Discrete Math - Day 17

Today we got back to discrete probability. I posted 3 problems based on cards games. Students were to find the probability for each event.

This was another struggle for students. The tendency is for students to want to add things together. For finding the probability of being dealt a "blackjack" they wanted to add 4/52 + 16/52. I kept reminding them to look back at how we were counting outcomes on other problems. The same could be said for the other two problems.

I finally had to step in to help get things re-focused for the class. I first asked how many ways could I be dealt two cards from a deck of 52? After some thought I got responses that there were ₅₂C₂ ways. I told them this was the value of the denominator; it reflected all the possible outcomes that could occur.

I then asked what the numerator should be. There was a lot of mumbling but no responses. I focused on the first card and asked how many different aces could I be dealt. The response was four. I then asked how many ways could I be dealt a face card or 10. The response was 16. I then asked how many ways could the 4 aces be mixed and matched with the 16 face cards/10s. Some students responded there were
4 x 16 = 64. I related this scenario to mixing and matching pants and tops.

I then asked students what if we were dealt a face card/10 first, how many ways would be match this situation with an ace. Students readily saw there were 16 x 4 = 64 ways for this to happen. The question is whether the desired outcomes were 64 or 128. This can be related back to the basketball teams problem. Once we have an ace and a 10 in our hand, does it matter whether the ace or the 10  was dealt first. The resultant probability is

     P("blackjack") = 64 / ₅₂C₂ = 64 / 1326 = .048265

So, approximately 1 in 20 hands results in a blackjack.

For being dealt a flush, students now realized the denominator would need to be ₅₂C₅. The issue was how many flush hands could be dealt. I asked the class how many cards do we have to draw from to form a flush. Students realized there were 13 cards to draw from and we were picking 5 of these, irrespective of the order. So, we have ₁₃C₅ ways to draw 5 cards of the same suit. But this is only one suit and we have four. Therefore, there are 4 x ₁₃C₅ = 4 x 1287 = 5148 possible flush hands. Our probability is

     P(flush) = 4 x ₁₃C₅ / ₅₂C₅ = .001981

or about .2%.

The last problem is one that confuses students because they are now primed to look at hands dealt and ways to get a single card from 13. I like to present this problem third for just that reason. Students can approach it from this perspective and they will reach the correct solution but a little thought will simplify things greatly.

In the game of spades, with four players, each player receives 13 cards. The queen of spades can only end up in one of those 4 hands, so the probability of getting the queen of spades dealt to you is 1/4. It is instructive to allow students to work through this problem the long way and come to the realization that the result is 1/4.

The long way first looks at how many ways a hand of 13 cards can be dealt, which is ₅₂C₁₃. The next issue to address is how many ways can the queen of spades be dealt in a hand of 13 cards. This means we have 12 cards that can be selected from 51 cards, since we are assuming our hand would have to contain the queen of spades. This means we now have ₅₁C₁₂ options. Finally,

P(holding queen of spades) = ₅₁C₁₂ / ₅₂C₁₃ = .25

There is usually a moment of shocked surprise at this result. Then students realize that this makes perfect sense.

Class concluded with students recording which probability rules are confusing to them and things they can do to help remember what to do.

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