four to a.....

lunapark · #1 01-22-2007, 12:59 AM

Hi...What are the odds that the board (after the river) would have four cards to a flush?

How about four cards to a straight (they dont have to be consecutive)?

It just seems that this occurance happens a lot more than I thought it would and was curious as to the real percentage.

jay_shark · #2 01-22-2007, 02:04 PM

Suppose we wish to know the number of boards that contain 4 clubs given 5 board cards . Once we've computed the total for clubs , we simply do the same for the other suits .

13C4*39C1/52C5=0.01 or 1% .

Since there are three other suits , we simply add 1% three more times to give you 4 % .

4 cards that fit into a straight that don't have to be consecutive :

case 1 (4 cards from a-5)

a234,a235,a245,a345,2345 and the fifth card is an x which may or may not belong to {a,2,3,4,5} but it cannot complete a straight . For instance , for {a234}, x=/ 5 or 6 since we would have a higher straight for x=6 and we don't want to overcount .

a234x : 4^4*28 + 4c2*4^3*4=8704
Now we do the same for all the other cases from using 4 cards from a-5.

8704*5 = 43520 (we have mutually exclusive events )

But this is also true if we pick 4 numbers from 2-6, 3-7, ...10-A

Simply take 43520*10=435200 .

Now we divide this number by 52C5 to give you 16.7%

jay_shark · #3 01-22-2007, 02:31 PM

One minor correction .

It is 43520*9 since we're summing from a-5, 2-6 , ...9-k and then we work out the last case from 10-A seperately .

5*[4^4*32 +4C2*4^3*4] = 5*9728

43520*9 +48640 = 440,320

The answer is 16.94%

lunapark · #4 01-22-2007, 04:42 PM

Thank you very much....Im suprised the number is that high for four to a straight.

So as a follow up how about the odds of either four to a straight or four to a flush.

I appreciate your help.

jay_shark · #5 01-22-2007, 06:18 PM

We add them and then we subtract the intersection for when you have 4 to a straight and with the cards being from the same suit .

a234x suppose a234 are all of one suit (4 suits) x=(7,8...k)

5C4*4*28 =560 .

Now we multiply this number by 5 since we are choosing 4 numbers from the set {a2345}

560*5=2800 .

2800*9 =25200 since we sum from a-5 , 2-6 , ...9-k .

The last case is when we choose 4 numbers from 10-A . X is any card other than the card that completes the straight .

5C4*4*32*5 = 3200

28900+3200= 32100

The probability is 32100/52C5= 0.0123

4%+16.94% - 1.23% = 19.71% or about 1 in 5 .