I'm awful at maths, so I thought I'd post this one here to get a
definitive answer. This isn't homework - just something I'd be
interested to know.

The UK National Lottery is drawn by picking six numbered balls
from a possible 49. The odds of any given six numbers coming up are,
I'm told, 14 point something million to one.

My question is this: if a second set of identical balls were put into
the draw machine, so that there were two 1 balls, two 2 balls, two
3 balls and so on, what difference would it make to the odds of any
six (different) numbers coming up?

I suspect that it would lengthen the odds of a jackpot - but I would
like to know whether it would be a significant change or more or less
the same probability.

As I mentioned, I'm a complete dunce at maths, so if you could 'show
your working' and explain it a little I'd be very grateful. Thanks!

Hi grimace-ga,

You're right about the odds of the lottery; thi probability is
calculated by counting the number of combinations that can result by
choosing 6 balls out of 49. The general formula for this is:

(n!)/((m!)(n-m)!)

where n is the total number of balls and m is the number of balls you
are choosing. Plugging into the formula:

(49!)/((6!)(43)!)

If you are unfamiliar with the "!" symbol, this actually represents
the factorial, which is the product of the integer and every integer
less than it and greater than one.
For example:

3! = 3*2*1 = 6

Now to answer your question about having two sets of balls (ie. two of
each number), I immediately saw a problem. What happens if both #1
balls are selected? To keep things consistent, I would think that if
the "twin" of a ball that has already been drawn is also drawn, then
this ball would be discarded.

By making this assumption, it is then clear that once a ball has been
drawn, its duplicate is eliminated from consideration. Lets use an
example to explain this situation:

First ball: (choosing 1 out of 98) #4 --> if #4 "twin" is drawn later
it won't count
Second ball: (choosing 1 out of 96) #22
and so forth....

What you can see from the above example is that implicitly, each draw
results in the removal of two balls (one right away and the other is
discarded if drawn later).

Original odds : (1/49)(1/48)(1/47)(1/46)(1/45)(1/44) = 1/13983816
New odds: (2/98)(2/96)(2/94)(2/92)(2/90)(2/88)
        = (1/49)(1/48)(1/47)(1/46)(1/45)(1/44) = 1/13983816

Conclusion --> the odds are the SAME in both draws!

Now if you want to consider two identical balls being drawn as being a
valid option, this is really the same as 98 choose 6, or:
98!/((6!)(92!)) = 1052618392 possible combinations

I hope that answer your question...

Cheers!

answerguru-ga

---

Request for Answer Clarification by grimace-ga on 12 Jul 2002 14:03 PDT

Thanks, answerguru. You explain everything beautifully.

Just to clarify, though - I was indeed thinking of a model where the
duplicate ball is *left in* rather than removed. I'm not quite so dim
as to miss the fact that the odds would be the same if duplicates were
removed! The duplicates are the element which complicate the puzzle
for me.

The model I was thinking of was exactly the same as the standard
lottery draw, except with duplicates. Thus, if "1" is the first ball
to be drawn, it is removed from the draw, but its duplicate is not.
Thus, "1 1 2 2 3 3", might be a possible draw. What I'm looking to
find out is the odds of matching six different numbers, even if there
is a possibility of numbers being duplicated.
I'm sorry I didn't make this clear from the start. 

Doesn't your final solution, '98 choose 6', imply that no balls are
removed from the draw at all?

---

Clarification of Answer by answerguru-ga on 12 Jul 2002 18:09 PDT

Hi again grimace-ga,

If we consider the draw (1,1,2,2,3,3) as being feasible, then we have
to give the people buying tickets the option of choosing the same
number twice. Do you see what I mean?

If you really think about it, this is the same as a draw with 98
distinct numbers and 6 are chosen from those. So your question won't
really have a valid answer because if you do draw a duplicate at any
point, everyone will lose automatically! Who wants to buy into a
lottery like that? :)

Anyways, I'm sure you still want the answer to your clarification and
I have no problem  with that..here it goes:

If you assume that tickets bought cannot contain duplicates then any
draw containing duplicates is a 'loser'. Thus the total # of losers
is:
total # of draws (98 Choose 6) -  total # of tickets (49 Choose 6)
  = 1038634576 'loser' draws

Then you have your valid draws (49 Choose 6) so the total odds in your
scenario is: 1/(98 Choose 6)

To answer your concern about the balls not being removed...they
actually are being removed. If you look back at the combinations
formula you'll see that it simplifies to:

98*97*96*95*94*93

This means that there are 98 choices when the first ball is taken out,
97 choices when the second ball is taken out....and so forth until the
six balls have been removed.


answerguru-ga

---

Request for Answer Clarification by grimace-ga on 13 Jul 2002 02:28 PDT

Thanks, answerguru. I don't want to be a difficult customer, but your
solution bothers me still.

