The only thing that is different for each run is the # of cards available in the deck. I have a feeling that the lower the # of cards available (regardless of what those cards are), the more favoured that AA becomes (by a tiny amount).
The only thing that is different for each run is the # of cards available in the deck. I have a feeling that the lower the # of cards available (regardless of what those cards are), the more favoured that AA becomes (by a tiny amount).
/g2
Sigh.
Let's try this again.
Let's say you have AsAh and your friend has QhQd
You agree to deal out 9 sets of Flop/Turn/River with the remaining cards.
(naming the sets of 5 cards A, B, C, etc.)
Every possible unique set of 5 cards will show up in column A as many times as it does in column B, or column C, etc. over the course of dealing out every possibility.
Since the best hand that would be made from the combination of your hole cards and the five cards from each column would not change, then the EV would not change from column to column. It is the VARIANCE that might change.
No, that would be like saying that the number of players in the table will magically affect how much AA is favoured over QQ since there are fewer cards available in the deck. Given that all the other cards are UNKNOWN prior to any runs, AA is a ~81.5% favourite over QQ and has an equity of ~63%. It does not matter if you agree to run it once or a trillion times, it does not matter if you will reshuffle or not, it does not matter if you take out 5, 8 or 47 UNKNOWN cards from the deck before dealing out the board; it does not matter if you do a happy dance around the table! Of course if you peek at the flop or make assumptions on what the first run will be, then the cards become known and the probabilities and EV will change AT THAT POINT.
THE EV REMAINS THE SAME prior to seeing any of the UNKNOWN cards. No matter what the players agree to do, AA is still a 81.5% favourite and has an equity of 63% over QQ.
Even if you do not understand the poker math or logic behind running it x times, the important thing to remember if somebody offers it to you is that it does not matter for EV, only variance will be affected. If you have brought sufficient bankroll and rebuying is not a problem, you can always decline running it more than once.
The only thing that is different for each run is the # of cards available in the deck. I have a feeling that the lower the # of cards available (regardless of what those cards are), the more favoured that AA becomes (by a tiny amount).
You fuckers! Quit repeating the same thing over and over and actually show some data, or reference some data, proving the EV is EXACTLY the same when you run it k times... where k = 1 to 9. (9 being the max # of times you can run a full board w/out reshuffling).
Listen carefully...
The only thing you have convinced me of is the fact that the EV is the same on the FIRST run of k runs. My feeling tells me that the EV might be different on subsequent runs. Stop TELLING me I'm wrong and PROVE me wrong.
The only thing you have convinced me of is the fact that the EV is the same on the FIRST run of k runs. My feeling tells me that the EV might be different on subsequent runs. Stop TELLING me I'm wrong and PROVE me wrong.
Greg - you are right that the chance to win may change after the first run. However, you don't do 1 run and then agree to run it twice, you always decide how many times PRIOR to running it (once my aces hold up why would I then give you another chance to crack me?). The EV is set BEFORE the runs (1, 2, or a dozen) are done, so the % to win after that point (once the first river is shown) is irrelevant. No math there, but does that help?
You fuckers! Quit repeating the same thing over and over and actually show some data, or reference some data, proving the EV is EXACTLY the same when you run it k times... where k = 1 to 9. (9 being the max # of times you can run a full board w/out reshuffling).
You win in 12 of the 15 scenarios (80%), so your EV is 0.6 units ((12-3)/15) per hand.
Now you and your friend decide to "run it twice".
If the deck is shuffled each time then you would scoop the whole pot 64% (80% * 80%) of the time
If the deck is shuffled each time then you would split the pot 32% of the time (2 * 80% * 20%)
If the deck is shuffled each time then your opponent would scoop the pot 4% of the time
Your EV in this case is still 0.6 units (1 * .64 + 0 * 0.32 - 1 * 0.04)
BUT if you did not shuffle the deck then there are 90 scenarios:
NOTE: Every 'flop' from the first scenario is represented in each of first flops 6 times, and every one of the second flops 6 times.
This means that the EV is the same for the first flop as it is for the second flop. - This will be the same in poker scenarios.
