Questions re: Probability of Higher Flush by WoO
August 25th, 2010 at 5:17:20 AM
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Hi All!
New here, so please forgive any forum faux-pas...
I have a couple of questions about the following:
https://wizardofodds.com/ask-the-wizard/holdem-probabilities/
1. About halfway down the page, Mr. Shackleford responds to a question from Bob B from Scottsdale re: the
probabilities of his flush being beaten by a higher flush versus n opponents.
I would be very grateful if someone could please expand on the methodology used to generate these percentages, as I am getting the same results, but displaced by one player e.g. my percentage for 3 players matches Mr. Shackleford's result for 2, etc.
Am I missing something somewhere? Or is this maybe a typo in the table?
2. Mr. Shackleford goes on to explain that "...these probabilities assumed independence between hands, which is not a correct assumption, but the results should be a close estimate", which I assume to mean that the effect of card-removal on the percentages for player 2,3....n are simulated.
How accurate is this simulation method, and can it be applied generically to other scenarios re: percentages
chances of multiple opponents beating ones hand?
Or does anyone know of a method to evaluate/simulate this effect?
Many, many thanks for your time!
Cheers!
p
New here, so please forgive any forum faux-pas...
I have a couple of questions about the following:
https://wizardofodds.com/ask-the-wizard/holdem-probabilities/
1. About halfway down the page, Mr. Shackleford responds to a question from Bob B from Scottsdale re: the
probabilities of his flush being beaten by a higher flush versus n opponents.
I would be very grateful if someone could please expand on the methodology used to generate these percentages, as I am getting the same results, but displaced by one player e.g. my percentage for 3 players matches Mr. Shackleford's result for 2, etc.
Am I missing something somewhere? Or is this maybe a typo in the table?
2. Mr. Shackleford goes on to explain that "...these probabilities assumed independence between hands, which is not a correct assumption, but the results should be a close estimate", which I assume to mean that the effect of card-removal on the percentages for player 2,3....n are simulated.
How accurate is this simulation method, and can it be applied generically to other scenarios re: percentages
chances of multiple opponents beating ones hand?
Or does anyone know of a method to evaluate/simulate this effect?
Many, many thanks for your time!
Cheers!
p
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 25th, 2010 at 8:10:14 AM
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Quote: podski...probabilities of his flush being beaten by a higher flush versus n opponents.
... my percentage for 3 players matches Mr. Shackleford's result for 2, etc.
Boldface type is my emphasis.
I'm not going to attempt the math, but could it be that the Wizard's answer is for number of "opponents" while yours is for number of "players"?
August 25th, 2010 at 4:57:02 PM
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Hi Doc,
Many thanks for the response. Apologies for the ambiguity; I was trying to condense Bob B's question, and introduced the word 'opponents" without thinking.
From The Wizard's reponse to Bob B, "The following table shows the probability for 1 to 8 higher ranks and 2 to 10 players, including yourself", and the table is entitled "Probability of Higher Flush: Higher Ranks (down) by Total Players (across)"
For the combination in question, The Wizard's numbers are:-
Higher Ranks = 4
2 Players 4.03%
3 Players 5.98%
4 ...etc.
Mine are:-
Higher Ranks = 4
2 Players 2.04%
3 Players 4.03%
4 Players 5.99%...etc.
Any clues?
Many thanks for the response. Apologies for the ambiguity; I was trying to condense Bob B's question, and introduced the word 'opponents" without thinking.
From The Wizard's reponse to Bob B, "The following table shows the probability for 1 to 8 higher ranks and 2 to 10 players, including yourself", and the table is entitled "Probability of Higher Flush: Higher Ranks (down) by Total Players (across)"
For the combination in question, The Wizard's numbers are:-
Higher Ranks = 4
2 Players 4.03%
3 Players 5.98%
4 ...etc.
Mine are:-
Higher Ranks = 4
2 Players 2.04%
3 Players 4.03%
4 Players 5.99%...etc.
Any clues?
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 25th, 2010 at 7:01:44 PM
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Sorry. Doubt I could help. If I tried to do the math, I would likely screw it up far worse than you may have, if you did at all. I just thought I would try to point out a "Doh!" if that was what happened.
