Questions re: Probability of Higher Flush by WoO

Hi All!

New here, so please forgive any forum faux-pas...

I have a couple of questions about the following:

http://wizardofodds.com/askthewizard/holdem-probabilities.html

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

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"?

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?

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.

:)

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

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

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 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

'tastic!

Are you willing/able to explain how you generated these numbers?

Thanks for all your help, miplet.

Cheers!
p

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.