# Homework Help - Elementary School Math Problem

Darn, I’m not even smarter than a fourth grader. My son has this homerwork problem, and I’m not quite sure how to solve it, much less explain to him how to work it out. I’m sure many, if not most of you, can do this one with 1,000,000 synapses tied behind your backs.

Here goes:

You have a chance to win \$1 million* based on a cut of cards.

Ten card deck from 1-10. Draw a card. Return the card to the deck. Draw a second card from the complete deck.

What are the odds (or percent chance) that the two cards will add to nine?

By my thinking, the first draw has a 20% chance of giving a bad result, and the second draw has a 10% chance of matching up with the initial 80%. But I think that’s the wrong way to think about it and maybe there is an easier way, by doing a matrix of all possible outcomes.

Suggestions on how to most easily solve this?

• Shibb, feeling very dumb tonight.

There are a hundred possible results of the two cuts. Of those, eight combinations add up to 9 (8-1, 7-2, 6-3, 5-4, 4-5, 3-6, 2-7, 1-8). So the answer is 8%.

Your way also works. There is an 80% that getting a nine is still possible based on the first cut, and a 10% chance that you will actually get a nine once you make the second cut if your first cut was a number between 1 and 8. Multiply those together, and you get 8%.

Think of it this way - you have 100 (ten times ten) possible ways to draw two cards. I’ll write each possibility as first-second. (1-1, 1-2, 1-3, 1-4, …10-7,10-8,10-9,10-10).

How many of those will add up to nine? You could draw:

1-8
2-7
3-6
4-5
5-4
6-3
7-2
8-1

That’s 8 ways to add up to nine, and 8/100 = 8%.

Well, that’s certainly a fine way to do it. 80% chance of the first draw going acceptably, and, given that it does, a 10% chance of the second card matching up. That gives a total success probability of 80% * 10% = 8%. (ETA: Ah, ENugent made this same point. Somehow, I missed that.)

Thank you all, very simple, well explained answers. My son was able to grasp both methods, although he preferred the matrix. We did several others solutions out of ten and also tried it with 12 cards, to help it sink in.