# Thread: Lotto Probabilities for multiple entries

1. ## Lotto Probabilities for multiple entries

I am interested in two aspects of lotto probability:

- accurately calculating P(win) when entering multiple tickets.
- accurately calculating P(win) for any of Division 1 or 2 or 3.

Australian Gold Lotto draws 6 balls from 45, and 2 supplementary balls...so 8 all up.
Entries must fill out 6 of the 45 numbers. Those 6 are used to match the conditions of each Division. There are 5 divisions, but lets focus on the top three.

Division 1 requires 6/6 balls to match
Division 2 5/6 balls plus either supplementary
Division 3 5/6 balls

Using Microsoft Excel

Total unique combinations for drawing 6 balls from 45
= combin(45,6) = 8,145,060 [lets round that off to 8m for the sake of brevity)

Div 1 winning combinations
=combin(6,6) = 1.

Div 2 winning combinations are =((PRODUCT(COMBIN(6,5),COMBIN(2,1),COMBIN(37,0)))) = 12
that's the P(match 5/6) * P(match 1 of 2 supplementaries) * P(match none of the 37 losers)

Div 3
=((PRODUCT(COMBIN(6,5),COMBIN(2,0),COMBIN(37,1)))) = 222
that's the P(of matching 5 from 6) * P(picking none of the supps) * P(picking 1 of 37 losers)

Now, the following bit takes some genuine understanding of probability to ensure it is right. To work out P(win) for each division, it might be expected you can simply use the number of winning combos over the number of total combos.
Hence you get
Div 1 1/8m

Div 2 12/8m

Div 3 22/8M

Now there pretty low P(wins).

So where I come unstuck is working out P(wins) when you make multiple entries in the one draw of the lotto balls.

I presumed it was a matter of multiplying the P(win) of one entry times the the number of entries. However, this appears flawed because if I enter 4 million odd entries (for arguments sake), I wll get a P value >1. And Probabilities can only be a number between 0 and 1.

Further, I figured you cannot multiply the P values of each entry by each other, because this is not an example of dependent outcomes. And if I did mutliply Ps, I would get a P(win) with 4m entries of much less than 0.5. which seems wrong to me. Because if I enter 4m entries, it is the same scenario as expecting a 1 or 2 or 3 if I roll a dice. I am after the P(win) for half the possible outcomes in both cases.

And then things get really tricky if I try and work out the P(win Div1 or 2 or 3) with 4m entries.

Anyone who can help with this problem, I'd really appreciate it.
But I believe it is not as intuitive as you might first expect.

I would be even grateful if someone can confirm if I enter 4m entries, whether my P(win Div1) in one draw of the balls is 0.5.

Regards
Winston

2. Originally Posted by winstonw
I would be even grateful if someone can confirm if I enter 4m entries, whether my P(win Div1) in one draw of the balls is 0.5.
If these 4M entries are selected without replacement at random from the total of 8M distinct possible entries yes, as then you will have divided the 8M equally likely possible tickets into two equal ly likely sets and the winner must be in one of the sets.

RonL

3. Thanks Ron. That seems intuitively right to me as well.
So I would write the formula for P(win div 1) with 4m entries as

= P(win with 1 entry) * no. of entries
= 1/8,145,060 * 4,072,530 = 0.5.
OK, that's seems pretty set in concrete to me.

Now, if you are able to shed light on the next bit.
How do I work out the P(win Div 3 with 4m entries)?

If I use the same logic and formula as above, I would get a number greater than 1.
i.e.
= 222 / 8,145,060 * 4,072,530 = 111.0

And my final query is how do I work out the P( of winning Div 1, OR Div 2 OR Div 3)

Thanks in advance if you are able to help.