One-card poker

Originally posted by iamatiger
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%

If P2 sees raise he folds with jack (loses 1), calls 50/50 with queen (approx breaks even) and calls with king (wins 2). If he sees a pass then he calls and wins 2 with 100% certainty. Sounds like a clear win for P2.

Originally posted by Palynka
If P2 sees raise he folds with jack (loses 1), calls 50/50 with queen (approx breaks even) and calls with king (wins 2). If he sees a pass then he calls and wins 2 with 100% certainty. Sounds like a clear win for P2.

hmm, you are right, this is indeed a tricky problem! I was pretty sure that P2 would have to pass with the queen because P1 was more likely to have a king than a jack, but this is not the case.

Originally posted by PBE6
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 situatio ...[text shortened]... e other player is trying to confound you is precisely what makes the problem more difficult.

The key is what do you do if you have a queen. The other person might have a king or might have a jack. You won't know.

So, my solution is to always raise and always call. If you have a jack and you are raised, you should fold. You should come out even.

Originally posted by uzless
You should come out even.

If the other player is playing optimally, you won't.

aren't we assuming that after the first hand, player 1 plays 2nd and player 2 plays 1st, thus reversing the strategy?

Each player will play the same strategy on every other hand so they will end up even.

Originally posted by uzless
aren't we assuming that after the first hand, player 1 plays 2nd and player 2 plays 1st, thus reversing the strategy?

Each player will play the same strategy on every other hand so they will end up even.

I don't think we are assuming that, but you would be right in that case.

However, there would still be the question of which are the 2 optimal strategies (when you go first and when you go second).

I think there are five goals P1 can consider in his strategy
1) When he has a king, can he maximise his chance of P2 seeing a raise (benefit +$1 each time he manages this)
2) When he has a jack can he:
2a) Maximise the chance of P2 folding (benefit +$2 each time he manages this)
2b) Minimise the chance of P2 seeing a raise (loss +$1 each time)
3) When he has a queen can he:
3a) Maximise the times he holds when P2 has a jack and raises(benefit +$3)
3b) Minimise the times he folds when p2 has a king and raises (loss +$1)

The optimal strategy will maximise the sum of the (benefit of outcome * chance of outcome)

you need to mix up your betting... if you have the same tactics throughout you will become very predictable.

plus when someone bets $1 are we allowed to re-raise $1 as in normal poker?

whatever way you look at it though i can't find a reason for checking with a king... if your opponent has a queen xe will assume that you've just checked with a jack and therefore won't get paid with a bet so will also check. the only way you could get away with checking with a king is if your opponent has a jack and tries to steal the pot... but after seeing you do that once xe is unlikely to fall for it again.

Further thoughts:
If P2 has a queen, and hears you raise the possible outcomes for him are:
See, you hold king, loss= 2
See, you hold jack, gain = 2
Fold, you hold king, loss = 1
Fold, you hold jack, loss = 1

So we can see that seeing when you hold a jack benefits him $3 over folding, and folding when you hold a king benefits him only $1 over seeing.

Therefore, if you raise, and he has a queen, he should see you, unless you are 3 or more times likelier to be holding a king than a jack.

To see this in action, assume you are 60/40 on holding a king/jack for your raise, given the above he will see you and in 100 deals he will make:
60*$2 loss - 40*$2 win = net loss $40
(if he had always folded he would have lost $100)

alternatively, assume you are 80/20 on holding a king/jack for your raise
now he should fold, losing $100 over 100 deals, rather than seeing which loses him $120 overall.

the halfway point is 75/25, where he loses $100 overall whatever he does. And your maximum profit when he holds a queen and you raise is $100 in 100 deals, which you get by being 75/25 or better.

What about when he holds a queen and you don't raise?
If P2 has a queen, and you don't raise, he should pass also, because if he raises it gives you the chance to hold with a king, or fold with a jack.

So, we can fill in a few items in our ideal strategy description:

P1 strategy:
Hold King, raise with probability A
Hold Queen, Pass.
Hold Jack, raise with probability B
Hold queen, hear raise, call with probability C
Hold king, hear raise, call
Hold jack, hear raise, fold.

P2 strategy:
Hear raise, hold king, see
Hear raise, hold queen, fold if A/B >= 3, hold otherwise
Hear raise, hold jack, fold
Hear pass, hold king, raise with probability D
Hear pass, hold queen, pass.
Hear pass, hold jack, raise with probability E

Originally posted by trev33
you need to mix up your betting... if you have the same tactics throughout you will become very predictable.

plus when someone bets $1 are we allowed to re-raise $1 as in normal poker?

whatever way you look at it though i can't find a reason for checking with a king... if your opponent has a queen xe will assume that you've just checked with a jack and es to steal the pot... but after seeing you do that once xe is unlikely to fall for it again.

The allowed bidding in the problem does not include re-raises, it gets quite complicated enough without them!

