In *Rolemaster*, you roll 1D100 (1D10 for tens and 1D10 for units) and add your skill.

If you roll 96 or more, you reroll and add again that roll

If you roll 5 or less you reroll and substract that roll.

Effectively possible to kill a fully armored knight with a kitten attack…if the kitten is lucky enough…

Oh, for sure, that part isn’t particularly crazy. It goes back to pretty early in AD&D’s history, particularly for rolling for ability scores. However, the FFG version of *Legend of the Five Rings* also uses specialized dice with special symbols that do different things and interact with each other in particular ways. There *wasn’t* a “lowest” die to drop. You might get a die face that generates a Success, which is good, but also Strain, which is bad, while another die generates Advantage, which is good, but not generally as good as a Success, except in some situations extra Successes don’t do any good but extra Advantages do, and is the Success worth the Strain, and there are some special rules about which dice you can drop and which you have to keep…and…and…

FWIW, I ended up going with the Monty Hall problem: I started with a Brooklyn 9-9 clip where they describe the problem and the debate (skipping the sections of the clip where they descend into hilarious raunch), then gave the class a few minutes to talk about their predictions (one super-smart kid said, “I’d think there would be a 50-50 chance of getting the right door, but I also know that you’re up to something, so switching is probably the right answer”). They broke into pairs and used 3 playing cards and a recording sheet to simulate outcomes of the game when the player sticks with their original guess or switches to the new guess.

In 45 minutes, I not only had everyone convinced that switching was the right strategy, but most of them also understood *why* it was the best strategy.

Simulations are powerful!

A better simulation gives this

1: 1/36

2: 7/216

3: 49/1296

4: 343/6^5

5: 2401/6^6

6: 16807/6^7

7: 493/11664

.

.

.

```
>>> from sympy import *
>>> z = symbols("z")
>>> q = z*(z**6-1)/(z-1)/6
>>> q
z*(z**6 - 1)/(6*(z - 1))
>>> factor(q)
z*(z + 1)*(z**2 - z + 1)*(z**2 + z + 1)/6
>>> r = (q+q**2+q**3+q**4+q**5+q**6)/6
>>> expand(factor(r))
z**36/279936 + z**35/46656 + 7*z**34/93312 + 7*z**33/34992 + 7*z**32/15552 + 7*z**31/7776 + 77*z**30/46656 + 131*z**29/46656 + 139*z**28/31104 + 469*z**27/69984 + 889*z**26/93312 + 301*z**25/23328 + 4697*z**24/279936 + 245*z**23/11664 + 263*z**22/10368 + 691*z**21/23328 + 1043*z**20/31104 + 287*z**19/7776 + 11207*z**18/279936 + 497*z**17/11664 + 4151*z**16/93312 + 3193*z**15/69984 + 1433*z**14/31104 + 119*z**13/2592 + 749*z**12/15552 + 2275*z**11/46656 + 749*z**10/15552 + 3269*z**9/69984 + 4169*z**8/93312 + 493*z**7/11664 + 16807*z**6/279936 + 2401*z**5/46656 + 343*z**4/7776 + 49*z**3/1296 + 7*z**2/216 + z/36
>>>
```

Nice!

Maybe Bayes’s theorem is a good topic for a probability class; it is often quite useful.

The last old Let’s Make a Deal I saw had an alpaca as the other zonk. So if you’re a kid who wants an alpaca more than a 1970s AMC Gremlin, the correct strategy is not to switch.

I will admit, that is brilliant. The Monty Hall problem is complex enough and so non-intuitive that some of the most intelligent people in the world have had public arguments about it. (Even world-renowned mathematician Paul Erdős refused to believe the solution until shown a computer simulation proving it.) Convincing students that the correct answer is what it is, and even getting them to understand how it’s correct, that is no small feat. Bravo!

Give them two six-sided dice; one with sides numbered 1-2-2-3-3-4, the other numbered 1-3-4-5-6-8. Let them figure out the probabilities of the possible results.

They will discover that the results are identical to what you get from rolling two standard dice.

Introduce them to Penney’s Game

It’s a really simple system - that is nevertheless non-transitive

And to bring simple probability calculations into this, the odds of you losing on one die roll are 5/6 ≈ 83%. The odds of you losing in two die rolls is 5/6 * 5/6 = 25/36 ≈ 69%. The odds of you losing in three rolls is 5/6 * 5/6 * 5/6 = 125/216 ≈ 58%.

But I like the real world aspect you bring in your example.

Just like in the Monty Hall problem, Monty is restricted in his choices. He has to show you the goat. Your choice on the door is random. His isn’t. Straight probability calculations go right out the window when options are forced in some way. Almost every seemingly obvious equal-odds probability problem involves some sort of forcing of choice aspect that results in a counter-intuitive result.

