Today I’m going to walk you through the design of a divided die roll, one of my favorite RPG tools.

It used to be believed in Europe that there were no such things as Black Swans. Then the Dutch discovered them living quite happily in Australia. Scientists now sometimes refer to improbable events as evoking the Black Swan Effect. Image by Zhu Bing from Pixabay

Dividing one die roll by another creates a large low-result probability and a long tail of low-probability high results. I’ve written about them before in The Physics Of Uncertainty, and used them in Oddities Of Values: Recalculating the price of valuables as well as in the Trade In Fantasy mechanics.

I was doing final revision of the structure of the next chapter of Trade In Fantasy when I realized that I would need just such a die roll. Normally, I would throw an extremely condensed explanation into a sidebar and that would be that, but the more I worked on prepping for the article, the less confident I became that I would have it anywhere near finished in time for publication.

You see, last Tuesday I got caught by the rain and since Tuesday night I’ve had a massive head cold. Symptoms are now starting to moderate, but my thinking is still fuzzy and wrapped in marshmallow and cotton wool. It is literally taking me a lot longer to do anything, and anything that I would normally knock off in a couple of days is now more likely to take three or four.

So I decided that it would be better to take a really small topic that I could explain in detail in the time available rather than the big article that I probably wouldn’t finish in time – and, since I was going to need to design this die roll anyway, it seemed like an obvious choice.

I’m not going to go into the specifics of what it’s for, or why I need it; that would steal too much from the Trade In Fantasy chapter.

That’s why we are where we are. Let’s get on with it!

Divided Die Roll Characteristics

Divided Die Rolls can be specified by defining four values – the minimum, the maximum, the average, and the interval.

In this case, I want an average of 1, a minimum of 0, a maximum of around 4 or 5 but I can be a little flexible in that area, and an interval of 0.1.

The format of a divided die roll is [(AdX + B) x C + D / (EdY + F)] / C.

The C and D variables

In this case, D is a modifier that’s going to come from other mechanics, so I can presume that it’s zero at the moment and come back to it as the very last thing. C is the reciprocal of the interval – so if I want results to be every 0.1, then C is 10.

So I can simplify the above to [(AdX + B) x 10 / (EdY + F)] / 10.

But there are alternate configurations that need to be considered before settling on this. The /10, that’s settled, absolutely essential – that’s how you go from while number results to results that are 1/10th apart. But the x 10 – I could make that “x d10” or “x d20” or “x (d10+5)” and all I’m going to be doing is changing the roll requirements for E, Y, and F.

In fact, it’s essential that I pick something other than x 10 because when you x10 and then /10, you end up with the same number.

Because this is not supposed to be the primary driver of the randomness within the roll, I like the notion of using “x (d10+4.5)” – a slight variation on the third option listed. That gives this part of the roll a range of 5.5 to 14.5 and an average result of exactly 10.

Those are important numbers because this defines the range over which the actual random die roll takes place.

On a low roll on the d10, the range is going to be about 1/2 of what it normally is (55% to be exact), on an average roll, the range will be present in full, and on a high roll, it will be 145% of normal – so that the randomness will occupy a wider range of results.

With those items specified, I can reformat the whole to f(1) x (d10+4.5)/10, which makes this effect a whole lot clearer and shifts everything that hasn’t been decided yet into the umbrella f(1) function.

Eliminating D

I can further simplify things by eliminating D altogether and specifying B as the additive from the game mechanics – which effectively eliminates it for the time being, too. So now we have f(1) = AdX / (EdY + F).

Adjusting Parameters

The next thing I have to do, having isolated and defined the C “value”, is to adjust the desired minimum, average, and maximum results, because those are the parameters that will define A, X, E, Y, and F.

Minimum:
f(0) x (1+4.5) / 10 = 0
f(0) x 5.5 / 10 = 0
f(0) x 5.5 = 0 x 10
f(0) = 0 x 10 / 5.5 = 0

Average:
f(1) x (5.5 + 4.5) / 10 = 1
f(1) x 10 / 10 = 1
f(1) = 1 x 10 / 10 = 1

Maximum:
Three possible values that are acceptable, this is something that is a bit flexible. But let’s work all three:
f(2) x (10 + 4.5) / 10 = 3
f(2) x 14.5 / 10 = 3
f(2) x 14.5 = 3 x 10 = 30
f(2) = 30 / 14.5 = 2.0689655172413793103448275862069 = 2.07

f(3) x 14.5 / 10 = 4
f(3) x 14.5 = 4 x 10 = 40
f(3) = 40 / 14.5 = 2.7586206896551724137931034482759 = 2.76

f(4) x 14.5 / 10 = 5
f(4) x 14.5 = 5 x 10 = 50
f(4) = 50 / 14.5 = 3.4482758620689655172413793103448 = 3.45

There’s one more useful set of values that can be derived at the moment – (f(n) – f(0)) / i.