August 26th, 2010 at 12:24:45 AM
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:)
Well, thanks anyway, Doc - I appreciate you taking the time. It's alway useful to have the benefit of another person's viewpoint, oui?
Cheers!
p
Well, thanks anyway, Doc - I appreciate you taking the time. It's alway useful to have the benefit of another person's viewpoint, oui?
Cheers!
p
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 26th, 2010 at 2:38:47 AM
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Ok here is my attempt of only against one other player, subject to brainos and/or typos. There are 45 unknown cards. 52 (total cards) - 5 (board cards) - 2 (your hole cards)= 45. This means there are combin(45,2)= 990 posible hands for the other player. There are 8 flush cards left. 13 (total cards) - 3 (board cards) - 2 (your hole cards)=8. There are combin(8,2)=28 ways for the other player to have 2 flush cards. If there are 7 or 8 flush cards higher than yours, then all 28 ways will be better than yours, for a total probabilty of 28/990 or 2.828 %. If there are 6 flush cards higher than yours, then 27 ways will be better than yours. The only way theirs would be lower is if they are both lower. For 5 flush cards higher than yours there are 3 lower. combin(3,2)=3 . 28-3=25. 4 Flush cards higher: combin(4,2)=6 28-6==22. .... For 1 card higher than yours: combin(7,2)=21. 28-21=7. Here is a summery:
1 = 0.007070707070707
2 = 0.013131313131313
3 = 0.018181818181818
4 = 0.022222222222222
5 = 0.025252525252525
6 = 0.027272727272727
7 = 0.028282828282828
8 = 0.028282828282828
1 = 0.007070707070707
2 = 0.013131313131313
3 = 0.018181818181818
4 = 0.022222222222222
5 = 0.025252525252525
6 = 0.027272727272727
7 = 0.028282828282828
8 = 0.028282828282828
“Man Babes” #AxelFabulous
August 26th, 2010 at 3:39:57 AM
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miplet, lovely miplet....
I am in complete agreement with your numbers, although I arrive via (three slightly) different methods.
So...either we're both making the same error(s), or [gulp] The Wiz's table is wrong. Can it be true?
Thank you so much for your time on this. Verily, 'tis appreciated.
I am in complete agreement with your numbers, although I arrive via (three slightly) different methods.
So...either we're both making the same error(s), or [gulp] The Wiz's table is wrong. Can it be true?
Thank you so much for your time on this. Verily, 'tis appreciated.
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 26th, 2010 at 7:12:16 AM
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I did a quick sim. Here are the results. 1-8 overcards, 2-10 total players:
| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
|---|---|---|---|---|---|---|---|---|---|
| 1 | 0.69 | 1.41 | 2.12 | 2.83 | 3.54 | 4.25 | 4.96 | 5.66 | 6.36 |
| 2 | 1.28 | 2.6 | 3.91 | 5.21 | 6.48 | 7.77 | 9.05 | 10.31 | 11.56 |
| 3 | 1.79 | 3.61 | 5.41 | 7.19 | 8.92 | 10.68 | 12.41 | 14.11 | 15.79 |
| 4 | 2.19 | 4.4 | 6.58 | 8.75 | 10.85 | 12.94 | 15.01 | 17.04 | 19.04 |
| 5 | 2.5 | 5 | 7.47 | 9.92 | 12.28 | 14.63 | 16.94 | 19.21 | 21.43 |
| 6 | 2.71 | 5.4 | 8.06 | 10.69 | 13.23 | 15.74 | 18.23 | 20.65 | 23.03 |
| 7 | 2.81 | 5.61 | 8.36 | 11.08 | 13.7 | 16.3 | 18.86 | 21.37 | 23.82 |
| 8 | 2.81 | 5.61 | 8.36 | 11.08 | 13.7 | 16.3 | 18.86 | 21.37 | 23.82 |
“Man Babes” #AxelFabulous
August 26th, 2010 at 7:27:47 AM
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'tastic!
Are you willing/able to explain how you generated these numbers?
Thanks for all your help, miplet.
Cheers!
p
Are you willing/able to explain how you generated these numbers?