Extending my simulation out to a few hundred billion cases matches DPRK’s results, I think

1: 7776/6^7 = 1/36

2: 9072/6^7

3: 10584/6^7

4: 12348/6^7

5: 14406/6^7

6: 16807/6^7

7: 11832/6^7

8: 12507/6^7

9: 13076 /6^7

10 13482/6^7

11: 13650/6^7

12: 13482/6^7

13: 12852/6^7

14: 12897/6^7

15:12772/6^7

16: 12453 /6^7

17: 11928/6^7

18: 11207/6^7

19: 10332 /6^7

20: 9387/6^7

21: 8292 /6^7

22: 7101/6^7

23: 5880/6^7

24: 4697/6^7

25: 3612/6^7

26: 2667/6^7

27: 1876 /6^7

28: 1251/6^7

29: 786/6^7

30: 462/6^7

31: 252 /6^7

32: 126 /6^7

33: 56/6^7

34: 21 /6^7

35: 6 /6^7

36: 1/6^7

Here is a tool for exploring dice probabilities I heard about on a board gaming podcast. I have not used it.

Brian

Since we’ve veered into a discussion of dice games in general, here’s an interesting game I (maybe?) invented.

Two players, playing against each other, take turns. On a player’s turn, they secretly roll and look at a certain number of a certain kind of dice. (This is locked in for the whole game… for instance, a simple version of the game would use 2d6. A more complicated version would use 10d20, etc.)

Then the player who rolled starts to reveal the dice to their opponent, one at a time, in an order of their choosing. At any time, opponent can say “stop”, and then opponent takes the most recently revealed die, and gets that many points.

Take turns, play a certain number of rounds, highest score wins.

There’s interesting strategy in which order you reveal them in, and also when to say “stop”. (A more “fair” version would be to play it duplicated, so teams of 2 face off, each simultaneously playing the same rolls, but from opposite sides.)

That’s a great way to organize the investigation. Looking at actual data is powerful stuff!

Too late for this year, and I don’t know how you got into the question of dice probabilities to begin with, but here’s an activity that I’ve used many times that leads into an analysis of dice rolls (I was an elementary math specialist for many years, among other educational roles):

Tell the kids they’re going to play a dice game in partners. On one side of the board, write the numbers 2, 3, 4, 10, 11, 12. On the other, write 5, 6, 7, 8, 9. Explain that in each pair one kid gets the numbers on the left and the other the numbers on the right. Then they roll two dice and record whether the number is in the group on the left or the group on the right. Whoever has that number gets a point. Play till someone has 20 points.

Someone is almost sure to note that the game looks unfair because one player has six winning numbers and the other only five, but if they don’t then you can bring it up. Either way. “Yeah, it’s true that there are more numbers on the left, but hey, there are only 11 possible rolls, and 11 is an odd number, so…” Shrug. “It does look like the player with six numbers has an advantage, so the game might not be completely fair. Still, let’s give it a try.”

Your “super-smart” kid might suspect that something is up, and usually I have a couple of kids who’ve played a lot of board games who have a nagging feeling that all is not quite as it appears. But the majority of kids go off cheerfully to play the game, fair or not–and are surprised to discover that the player with 5-6-7-8-9 is the winner–and then shocked to discover that in every pair 5-6-7-8-9 was the winner. Which prompts the question, “How come? What’s going on?” Which in turn gets the mental gears moving and primes them for a more thorough look into what’s happening.

Anyway, too late for this year, but you might consider it for next year. As I say, I’ve done it a lot, and it has worked well. Good luck!

*In the interest of full disclosure, I did *once* have a game which was won by the “wrong” player. It was fine, didn’t matter, we still had 10 that came out with the 5-6-7-8-9 player winning and only one with the reverse. But it did happen. Given the personalities involved, one who liked a joke, one who was out to lunch, I’ve always suspected a certain degree of chicanery…“Let’s see, four and three, that’s five, a point for me! Here, you roll!”

One that I just remembered:

I briefly ran a campaign of “Dogs in the Vineyard,” one of the first really breakthrough “indie” RPGs. It had a very funky “high concept”: the PCs are essentially supernatural troubleshooters for the LDS Church in 19th Century frontier Utah.

Anyway, the game’s dice mechanic was built around each player having a dice pool which they roll, and then use the results of in a poker-style series of bids and raises.

Another that’s more crazy dice than crazy dice rolls:

The Ubiquity system, used in the *Hollow Earth Expedition* RPG, uses special d8s. There are white, red, and blue dice. The white dice are numbered {0, 0, 0, 0, 1, 1, 1, 1}. The red dice are numbered {0, 0, 1, 1, 1, 1, 2, 2}. The blue dice are numbered {0, 1, 1, 1, 2, 2, 2, 3}. So, rolling one red die is mathematically equivalent to rolling two white dice. And rolling one blue die is equivalent to rolling three white dice. The game uses a dice pool system, with the pool rating being the number of white dice you roll. Since the red and blue dice replace two or three white dice, you can roll large dice pools with a relatively small set of dice.

The game publishers also released a special set of “exploding” dice. They had eight “main” sides, numbered as above, but also had small faces on the vertices, resulting in a small chance of a die landing “cocked”. The vertex face had an explosion symbol; if the die landed on the vertex, it counted as the highest value on the die, and you rolled the die again and added the value.