But - Now, you win both hands 54 times out of 90 (60%), and split 36 times out of 90 (40%)
There is no scenario where your friend will win both hands - there is only 3 red cards and they would need 2 in both flops to win the whole pot.
So, your EV remains the same, but by running it twice you have taken all the risk of losing all of your money away.
I told you your friend was stupid.
It gets more complicated in poker because there are more cards and types of hands, but the same concepts apply - EV remains the same, but variance changes the more times you run the cards.
LOL... you guys suck. Seriously though, no one has indicated that they actually understand what I am talking about. I'm working on a simulation right now. If anyone wants to see my code or the results PM me.
Ok I'm sorry I didn't get here sooner to do the followup post, wow lots of replies
1) I don't mind taking big inevitable beats becuz it was a big hand, I call the inevitable ones good cuz we both played our hands perfect and Poker is entertaining and what goes around comes around in poker.
2) I ran it twice cuz I brought 1 buyin and I decided to protect myself cuz even if we chopped I'd still win the initial raisers chopped raise money
3) my buddy on the button (the inital raiser) isn't really that good of a player, so if he misses the flop I can bet him out very easily or if he hits his TP (hopefully AK $$$) I'd have all of his money
and yes I'm stupid as usual we did reshuffle the FT&R + the burns so it probably would've ended up a chop had we not reshuffled, but hey shit happens.
Does anyone understand my question yet? LOL... I almost don't understand it anymore.
i sure do!
what language/system are you using for the simulator? if it is 360 Assembler the i am all over it! ok, the reason i ask is, pokercalculator does most of what you want. it is free, written in java and comes with a bunch of 'class'es. i'm not familiar with java, just wondering if the classes could be reused for the bulk of your simulator.
what language/system are you using for the simulator? if it is 360 Assembler the i am all over it! ok, the reason i ask is, pokercalculator does most of what you want. it is free, written in java and comes with a bunch of 'class'es. i'm not familiar with java, just wondering if the classes could be reused for the bulk of your simulator.
I'm using C++... my second choice was 360 assembler (obviously, a close third was slitting my wrists). Java is great for developers... but it runs really slow.
it is free, written in java and comes with a bunch of 'class'es. i'm not familiar with java, just wondering if the classes could be reused for the bulk of your simulator.
The classes he released are almost useless for extension without the source (unless you got way too much time on your hands).
Really? What happened? I haven't used Java in about a year and a half... maybe 2.
I really don't want to hijack this thread, but you can google "java vs c++ performance" and PM me if you want to chat. In general, the new runtime JVM's are much better, so IF you don't program horribly, you can achieve great performance.
I really don't want to hijack this thread, but you can google "java vs c++ performance" and PM me if you want to chat. In general, the new runtime JVM's are much better, so IF you don't program horribly, you can achieve great performance.
Quote from my google search:
The results I got show that no one should ever run the client JVM when given the choice. (Everyone has the choice. To run the server VM, see instructions in the Using the Server JVM section below.)
That's all I needed to know. Thanks beanie. Now back to the thread.... umm yeah... losing the whole pot with AA vs QQ running it twice that sucks.
I am trying to prove the EV concept to you without the numbers (which you may or may not believe anyways).
Let's say that you and your friend have AA and QQ, and instead of dealing the flop off the top of the deck, you discard the first 8 cards and then do the burn, lay out a flop, burn, turn, burn river. Would you expect the EV to change because you got rid of the first 8 cards from the deck before doing the flop/turn/river?
Or, would you expect it to change if you dealt the flop, turn, and river off the bottom of the deck?
Each time you would start with a totally random deck.
I think where the confusion lies is in the fact that if you already knew the result of running it once and THEN you decided to run it again there would be a difference in your EV for the second run (to make up for the fact that you now know whether you were a winner or a loser after the first FTR), but as Beanie already stated you usually would set the number of times that you are going to run the cards before any of the cards are revealed.
The EV would be the same as long as the deck started with the same cards in it each time, regardless of where the flop/turn/river came from.
By discarding the top 8 cards and then doing the flop, turn, and river you have effectively created what the second Flop/Turn/River would be if you ran it twice. As we stated before, this second flop/turn/river creates the same EV as the frist flop/turn/river. Combining the 2 EVs would be an (x + x)/2 = x scenario.