Thanks for all your help, miplet.
Cheers!
p
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 26th, 2010 at 7:50:38 AM
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Quote: podski'tastic!
Are you willing/able to explain how you generated these numbers?
Thanks for all your help, miplet.
Cheers!
p
I took the 45 remaning cards and numbered them 1 to 45. Cards 1 to 8 represent the 8 cards that will make the flush. Randomly give 2 cards to each of the other 9 players. If they are both less than 9, and atleast 1 is equal or less than the number of over cards, that player would have a higher flush.
“Man Babes” #AxelFabulous
August 27th, 2010 at 4:22:49 AM
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luvvly jubbly.
Even allowing for a bit of divergence as x and y increase, my figures are never more than 0.8% out from yours.
So..back to my original question 1:- What is Mr Wizard doing to get his figures?
Oddly ("ODDly" haha!), if I regenerate my numbers post-flop (i.e. divide by combin(47,2)= 1081), I get identical results to the Wizard, but (as before) displaced by one column.
(If only I knew how to paste tables into this thread....)
Even allowing for a bit of divergence as x and y increase, my figures are never more than 0.8% out from yours.
So..back to my original question 1:- What is Mr Wizard doing to get his figures?
Oddly ("ODDly" haha!), if I regenerate my numbers post-flop (i.e. divide by combin(47,2)= 1081), I get identical results to the Wizard, but (as before) displaced by one column.
(If only I knew how to paste tables into this thread....)
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 27th, 2010 at 4:31:05 AM
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And the other question, of course, is "is this method generic enough to be applied across other scenarios?"
e.g. if i can calculate the heads-up (i.e. 1 opponent or 2 total players) probability of the other player having a better hand, can I then infer the probabilities for 2, 3....n other players?
Muchas Gracias for the input on this miplet!
e.g. if i can calculate the heads-up (i.e. 1 opponent or 2 total players) probability of the other player having a better hand, can I then infer the probabilities for 2, 3....n other players?
Muchas Gracias for the input on this miplet!
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 27th, 2010 at 10:05:05 AM
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Sorry for the late arrival.
To answer the question at hand, I don't remember how I came up with that table. My attempts to recreate those numbers have failed. So I think I made a mistake somewhere.
I just updated my table with the way I should have done it. If c is the number of overcards and p is the number of total players, including yourself, then the answer should been:
1-(1-(COMBIN(8,2)-if(c<=6:combin(8-c,2),0))/COMBIN(45,2))^(n-1)
As I noted, this is just an estimate. I would take Miplet's simulation results over my estimate.
Speaking of which, I ammended my answer to include your table Miplet. I hope that is okay. If you prefer another attribution other than "miplet," please let me know.
To answer the question at hand, I don't remember how I came up with that table. My attempts to recreate those numbers have failed. So I think I made a mistake somewhere.
I just updated my table with the way I should have done it. If c is the number of overcards and p is the number of total players, including yourself, then the answer should been:
1-(1-(COMBIN(8,2)-if(c<=6:combin(8-c,2),0))/COMBIN(45,2))^(n-1)
As I noted, this is just an estimate. I would take Miplet's simulation results over my estimate.
Speaking of which, I ammended my answer to include your table Miplet. I hope that is okay. If you prefer another attribution other than "miplet," please let me know.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 27th, 2010 at 4:58:29 PM
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Quote: WizardSorry for the late arrival.
To answer the question at hand, I don't remember how I came up with that table. My attempts to recreate those numbers have failed. So I think I made a mistake somewhere.
I just updated my table with the way I should have done it. If c is the number of overcards and p is the number of total players, including yourself, then the answer should been:
1-(1-(combin(8,2)-if(c<=6:combin(8-c,2),0)))/combin(45,2))^(n-1)
As I noted, this is just an estimate. I would take Miplet's simulation results over my estimate.
Speaking of which, I ammended my answer to include your table Miplet. I hope that is okay. If you prefer another attribution other than "miplet," please let me know.
Using my table is fine with me. "miplet" is prefered. My sim was a quick million hands using php's shuffle function, so I don't realy know how random it is. (I have 2 c++ books sitting around that I should read.)
