# The Lottery Problem – Solved

On Friday I asked you the following riddle. Assume there are 2 different types of lottery. Lotto A makes you choose 6 numbers out of a possible 75,and lotto B makes you choose 5 numbers out of a possible 60 and 1 ‘mega’ number out of a possible 40. Which lottery gives you the best odds? Original problem here.

I’m sure a lot of you said Lotto B. And this is how you probably came up with the answer:

• Lotto A: 75 numbers, choose 6, so in order to choose the first number we have 75 options, for the second number, we have 74 options (since we just used up one number), etc. So we get:
$$75 \cdot 74 \cdot 73 \cdot 72 \cdot 71 \cdot 70 = 144,978,876,000$$ possibilities
• Lotto B: Out of 60 numbers, choose 5, then out of 40 numbers, choose 1. So the same logic as in lotto A, we get:
$$60 \cdot 59 \cdot 58 \cdot 57 \cdot 56 \cdot 40 = 26,215,257,600$$ possibilities

Congratulations if you got this far! This is technically correct IF the lottery forced you to choose the numbers in a specific order. The lottery (luckily) doesn’t force you to get the order right though, you can choose the numbers in any order. So then what are the chances for each lotto? Let’s first look at an easier example.

Choose 2 numbers out of 1-10. The first number has 10 choices, the second number has 9 choices. So we get a total of 90 different options ($$10 \cdot 9 = 90$$). BUT again the order doesn’t matter. Pretend your numbers were 1 and 7. We’ll call this pair (1,7) for simplicity. Selecting number 1 and then selecting number 7 is the same thing as selecting number 7 and then selecting number 1! So for each pair […(1,7), (1,8), (1,9),(1,10), (2,2), (2,3)…] we have 2 different options. So we must divide 90 by 2 to get the real outcome of 45 different outcomes.

Now let’s take it one step further and pretend we picked a 3rd number. Let’s also pretend this number is the number 4. Earlier we saw that the outcome (1,7) could either show up as 1 then 7, or 7 then 1. Let’s suppose the first occurred (1 then 7), then by adding the number we get 3 different choices as to where to put it: 4 then 1 then 7, 1 then 4 then 7, 1 then 7 then 4. [Note that the same thing can be said if we had chosen 7 then 1]. So this time instead of having $$10 \cdot 9 \cdot 8 = 720$$ possibilities, we have $$\frac{720}{6} = 120$$ possibilities. [We get 6 by multiplying 2 with 3. 2 since we can have 1 then 7 and 3 since we can put the number 4 first, middle, or last].

Now we can see that if we added a 4th number we would have $$\frac{10 \cdot 9 \cdot 8 \cdot 7}{2 \cdot 3 \cdot 4} = \frac{5040}{24} = 210$$.

The lottery follows this same principle. So applying this logic, lets reexamine our lottery:

• Lotto A: Since 6 numbers were chosen we must take our original number and divide appropriately:
$$\frac{75 \cdot 74 \cdot 73 \cdot 72 \cdot 71 \cdot 70}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} = \frac{144,978,876,000}{720} =201,359,550$$ possibilities
• Lotto B: Since 5 numbers were chosen the first time, and only 1 the second time we must take our original number and divide appropriately:
$$\frac{60 \cdot 59 \cdot 58 \cdot 57 \cdot 56 \cdot 40}{2 \cdot 3 \cdot 4 \cdot 5} = \frac{26,215,257,600}{120} =218,460,480$$ possibilities

So you have to buy over 17 million more tickets for Lotto B to give you a guaranteed winning ticket! So if you chose Lotto A, you just won the lotto!