Post

_TKO_ · #1 11-23-2006, 12:35 AM

The game is called Post.

Every player antes $1.
The action starts with the player left of the dealer.
The dealer deals out two cards. If the first card is an A, the player may choose whether it acts as a high card or low card.
The player must wager anywhere from $1 to the size of the pot that the next card off the deck will be between the two initial cards (if the cards are equal or connected, there will be a redeal).
The player is paid even money if the card is in fact in the middle. If the card is outside the range, the player's wager is added to the pot. If the card is equal to either of the cards, then the player must forfeit double his wager to the pot.
The game ends when one player takes the entire pot.

I'm trying to determine the optimal strategy for this game. I think that your edge comes from other players not trying to scoop the pot when they have an edge. Also, you have to know when the odds are in your favour, so you know when to go for the scoop.

This game is such a gambler's game, that you could probably convince people to play it with you. I would love to play it knowing I have an edge.

Okay, so I'm trying to calculate when the odds are in your favour. It's dependent on the spread of two cards that you see. The possible spreads range from 2 up to 12. There are 4 cards per card within the spread, 4 outisde the spread, and 3 for each post. When the first two cards are up, there are 50 cards remaining.

So, the probability of hitting a winning card is W = 1/50 * 4(x-1), where x is the size of the spread (the -1 is included since the range is non-inclusive). The probability of losing is L = 1/50 * 4(13-x-1) = 1/50 * 4(12-x). The probability of hitting a post is P = 6/50. Since you win 2 in the first case, lose 1 in the second, and lose 2 in the third, your EV per spread is:

2W - L - 2P = 2/25 * (3x - 17)

Is this logic correct so far?

Anyway, once we calculate the EV per spread, we need to multiply each by the frequency of the spread and sum the results the get our total EV. This is the part I need help with.

Now, I'm not sure how to factor in the ante and the likelihood of us not getting a chance to play (ie it's better to be in early position rather than late position). Also, when I played this last night, we dealt through 3/4 of the deck, so counting cards also adds to our advantage per turn.

_TKO_ · #2 11-23-2006, 02:13 AM

Okay, I've got the frequencies now and I've calculated your EV if you go first and bet the pot if it's in your favour and bet the minimum if it's not. The pot size is based on 2 players. I played around with the numbers a bit and found that each additional dollar in the pot increases your EV by about 0.1892.

Code:<hr /><pre> Spread Win Loss Post EV Wager Frequency Total EV
12 1.76 0 -0.24 1.52 2 1.21% 0.03668175
11 1.6 -0.08 -0.24 1.28 2 2.41% 0.061779789
10 1.44 -0.16 -0.24 1.04 2 3.62% 0.075294118
9 1.28 -0.24 -0.24 0.8 2 4.83% 0.077224736
8 1.12 -0.32 -0.24 0.56 2 6.03% 0.067571644
7 0.96 -0.4 -0.24 0.32 2 7.24% 0.046334842
6 0.8 -0.48 -0.24 0.08 2 8.45% 0.013514329
5 0.64 -0.56 -0.24 -0.16 1 9.65% -0.015444947
4 0.48 -0.64 -0.24 -0.4 1 10.86% -0.043438914
3 0.32 -0.72 -0.24 -0.64 1 12.07% -0.077224736
2 0.16 -0.8 -0.24 -0.88 1 13.27% -0.116802413
1 0 0 0 0 0 14.48% 0
0 0 0 0 0 0 5.88% 0
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|TOTAL: 0.125490196
</pre><hr />

So this table shows the EV of going first, assuming the deck is reshuffled before all new draws.

Can anyone help me determine the value of position assuming all opponents are playing perfectly?

Let's first assume we're playing heads up. If we are second to act, then our EV is -1 for every time our opponent scoops. Of the times he doesn't scoop, we are interested in how many times he attempts to scoop and misses (our EV becomes 0.1255+0.1892*2), how many times he doesn't attempt to scoop and misses (our EV becomes 0.1255+0.1892), and how many times he doesn't attempt to scoop and hits (our EV drops to 0.1255-0.1892).

_TKO_ · #3 11-23-2006, 02:15 PM

Here's my stab at it.

