We often play $40/80 stud with a $10 ante a start of a mix
Game. The time is $7 a down and we play time pots for all games, if the first had of the down is a stud variant we all ante $15 instead of $10 and the dealer takes the time from the pot....

We also play with a kill, if you are leaving at the end of the down and scoop a pot you have to play the next hand but you do not ahve to pay time (can only pay your kill).

A player scoops the last hand of stud and owes a kill, he announces he is leaving but has to play one last hand which is now a time pot. Player A think he owes a $8 ante and player B thinks he owes a &10 ante. Which is correct?

Obviously the actual money doesn't matter but both sides feel pretty strongly they are correct

I don't get why you would ante less in the time pot. Why not just have people post 17, have the dealer pull in 7 for time and ten into the pot for each? The way you have it the dealer still has to make change etc so I don't see how it saves time (might be missing something though). Ot just have someone put up the time for everyone and collect the time back from the winners of the first two pots.

They ante more in the time pot, not less. They ante extra (but not $7 extra) so there's still some ante money to play for in the time pot. It's not to save time.

Thats a really interesting question actually. My initial thought was $10 but technically everyone is anteing $8 that hand, so sure, have him ante $8.

Couldn't everyone just agree to have this hand not be the time pot?

yeah wtf?? the guys is leaving but has to pay his share of the time for the next down?

Gotta ask: What if players are walking when this happens?

there must be a misunderstanding. The time for the down is $7. if one player is walking the dealer will take $42 (or $35 in this case) from the pot and then $7 from each player.

In this case there were 7 players dealt in, 6 people anted $15, the 7th $8 or $10 and the dealer took $42 from the pot (as opposed to standard $49).

If the time is $7/plAyer and they are taking $35 from the pot ($5/player) doesn't each player owe only $2 each?

The winner of the pot pays the time charge -- 7$ for each active player. Of this $7, $2 is from the antes and $5 from a surcharge (now called an upcharge at my local movie theater). I would say that the Killer can avoid the surcharge, but must ante. The ante is $10. No reason he should take $2 back.

And a walking player gets a button and pays $7 when he returns if he wishes to continue playing.

jmo

ah. Yeah, I agree with this, but I can see both sides of the argument. A player leaving shouldn't have to pay the time, but should have to pay the ante given he is the kill.

FWIW this was my exact view as well, but I can see both sides of the argument. The $8 argument is that the $15 is divided between a $7 rake and $8 ante
The $10 argument is that the $15 (which is an arbitrary number) is broken down as a $10 ante, $5 towards the time and then the winner of the pot pays $2 for each player....

Basically when we are 7 handed and playing a draw game the winner of the pot pays $49. When we are 7 handed and playing stud the winner of the time pot pays $14

Additionally, when we play a time pot at 100-200 its a $25 ante in normal pots and time pots alike, so the $15 is def an arbitrary number

So if you're always doing time pots anyway, why are you anteing extra if it happens to be a stud hand?

Less action/smaller pots on the stud rounds?

Put me in the $8 camp. Everyone is only paying an $8 ante, the $7 is for time. Which he doesn't owe.

I am fairly certain I have seen this at the 20-40 OE game where we ante $10 in time pots and the person leaving only had to ante $3.

This is the argument the other side keeps making but I dont understand it. Lets sat you steal a normal pot, you win $80 (antes + bring in). Now lets assume we steal a time pot, you win $66. Whoever wins the pot eat the $14 which is $2 per player and supports the break down like I said earlier.

Its a $10 ante, $5 towards time and the winner pays $2 time for each player. So in this scenario the player antes $10, doesn't have to pay his $5 and the time pot winner will save $2 as well.

Bit lots of people disagree with me, so its def not a clear cut issue

This just seems wrong to me. Let's say the killer wins the pot. Why should he have to pay $2 for each player if he wins while they wouldn't have to pay $2 for him if they won?

Having to pay $2 for each player the next hand is a built in cost of playing the last hand of the previous down.

To me, the bigger antes don't seem very optimal. When you keep the antes normal, it usually ends up with the fun players paying the bulk of the time for everyone.

I think it's pretty clear he owes 10 here fwiw.

But only if you happen to win that hand and kill it? It doesn't seem like having to kill the hand should have anything punitive other than the fact you have to kill it.

people aren't anteing $10 this hand. they're anteing $8.

They're anteing 10 and putting 5 extra in to help generate more action in the time pot.
If they were anteing 8, and putting 7 in for time, they might as well all just pay the time charge individually. I think 10s the right number here, but I think there's a better argument that he has to go the full 15 than there is for him to ante 8. The extra ante money is solely to loosen up play on that particular hand. The winner of the pot pays all the time.

No, they are anteing $15.

What do you do if someone isn't leaving, but just doesn't want to be in the time pots? Isn't that allowed?

The plaque next to the dealer says $10 ante. The game being spread. If they put in more, wheee gamble gamble. If they put in less, they don't get a hand.

I guess I also still don't exactly understand how it is working now. Regardless of how much he antes, what happens if the killer wins the pot? If they take time out, he is getting extra screwed no matter what ante he paid.

Seems to me like the only fair way to do it is to wait until the next hand to do the time pot, or whatever the first hand is that everyone who owes time is in, and no one else.