Hi there, small marital tiff going here, and we could use some help! :D

I am looking for a formula that will determine the likelihood of any
number being drawn for a 6/49 lotto, over X number of drawings. To
say, there have been 1721 lotto drawings for my state's 6/49 game. I
would like to know how many times my favorite number (15) should have
been drawn by now.

My idea was that there have been 1721 games, and 6 numbers are drawn
each time, totalling 10326 drawn numbers. Divided by the 49 numbers,
each should have come out around 210 times. My formula is (X*6)/49.

My DH says that the odds aren't that even <grin>. That as each number
is drawn, the odds of a remaining number being drawn increases by 1,
and this is carried through all the games. By his thinking, the number
15 should have come out around 222 times. I'm not sure about the
formula, so here's a working example:

First draw	1721/49 =  35.12244898
Second draw	1721/48 =  35.85416667
Third draw	1721/47 =  36.61702128
Fourth draw	1721/46 =  37.41304348
Fifth draw	1721/45 =  38.24444444
Sixth draw	1721/44 =  39.11363636
                  Sum   = 222.3647612

Please help us out with the correct formula and a working example
using the 1721 drawings for the 6/49 game.

Thanks!

Dear looking4lotto,

Your formula is correct, although the reasoning you give does not
quite justify it. The 10326 individual number drawings that have taken
place over 1721 games are not independent events. The probability that
the very first of the 10326 values will be a 15 is 1/49, since there
are 49 values from which to choose. If the first value turns out not
to be a 15, then the odds that the second value will be a 15 are 1/48.
But if the first value is indeed a 15, the probability for the second
value is now 0. In general, then, we may not take the number of
drawings, divide by the total number of available values, and say that
this is the expected number of occurrences of any one value.

Such a calculation works only in cases where the total number of
values is divisible by 6, since we dealing with a sequence of lottery
drawings in which the probabilities are reset after every 6 values.
Within these intervals, the reasoning I outlined in my Comment below
is the correct one. To recapitulate, the odds that a given value will
appear in one lottery game are

     48*47*46*45*44 / (5*4*3*2)        6
  -------------------------------  =  -- .
  49*48*47*46*45*44 / (6*5*4*3*2)     49

In other words, a given value is expected to occur 6/49 times in one
lottery game. In two games, it is expected to occur 2*6/49 times. In
49 games, it is expected to occur 49*6/49 = 6 times, and in 1721
games, the expected number of occurrences is

  1721.0*6/49 = 210.73469387755102 .

As for the "working example" given by your Designated Hitter, it seems
to be based on the premise that the value 15 can never be drawn, hence
the number of available values should be strictly decremented with
each draw. This is a mistaken premise, for it fails to take into
account the probability that a 15 can indeed be drawn.

I earnestly hope that I have made some small contribution to the
peaceful resolution of your tiff.

If you feel that my answer is incomplete or inaccurate in any way,
please post a Clarification Request so that I have a chance to meet
your needs before you assign a rating.

Regards,

leapinglizard

Fabulous! And not just because I'm right, because I'm not totally
right...right answer, faulty logic. Thanks for taking the time to
explain it thoroughly and provide an example.

The number of ways to select 6 numbers among 49 is written C(49, 6),
the value of which is

  49*48*47*46*45*44 / (6*5*4*3*2).

If we fix one choice among the 6 numbers at, say, your favorite number
15, then the number of ways to select the other 5 numbers among the
remaining possible 48 is C(48, 5), or

  48*47*46*45*44 / (5*4*3*2).

You can plainly see that C(48, 5) divided by C(49, 6) is just 6/49. In
other words, the probability that one drawing of the 6/49 lottery will
contain the number 15 is 6/49.

Another way to put this is that the average, or expected, number of
times the number 15 will appear in one lottery drawing is 6/49. To
calculate the expected number of times that 15 will appear in X
drawings, we multiply 6/49 by X. Thus, you can tell your DH -- what
exactly is that? your designated hitter? -- that your formula is
absolutely correct.

leapinglizard

More could be said about why your husband's reasoning is incorrect,
with reference to the Poisson distribution and other such statistical
concepts, but perhaps another Researcher would like to attend to this.

leapinglizard

Another way to see it is to calculate the probability that none of the
numbers are 15.

The probablility that the first number is not 15 is 48/49, because 48
of the 49 choices are not 15.
The probablitity that the second number is not 15, given that the
first number was not 15, is 47/48.

And so on.

So, the probability that none of the six numbers is 15 is:

48/49 * 47/48 * 46/47 * 45/46 * 44/45 * 43/44

The 45's, 46's, 47's, and 48's cancel, leaving 43/49.

So the probability that at least one of the numbers is 15 is 1 - 43/49, or 6/49.

The 44's cancel too...

Annnd, my DH = My Dear Husband...which he is, though I was typing a
little tongue in cheek at the time :)

It seemed like such a simple thing at first. We were gabbing about
favorite numbers and I said that my favorites never come out...which
he said couldn't be true, which led to finding the state lottery
archives and tossing the numbers into a little database and tallying.
I sorted the numbers by occurrence, decreasing.  There was my 15,
sitting near the middle of the list at 212. Based on the curve, it
looked about average, but I wanted to know ~how~ close to average it
was...

Yeah, I know, we need to get out more, but it was a rainy, blustery day. :D

Thanks again, LL, for the examples and all the extra thoughts and
ideas about playing with the numbers!

And it will be a most peaceful resolution: neither of us was right,
and I wasn't fully wrong!

Not at all small, your contribution...