First I calculated the probability of each situation occurring. I then calculated the new pot size assuming villian is playing perfect strategy.

Code:<hr /><pre> Spread Chance of hitting Chance of losing Chance of posting
12 0.88 0 0.12
11 0.8 0.08 0.12
10 0.72 0.16 0.12
9 0.64 0.24 0.12
8 0.56 0.32 0.12
7 0.48 0.4 0.12
6 0.4 0.48 0.12
5 0.32 0.56 0.12
4 0.24 0.64 0.12
3 0.16 0.72 0.12
2 0.08 0.8 0.12

</pre><hr />

Code:<hr /><pre> Spread New pot: Villian hits / Villian Loses / Villian Posts
12 0 4 6
11 0 4 6
10 0 4 6
9 0 4 6
8 0 4 6
7 0 4 6
6 0 4 6
5 1 3 4
4 1 3 4
3 1 3 4
2 1 3 4
</pre><hr />

I then multiplied the first table by the frequency of each spread occurring. This gave me the probabilities of each pot size being the new pot size.

Code:<hr /><pre> Spread Chance of new pot existing: Villian hits / Villian Loses / Villian Posts
12 0.010618401 0 0.001447964
11 0.019306184 0.001930618 0.002895928
10 0.026063348 0.005791855 0.004343891
9 0.030889894 0.01158371 0.005791855
8 0.033785822 0.019306184 0.007239819
7 0.034751131 0.028959276 0.008687783
6 0.033785822 0.040542986 0.010135747
5 0.030889894 0.054057315 0.01158371
4 0.026063348 0.069502262 0.013031674
3 0.019306184 0.086877828 0.014479638
2 0.010618401 0.106184012 0.015927602
</pre><hr />

I determined the EV of each new pot by using the formulas stated in my previous post. A pot of zero size has an EV of -1, since you paid one unit that you will never get back. Every other pot has an EV of (-0.0637 + (P - 1)*0.1892), where P is the size of pot.

Code:<hr /><pre> Spread EV of new pot: Villian hits / Villian Loses / Villian Posts
12 -1 0.503892196 0.882294196
11 -1 0.503892196 0.882294196
10 -1 0.503892196 0.882294196
9 -1 0.503892196 0.882294196
8 -1 0.503892196 0.882294196
7 -1 0.503892196 0.882294196
6 -1 0.503892196 0.882294196
5 -0.063710804 0.314691196 0.503892196
4 -0.063710804 0.314691196 0.503892196
3 -0.063710804 0.314691196 0.503892196
2 -0.063710804 0.314691196 0.503892196
</pre><hr />

Finally, we multiply each of these EV's with their probabilities of occurring. We then sum the results to get our total EV.

Code:<hr /><pre> Spread EV
12 -0.010618401 0 0.00127753
11 -0.019306184 0.000972824 0.00255506
10 -0.026063348 0.002918471 0.00383259
9 -0.030889894 0.005836941 0.00511012
8 -0.033785822 0.009728235 0.00638765
7 -0.034751131 0.014592353 0.00766518
6 -0.033785822 0.020429294 0.00894271
5 -0.00196802 0.017011361 0.005836941
4 -0.001660517 0.02187175 0.006566559
3 -0.001230013 0.027339688 0.007296177
2 -0.000676507 0.033415174 0.008025794
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|TOTAL: 0.022876744
</pre><hr />

TOTAL EV OF ACTING SECOND: 0.022876744

Unfortunately, this result seems counterintuitive, so maybe someone can help me interpret it. How can both players have a positive expectation here? I suppose this isn't a zero-sum game, since other players are adding to the pot before your turn.

I would assume that adding more players doesn't change your expectation to negative, but it may increase your overall EV. I still think that you will have the worst EV if you are last to act.

VBCurtis · #4 11-25-2006, 02:12 PM

I didn't analyse the calculations closely, but I imagine you failed to take the ante into account on each EV calculation-- for instance, when you bet pot and win going first, you only really win $1, since the other dollar is your own ante. Such errors would lead to a +EV answer for both players, which clearly can't be true.
-Curtis

_TKO_ · #5 11-25-2006, 03:25 PM

Thanks, I'll take a look at that situation.