This die defines your luck in life, and is rolled by your god of choice at every important event: roll a (0) you fail miserably, (6) you succeed briliantly, (4) you pass, barely.
Would you rather have:
A: a 000666 die?
B: a 111555 die?
C: a 222444 die?
D: a 333333 die?
Option C “222444”.
I coded successes as positive values and failures as negative values. I arbitrarily used a doubling for each greater success/failure level and came up with the following value coding:0 1 2 3 4 5 6 -8 -4 -2 -1 +1 +2 + 4 This results in the following expected values for the offered dice:
A: -2
B: -1
C: -1/2
D: -1All dice are bad, option C is the least bad. And this kinda makes sense. For option A, you may have a fantastic success, but you are also just as likely to complete crash out. And a “crash out” should happen after very few rolls. Option B is a slightly less extreme version of this, but any gains from the 5 results should be more than wiped out by the 1 results. And those should be happening with similar frequency. Option C is again the same thing, but with a slower circling of the drain. 4 results let you recover some, but the 2 wipes out that 4’s benefits and more resulting in a slow decline. And option D is just straight out bad, every result is a failure.
It seems that the only good choice is not to play. ;-)
EDIT: I realized, I made a mistake in my original numbers, I forgot to divide by 6. And this is why coffee should come before math. The conclusions are still the same, but the numbers are different. I’ve corrected those.
That’s so cool. If you could design a die with average face value of 3, min face value of 0, max face value of 6, what would be the best die?
Do note that I made a mistake in the original post, but the conclusion was still the same. I forgot to divide the Expected Value (EV) for all dice by 6 (the number of faces).
If you could design a die with average face value of 3, min face value of 0, max face value of 6, what would be the best die?
I’m not sure how to prove this empirically, but playing with it on my whiteboard I get a sense that the die 444222 is going to have the best EV, under the given constraints and my value assignments. The real kicker is “average face value of 3”. Given that constraint, you will never be able to create a die with a positive or even zero EV using my values. Consider die 333333 and each face’s value:
3 3 3 3 3 3 -1 -1 -1 -1 -1 -1 This die has an average face value of 3 ( (3 * 6) / 3) and we can consider changing any face up or down. But, in order to keep the average a 3, moving one face up one number requires we move a different face down one number and vice-versa. For example, if we push one face from a 3 to a 4, we must also pull one face from a 3 to a 2 to balance out the average. And because the value for positive value numbers (4, 5, 6) starts off one doubling behind the values for the negative value numbers (3, 2, 1, 0), going any further than 4 in the positive direction on a face means that another face will be pushed down far enough to cancel out the benefit of going to a 5 or beyond.
To look at it another way (the way I did on my whiteboard), let’s just consider a two sided die (a coin flip). Using the same values for each number, we can consider a 33 coin. This has an EV of -1 ( (-1 * 2) / 2) and an average of 3 ( (3 * 2) / 2 ). Now, move the numbers, but keep the same average of 3. Moving to a 42 coin changes the EV to -1/2 ( (+1 + (-2)) / 2 ) and the average is still 3 ( (4 + 2) / 2 ). The EV got better. So, let’s take another step in each direction. We get a 51 coin with an EV of -1 ( (+2 + (-4)) / 2) and the average is unchanged at 3 ( (5 + 1) / 2 ). And going to a 60 coin takes us to an EV of -2 ( (+4 + (-8)) /2 ) with a average of 3 ( (6 + 0) / 2 ). This means that the best coin for this scenario is a 42 coin. Taking that coin idea back to the die, you can think of the die as a bunch of linked coins. If you want one face to be a 5 the one face must be a 2, which would be worse than having the pair of faces be a 4 and a 2. So, to maximize the EV, you want to create a bunch of 42 pairs.
Of course, we could fiddle with multiple faces at once. What about a 622233 die. Well, it gets worse. EV is -2/3 ( +4 + (-2) + (-2) + (-2) + (-1) + (-1))/6).
Maybe a 522333, EV is -5/6 ( (+2 + (-2) + (-2) + (-1) + (-1) + (-1)) / 6). Again, since lower numbers get a more negative valuation faster than higher numbers get a positive valuation, you just really don’t want to let numbers get any lower than necessary. The 42 paring just happens to hit a sweet spot where that effect isn’t yet pronounced enough to cause the EV to drop off.So ya, while I don’t know the maths to prove it. I’m gonna say that the 444222 probably maximizes the EV under the given model.
I’m going to go with option C with the following reasoning:
I’m going to assigned (somewhat arbitrarily) the following values to each outcome:0 | 1 | 2 | 3 | 4 | 5 | 6
- | - | - | - | - | - | - -8 | -4 | -2 | -1 | +1 | +2 | +4
This codes failure outcomes as having a negative value and success outcomes as having a positive value, with the value doubling for each increase in success/failure. So, the expected value for the 4 options are:
A: -12 B: -6 C: -3 D: -6
Meaning all of the options are bad, but the least bad is option C. And this makes some intuitive sense. You have an equal chance of success or failure and while no success will be all that spectacular, you will also never suffer a spectacular failure. Die A seems like an interesting choice, but you would expect to suffer a catastrophic failure about half the time and that may end your ability to keep rolling. Die B is a slightly less bad version of die A, and may be an ok choice, if a 1 result doesn’t result in you no longer being allowed to roll. Though, if you are not able to stop rolling whenever you want, a 5 outcome is likely to be wiped out fairly soon. Die D is just straight out bad. It always results in a failure; so, there is no point playing.
A. the 000666 dice.
Failure is capped at rock bottom, whereas the possibilities for success are infinite, so the brilliant success should outweight the miserable failure.
I assume this is the dice for people like Trump. Fail miserably and get convicted in your court case, then get elected president of the united states anyway.
It’s important that you know: the singular of “dice” is “die”.
Thanks, corrected it
That’s four choices, let me go roll my D4
Considering barely passing is at 4, the 333333 die has no appeal. I would go with 000666 because it’s a 50/50 to kick ass the most possible.
Would still rather the standard 123456 of everyday life (lol more like 500001)
(4) you pass, barely
That implies that 1, 2, and 3 are all failing (perhaps with different degrees of embarrassment). If all failure is equivalent in practice, you might as well maximize the non-failing outcomes and go with A.
(perhaps with different degrees of embarrassment)
Exactly. Roll a (0) you’re paralysed for life, roll a (3) you break an ankle, roll a (6) you perfectly execute the salto on your first try.
That seems pretty arbitrary
0-2 would indicate negative, 0 being extremely bad and life changing and 2 pretty bad (worse than broken bones)
3 which should be neutral is broken bones
4-6 indicates success, but 4 seems to be just barely passing (e.g. landing it but with a fumble and falling) which doesnt seem to balance out a broken bone for 3 or something worse at 2
That seems pretty arbitrary
The entire question is?
