Two players each get dealt a real number between zero and one. They each bet the size x pot or check. Both bets are revealed at the same time. High number wins the 3x pot if they both bet. If only one bets, he wins. If neither bets they both have a second chance to bet the pot with the knowledge that the first round was two checks. (Also on the first round both players know that there might be a second round.) If they still both check they get a refund.

Who is paying the extra when they both bet and one scoops 3x the pot?

You didn't describe any antes or blinds so when one bets and the other checks, the pot that he wins is composed entirely of his own bet, i.e. just gets his money back?

It doesn't matter. There is an initial pot of x.

By "If they still both check they get a refund.", do you mean that they split the pot x?

Tried to find a solution using indifference criteria and produced an incomplete set of equations.

I assumed the form:

For numbers 0-a, bet as a bluff (B)
For numbers a-b, check, then bet the next street as a bluff (CB)
For numbers b-c, check, then check the next street (CC)
For numbers c-d, check, then bet the next street for value (CB)
For numbers d-1 (easier to work with than 100), bet for value (B)

The problem is that the EV equation for indifference with c is the same as with b
i.e.:

the EV of CC with b = the EV of CC with c
(as we only win/lose a non-zero amount if the other player checks as well, which they do with a frequency c-b, and this is independent of our hand)

and...

the EV of CB with b = the EV of CB with c
(as we win x when the opponent CCs, win 3x when they CB as a bluff [we beat all CB bluffs with any hand above b] and lose x when they CB for value [we lost to call CB value with any hand below c])

I guess this form must be wrong - don't have time to try another though - hope that is of some use at least.

As for a better form - given that the strength of our hand doesn't matter at all for the CC line, I guess we should actually use this for the bottom of our range and therefore use the hands just below our value CB range as CB bluffs, which should hopefully give some more useful equations - maybe try that?

There are three possible moves:
(1) Bet
(2) Check, Bet
(3) Check, Check

My guess is that this will be arranged linearly. There is no advantage that I can see in "bluffing" with bad hands.

So, if we are given card with value C, then we choose two values 'a' and 'b' with:
b <= C <= 1: Bet
a <= C < b: Check, Bet
0 <= C < a: Check, Check

I've yet to work on figuring out exact EVs and maximizing/etc, but the problem seems straightforward enough.

Oops... Guess I was wrong. Bluffs do seem necessary. I was hung up on the fact that they split the pot regardless of holdings when they both check. But if it's only one street the there should be no benefit in betting your worst card as opposed to betting the bottom of your "value" range. In two streets, it's advantageous to save good hands for the next street.

Therefore, no bluffs on second street... I think?

So I think it should be
c <= C <= 1: value bet
b <= C < c: Check, bet
a <= C < b: Bluff bet
0 <= C < a: Check, check

Yeah, agreed. Though the distinction between bluffs and value bets in a merged betting range is not that meaningful.

Having some stronger hands on the second street seems logically to be important to prevent your opponent from betting with impunity.

I think this is actually going to require a mixed strategy. I just can't figure out a way to organize ranges in a way that yields a concrete solution.

So far I've figured out that:
(1) We should bet 2/5 of the time on the first street, and check 3/5 of the time.
(2) We should bet 1/3 of the time on the second street (the 1/3 top part of our range that makes it there).
(3) We should bet C=4/5 and up 100% of the time. At C=4/5, we become indifferent to a check/bet line.

I think between C=2/5 and C=4/5, we need to mix betting with checking on the first street, and we always bet for C>2/5 on the second street. I'm not yet sure if we can mix linearly (e.g., so that C=3/5 we're betting the first street exactly 50% of the time).

From C=0 to C=2/5, we should check both streets 100% of the time.

Final answer....

4/5 <= card < 0... Bet
2/5 <= card < 4/5... Bet 50% of the time. Check 50% of the time. Always bet second street.
0 <= card < 2/5.... Check both streets

EV(card = 1) = 7/5
EV(card = 4/5) = 4/5
EV(card = 3/5) = 1/2
EV(card = 2/5) = 1/5
EV(card = 1/5) = 1/5
EV(card = 0) = 1/5

Im not sure if we need mix strategy when we have infinitely many cards.
Imo hero probably needs to bet some top % of his best hands and some % of his worst hands on first streets and then only bet some % of best hands on second street because value of check is same for every hand.So we need to find 3 numbers
X-if card is bigger is then x then bet first street
Y-if card is lower then Y then bet first street
Z-if card is bigger then Z then check first street and bet second one if you get there and if its lower then Z check down.
Then we just need to find number for what bet and check on first street has same EV and then we for what check check and check bet line has same EV that should give us 3
equations form which we can fine our points of indifferent.

