Question

HURRY PLEASE

To win a state "A" lottery, one must correctly select 5 numbers from a collection of 50 numbers (one through 50). The order in which the selection is made does not matter. The neighboring state "B" has a lottery where one must correctly select 6 numbers from a collection of 60. Which lottery would you rather play? Support your choice with mathematical calculations.

Answer 1

To compute how many groups of $k$ elements we can extract out of a group of $n$ elements, we can use the binomial coefficient:

$\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$

where the factorial is defined as

$n! = n(n-1)(n-2)(n-3)\ldots 3\cdot 2$

So, this is the number of possible quintuplet out of 50 numbers:

$\binom{50}{5} = \dfrac{50!}{5!45!}= \dfrac{50\cdot 49\cdot 48 \cdot 47 \cdot 46}{5\cdot 4\cdot 3\cdot 2} = 2118760$

Similarly, this is the number of possible groups of six numbers out of 60:

$\binom{60}{6} = \dfrac{60!}{6!54!}= \dfrac{60\cdot 59\cdot 58 \cdot 57 \cdot 56 \cdot 55}{6\cdot 5\cdot 4\cdot 3\cdot 2} = 50063860$

So, you win with a combination out of 2118760 in state A, and with a combination every 50063860 in state B this means that winning in state A is easier with a ratio of

$\dfrac{50063860}{2118760} \approx 23$

Whichi means that winning in state B is 23 times harder