Discrete Math - Day 15

Today we started to look more closely at discrete probability. I re-visited the idea that we are counting desirable outcomes and total possible outcome. We then divide the desirable count by the total count to determine the probability.

To start things off I asked students to calculate the probability of being dealt a four of a kind (four cards of the same rank) in poker. This boils down to drawing 5 cards and having four of the same rank.

Students were basically stumped on this problem. It took a lot of guiding questions, asking students to look back at what we had done before, think about connections to the flag problems we had worked on, etc.

Even the denominator, which we had determined yesterday, was somewhat of a struggle. Students were thinking it would be ₅₂C₅, but they had no confidence that their thinking was correct.

After this, students were thinking that there would be 13 choices for the rank of the 4 of a kind. Then they were stuck again. I tried to get students to think of the other problems we worked on and what was happening with the cards. Finally a few students started considering the five cards in the hand and realized they could focus on the fifth, non-matching card. They determined there were 12 ranks this card could take on and then 4 suits. I had these four students go around to the other groups to discuss their idea. Afterward, I went around to the groups and asked them to communicate what they had learned. It appeared this was effective in helping the other students to make sense of the problem.

We finally arrived at a result of P(4 of a kind) = 13 x 12 x 4 / ₅₂C₅

I  then had students review what probability that they knew. It was limited to very vague concepts and a few considerations as to what valid probability values look like and must sum to.

I went through a few basic rules and then asked students, for homework, to produce examples of these rules.

The rules covered were:

1. 0<= P(A) <=1     all probabilities fall between 0 and 1, inclusive
2. Sum of all probabilities equals 1
3. P(A) + P(not A) = 1     probability of an event and its compliment are 1 (this can be used to calculate P(A) by using 1 - P(not A))
4. P(A or B) = P(A) + P(B)     for mutually exclusive (non-overlapping) events
5. P(A and B) = P(A) x P(B)     for independent events (when one event does not influence the outcome of the second event)

Visit the class summary for a student's perspective of the class and to view the lesson slides.