Lottery Quesiton

If I purchase at random 3 million tickets in a 49/6 lottery (just under 14 mil combinations) on average how many duplicate tickets will I have?

Quote: hodag

49/6 lottery

Is that 49 balls, where 6 are drawn without replacement?

Correct. Typical lottery setup.

My naive ansatz would be: The probability of N tickets for any specific combination will be (3e6 chose N)*p^N*(1-p)^(3e6-N), with p = 1/14e6. Since p is low and N large, a Poisson distribution should be a good enough approximation with lambda = <N> = 3e6/14e6.

Probability of 0 tickets will then be exp(-lambda), for 1 ticket it will be lambda*exp(-lambda). For your numbers, probability of 0 or 1 tickets will be P=98%.
So for a specific combination, probabiltiy of 2 or more tickets is 2%.

This part I'm not quite sure: If you have 14e6 possible combinations, I would estimate 14e6*2% = 280,000 combinations with multiple tickets.

Quote: hodag

If I purchase at random 3 million tickets in a 49/6 lottery (just under 14 mil combinations) on average how many duplicate tickets will I have?

My try
this looks just like a Roulette question about the number of repeat and uniques over N spins
so I show how I work it out for Roulette and we can see if it matches my answer for your Lottery

38 slot wheel and spin 29 times.
N=38
S=29
how many repeats and uniques.
I think Roulette players know the answer to this (oh, they may only know the 38 spin value. hehe)
how about working it out and I could even show the distribution and the formula (function) that can make it
but I think I have done this before. So I find the link to it.

where were we
another skittle

One formula for the expected number of repeats in N trials is =S-(N*(1-((N-1)/N)^S))
very simple once we break it down

One formula for the expected number of uniques in N trials is =N*(1-((N-1)/N)^S)
((N-1)/N)^S
well (37/38)^29 is the probability of NO hit for 1 number in 29 spins
so 1- is at least 1 hit
we times that by 38 because you know why = 20.46 = # of unique numbers

We have 29 spins so 29 - 20.46 = 8.54 must have repeated. love averages!
we do not know the distribution of those, but it can also be figured out too)

now try your question
N=C(49,6)=13,983,816
S=3,000,000

I get in Excel
299,971.24
another estimated 280,000
How close is Excel?


do the two answers given so far feel right?
hmmm

how about my Roulette answer - seem right? (that can easily be simulated)

how many did you think would be repeats on average?

Sally

Quote: hodag

If I purchase at random 3 million tickets in a 49/6 lottery (just under 14 mil combinations) on average how many duplicate tickets will I have?

A lot.

My estimates (based on simulation):
-Duplicated numbers (sets of numbers which are on 2 or more tickets): 279,300
-Tripled (or more) Numbers (sets of numbers which are on 3 or more tickets): 19,610
-Duplicated tickets (total tickets which duplicate each other): 579,300

Mango's use of the a Poisson Approximation is excellent in this case.