run it 'n' times - calling all math geeks
ok, i just love beating a dead horse...
you have QQ, villain has AA
flop comes Q75 rainbow
run it once, you are approx 91.5% fav. in other words you have an 8.5% chance of going broke. run it more than once, people claim you still have an EV of 91.5% but your risk of going broke drops exponentially. we've had a lot of debate about that. i'm getting close to believing the EV stays the same (give or take...) but I believe your risk of going broke drops faster than they claim.
you can run this 11 times if you are a glutton for punishment. you will have 22 face up cards and 23 burned cards.
someone tell me the exact odds of where the Aces will end up - both burned, both face up, or one of each. i'm thinking just over 25, just under 25 and just under 50...
* if they are both burned, AA has an EV of 0%.
* if they are both face up, AA can win 2, 1 or 0 of the runs, depending on whether they both come on the same hand or in a hand with the case Q. anyone care to work out the percentages?
* if one of them is in the face up group, it can win 1 or 0 hands again depending on where the other Q ends up.
i would expect AA to win slightly less than 1 of the 11 runs over the long haul. i can't say for certain if that puts it higher or lower than 8.5% but my gut says slightly lower. can someone show me the math on that?
i say your chance of going broke is 0 if you do at least 3 runs, not 0.085^n. any arguments?
let's see if we can get 7 or 8 pages of discussion on this...
you have QQ, villain has AA
flop comes Q75 rainbow
run it once, you are approx 91.5% fav. in other words you have an 8.5% chance of going broke. run it more than once, people claim you still have an EV of 91.5% but your risk of going broke drops exponentially. we've had a lot of debate about that. i'm getting close to believing the EV stays the same (give or take...) but I believe your risk of going broke drops faster than they claim.
you can run this 11 times if you are a glutton for punishment. you will have 22 face up cards and 23 burned cards.
someone tell me the exact odds of where the Aces will end up - both burned, both face up, or one of each. i'm thinking just over 25, just under 25 and just under 50...
* if they are both burned, AA has an EV of 0%.
* if they are both face up, AA can win 2, 1 or 0 of the runs, depending on whether they both come on the same hand or in a hand with the case Q. anyone care to work out the percentages?
* if one of them is in the face up group, it can win 1 or 0 hands again depending on where the other Q ends up.
i would expect AA to win slightly less than 1 of the 11 runs over the long haul. i can't say for certain if that puts it higher or lower than 8.5% but my gut says slightly lower. can someone show me the math on that?
i say your chance of going broke is 0 if you do at least 3 runs, not 0.085^n. any arguments?
let's see if we can get 7 or 8 pages of discussion on this...
Comments
cards win %win lose %lose tie %tie EV
Qs Qc 905 91.41 85 8.59 0 0.00 0.914
Ad Ah 85 8.59 905 91.41 0 0.00 0.086
With your scenario, AA can only win a maximum of two times. I agree that QQ cannot go broke if you run it three or more times.
Edit#2 - yeah, my bad... I thought I was onto something, but no. Stupid 2, why couldn't you be a 1?!?!?
Edit #3 - Sorry for the confusion, but I ended up figuring out that the EV is the same whether you run it 1 to 5 times for the following situation. AA vs QQ on a board of Q752 rainbow. This is nothing definitive, but it has taken a bite out of my hopes of being right. If anyone cares to see my calculations I'll share my spreadsheet, it's probably a good start for calculating the EV when there are 2 cards to come.
/g2
also "* if they are both face up, AA can win 2, 1 or 0 of the runs, depending on whether they both come on the same hand or in a hand with the case Q. anyone care to work out the percentages?"
if they are both face up AA wins atleast one of the runs, because if it comes on one run turn A river Q, that implies that the other A will be somewhere and the QQ will lose that run.
Correct.
How would one be possible given the board?
/g2
right, misread the "rainbow" part
Let's start with 'running it once'....
(I will use TR to indicate Turn/River)
2 Aces come on the TR 0.10 % of the time
1 Ace comes on the TR, with no Queen 8.48 % of the time
1 Ace comes on the TR, with a Queen 0.20 % of the time
No Ace comes on the TR 91.21 % of the time
Thus, the Queens win 91.41% of the time
But, as pkrface indicated, this means that you go broke 8.59% of the time
Let's look at running the TR 11 times...
Cases where the Queens win it all
=========================
No Aces in any of the TRs 25.56%
1 Ace in a TR, but a Queen in the same TR 1.19%
Sub-Total - 26.75%
Cases where the Queens win 10 out of 11 hands
====================================
1 Ace in a TR, no Queen showing in any TR 26.15%
1 Ace in a TR, Queen showing in a different TR 23.77%
2 Aces in the same TR 1.11%
2 Aces in different TRs, Queen in one of those TR 1.03%
Sub-Total - 52.06%
Cases where the Queens win 9 out of 11 hands
===================================
2 Aces in different TRs, Queen in different TR
2 Aces in different TRs, Q in another TR 9.30%
2 Aces in different TRs, Q in burn cards 11.89%
Sub-Total - 21.19%
So, our final results are....
Win 11/11 TRs... 26.75%
Win 10/11 TRs... 52.06%
Win 9/11 TRs... 21.19%
EV = 26.75% * (11/11) + 52.06% * (10/11) + 21.19% * (9/11) = 91.41%
Thus, our EV is the same even though we have run it 11 times, but the variance has gone way down. As pkrface said, the player with Queens cannot go broke if the hand is run 3 or more times. In the case of running it 11 times the player cannot lose more than 18.2% of the pot!
Can you PM or email me the details of how you came up with all these numbers?