Minimum – zero divided by anything is still zero.
Average:
f(5) = [f(1) – f(0)] / i = [1 – 0] / 0.1 = 10

Maximums:
f(6) = [f(2)-f(0)] / i = [2.07 – 0] / 0.1 = 20.7
f(7) = [f(3)]-f(0)] / i = [2.76 – 0] / 0.1 = 27.6
f(8) = [f(4)-f(0)] / i = [3.45 – 0] / 0.1 = 34.5

E, Y, and F

The way divided die rolls work is to pair a set of A and X with a set of E, Y, and F to give the correct results.

The width of the denominator makes a big difference. Below are several probability charts, all based on 6d10 / Ed6, and all to more-or-less the same scale:

At the top, the red curve is 610/d6. Note the hump before the curve finally flat-lines. Below that is 6d10/2d6. Notice how the probability is considerably higher for the average result. The purple graph is 6d10/3d6, and it’s higher again. The blue line on the right is 6d10/4d6, and the increase in probability of the 1 result is starting to get smaller. The light blue shows 6d10/5d6, and any result over 2 is now extremely unlikely. The dark and light green that start the bottom row are 6d10/6d6 and 6d10/7d6, respectively, and the probability of a 1 result peaks. That’s followed by yellow and gold, showing 6d10/8d6 and 6d10/9d6, respectively. Notice that the probability of a zero result is growing and has in fact started eating into the probability of a 1 result. With the last curve, in orange, the probability of a zero result is now higher than the probability of a 1.

AnyDice is very useful, but it’s a die rolling program and it expects integer outcomes. It doesn’t do fractions – it rounds.

Another trick that’s really useful is to add an ordinary NdZ die roll to the results.

For the next set of graphs, I’ve taken the maximum result and added 2d-something – for the first chart, 2d10; for the second, 2d8; for the third, 2d6; for the fourth and fifth, 2d4; for the rest, 2d3. These sizes have been chosen because they are roughly 1/4 of the maximum result on the divided die roll. Results and probabilities have been scaled to be the same, and the same colors have been used.

Wow, what a difference! The “+2dN”.clearly dominates the results at lower values, but the long tail continues to exist, low probability outcomes significantly greater than the base range.

So the combination of these techniques gives a huge level of control over the end probability curve, and the shape of that curve is dependent on what’s on either side of that division mark.

Maximum Over Minimum

The other thing to notice is the number of results. The maximum outcome occurs when the maximum is rolled above the division line and the minimum below it.

6d10 has a maximum outcome of 60. 1d6 has a minimum of 1, so the maximum result of dividing the first by the second is 60.

With 2d6 as the divisor, the minimum roll becomes 2, and the maximum result overall becomes 60/2=30.

With 3d6 as the divisor, the minimum roll becomes 3, and the maximum overall result becomes 60/3=20. We would get the same divisor minimum with 2d6+1, but the shape of the curve would be different – more like the 2d6 curve, but scrunched into 2/3 the size, a compromise between the two.

Relationship to Parameters

Our maximum parameter is 20.7, or 27.6, or 34.5 – or somewhere in that vicinity. So we want around 30 results. We could get there with 3d10, or 5d6, or even d10+d20. These are going to have the same maximum result, but different minimums – 3, 5, and 2, respectively. Another option is 3d12, with it’s peak of 36, or 2d12+d8, or 2d12+d6, with peaks of 32 and 30, respectively.

Every additional die or +1 in the denominator multiplies these options by the total +1, because we still want the same end result.

The problem is that lots of dice take time to add up when you roll them. d6 work well because it’s easy to group them into sets totaling 10. d8s and even d10s can be handled using the same technique, but because there are fewer of them, you’re less likely to get convenient sets. d12 and d20…. not so much. That means that the first group of possible rolls is preferable to the second.

