I’m going to post a question here, and I’m not going to mention it in class. I want one of my students to post a solution as a comment or to otherwise send me a solution digitally (and I’ll post it here). I want to see if anyone is reading this, and I want to challenge you a bit. :)

# The Question

When playing Delve: The Dice Game you notice that the Rogue’s **Crippling Strike** can be a very powerful attack against a single opponent (depending on the value on the sixth die). Crippling Strike requires a **full house**, which is three of one value and two of another value (a roll of 2-2-5-5-5-4 would be a Crippling Strike dealing 4 damage). Assuming you roll all six dice at once (e.g. on the final roll of a turn), how many different combinations are there which will result in a successful Crippling Strike?

There are six die being rolled, with one pair of two and one pair of three die being identical, with a third being random. Two patterns may be formed this way, (AA-BBB-C) and (AAA-BB-C), both are potentially valid, if that is how the game goes. Assuming it does, that makes for two very similar (the same) cases. In the first case, AA-BBB-C, A may have 6 valid choices, B can only have 5, as one is taken, and the third is (assumed to be) again 6. Thus, the formula goes 6 choose 1 times 5 choose 1 times 6 choose 1 (even though there are two/three #’s, they are both the same, so one number is chosen for both/all three), all divided by two, to remove repetitions. This simplifies to 6x5x6 over 2, which equals 90. The same process must be repeated for case 2, AAA-BB-C, which again equals 90. Take the two results and add them together, which equals a sum of 180. Therefore there are 180 different combinations that result in a successful crippling strike. Assumptions made are that AA-BBB-C is as valid as AAA-BB-C and C can be any number (1-6)..

Hi Erik,

Thanks very much for responding! I’ll give you a little feedback on your response.

Because the order of the dice doesn’t matter, AA-BBB-C and AAA-BB-C are really the same case; they shouldn’t be counted separately.

The case to watch that you don’t do some double-counting is for AAA-BB-C where B=C. For example, if you have counted 111-22-2, you don’t want to also count 222-11-1 as a separate case (both are 111222).

That’s some great thinking you have there, and I’m happy with how clear your response is when you’re typing in a text area! Thanks!