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Expert: - 1/9/2011


Question
In a 6 in 49 lottery, if I were to choose only odd numbers, am I equally likely to win the lottery as if I picked odd and even numbers?

I think that the answer is no as there are 13983816 possible combinations and only 177100 possible combinations with just odd numbers but how do I prove it?

Answer
1st, I will say that C(a,b) = a!/(b!(a-b)!).
The chance of them all being odd is C(49,25).
The chance of them all being even is C(49,24).

For all odd, since there are 25 odd numbers, it is (49*48*47*46*45*44)/(25*24*23*22*21*20).
That gives the 13,983,816.

For all even, it is (49*48*47*46*45*44)/(24*23*22*21*20*19).
That gives the 177,100.

However, once the numbers are picked, no matter what there are, there are C(49,6) ways of picking them and only 1 of them is right.

So with the number all being odd, there is a slightly higher probability of this occuring,
but that does not give any advantage to picking the numbers.


For example, let's look at picking 2 numbers out of 1, 2, 3, 4, and 5.
There is only 1 way they could all be even, but 3 ways they could be odd.
There are 10 ways total, which means 6 of them contain both even and odd.
The numbers chosen have a higher chance of all being odd.

To buy all of the odd combinations would cost the price of three tickets and
the price would be 3 units.

There would only be one ticket to buy to cover all of the evens,
and the price would be 1 unit.

What this comes down to is spending 3 units gives you 3 times the chances of winning,
but cost 3 times as much, so there is no gain.

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