James Ernest and Glen Barnett
Here’s the situation. It’s your turn. You’re playing against a larger character, let’s call him Hammer, who’s got two dice left: (3/4, 17/20). Your dice lie (5/6, 10/10). You have already figured out that Hammer needs to keep his 20 to win. How do you get your best chance of taking it?
If you’re playing to keep the highest number on board, as you usually do, you’d do 5-takes-4, and reroll the 6-sider. But that’s going to give you an expected value of 3.5 on your lowest die (an even chance of 1 through 6, averaged). Hammer is obviously going to take your highest number next turn, to have the best shot at protecting his 20. So you want to come out of this turn with the best possible value on the lower of your two dice.
With the hand above, rolling the 10-sider is the better choice. Half the time, the 10 will roll higher than the 5/6, leaving 5 as your lower number. In fact, the average expected value of the lower number in this roll is 4.0.
Unfortunately, it’s not as simple as always rolling the larger die. If your 6 were showing a 3 rather than a 5, you’d be better off rolling the 6-sider. Rolling the 10 in this case only gives you an average lower number of 2.7.
And now, before you read further, figure out which die you’d roll if you were showing (4/6, 5/10.)
The Best Lower Number
There’s an easy way to calculate the expected value of the second highest number when you have a choice of which of two dice to roll.
First, some preliminaries. Triangular numbers are the number of dots you get when you lay dots out in triangles, like so:
A quick way to work out these numbers is to write down consecutive numbers from 1 on up (to 12 will do), and then underneath write a 1 under the 1, and then under each other number write the sum of the number to its left and the number above it, so the number under 2 is 1+2 = 3, and so on:
1 2 3 4 5 6 7 8 9 10 11 12
1 3 6 10 15…
The next few triangular numbers are 21, 28, 36, 45, 55 and 66. These numbers are important; if you know them already, great. If not, they are easy to work out.
Let’s say you have a 4 on a d6 (six-sided die) and a 5 on a d10, and your opponent has a 3 on a d4 and a 17 on a d20 – you want to know what to take the d4 with to maximize your chances of getting that d20. As James points out, you need to maximize the average value of the lower of your two numbers after the roll (since your opponent will take the higher of the two). Maximizing the lowest number’s average is the same as maximizing the probability that your rival will roll under your lower die.
Here’s how to find the average value of the lower number when rerolling a given die:
Take the face-up number on the *other* die. Now subtract the previous triangular number divided by the number of sides on this die. For example, if you reroll the d10, then the number on the other die is 4. Subtract the previous (ie third) triangular number (which is 6) divided by 10: 4 – 6/10 = 3.4. That’s all there is to it! (That’s not quite true – sometimes you do something that’s even easier.*)
Now what if we reroll the d6? The average lower number is 5 – 10/6 (i.e. 5 on other die – 4th triangular number divided by 6 for the d6), which gives 3 1/3. So you’d better take with your d10.
*That formula works when the number on the other die is smaller than the number of sides on the die you’re rerolling. If it is bigger, just take the average for the die you’re rerolling (since it will always be smaller than the other die, right?).
You don’t generally have to calculate the values to this accuracy -typically you only need to go far enough to figure out which one is bigger. Let’s look at another case: 3 on a d6 and 7 on a d10. Rerolling the d6: 3.5 (since 7 is bigger than 6) d10: 7 – 21/10
You don’t need to work out 7 – 21/10 if you can see it is bigger than 3.5 (which it is). You can say “21/10 is just over 2, so 7 minus something less than 3 is more than 4”. Neat, huh?
“Help, what’s the 8th triangular number?” If you don’t have the numbers to hand, and you don’t want to work the whole list out to get just one of them, here’s how to get a particular triangular number. You take that number, multiply it by the next number, and halve. So the 8th triangular number is 8 x 9 / 2 = 36. If you don’t know your multiplication tables, it only takes a moment to write down those two rows of numbers I mentioned before (if you can’t /add/, pick a different game to Button Men).
Here’s the same information in a table. The column headed “f” is the face value on the other die. The column headed “V” is the formula for the values in the table, unless we have gone past the number of sides on the die already.
Table: Average value on lowest die. (4 through 12 represent sides on rolled dice)
Note that we don’t need to continue the table to 20, since all but the last column don’t change, the rest of the values in the d20 column are bigger than all the values in the table, and if they are both d20’s you obviously just reroll the lower one.
You can notice some interesting stuff right away. For example, one rule you can see is that if the die with the fewest number of sides is showing 8 or more, reroll the other die. Another pretty obvious rule is that if you work out a lowest number value for one die that’s already above the other die’s ordinary average, it must be above the other die’s reroll value. Also, if the die with the larger number of sides is showing the lower number, reroll it. There are lots of other things to notice that I’ll leave up to you.