend the Universe
11 Mar '10 17:141 edit
I ran a simulation where each player uses a fixed strategy (i.e. always makes the same play in the same situation).

I found that player 1 has an advantage with a worst case EV of \$0.33/hand for each combination of strategies. The best case EV is \$0.67/hand if player 1 always checks his King and player 2 always bets his Jack.

A mixed strategy may be able to improve upon this some, but I would guess there is a true advantage to being player 1 since player 2 doesn't have the option of raising a bet, only calling.

I would also think that identifying the play style of the person you are playing would be more important that sticking to any sort of fixed or mixed strategy.

*edit* This post is entirely incorrect, there was a mistake in my simulation output. I'll fix it and get back
10. Palynka
Upward Spiral
11 Mar '10 17:451 edit
Originally posted by forkedknight
I ran a simulation where each player uses a fixed strategy (i.e. always makes the same play in the same situation).

I found that player 1 has an advantage with a worst case EV of \$0.33/hand for each combination of strategies. The best case EV is \$0.67/hand if player 1 always checks his King and player 2 always bets his Jack.

A mixed strategy ma n you are playing would be more important that sticking to any sort of fixed or mixed strategy.
Player 2 has the option of raising after 1 checks. In fact, intuitively I feel player 2 has an advantage here because he can gain some information from 1's initial move.
11. forkedknight
Defend the Universe
11 Mar '10 18:06
Originally posted by Palynka
Player 2 has the option of raising after 1 checks. In fact, intuitively I feel player 2 has an advantage here because he can gain some information from 1's initial move.
You are correct; if p1 chooses a fixed strategy, p2 can always counter with a fixed strategy that will result in an advantage for p2.

In this case, p1's best strategy is check/folding a J, check/calling a Q, and betting a K.
p2 will counter by fold/checking a J, fold/betting or fold/checking a Q, and betting a K.
The overall result of this scenario is an EV of -\$0.33/hand for p1
Against a random p2 strategy, this strategy for p1 results in an EV of \$0.166

However, if p2 chooses a random fixed strategy each hand, p1 can achieve an advantage by check/folding a J, check/calling a Q, and check/calling a K.

The result is an EV of \$0.25/hand in favor of player 1.
12. forkedknight
Defend the Universe
11 Mar '10 18:14
Originally posted by forkedknight
You are correct; if p1 chooses a fixed strategy, p2 can always counter with a fixed strategy that will result in an advantage for p2.

In this case, p1's best strategy is check/folding a J, check/calling a Q, and betting a K.
p2 will counter by fold/checking a J, fold/betting or fold/checking a Q, and betting a K.
The overall result of this scenario ...[text shortened]... /calling a Q, and check/calling a K.

The result is an EV of \$0.25/hand in favor of player 1.
In addition, the ideal scenario for p1 is:
p1 strategy: check/fold J, check/call Q, check/call K
p2 strategy: fold/bet J, call/bet Q, bet K
This results in an EV of \$1.33/hand for p1

The ideal scenario for p2 is:
p1 strategy: bet Jack, check/fold Q, bet K. This is the worst strategy against a random opponent as well.
p2 strategy: bet Jack, call/check Q, bet K
This results in an EV of -\$1/hand for p1
13. 11 Mar '10 21:212 edits
Originally posted by forkedknight
In addition, the ideal scenario for p1 is:
p1 strategy: check/fold J, check/call Q, check/call K
p2 strategy: fold/bet J, call/bet Q, bet K
This results in an EV of \$1.33/hand for p1

The ideal scenario for p2 is:
p1 strategy: bet Jack, check/fold Q, bet K. This is the worst strategy against a random opponent as well.
p2 strategy: bet Jack, call/check Q, bet K
This results in an EV of -\$1/hand for p1
If, by fixed strategy we mean "in situation X I do Y". I'm pretty sure such a strategy can always be countered by the other player. however I would think P1 can at least break even by a strategy such as "in situation x I do y with probability A and do Z with probability 1-A"

Such a strategy is also "fixed" in that it does not change over time, however shall we call it a "probabalistic" strategy?
14. 11 Mar '10 22:49
My calculations seem to show that P1 can do not better than break even, which he achieves for example by:

Raising when he has a king
Raising when he has a queen
Raising 99.99% of the time he has a jack.

I don't think P2 can beat this strategy to get better than 50%
15. Palynka
Upward Spiral
11 Mar '10 23:20
Originally posted by iamatiger
Such a strategy is also "fixed" in that it does not change over time, however shall we call it a "probabalistic" strategy?
The technical terms in game theory are pure and mixed strategies.