“Man Babes” #AxelFabulous
August 27th, 2010 at 5:23:15 PM
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Ah, Mr. Bond; I've been expecting you....
:)
Thanks very much for the input, Wizard. Most helpful.
(BTW - you also need to edit the text that goes with the table, as "In the case of your example of 4 higher ranks and 9 total players, the probability is 16.89%" still refers to the old calculation. The value in the updated table is 16.45%.)
Cheers!
p
:)
Thanks very much for the input, Wizard. Most helpful.
(BTW - you also need to edit the text that goes with the table, as "In the case of your example of 4 higher ranks and 9 total players, the probability is 16.89%" still refers to the old calculation. The value in the updated table is 16.45%.)
Cheers!
p
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 28th, 2010 at 4:08:03 AM
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Quote: Wizard
If c is the number of overcards and p is the number of total players, including yourself, then the answer should been:
1-(1-(combin(8,2)-if(c<=6:combin(8-c,2),0)))/combin(45,2))^(n-1)
Hi Wiz!
Apologies if I seem to be turning into your NetNemesis, but that formula isn't right:-
* shouldn't it be combin(45,2))^(p-1), if p is the number of players? and
* you've got one too many/too few brackets
Cheers!
p-1 (hee hee! the leetle joke I am makings!)
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 28th, 2010 at 8:14:31 AM
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Quote: mipletUsing my table is fine with me. "miplet" is prefered. My sim was a quick million hands using php's shuffle function, so I don't realy know how random it is. (I have 2 c++ books sitting around that I should read.)
I'm not sure how good it is either, but in general don't trust off the shelf RNG's very much. I highly recommend a compiled langauge for big simulations, and a Mersenne Twister for an RNG. Once you know one language picking up others is easy. It also looks great on a resume to list as many computer languages and skills as possible.
Anyway, thanks for the use of the table. If you ever redo it, let me know and I'll update the table.
Quote: podskiHi Wiz!
* shouldn't it be combin(45,2))^(p-1), if p is the number of players?
Dang, I knew I should have double checked that. Thanks also for the correction on the example.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 28th, 2010 at 10:56:49 AM
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Quote: WizardDang, I knew I should have double checked that. Thanks also for the correction on the example.
Really, don't mention it, Sir.
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 28th, 2010 at 11:26:13 AM
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So...Wizard 'n' miplet - any thoughts on the other bit:
Thanling You most politely,
p
Quote: podskiis there a method generic enough to be applied across all other scenarios?
e.g. if i can calculate the heads-up (i.e. 1 opponent or 2 total players) probability of the other player having a better hand than me given my pocket cards and the Flop/Turn/River, can I then infer/simulate the probabilities for 2, 3....n other players?
Thanling You most politely,
p
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 28th, 2010 at 11:27:30 AM
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Quote: podskiThanling You most politely,
as in "Thanking You", obviously
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 28th, 2010 at 12:36:22 PM
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Quote: podskiSo...Wizard 'n' miplet - any thoughts on the other bit...
If we incorrectly assume that each player is dealt cards from a separate deck, then if the probability that one player beats you is p, then the probability that at least one player out of n will is 1-(1-p)^n. My table is based on that assumption.
For example, the probability of rolling at least one six in 10 rolls of a die is 1-(1-(1/6))^10 = 1-(5/6)^10=0.838494417.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 28th, 2010 at 1:43:07 PM
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Quote: Wizard...the probability that at least one player out of n will [beat you] is 1-(1-p)^n. My table is based on that assumption.
So, I believe, is mine.
Do I take it then that factoring in the effect of card removal on the chances each opponent after the heads-up calculation is too onerous?
Or is it just that it makes bugger-all difference?
Sorry for being a pest; my thirst for knowledge gets the better of me sometimes.
Cheers!
p
We Have Assumed Control, We Have Assumed Control, We Have Assumed Control
August 28th, 2010 at 2:54:42 PM
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Quote: podski
Do I take it then that factoring in the effect of card removal on the chances each opponent after the heads-up calculation is too onerous?
Yes. If you wish to factor that in, then a simulation is the easiest way to go.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