Let me try and express what I see as the flaw:

"If you assume that tickets bought cannot contain duplicates then any
draw containing duplicates is a 'loser'. Thus the total # of losers
is:
total # of draws (98 Choose 6) -  total # of tickets (49 Choose 6) 
  = 1038634576 'loser' draws"

This seems to be flawed. It doesn't seem to take into account the fact
that for any ticket there is more than one winning draw. Thus, if my
ticket is "1 2 3 4 5 6", the combinations (where A is the first set of
balls and B is the second)

1A 2A 3A 4A 5A 6A
1A 2A 3A 4A 5A 6B
1A 2A 3A 4A 5B 6B
etc.

are possible.

Wouldn't a draw of 49 numbered balls and 49 blank balls also give
1038634576 'loser draws'?

I like the methods the commenters have used to get at the answer, and
they have provided two very similar solutions to set against your very
high one. Any thoughts on unravelling this knot?

---

Clarification of Answer by answerguru-ga on 13 Jul 2002 08:26 PDT

Hi again grimace-ga,

Yes upon further investigation I have to say that the comments
provided by calub and charles consider your situation...my apologies.


answerguru-ga

Okay - answerguru did his best, but never quite hit the mark here. I
learnt a lot about ways of solving a problem like this, both from the
answer and the comments; but in the end, it was the commenters who
came up with the correct solution.

I feel like a swine giving a low rating, since I'm a researcher
myself. but with questions like this there's clearly a right and a
wrong answer.

Think of the problem this way :

Probability of winning = (# of ways in which you can win)/(# of
different outcomes).

Suppose you have 49 numbered red balls and 49 numbered blue balls.

The total number of different lottery drawings is 98!/92!. If you had
to pick the numbers AND the colors, the number of ways in which you
could win would just be 6!. So in this case, the probability of
winning is 1 over 98!/(92!6!).

However, you don't care about color. Red 3, Blue 5, 7, 8, 9, 10 is the
same as Blue 3, Red 5, 7, 8, 9, 10.

You have 12 shots at getting the first ball out of a pot of 98. You
don't care which of your 12 "winning" balls it is. You then have 10
shots at getting one of your remaining 5 numbers out of a pot of 97.
Likewise 8 shots from 96...

The odds on winning this jackpot are :

(12/98) * (10/97) * (8/96) * (6/95) * (4/94) * (2/93) = (2^6)6!92!/98!
= 64/1052618392 = 1/16447162.38

Because there are a lot of combinations where nobody wins the lottery
(multiple balls with the same number) - you do not get a nice 1/"round
number".

So while Answerguru answered the question, I think this is the
question you were trying to ask. I could be completely wrong here :)

-calebu2-ga

calebu - I like your answer. The red/blue analogy works well, and I
hadn't thought of the first ball being 12/98 rather than 2/98. That
seems to make sense. However, it bothers me that answerguru's
calculation of the original odds:

(1/49)(1/48)(1/47)(1/46)(1/45)(1/44) = 1/13983816

comes out with the 'right' - i.e. published and widely quoted - odds
for a jackpot on a conventional draw.

Using your method of calculation, wouldn't

(6/49)(5/48)(4/47)(3/46)(2/45)(1/44) 

give much shorter odds?

I'm still confused, I'm afraid.

Hi grimace,

I figured the odds using a different method, and they come out close
to calebu2's. I think my method might be less precise, but it gives
you a clearer picture of how the odds are affected.

If you look at each draw:
1st draw: odds aren't affected - 6/49 (or 12/98) of picking one of
your numbers
2nd draw: odds are the same + 1/97 chance of picking the duplicate
ball
3rd draw: " " + 2/96 chance of picking one of the two duplicates
4th draw: " " + 3/95
5th draw: " " + 4/94
6th draw: " " + 5/93

I added up the increased odds on the 5 draws that are affected, which
comes to .1590379 on my not-very-precise hand calculator. Multiplying
1.1590379 x 13,983,816 comes out to 16,207,769. That's less than 1.5%
off from calebu's calculation.

Hope this helps.

grimace-ga: you are correct. The formula in the answer involving 98
choose 6 - 49 choose 6 is incorrect, and for the reason you give.