But, as I stated earlier, your variability fluctuates much less when you run it x times.
I am trying to prove the EV concept to you without the numbers (which you may or may not believe anyways).
Let's say that you and your friend have AA and QQ, and instead of dealing the flop off the top of the deck, you discard the first 8 cards and then do the burn, lay out a flop, burn, turn, burn river. Would you expect the EV to change because you got rid of the first 8 cards from the deck before doing the flop/turn/river?
Or, would you expect it to change if you dealt the flop, turn, and river off the bottom of the deck?
Each time you would start with a totally random deck.
I think where the confusion lies is in the fact that if you already knew the result of running it once and THEN you decided to run it again there would be a difference in your EV for the second run (to make up for the fact that you now know whether you were a winner or a loser after the first FTR), but as Beanie already stated you usually would set the number of times that you are going to run the cards before any of the cards are revealed.
The EV would be the same as long as the deck started with the same cards in it each time, regardless of where the flop/turn/river came from.
By discarding the top 8 cards and then doing the flop, turn, and river you have effectively created what the second Flop/Turn/River would be if you ran it twice. As we stated before, this second flop/turn/river creates the same EV as the frist flop/turn/river. Combining the 2 EVs would be an (x + x)/2 = x scenario.
But, as I stated earlier, your variability fluctuates much less when you run it x times.
I understand that the EV of each run will be the same (x+x)/2=x... what I am unsure about is that the x is independent of the # of cards in the deck. If it is dependent on the # of cards in the deck, then the EV will be different based on how many times you decide to run it.
Al, thanks for all your hard work trying to convince me Best of luck finishing the job!
Yes, EV is independent of the # of cards in the deck since the cards are unknown. Extending DataMn's example, let's say you and I agree to run AA vs. QQ two times. So we split the pot into two, e.g., $200 / 2 = $100. As an experiment, I lay face down six boards in sequence (5 + 3 burn cards for each board), then each of us takes turns eliminating four of the boards so that only two boards will be left for the two runs. Since the cards are unknown, each of those six boards has the same EV as all the other boards. In effect, we have eliminated 32 cards from the deck but the overall EV has not changed. In other words, EV is independent of the # of cards in the deck.
Let's say that just before turning over the first card, you change your mind and propose that we run it six times instead of two. Since I know that all six unknown boards have the same EV as each other and the overall EV will not be affected, I agree and we split the pot into six for each of the runs, i.e., $200 / 6 = $33.33. So before seeing the cards, whether we agree to use 8, 16 or all of the cards I have laid out, the EV is the same at that point. Only our variance/risk is reduced.
Comments
/g2
Sigh.
Let's try this again.
Let's say you have AsAh and your friend has QhQd
You agree to deal out 9 sets of Flop/Turn/River with the remaining cards.
(naming the sets of 5 cards A, B, C, etc.)
Every possible unique set of 5 cards will show up in column A as many times as it does in column B, or column C, etc. over the course of dealing out every possibility.
Since the best hand that would be made from the combination of your hole cards and the five cards from each column would not change, then the EV would not change from column to column. It is the VARIANCE that might change.
/g2
THE EV REMAINS THE SAME prior to seeing any of the UNKNOWN cards. No matter what the players agree to do, AA is still a 81.5% favourite and has an equity of 63% over QQ.
Even if you do not understand the poker math or logic behind running it x times, the important thing to remember if somebody offers it to you is that it does not matter for EV, only variance will be affected. If you have brought sufficient bankroll and rebuying is not a problem, you can always decline running it more than once.
Listen carefully...
The only thing you have convinced me of is the fact that the EV is the same on the FIRST run of k runs. My feeling tells me that the EV might be different on subsequent runs. Stop TELLING me I'm wrong and PROVE me wrong.
/g2
So.. AA lost twice. Did you contact Ripleys Believe it or not? You could be in line to be on TV with a beat that bad..
let's run it again....
WHO CARES
and then re-shuffle.....
WHO CARES...