If they are playing gto the ev of every match should be 0.5pot. Am I right?

No we do need a mixed strategy... I tried to solve it in various ways with pure strategies, and it doesn't work.

Here is how I solved it.

First consider the second street alone. Using indifference criteria and the assumption we bet top X% and check bottom, this turns out to be easy to solve. We find that the solution is to bet the top 1/3 of our range and check the bottom 2/3.

Now go back to the first street. Consider we bet Y% of the time on the first street. Consider the EV of the worst card we bet with. We will win (1-Y) pots, and we will lose (Y) pots. So our EV is 1-2Y. But opponent will check both streets (1-Y)*2/3. So our EV from checking twice is (1-Y)*2/6 (our EV from two checks is 1/2)... We are indifferent between checking twice and betting first if 1-2Y = (1-Y)*2/6... 6-12Y = 2-2Y... Y = 4/10 = 2/5.

Note that if we aren't indifferent between checking twice and betting first, we would "bluff" the bottom of our range (if betting is higher EV), or just check/check the bottom of our betting range (if check/checking is higher EV).

So we should bet 2/5 of the time on the first street.

Next, we find the point at which we become indifferent from betting versus check, betting (the top of our check/bet range). Let's call that point card = Z.

EV(bet Z) = 3/5 + 2*(2/5 - (1-Z)) - (1-Z) = 7/5 - 3 + 3Z = -8/5 + 3Z
EV(check, bet Z) = 2 * 3/5 * 1/3 + 3/5 * 2/3 = 4/5

To make us indifferent, set these equal, and we have -8/5 + 3Z = 4/5... 3Z = 12/5...

Z = 4/5.

So far we know:
* We bet 2/5 of the time on the first street, and check 3/5
* We bet 1/3 of the time on the second street, and check 2/3
* We bet 100% of the time if dealt a card >= 4/5.
* At card = 4/5, we are indifferent between a bet versus a check/bet line.

Ok now we want to find the probability we bet when we're below 4/5. Let's suppose we're evaluating what we should do with card Q, slightly below 4/5. Let's suppose that we bet P percent of the time from Q through 4/5 (and check/bet (1-P) percent of the time).

Then:
EV(bet Q) = 3/5 + 2 * (1/5 - (4/5 - Q)*P) - 1/5 - (4/5 - Q)*P
= 3/5 + 2/5 - 8/5*P + 2*PQ - 1/5 - 4/5*P + PQ
= 4/5 - 12/5*P + 3*PQ
EV(check, bet Q) = 3/5*2/3 - (1-P)*(4/5 - Q) + 2*(3/5*1/3 - (1-P)*(4/5 - Q))
= 2/5 - 4/5 + 4/5*P + Q - QP + 2/5 - 8/5 + 8/5*P + 2*Q - 2PQ
= -8/5 + 12/5*P + 3Q - 3PQ

To make us indifferent, set these equal:
4/5 - 12/5*P + 3*PQ = -8/5 + 12/5*P + 3Q - 3PQ
4 - 12*P + 15*PQ = -8 + 12*P + 15Q - 15PQ
12 - 15*Q = 24*P - 30*PQ
12 - 15 * Q = P(24 - 30*Q)
P = (12 - 15*Q) / (24 - 30*Q)
P = 1/2

(Note: There's probably a much simpler way to do that....)

So basically, for the portion of our range where we're indifferent between betting and check/betting, we bet with probability 50% and check/bet with probability 50%.

This yields my solution in my previous post above.

Before they're dealt cards, yes. This requires the assumption that they split pot when they both check, which wasn't entirely clear in the op, but I did make this assumption in solving the problem.

Seems right. So how bout if the best hand wins if there is no bet both times?