Deciding The Denominator

Because of the multiplication factor, the smaller and simpler the denominator, the better. But using one dice produces strange interactions that produce a second peak in the probabilities. Here’s a look at 5d6/d6, stretched a little vertically to make the bulge that results a little more obvious:

Using 2 or more dice in the denominator eliminates this almost completely. Here’s 10d6 / 2d6 for comparison:

Using 3d4 instead of 2d6 would be even better, but d4s (the traditional tetrahedron design) don’t roll very well. Using d3s would solve that – at the price of added inconvenience.

So 2d6+0 is the denominator of choice, at least for the moment.

Peak probability

The other problem I have is with the average result. I very much want it to center around 1, with a range down to 0 and up to 2 being the dominant probability outcome. Taking the desired intervals into account, that’s 0 to 10 to 20. I can achieve that, no problem, just by adding 2d10, or 4d5, or maybe even 4d6-4.

But that maximum of 20 then has to come off our range of results from the divided part. So, instead of 20.7, or 27.6, or 34.5, or similar, we’re now talking 0.7 or 7.6 or 14.5.

The 0.7 is so small as to be trivial. There might as well not be a divided roll. The other two remain viable choices. The question is how to get to them?

Since the denominator is to be 2d6, I can go ahead and double these to 15.2 and 29. Two options immediately come to mind: d10+d6-1 and 3d10.

But these will shift the average result, so whatever the overall average works out to be, I have to deduct that from the roll.

d10+d6-1: average is 5.5+3.5 = 9. So the -1 has to become a -8.

The alternative is 3d10. Average is 5.5 x 3 = 16.5. So the +0 has to become a -16.5. Or add 4 and subtract 20. I don’t like this roll as much as the first.

The third alternative, not mentioned so far, is to apply the average divisor, and shift the adjustment modifier to the other part of the calculation, shown in the first paragraph. The average denominator roll of 2d6 is 7. 9/7 = 1.2857142857142857142857142857143 = 1.286. That’s a bit of a pain; it’s easy with a calculator but otherwise user-unfriendly.

16.5/7 = 2.3571428571428571428571428571429 = 2.36. That’s one decimal place less, which is a good thing, but it’s still not ideal.

But I have one more trick up my sleeve: multiplication-adjustment-division.

The perfect solution would be 7 x (added roll), apply 7 x modifier, divide result by 7. But 7 is a pain in the neck when it comes to multiplication and division. So let’s look at alternatives and see whether or not there’s something close enough to be more convenient.

9/7 first:

2 x 9/7 = 18/7 = 2.57 = 2.6. Better, but not good enough.
3 x 9/7 = 27/7 = 3.857 = 3.86. No.
4 x 9/7 = 36/7 = 5.143 = 5. A contender.
5 x 9/7 = 45/7 = 6.4286 = 6.43. No.
6 x 9/7 = 54/7 = 7.718 = 7.2. No.
7 x 9/7 = 9. Obvious.
8 x 9/7 = 72/7 = 10.286. Rounding to 10 – a contender, but marginal.
9 x 9/7 = 81/7 = 11.57. No.
10 x 9/7 = 90/7 = 12.86. No.
20 x 9/7 = 180/7 = 25.714. No.

No other multiplication looks easy enough. But one serious contender.

16.5/7 second:

2 x 16.5 / 7 = 33/7 = 4.714. No.
3 x 16.5 / 7 = 49.5/7 = 7.07 = 7? A contender.
4 x 16.5 / 7 = 66/7 = 9.423. No.
5 x 16.5 / 7 = 77.5 / 7 = 11.07 = 11? A contender.
6 x 16.5 / 7 = 99/7 = 14.143. No.
7 x 16.5 / 7 = 148.5 / 7 = 21.214. No.
10 x 16.5 / 7 = 165/7 = 23.57. No.
20 x 16.5 / 7 = 330/7 = 47.143. No.

I’ve rejected options with smaller rounding errors than the 8, 10 option based on 9/7, so I’m tempted to eliminate it – but subtracting ten is such an easy calculation that I think it has to stay in the mix. So I’ve now got four contenders.

Combinations

I’ve ended up with six combinations for the divided part and three for the added part, or 18 combinations in all.

