Quote:
Originally Posted by answer20
While I figuratively missed the boat on this one .. it's because I was literally on a boat and off the grid!
This RIT quandary is very common in the home games in our area. It doesn't matter if there are side pots or not. It's a 'simple' if/then spot for the Dealer. If the non-RIT Player(s) win the first Board then there's no 2nd Board for the impacted pots .. otherwise it becomes a split pot based on two Boards.
It's extremely common for a short stack to only want to run the Main Pot once even if they regularly RIT in most pots.
Lately I have leaned towards only running it once in pots less than $X, but I don't care if other Players want to split up the pot if I lose the first Board.
AND .. IMO .. a Player's EV automatically changes when you RIT since you are now running split pots. It's cut in half and you just happen to be running the same EV twice in a row. Your probability doesn't change by much, but your EV is reduced. EV is chips, probability is outs. Variance is the spread between data points, so when you reduce your EV you have reduced the potential data points on your results graph. Same impact if you win or lose both, less impact if you split. When you RIT are you supposed to enter one or two data points? One 'net' data point or two 'actual' data points?
Yes, Player A's spot is a 'loop hole' since they have opened up the door to a potential win when they otherwise were drawing dead.
Some Players get caught up on the split pot thing .. I just tell them it's just like playing two pots the same size 'very fast' instead of 20-30 minutes apart, which rarely happens anyway. GL
In a normal RIT spot, EV does not change. That is what brought up the subject of this post in the first place - this is not a normal RIT situation, and the EV does indeed change for the players who agreed to RIT. EV is just the probability of each outcome multiplied by the value of that outcome, with a sun over all possible outcomes. To make the point concrete, assume two players all in on the turn. Player A has AKo, player B has AsQs. Board is Ah Ks 9s 4d. Obviously player B has exactly 9 outs. (any remaining spade). For mathematical simplicity let the pot be $1892 (after all action is settled and rake taken).
Now if they run it once, player A wins 1892 with probability 35/44. Player A therefore has an EV of 1892 x 35/44 = 1505. Player B has an EV equal to the remainder of the pot, or 387 (which of course could also be found from 1892 x 9/44).
If they run it twice the calculation is a bit less simple. Player A will scoop with probability 35/44 x 34/43 = 1190/1892. Player B scoops with probability 9/44 x 8/43 = 72/1892. The remaining times will be a split, so the split probability must be 630/1892 (since 1190+630+72 = 1892). Now the EVs are easy to calculate. Player A wins 1892 for a scoop and 946 for a split, so his EV is 1892 x 1190/1892 + 946 x 630/1892 = 1190+315 = 1505, exactly the same as when they ran it once. Likewise if player B will have an EV equal to the rest of the pot or 387, again identical regardless of how many times they run it.
RIT does reduce variance though. To see this calculate player AÂ’s variance in the above spot. His EV is 1505, and he actually wins 1892 with probability 35/44 and zero with probability 9/44. His variance is therefore 9/44 x1892^2 + 35/44 x 387^2 which equals 851,388. For RIT he wins 1892 with probability 1190/1892, 946 with probability 630/1892 and zero with probability 72/1892. Variance is then
1892^2 x 72/1892 + 559^2 x 630/1892 + 387^2 x 1190/1892, which is indeed lower equaling 334,473.
Last edited by stremba70; 04-17-2023 at 12:10 PM.
