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Dave H.
11-23-2004, 03:11 PM
Two different sites and I quote:

[ QUOTE ]
N.B. If you take two sutied cards to the river, you have a 15/1 (6.4%) chance of making a flush in your suit by then.

[/ QUOTE ]

[ QUOTE ]
Finally for all the possibilities if you start suited and stay to see all seven cards (your two and the five board cards) the probability that you will make a flush is 5.77%. The odds against you are 16.3:1

[/ QUOTE ]

Which of the above is correct, please?

Also, I thought I read (and I searched for this a long time) that there was something like a 60% probability that the board could be exactly 2 suited. Can that be??

Thank you!

Lost Wages
11-23-2004, 04:25 PM
Question 1)
If you start with 2 suited cards then you will make a flush (or straight flush):

C(11,3)*C(47,2)/C(50,5) = 8.42% = 10.9:1 against

However, that includes the times that the board contains 4 or 5 of your suit which you generally would not prefer.

The probability of making exactly a 5 card flush (or straight flush) is:

C(11,3)*C(39,2)/C(50,5) = 5.77% = 16.3:1 against

Question 2)
Not sure what you mean by "exactly 2 suited". Can you give an example?

Lost Wages

Stork
11-23-2004, 09:46 PM
I'm pretty sure the second one is correct. I just remember hearing that you make the flush slightly less than 6% of the time.

gaming_mouse
11-24-2004, 09:42 AM
5.77% is the chance of having a board with EXACTLY 3 of your suit.

6.4% is the chance of EXACTLY 3 OR EXACTLY 4.

Lost wages gives the correct calculation for having a flush made any way at all (including a str8 flush).

gm

EDIT: Actually, LostWages calculation is not quite right for making a flush any way at all. It counts certain hands multiple times. The correct answer is 6.4%, because having a board of 5 suited cards does not even affect the calculation by as much as 1 digit.

gaming_mouse
11-24-2004, 10:03 AM
Dave,

In case you want to see the math, here it is:

ncr(11,3)*ncr(39,2)/ncr(50,5)=.0577 (EXACTLY 3)
ncr(11,4)*ncr(39,1)/ncr(50,5)=.0060 (EXACTLY 4)
ncr(11,5)/ncr(50,5)=.0002 (EXACTLY 5)

gm

Lost Wages
11-24-2004, 10:30 AM
Good catch, I fudged it.

Lost Wages

Dave H.
11-24-2004, 12:03 PM

By 2 suited, I meant two and ONLY two suits on the board, i.e. 2 diamonds and 3 spades, 4 hearts and 1 club, etc.

Dave H.
11-24-2004, 12:05 PM
Yes, I absolutely wanted to see the math. Thank you very much!

gaming_mouse
11-24-2004, 02:29 PM
[ QUOTE ]
By 2 suited, I meant two and ONLY two suits on the board, i.e. 2 diamonds and 3 spades, 4 hearts and 1 club, etc.

[/ QUOTE ]

ncr(4,2)* (ncr(13,1)*ncr(13,4)*2 + ncr(13,2)*ncr(13,3)*2)/ncr(52,5)

This works out to 14.5%. So yes, the 60% is way off.

gm

Lost Wages
11-24-2004, 02:53 PM
GM has it right. Link (http://www.math.sfu.ca/~alspach/comp3.pdf) to the solution for every possible suit distribution.

Lost Wages

semipro
11-24-2004, 04:32 PM
Pardon my ignorance,
What is ncr? And please explain the math.

Thank you.

gaming_mouse
11-24-2004, 06:09 PM
nCr(n,r) is the number of ways you can choose r things from a total of n things.

Thus, for example:

nCr(2,2) = 1
nCr(3,1) = 3
nCr(4,2) = 6

The formula is given by:

nCr = n!/( (n-r)! * r!)

where n! = n*(n-1)*(n-2)*...*3*2*1

If you are not familiar with these concepts, I won't be able to explain the math to you.

gm

mannika
11-24-2004, 10:20 PM
The 60% probability probably refers to the probability of the flop being two-suited, not the entire board.

Probability of flop being two suited = 1 - prob(rainbow) - prob(single suit)

prob(rainbow) = (52/52)*(39/51)*(26/50) = 0.3976
prob(single suit) = (52/52)*(12/51)*(11/50) = 0.0518

prob(two-suited) = 1 - 0.3976 - 0.0518 = 0.5506

Which is fairly close to your 60% figure. Not sure if this is what was intended when you heard that remark.

Dave H.
11-26-2004, 10:01 AM
That may very well be what I read...thanx much!

Dave H.
11-26-2004, 10:38 AM
[ QUOTE ]
ncr(4,2)* (ncr(13,1)*ncr(13,4)*2 + ncr(13,2)*ncr(13,3)*2)/ncr(52,5)

This works out to 14.5%. So yes, the 60% is way off.

[/ QUOTE ]

gm,

I'm struggling with this...can you help?

1. ncr(4,2) would be the number of ways to combine 4 suits two ways...CORRECT?

2. ncr(13,1)*ncr(13,4) refers to the number of ways to get 1 card of one suit and 4 cards of another suit...CORRECT?

3. ncr(13,2)*ncr(13,3) refers to the number of ways to get 2 cards of one suit and 3 cards of another suit...CORRECT?

4. the ncr(52,5) term is obvious and the '+' sign is obvious

5. What is the *2 factor that's used with both terms in the parentheses?

Thank you.

gaming_mouse
11-26-2004, 01:26 PM
ncr(4,2) would be the number of ways to combine 4 suits two ways...CORRECT?

Correct.

ncr(13,1)*ncr(13,4) refers to the number of ways to get 1 card of one suit and 4 cards of another suit...CORRECT?

Correct.

ncr(13,2)*ncr(13,3) refers to the number of ways to get 2 cards of one suit and 3 cards of another suit...CORRECT?

Correct.

What is the *2 factor that's used with both terms in the parentheses?

Take two particular suits, hearts and clubs. We can have:

1 heart, 4 clubs
2 heart, 3 clubs
3 heart, 2 clubs
4 heart, 1 clubs

But the # of combos for "1 heart, 4 clubs" = # of combos for "4 clubs, 1 heart". Similarly for 2,3. So we just multiply each by 2.

gm