Because I’ve already used A through F, in general definitions, as well as X and Y, I’ll label these elements starting with G.

1. G1+K(g1) = (d10+d6-1) / 2d6 + (4 x 2d10 -5) / 4
2. G1+L(g1) = (d10+d6-1) / 2d6 + (4 x 4d5 – 5) / 4
3. G1+M(g1) = (d10+d6-1) / 2d6 + (4 x 4d6 – 21) / 4
4. G2+K(g2) = (d10+d6-1) / 2d6 + (8 x 2d10 -5) / 8
5. G2+L(g2) = (d10+d6-1) / 2d6 + (8 x 4d5 -5) / 8
6. G3+M(g2) = (d10+d6-1) / 2d6 + (8 x 4d6 -37) / 8
7. H1+K(h1) = 3d10 / 2d6 + (3 x 2d10 -7) / 3
8. H1+L(h1) = 3d10 / 2d6 + (3 x 4d5 -7) / 3
9. H1+M(h1) = 3d10 / 2d6 + (3 x 4d6 -19) / 3
10. H2+K(h2) = 3d10 / 2d6 + (5 x 2d10 – 11) / 5
11. H2+L(h2) = 3d10 / 2d6 + (5 x 4d5 – 11) / 5
12. H2+M(h2) = 3d10 / 2d6 + (5 x 4d6 – 31) / 5
13. I+K = (d10+d6-8) / 2d6 + 2d10
14. I+L = (d10+d6-8) / 2d6 + 4d5
15. I+M = (d10+d6-8) / 2d6 + 4d6-4
16. J+K = (3d10 + 4 – 20) / 2d6 + 2d10
17. J+L = (3d10 + 4 – 20) / 2d6 + 4d5
18. J+M = (3d10 + 4 – 20) / 2d6 + 4d6-4

But I have one more trick up my sleeve – instead of multiplying the result, I can multiply the number of dice to be rolled. This produces an additional set of 12 options to consider:

19. G1+K(g1a) = (d10+d6-1) / 2d6 + (8d10 -5) / 4
20. G1+L(g1a) = (d10+d6-1) / 2d6 + (16d5 – 5) / 4
21. G1+M(g1a) = (d10+d6-1) / 2d6 + (16d6 – 21) / 4
22. G2+K(g2a) = (d10+d6-1) / 2d6 + (16d10 -5) / 8
23. G2+L(g2a) = (d10+d6-1) / 2d6 + (32d5 -5) / 8
24. G3+M(g2a) = (d10+d6-1) / 2d6 + (32d6 -37) / 8
25. H1+K(h1a) = 3d10 / 2d6 + (6d10 -7) / 3
26. H1+L(h1a) = 3d10 / 2d6 + (12d5 -7) / 3
27. H1+M(h1a) = 3d10 / 2d6 + (12d6 -19) / 3
28. H2+K(h2a) = 3d10 / 2d6 + (10d10 – 11) / 5
29. H2+L(h2a) = 3d10 / 2d6 + (20d5 – 11) / 5
30. H2+M(h2a) = 3d10 / 2d6 + (20d6 – 31) / 5

Thirty options. There’s no way to test them all at the same time, even using something like AnyDice. But, just looking at them, I have one clear favorite already: H1+K(h1a) is much simpler ad uses a lot less dice than any of the alternatives. I also have a clear alternate preference, I+K, for similar reasons. H2+K(h2), H2+L(h2), J+K and J+M are third choices. Most of the others involve too many dice to be convenient.

Time for another set of compounded graphs looking at those 6 choices. A mistake made through the fuzzy thinking mentioned earlier has put them in reverse order, but oh well.

Analysis

J+M (red)

1. The shape is more or less right.
2. The average is right.
3. The tail isn’t long enough.
4. The minimum comes up as -6. That can be easily corrected but doing so wrecks the average. That can also be fixed but doing so makes the tail – which would have been improved by the first fix – too short again.

Rejected.

J+K (purple)

1. The shape is dominated by the addition.
2. The average is right.
3. The tail isn’t long enough.
4. The minimum comes up as -4. That can be easily corrected but doing so wrecks the average. That can also be fixed but doing so makes the tail – which would have been improved by the first fix – too short again.

Rejected.

H2+L(h2) (green)

1. The shape is more or less right.
2. The average is right.
3. The tail is long enough; the maximum comes up as 32.
4. The minimum comes up as 1. Not perfect but close enough.