I will provide actual numbers soon, but I am going to use the example of a simpler game to explain the concepts that I have been trying to portray:
You and your friend are going to play a game.
Your friend isn't very bright, so the game will be tilted in your favour.
You pull the Ac, Kc, Qc, Jc, Kd, Qd, Jd from the deck
The Ac will be a communal card. The rest of the cards will be shuffled and the top 2 cards will be revealed.
If the majority of the 3 cards are black then you win, otherwise you lose.
There are 15 ways the 'flop' can come down (I have taken order out for simplicities sake):
KcQc KcJc KcKd KcQd KcJd
QcJc QcKd QcQd QcJd JcKd
JcQd JcJd KdQd KdJd QdJd
You win in 12 of the 15 scenarios (80%), so your EV is 0.6 units ((12-3)/15) per hand.
Now you and your friend decide to "run it twice".
If the deck is shuffled each time then you would scoop the whole pot 64% (80% * 80%) of the time
If the deck is shuffled each time then you would split the pot 32% of the time (2 * 80% * 20%)
If the deck is shuffled each time then your opponent would scoop the pot 4% of the time
Your EV in this case is still 0.6 units (1 * .64 + 0 * 0.32 - 1 * 0.04)
BUT if you did not shuffle the deck then there are 90 scenarios:
(KcQc JcKd) (KcJc QcKd) (KcKd QcJc) (KcQd QcJc) (KcJd QcJc)
(KcQc JcQd) (KcJc QcQd) (KcKd QcQd) (KcQd QcKd) (KcJd QcKd)
(KcQc JcJd) (KcJc QcJd) (KcKd QcJd) (KcQd QcJd) (KcJd QcQd)
(KcQc KdQd) (KcJc KdQd) (KcKd JcQd) (KcQd JcKd) (KcJd JcKd)
(KcQc KdJd) (KcJc KdJd) (KcKd JcJd) (KcQd JcJd) (KcJd JcQd)
(KcQc QdJd) (KcJc QdJd) (KcKd QdJd) (KcQd KdJd) (KcJd KdQd)
(QcJc KcKd) (QcKd KcJc) (QcQd KcJc) (QcJd KcJc) (JcKd KcQc)
(QcJc KcQd) (QcKd KcQd) (QcQd KcKd) (QcJd KcKd) (JcKd KcQd)
(QcJc KcJd) (QcKd KcJd) (QcQd KcJd) (QcJd KcQd) (JcKd KcJd)
(QcJc KdQd) (QcKd JcQd) (QcQd JcKd) (QcJd JcKd) (JcKd QcQc)
(QcJc KdJd) (QcKd JcJd) (QcQd JcJd) (QcJd JcQd) (JcKd QcJd)
(QcJc QdJd) (QcKd QdJd) (QcQd KdJd) (QcJd KdQd) (JcKd QdJd)
(JcQd KcQc) (JcJd KcQc) (KdQd KcQc) (KdJd KcQc) (QdJd KcQc)
(JcQd KcKd) (JcJd KcKd) (KdQd KcJc) (KdJd KcJc) (QdJd KcJc)
(JcQd KcJd) (JcJd KcQd) (KdQd KcJd) (KdJd KcQd) (QdJd KcKd)
(JcQd QcKd) (JcJd QcKd) (KdQd QcJc) (KdJd QcJc) (QdJd QcJc)
(JcQd QcJd) (JcJd QcQd) (KdQd QcJd) (KdJd QcQd) (QdJd QcKd)
(JcQd KdJd) (JcJd KdQd) (KdQd JcJd) (KdJd JcQd) (QdJd JcKd)
NOTE: Every 'flop' from the first scenario is represented in each of first flops 6 times, and every one of the second flops 6 times.
This means that the EV is the same for the first flop as it is for the second flop. - This will be the same in poker scenarios.
But - Now, you win both hands 54 times out of 90 (60%), and split 36 times out of 90 (40%)
There is no scenario where your friend will win both hands - there is only 3 red cards and they would need 2 in both flops to win the whole pot.
So, your EV remains the same, but by running it twice you have taken all the risk of losing all of your money away.
I told you your friend was stupid.
