lottery

We have this problem in my class, and none of us have been able to get a right answer for any except the first. Can someone walk me through the solutions to at least one of these?

In the "Make More Money Game," six different winning numbers are selected at random from, 1 to 33. A player also selects six different numbers, each from 1 to 33. There are different payoffs for different numbers of matches. Note that order does not matter for any of these matches. In each case, tell the odds of the player winning.

all 6 numbers - 1 in 1107568
exactly 5 of 6 - 1 in ?
exactly 4 of 6 - 1 in ?
exactly 3 of 6 - 1 in ?

Hello, jangalinn!

In a lottery, six different winning numbers are selected at random from 1 to 33.
A player also selects six different numbers, each from 1 to 33.
There are different payoffs for different numbers of matches.
Note that order does not matter for any of these matches.
In each case, find the probability of the player winning:

all 6 numbers: 1 in 1107568
exactly 5 of 6: 1 in ?
exactly 4 of 6: 1 in ?
exactly 3 of 6: 1 in ?

Click to expand...

There are: .\(\displaystyle {33\choose6} \:=\:1,\!107,\!568\) possible outcomes.

There are: .6 Winners and 27 Others.


All 6 Winners

There is one way to get all 6 Winners.

\(\displaystyle P(\text{6 Winners}) \:=\:\dfrac{1}{1,\!107,\!568} \quad\Rightarrow\quad \text{1 in 1,107,568}\)


Exacty 5 Winners

We want 5 Winners and 1 Other.

There are: .\(\displaystyle {6\choose5}{27\choose1} \:=\:6\cdot27 \:=\:162\) ways.

\(\displaystyle P(\text{5 Winners}) \:=\:\dfrac{162}{1,\!107,\!568} \:\approx\:\dfrac{1}{6836.84} \quad\Rightarrow\quad \text{1 in 6,837}\)


Exactly 4 Winners

We want 4 Winners and 2 Others.

There are: .\(\displaystyle {6\choose4}{27\choose2} \:=\:15\cdot351 \:=\:5265\) ways.

\(\displaystyle P(\text{4 Winners}) \:=\:\dfrac{5,\!265}{1,\!107,\!568} \:\approx\:\dfrac{1}{210.36} \quad\Rightarrow\quad \text{1 in 210}\)


Exactly 3 Winners

We want 3 Winners and 3 Others.

There are: .\(\displaystyle {6\choose3}{27\choose3} \:=\:20\cdot2925 \:=\:58,\!500\) ways.

\(\displaystyle P(\text{3 Winners}) \:=\:\dfrac{58,\!500}{1,\!107,\!568} \:\approx\:\dfrac{1}{18.93} \quad\Rightarrow\quad \text{1 in 19}\)