Consider Further.

H2+K(h2) (blue)

1. The shape is dominated by the addition. It’s less of a bell curve.
2. The average is right.
3. The tail is the longest of all the contenders; the maximum result is 33.
4. The minimum comes up as -1. The formula can be tweaked to fix that, but it would throw the average out. That can also be fixed, at the cost of a fiddly little bit of arithmetic – every time the roll has to be made.

Rejected, but only because I already have a better candidate.

I+K (yellow)

1. The shape is dominated by the addition, even more than the other examples where that was the case.
2. The average is right.
3. The tail isn’t anywhere near long enough.
4. The minimum comes up as -1. That can be easily corrected but doing so wrecks the average. That can also be fixed but doing so makes the tail – already too short – even shorter, and adds some fiddly maths to the roll.

Rejected.

H1+K(h1a) (orange)

1. The orange is very similar to the green in shape; the peak is a little steeper.
2. The average doesn’t quite look right – it’s closer to 10.3 or something like that, even though 10 is marginally higher in probability, 11 is almost as high. If they were equal, I’d list the average as 10.5; since they aren’t, it has to be a little less than that.
3. The tail is perfectly satisfactory, though a little flatter and lower in probability compared to some of the other options trialed.
4. The minimum is shown as 0.

Rejected because of the average being off.

The wash-up: while several of the choices came close, there was only one that passed every test – H2+L(h2) (green), better known as 3d10 / 2d6 + (5 x 4d5 – 11) / 5.

Further Tweaking?

The maximum result of 32 is a little on the low side. It can be boosted by increasing the size of the roll on the top of the divisor to 4 or 5 d10. This would take the average from 16.5/7 to 22/7 or 27.5/7, respectively. Compensating for this would require further adjustment of the addition part to bring the average back down.

The bigger any such change, the harder this is to do. Things are already at the point where I was contemplating applying the x 5 multiplier to the divided roll so that the division can be handled more easily.

There is still the interval adjustment to incorporate, a further division by 10 once the roll has rounded off to whole numbers.

To be honest, though, my inclination at the moment is to leave well-enough alone and live with the smaller-than-desired maximum.

The Ultimate Modifier I

This formula – a process really – was worked out after simplifying the question by leaving out B. That has to get put back into the process, as follows:

(3d10 + Bias Modifier) / 2d6 + (5 x 4d5 – 11) / 5.

I was thinking that this modifier would be something like plus or minus 10. It’s intended to bias the results based on external non-random factors. But I could easily boost it to a higher plus or minus. This matters a bit because 1/7th of it will add to the average result (desired and intentional) and the whole of it adds to the maximum (also desired and intentional).

This is a factor that I was taking into consideration when mooting no change to the current die roll.

But, at the end of the day, there is a further option to keep up my sleeve in case I need it.

The Ultimate Modifier II

As well as applying the modifier to the divided roll, I could also apply a fraction of it to the end result IF the roll is above a threshold. Or even universally, to all rolls. The latter would take some of the randomness out of the roll and increase the effect of the non-random factors.

I don’t have to decide right now; I can leave that choice for when working on the system mechanics.

A further option to keep in that back pocket is not to base this adjustment on the non-random factors, but on the roll as a whole – if it were >2.0, and the net maximum got up to around 4 or 4.5, adding 0.1 to the result for every 0.5 or part thereof would increase the maximum considerably, and would be a lot simpler in terms of the overall mechanics.

I can even have it both ways – if the bias is above a certain threshold (to be determined, but probably +10 or better), I could use the bias as the basis of the bonus; if not, increasing the significance of the random element by basing the bonus on the overall roll as described.

In fact, I like this option because the bias, if it’s really significant, should mitigate the randomness. It would be a closer simulation of the real world.

Closing out this article

I thought it might be worth going right back to the basics and pointing out that not all divided die rolls have to be this complicated. The best way to illustrate that is by considering a much simpler one or two, just for the heck of it.

2d6/d6

This is one of the simplest you can imagine. Three dice.

Minimum: 2 on the 2d6. If you get a 1 on the 1d6, that’s an overall roll of 2. If you get a 2 on the 1d6, that’s 1. If anything more, that’s 0 – if you round down..Rounding matters because it greatly simplifies the maths of the division, enough that you can generally do it in your head. If, for example, you chose to round off instead of down, then results of 2, 3, or 4 on the single d6 give you a net result of 1.