Even for the single street version of this game, I think this will require a mixed strategy... It's clear that betting the top of our range is always best (if we check, we win when our opponent checks; if we bet, we win when our opponent checks OR when he bets). However, we prefer moving the bottom of our betting range down to 0 so that we have stronger checks.

We should therefore be indifferent between betting and checking the very bottom of our range. Since checking the bottom of our range has an EV of 0, we should bet exactly half of the time to get an EV of 0 at the bottom of our range when we bet.

If we're betting 50% of the time total, and the top X% always, we become indifferent between betting and checking when our CARD = 2/3.

This means
* For 2/3 <= CARD <= 1: Bet always
* For 0 <= CARD < 2/3: Bet 25% of the time... With CAVEAT...
Caveat: I don't think we can just bet 25% of the time regardless of what we have. It seems like we will bet more often when we have a better card, and less often for bad cards...

I think for the two street game, there will be a big chunk of our range that we'll have to mix between betting, check/betting, and check/checking.

Interesting problems. That's all I've got on this for now.

Seems like betting 2/3 or better along with 1/3 thru 1/2 can not be exploited by someone who deviates from that strategy (in the one round simultaneous pot size bet game where the best hand wins if they both check.)

If he bluffs his hand zero he gains or loses nothing since it is 50-50 his risk of one bet will succeed in winning the pot. If he bluffs with slightly better his risk is the same but his reward is less.

If he checks some of the worst hands in the top third he saves a bet a third of the time but costs himself two bets the sixth of the time that he faces the demibluff hands and blows the pot.

If he bets slightly more than the top third for value, he saves a two bet swing against the demibluff but costs himself a bet against the top third.

If he doesn't bet the worst of his demibluffs he saves a bet a third of the time but misses out on a two bet swing a sixth of the time.

Am I missing something?

If opponent simply bets top half and checks bottom half, then his strategy wins. Therefore it appears your suggested strategy is exploitable.

The differences are:

* When we have .5-(2/3) and face (2/3)-1, then we are saving PSB by checking.
* When we have (1/3)-.5 and face .5-1, then we are losing PSB by betting.

Since the frequency of the latter scenario is larger than the former, it results in a net loss.

For the single street version of the game where high card wins when both players check...

We cannot just split our range into a pure strategy. If we have a "high" betting range and a "low" betting range, with checks in between, then betting the bottom of the high part has the same EV as betting the top of our low part. However, checking at those points will always have a higher EV with the better card (bottom of high) than with the low card (top of low). So we will not be indifferent at one of the two points.

It seems like opponent could exploit this by betting top 2/5 on first street and checking down second street every time.

I got lazy. I looked only at single changes that no longer had you betting 50%.

Are you sure? For the strategy I listed, between 2/5 and 4/5, against the strategy you propose, it would lose EV when betting (since opponent bets a stronger range), but it would gain EV when check/betting (our EV from check/betting would be an automatic 3/5) as well as check/checking (EV increases from 1/5 to 3/10 at the bottom of our range).

I don't have the time right now to figure out the overall EV from both strategies, but I have a feeling you've gotta be wrong; since I did particularly look at the strategy of betting the top 2/5 and had already found that that was dominated by strengthening our check/betting range with a mixed strategy.

You are right, I found an error. I now show that the two strategies are even against each other (yours is -.02, then +.02), but I would guess your strategy would be less exploitable against outside strategies (i.e., more GTO).

This seems to be ahead of the 1/5 B, 2/5 50-50/B, 2/5 CC strategy:

2/3 =< card =< 1 : Bet
1/2 =< card =< 1 : 50% Bet, 50% Check/Bet
card =< 1/2 : Check/Check

If someone would please check my work and/or find a strategy that exploits it for +EV.

This strategy bets (first street) 1/3 + 1/6*1/2 = 5/12 of the time.

Consider the EV of betting the worst card we bet with, and our EV of check/checking that card (strategy against itself):
EV (bet bad card) = 7/12 - 5/12 = 1/6
EV(check/check bad card) = 1/2*1/2 = 1/4

So this strategy playing itself would prefer to bet less on first street, finding that it's sometimes more profitable to take a check/check line. Therefore this strategy cannot be GTO, and some other strategy can exploit it. Specially a strategy that exploits it would be one that mimics this strategy, but moves 1% of its betting range (from bottom part) to a check/checking range.