It gets more complicated in poker because there are more cards and types of hands, but the same concepts apply - EV remains the same, but variance changes the more times you run the cards.
(I'll read the rest of your post when I get home from work tonight)
/g2
The EV before the first flop is delt for the overall scenario is a constant.
The EV after the first flop when you don't reshuffle changes because you have more information (a card that can't occur again).
Is that EV independent of the number of times you're going to run it?
/g2
BY definition, yes. EV is the 'expected value' if you ran the trial an infinite number of times.
I meant "run it once", "run it twice", ... "run it k times"
I.e. Is the EV exactly the same when you run it once, as when you run it twice... etc.
I understand that EV is the expected value if you were in the same situation an infinite # of times.
Does anyone understand my question yet? LOL... I almost don't understand it anymore.
/g2
IF YOU DON'T RESHUFFLE, THEN IT IS NOT.
If you want more, go back to highschool and chat up your math teachers. Or maybe go beat them up because you got gypped on your education.
/g2
1) I don't mind taking big inevitable beats becuz it was a big hand, I call the inevitable ones good cuz we both played our hands perfect and Poker is entertaining and what goes around comes around in poker.
2) I ran it twice cuz I brought 1 buyin and I decided to protect myself cuz even if we chopped I'd still win the initial raisers chopped raise money
3) my buddy on the button (the inital raiser) isn't really that good of a player, so if he misses the flop I can bet him out very easily or if he hits his TP (hopefully AK $$$) I'd have all of his money
and yes I'm stupid as usual we did reshuffle the FT&R + the burns so it probably would've ended up a chop had we not reshuffled, but hey shit happens.
what language/system are you using for the simulator? if it is 360 Assembler the i am all over it! ok, the reason i ask is, pokercalculator does most of what you want. it is free, written in java and comes with a bunch of 'class'es. i'm not familiar with java, just wondering if the classes could be reused for the bulk of your simulator.
/g2
No longer true, but that's a topic for another thread
/g2
"read a card" "punch a card" "print a line"? Is there still a market for my programming skills?
The results I got show that no one should ever run the client JVM when given the choice. (Everyone has the choice. To run the server VM, see instructions in the Using the Server JVM section below.)
That's all I needed to know. Thanks beanie. Now back to the thread.... umm yeah... losing the whole pot with AA vs QQ running it twice that sucks.
/g2
I am trying to prove the EV concept to you without the numbers (which you may or may not believe anyways).
Let's say that you and your friend have AA and QQ, and instead of dealing the flop off the top of the deck, you discard the first 8 cards and then do the burn, lay out a flop, burn, turn, burn river. Would you expect the EV to change because you got rid of the first 8 cards from the deck before doing the flop/turn/river?
Or, would you expect it to change if you dealt the flop, turn, and river off the bottom of the deck?
Each time you would start with a totally random deck.
I think where the confusion lies is in the fact that if you already knew the result of running it once and THEN you decided to run it again there would be a difference in your EV for the second run (to make up for the fact that you now know whether you were a winner or a loser after the first FTR), but as Beanie already stated you usually would set the number of times that you are going to run the cards before any of the cards are revealed.
The EV would be the same as long as the deck started with the same cards in it each time, regardless of where the flop/turn/river came from.
By discarding the top 8 cards and then doing the flop, turn, and river you have effectively created what the second Flop/Turn/River would be if you ran it twice. As we stated before, this second flop/turn/river creates the same EV as the frist flop/turn/river. Combining the 2 EVs would be an (x + x)/2 = x scenario.
But, as I stated earlier, your variability fluctuates much less when you run it x times.
Al, thanks for all your hard work trying to convince me
/g2
Let's say that just before turning over the first card, you change your mind and propose that we run it six times instead of two. Since I know that all six unknown boards have the same EV as each other and the overall EV will not be affected, I agree and we split the pot into six for each of the runs, i.e., $200 / 6 = $33.33. So before seeing the cards, whether we agree to use 8, 16 or all of the cards I have laid out, the EV is the same at that point. Only our variance/risk is reduced.
I'll try not to sleep so much in stats class from now on.
/g2 <
DONK!!!!!!!!!!!!!