From the week of December 11 to December 13 we chose Wednesday's problem (that you can find here):

*Spence is playing a card game using a standard $52$-card deck. During the game he is dealt $4$ cards and he wins if he gets $4$ cards of the same rank. In this game the cards of rank $A$ act as a wildcard, so an $A$ can be used as any other card. How many winning hands are there for this game?*

17% of the students who tried the problem got it right on their first attempt.

This game Spence is playing wants him to get a *poker* when dealt only $4$ cards. That is usually a difficult task even with more than $4$ cards, however, this time the $A$ cards act as wildcards, so even if he is missing some cards to get $4$ of the same rank, he can get a little help.

To count this hands easily we want to take advantage of the* Sum R**ule*. We can first find the possible separate cases that could happen, count each of them, and then add together the number of ways each of them could happen. Since in the game Spence is dealt $4$ cards, this are the only $4$ things that could happen:

- Case 1: All $4$ cards are of the same rank (without using any wildcards).
- Case 2: $3$ cards are of the same rank, and the $4^{th}$ card is a wildcard.
- Case 3: $2$ cards are of the same rank, and the other two are wildcards.
- Case 4: $3$ wildcards and any other card.

The order in which the cards are dealt is not important, and all four cases could be counted in a similar manner. Note we can follow some steps to count each of the cases, so we can use the *Product Rule* to count each of them:

- Choose which rank is the one that will make that a winning hand.
- Choose the cards of that rank that we are using
- Choose the wildcards that we are using.

Did you think of a different way of counting the winning hands of this game?

Share your thoughts and questions below!