> 4x48x47x46x45 divided by 24 (which is 4 factorial, 4x3x2x1)
=778,320.
> As long as I did the math right this is how many combinations
that
> contain a single deuce out of the 2,598,960 combinations that one
can
> be dealt on the initial deal. By dividing 2,598,960 by 778,320
the
> result is seeing a single deuce on the flop an average of every
3.4
> deals.
Let's push it even further. How often would we see a hand that
contains two deuces on the flop?
Six combinations make a pair of deuces. Any one of four times any
one
of three divided by 2 factorial (4x3 divided by 2x1) = 6.
So the equation is 6x48x47x46 divided by 3 factorial(3x2x1) =
103,776
combinations. 2,598,960 divided by 103,776 = 25.04. So we'll see
two
deuces on the flop about every 25 games.
How often will we see three deuces on the flop? There are four
combinations that make three deuces. Any one of four times any one
of
three times any one of two divided by 3 factorial (4x3x2 divided by
3x2x1) = 4
So the equation is 4x48x47 divided by 2 factorial (2x1) = 4224
combinations. 2,598,960 divided by 4,224 = 615.28. So we'll see
three
deuces on the flop about every 615 games.
How often will we see four deuces on the flop? One combination
makes 4
deuces. Any one of four times any one of three times any one of
two
times any one of one. So the equation is 1x48. 2,598,960 divided
by
48 = 54,145. We'll be dealt four deuces every 54,145 games.
So now let's add some combination up-
Combinations containing one deuce, 778,320, plus combinations
containing two deuces, 103,776 = 882,096. 2,598,960 divided by
882,096 = 2.95. So we're gonna see a hand containing either one or
two
deuces about every three hands we're dealt.
Adding in the rest of the combinations has a negligible effect.
We'll
see a hand containing one or more deuces every 2.93
deals.
If one were seeing deuces at twice the rate of the true odds they
would
have to see one deuce or more in two out of every three hands
dealt.
Thanks Brian, for the confirmation and thanks, Harry. Now, I must
admit a little boo boo. It's statistically insignificant overall but
a mistake nonetheless. After Harry gave me the clue about non-deuce
hands I went straight to the calculator and punched in 48x47x46x45x44
divided by 120 (5 factorial) and came up with 1,712,304 combinations
that do not contain a deuce. I then subtracted that number from
2,598,960 for 886,656 combinations that contain one or more deuces.
So then I went to my work above and totaled the combinations. It
came to just 886,368. I knew I had made a mistake somewhere. So I
found it in the hands that contain three deuces. I got the equation
right 4x48x47 but punched it into the calculator as 4x48x44. So I
got the number of combinations wrong. It should be 4512 instead of
4224, a difference of 288 combinations and the chances of getting
three deuces on the flop is 576 instead of 615.
So I then added 288 more combinations to 886,368 for a total of
886,656. Adding this number to 1,712,304 hands that don't contain a
deuce totals to 2,598,960. So now we're johnny on the spot.
Incidentally, Harry, I'm living proof that 8th grade math works.
That was the last time I took a math class. Good luck.
···
--- In vpFREE@yahoogroups.com, "mickeycrimm" <mickeycrimm@...> wrote:
--- In vpFREE@yahoogroups.com, "mickeycrimm" <mickeycrimm@> wrote:
The