/g2
I'll explain it here (if possible) to convince others...
WARNING - If you don't like mathematical explinations move on to the HBK vs Cena thread in Off Topic Lounge.
Let's go through each scenario, math explinations will be under each sentence in square brackets.
Let's start with 'running it once'....
(I will use TR to indicate Turn/River)
2 Aces come on the TR 0.10 % of the time
[There are 45 Cards remaining. An ace will come as the first card 2 out of 45 times, and will come as the second card 1 out of 44 times. (2/45) * (1/44) =
0.10101%]
1 Ace comes on the TR, with no Queen 8.48 % of the time
[Once again an ace will come as the first card 2 out of 45 times, and a queen or ace will not come as the second card 42 out of 44 times. We then multiply this by 2 to figure into the equation the fact that the ace could come as the first card or the second card. (2/45) * (42/44) * 2 = 8.48485%]
1 Ace comes on the TR, with a Queen 0.20 % of the time
[Once again an ace will come as the first card 2 out of 45 times, and a queen will come as the second card 1 out of 44 times. Once again we multiply by 2 to figure in the cards coming as A then Q or Q then A. (2/45) * (1/44) * 2 = 0.20202%]
No Ace comes on the TR 91.21 % of the time
[No ace will come as the first card 43 out of 45 times, and no ace will then come as the second card 42 out of 44 times. (43/45) * (42/44) = 91.21212%]
Thus, the Queens win 91.41% of the time
[OK if you want 5 decimal places it is 91.41414%]
Cases where the Queens win it all
=========================
No Aces in any of the TRs 25.56%
[The first ace will be in the burn cards 23 out of 45 times, and if so the second ace will be in the burn cards 22 out of 44 times. (22/45) * (21/44) = 25.55556%]
1 Ace in a TR, but a Queen in the same TR 1.19%
[Let's first explain the likelihood of one Ace being in one of the turn/rivers. If we choose one of the aces, it will be in a TR 22 out of 45 times. The other ace will then be in the burn cards 23 out of 44 times. We multiply by 2 to account for the fact that the aces could be reversed in their positions. (22/45) * (23/44) * 2 = 51.11111% Once we determine this percentage, we can then say that there are 43 remaining positions that the Queen can be in (22 remaining burn cards, and 21 remaining positions in TRs). Only in one of these positions would this put the Queen in the same TR as the revealed ace. 51.11111% * (1/43) = 1.18863%]
Sub-Total - 26.75% [New value - 26.74419%]
Cases where the Queens win 10 out of 11 hands
====================================
1 Ace in a TR, no Queen showing in any TR 26.15%
[Using the 51.11111% for the likelihood of there being 1 Ace in the TRs (see above), the likelihood of the Queen being in the burn cards would then be 22/43. 51.11111% * (22/43) = 26.14987%]
1 Ace in a TR, Queen showing in a different TR 23.77%
[Once again using the same 51.11111% for 1 Ace in TRs, the likelihood of a Queen showing in a different TR from that Ace would be 20/43. 51.11111% * (20/43) = 23.77261%]
2 Aces in the same TR 1.11%
[Let us first figure out the likelihood that there will be 2 aces showing in the TRs. The first ace will show up 22/45 times, and the second ace will then show up 21/44 times. (22/45) * (21/44) = 23.33333% When both aces do show up they will be in the same TR once out of 21 times. (Imagine you planted the once ace in a fixed spot. There are 21 remaining positions for the second ace to be placed, but only one would put it in the same TR.) 23.33333% * (1/21) = 1.11111%]
2 Aces in different TRs, Queen in one of those TR 1.03%
[This one really gets complicated. We already figured out that the likelihood of there being 2 Aces in the TRs being 23.33333%. 20 out of 21 times the aces will not be in the same TR. If they are not in the TR, then there are 2 spots out of the 43 remaining locations (20 on the board, 23 in the burns) that would put the queen in the same TR as an ace. 23.33333% * (20/21) * (2/43) = 1.03359%]
Sub-Total - 52.06% [New total - 52.06718%]
Cases where the Queens win 9 out of 11 hands
===================================
2 Aces in different TRs, Q in another TR 9.30%
[Using the 23.33333% we already figured out for 2 Aces being in the TRs, and the 20 out of 21 times that those Aces will not be paired in the same TR, there are 18 locations out of the 43 remaining (20 on the board, 23 in the burns) that would put the Queen in a different TR than an ace. 23.33333% * (20/21) * (18/43) = 9.30233%]
2 Aces in different TRs, Q in burn cards 11.89%
[Using the 23.33333% we already figured out for 2 Aces being in the TRs, and the 20 out of 21 times that those Aces will not be paired in the same TR, there are 23 locations out of the 43 remaining (20 on the board, 23 in the burns) that would put the Queen in the burn cards. 23.33333% * (20/21) * (23/43) = 11.88630%]
Sub-Total - 21.19% [New total - 21.18863%]
Win 11/11 TRs... 26.74419%
Win 10/11 TRs... 52.06718%
Win 9/11 TRs... 21.18863%
EV = 26.74419% * (11/11) + 52.06718% * (10/11) + 21.1863% * (9/11) = 26.74419% + 47.3338% + 17.33425% = 91.41224
Now, before you get your panties in a bunch g2... this minute difference is due to rounding error.
I've got to get going to Bristol now. If you need any further explination I can do it tomorrow.
Allen
BMath, University of Waterloo, 1997
can we generalize for other situations that the risk of going broke drops faster than exponentially the more times you run it?
um, see above.
Yes.
I'm just watching the clock as the session fee comes due.......