Average: 7 on the 2d6. If you get a 1 on the 1d6, that becomes a result of 7. If you roll a 6, it becomes a 1. If you roll the average (3.5), it becomes a 2. If you measure the probability, that says that the peak probability is a 2, and that the range 1 to 2 is equal in overall probability to the range 2 to 7 – so the low results are going to be roughly 5 times more frequent than the higher ones.

Maximum: This is always when divided die rolls get interesting. Maximum on the 2d6 is 12. If you roll a 1 on the single d6, that’s an absolute peak result of 12. If you roll a 2, the overall result is a 6. If you roll average (3.5), the answer is 12/3.5 = 3.4, which becomes a 3. If you roll maximum on the denominator die, the overall result falls back to 2.

A good way to think about what’s happening is that the shape of the die roll being divided is changing with each different outcome on the single d6.

4d6/d6

Minimum: 4 on the 4d6. A roll of 1 on the denominator d6 yields an overall result of 4; a roll of 2 gives 2; a roll of 3 or 4 gives 1; what happen on a 5 or 6 depends on what rounding you mandate.

Average: 14 on the 4d6. A roll of 1 on the denominator gives a result of 14; a roll of 2 gives 7; a roll of 3 gives 4 2/3, which might be 5 or 4 depending on rounding; a roll of 4 gives 3; a roll of 5 gives 2 or 3 (rounding again); a roll of 6 gives 2 1/3 which rounds to 2. Peak probability is 14/3.5 = 4.

Maximum: 24 on the 4d6. A roll of 1 on the denominator gives the absolute maximum result of 24, and this will happen once in 6^(4+1) rolls, once in 7776 rolls. A roll of 2 gets 12. A roll of 3 gets 8. A roll of 4 gets 6. A roll of 5 gets 4.8 which could round to either 4 or 5. A roll of 6 gets a result of 4. The average result if you have rolled a maximum on the 4d6 is 24/3.5 = 6..86, which either rounds to 6 or 8.

It’s like rolling a d6 that can sometimes explode into 2, 3, or even 4d6 – if that d6 were weighted to produce low rolls.

One more.

2d6/d3

These results will look very familiar.

Minimum: 2 on the 2d6. A roll of 1 on the denominator d3 yields an overall result of 2; a roll of 2 gives 1; and a roll of 3 gets either 0 or 1 depending on rounding.

Average: 7 on the 2d6. A roll of 1 on the denominator gives a result of 7; a roll of 2 gives 3 or 4; and a roll of 3 gives 2 1/3, which rounds to 2. Peak probability is 7/2 = 3.5.

Maximum: 12 on the 2d6. A roll of 1 on the denominator gives the absolute maximum result of 12, and this will happen once in 3×6^(2) rolls, once in 108 rolls. A roll of 2 gets 6. A roll of 3 gets 4.

In other words, the result is effectively a d6 that has a 1 in 3 chance of doing double effect, and a 1-in-3 chance of being a weighted-low d4 – and both results will come up often enough to be noticed.

And this scales – you can replace any d6 with this divided roll.

This option for rolling damage ups the excitement of combat no end. Or you could replace the 3d6 used to roll stats with this.

Let’s look at that last one. The odds are that you’ll have one of each of the possible denominators – so 2d6/1, 2d6/2, and 2d6/3. There are enough dice to make this a genuine dumbbell probability. Your overall average is probably 7+3.5+2 = 12-13. But, should the denominators smile upon you, you could get a stat of 36; and, should they frown, a stat of 2. Is the risk worth it? A stat of 36 is awfully tempting…

In effect, the stat roll is a 6d6/3d3 divided roll. But you can take some of the wildness out – use d6 + 4d6/2d3, or 2d6 + 2d6/d3. The latter gives you a chance of effectively getting to roll 4d6 (4-14-24) instead of 3 (3-10.5-18) – but at the price of possibly rolling 2d6+1 or 2 (3,4-8,9-13,14).

These options are always food for thought.

Well, I made it to the end – but it took three times as long to write this post as it should have. The good news is that my head cold is substantially improved over the way it was when I started on this post. I think taking this option was the right choice to make.