If you always raise with a king, then the opp knows that any pass by you is a queen or jack. Therefore holding the jack he will be more able to put you on the spot by raising, in the knowledge that you don't have a totally safe take. This lets him win some of his jacks against your queens.

When you have a queen as P1 and you pass and hear a raise:
You gain $2 if you see the raise and P2 has a jack
You lose $2 if you see the raise and P2 has a king.
You lose $1 if you fold and P2 has a jack
You lose $1 if you fold and P2 has a king.

Therefore your "profit" when P2 has the king is reduced by $1 if you see.
Your "profit" when P2 has the jack is increased by $3 if you see.

So you should only fold if P2 is >=3 times more likely to have a king than a jack

This brings us down to only 4 probabilities to determine.


P1 strategy:
Hold King, raise with probability A
Hold Queen, Pass.
Hold Jack, raise with probability B
Hold queen, hear raise, fold if C/D >= 3, call otherwise.
Hold king, hear raise, call
Hold jack, hear raise, fold.

P2 strategy:
Hear raise, hold king, see
Hear raise, hold queen, fold if A/B >= 3, hold otherwise
Hear raise, hold jack, fold
Hear pass, hold king, raise with probability C
Hear pass, hold queen, pass.
Hear pass, hold jack, raise with probability D

Assuming A and B have been chosen in a way that makes P2 fold with the queen (A/B >= 3)
The expected profit when P1 raises can be worked out as follows:
a/(a+b) of the raises are when P1 has the king
b/(a+b) of the raises are when P1 has the jack
in a/(2(a+b)) of the raises P1 has the king, P2 has the queen and folds, +1 to P1
in a/(2(a+b)) of the raises P1 has the king, P2 has the jack and folds, +1 to P1
in b/(2(a+b)) of the raises P1 has the jack, P2 has the king and sees, -2 to P1
in b/(2(a+b)) of the raises P1 has the jacj, P2 has the queen and folds,+1 to P1
So the expected profit to P1 when he raises is a/(a+b) - b/(a+b) = (a-b)/(a+b)

The expected profit when a2 does not raise can be worked out as follows:
(1-a)/(3-a-b) of the passes are with the king
1/(3-a-b) of the passes are with the queen
(1-b)/(3-a-b) of the passes are with the jack

in 1/2 of the king passes, p2 has the queen and folds, +1 to P1
in d/2 of the king passes, p2 has the jack and raises, +2 to P1
all of the jack passes are -1 to P1
assuming that D and C have been worked out to make P1 fold with the queen (c/d >= 3):
in 1/2 of the queen passes, P2 has the king, -1 to P1
in d/2 of the queen passes , P2 has the jack and raises, -1 to P1
in (1-d)/2 of the queen passes, P2 has the jack and folds, +1 to P1
so the expected profit when P1 does not raise is:
(1-a)/(3-a-b)*(d + 0.5) - d/(3-a-b) - (1-b)/(3-a-b)
= ((1-a)(d + 0.5) - d + b - 1)/(3-a-b)
=( -0.5 -ad -a/2 +b)/(3-a-b)

So, interestingly profit is independent of c except for the inequality (c/d >= 3).
the larger d gets, the smaller P1s profit, so P2 will maximise D within this inequality ie:
c = 1
d = 1/3

therefore the expected profit when P1 does not raise is equal to:
( -0.5 -a/3 -a/2 +b)/(3-a-b)

To get the final expected profit we need the probability of P1 raising
this is
(a/3 + b/3) = (a + b)/3

Bringing everything together, P1's expected profit in this game is:
(a-b)/3 + 2*(a + b)*( -0.5 -a/3 -a/2 +b)/(3*(3-a-b))

Which, graphed in excel seems to have a maximum around:
a = 0.4 b = 4E-07 P1 profit = +0.047863057 per game

Assuming A and B have been chosen in a way that makes P2 fold with the queen (A/B >= 3)
The expected profit when P1 raises can be worked out as follows:
a/(a+b) of the raises are when P1 has the king
b/(a+b) of the raises are when P1 has the jack
in a/(2(a+b)) of the raises P1 has the king, P2 has the queen and folds, +1 to P1
in a/(2(a+b)) of the raises P1 has the king, P2 has the jack and folds, +1 to P1
in b/(2(a+b)) of the raises P1 has the jack, P2 has the king and sees, -2 to P1
in b/(2(a+b)) of the raises P1 has the jacj, P2 has the queen and folds,+1 to P1
So the expected profit to P1 when he raises is a/(a+b) - b/(a+b) = (a-b)/(a+b)

The expected profit when a2 does not raise can be worked out as follows:
(1-a)/(3-a-b) of the passes are with the king
1/(3-a-b) of the passes are with the queen
(1-b)/(3-a-b) of the passes are with the jack

in 1/2 of the king passes, p2 has the queen and folds, +1 to P1
in d/2 of the king passes, p2 has the jack and raises, +2 to P1
all of the jack passes are -1 to P1
assuming that D and C have been worked out to make P1 fold with the queen (c/d >= 3):
in 1/2 of the queen passes, P2 has the king, -1 to P1
in d/2 of the queen passes , P2 has the jack and raises, -1 to P1
in (1-d)/2 of the queen passes, P2 has the jack and folds, +1 to P1
so the expected profit when P1 does not raise is:
(1-a)/(3-a-b)*(d + 0.5) - d/(3-a-b) - (1-b)/(3-a-b)
= ((1-a)(d + 0.5) - d + b - 1)/(3-a-b)
=( -0.5 -ad -a/2 +b)/(3-a-b)

So, interestingly profit is independent of c except for the inequality (c/d >= 3).
the larger d gets, the smaller P1s profit, so P2 will maximise D within this inequality ie:
c = 1
d = 1/3

therefore the expected profit when P1 does not raise is equal to:
( -0.5 -a/3 -a/2 +b)/(3-a-b)

To get the final expected profit we need the probability of P1 raising
this is
(a/3 + b/3) = (a + b)/3

Bringing everything together, P1's expected profit in this game is:
(a-b)/3 + 2*(a + b)*( -0.5 -a/3 -a/2 +b)/(3*(3-a-b))

Which, graphed in excel seems to have a maximum around:
a = 0.4 b = 4E-07 P1 profit = +0.047863057 per game

Great work! A few small algebra issues, though.

When p1 raises you have:
n a/(2(a+b)) of the raises P1 has the king, P2 has the queen and folds, +1 to P1
in a/(2(a+b)) of the raises P1 has the king, P2 has the jack and folds, +1 to P1
in b/(2(a+b)) of the raises P1 has the jack, P2 has the king and sees, -2 to P1
in b/(2(a+b)) of the raises P1 has the jacj, P2 has the queen and folds,+1 to P1
So the expected profit to P1 when he raises is a/(a+b) - b/(a+b) = (a-b)/(a+b)

But this should be a/(a+b) - 2*b/(2(a+b))+b/(2(a+b)) = a/(a+b)-b/2(a+b)

And also the prob for the last term is incorrect. You pass with prob 1-(a+b)/3 = (3-a-b)/3. However, when I do both these corrections I get a value function of a/18+b/6-1-6, which has a degenerate maximum at a=1,b=1... Which can't be right. Maybe I'm doing a mistake myself. I can't see much fault in anything else, though...

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 jack and a raise by P2, fold.

P2:
*after raise by P1:
. see with king
. see with prob 50% with queen
. pass with jack
* after pass by P1
. raise with king
. pass with queen
. raise with prob 1/3rd with jack

These are only stable, optimal strategies if each player has full knowledge of the other's strategy, if such knowledge is unavailable players then will attempt to increase their profit by fiddling with the probability related actions. For instance:

If P2 tries raising with the jack more often than 1/3 then P1 should switch to always seeing with the queen (and P1's profit will increase). Until he realizes that he should do this P2 will increase his profit.

If P2 tries seeing seeing raises less with the queen, P1 should increase the % of times he raises with a jack, again, P2's profit will increase until P1 realises he must do this.

So, what is the profit for each player with these strategies?:
KQ - P1 raises , P1 gains $1.5 on average
KJ - P1 raises, P2 folds and P1 gains $1
QK - P1 passes, P2 raises and P1 loses -$1.5 on average
QJ - P1 passes:
* 2/3rds of the time P2 passes and P1 gains $1
* 1/3rd of the time P2 raises
* . On half of those raises, P1 sees the raise and gains $2
* . On half of those raises, P1 folds and loses -$1
this gives P1 an average gain of 2/3 + 1/3 -1/6 = +$(5/6)
JK - 2/3 of the time P1 passes, P2 raises and P1 folds, losing -$1
* . 1/3 of the time P1 raises, P2 sees and P1 loses $2
this gives P1 an average loss of -$(4/3)
JQ - 2/3 of the time P1 passes, whether P2 raises or not P1 loses -$1
* . 1/3 of the time P1 raises:
* .. on half of those raises, P2 sees the raise and P1 loses -$2
* .. on half of those raises, P2 folds and P1 gains $1
this gives P1 an average loss of -2/3 -1/3 + 1/6 = -$(5/6)

So most of the deals cancel out:
KQ = -QK
QJ = -JQ

However asymmetry remains in the KJ/JK deals where P1 gains $1 when he has the kings but P2 gains $(4/3) when he has the king. We can explain P2's advantage because he takes action after knowing P1's bid, which lets him more optimally fold with a jack than P1 can achieve.

So the expected loss by P1 is (1/6)*(-1/3) = minus one eighteenth of a dollar per deal.

Absolutely spectacular work! 😀

For reference, here's the problem site I took the problem from (No. 203):
http://mathproblems.info/