# Dice Game: 6-sided die vs 20-sided die

• Master1022
In summary: Both set of dice average 10.5. If the goal is to get the highest sum, then they are equal.However, would the variance come into consideration if we had to choose one set with which to play the game?Not really. It would only indicate if you lost (won) a roll by a lot or a little.

#### Master1022

Homework Statement
There is a game where one person rolls a 20-sided dice and another person rolls three 6-sided dice. They both roll their dice and whoever gets a sum of numbers wins the game. Which person would you rather be? Is it a fair game?
Relevant Equations
Expectation and variance
Hi,

I was taking a look at the following question.

Question: There is a game where one person rolls a 20-sided dice and another person rolls three 6-sided dice. They both roll their dice and whoever gets a higher number (for 20-sided die) OR sum of numbers (for three 6-sided dice) wins the game. Which person would you rather be? Is it a fair game? Assume the dice are all fair

Attempt:
I think the general concept is about looking at expectation and variance for both cases and then make a decision from there.

So for the person with the 20-sided die, the expectation and variance are given by:
$$E(\text{20 sided die}) = \frac{n + 1}{2} = \frac{21}{2} = 10.5$$
$$Var(\text{20 sided die}) = \frac{n^2 - 1}{12} = \frac{(20)^2 - 1}{12} = \frac{399}{12}$$

For the person with the three 6-sided dice:
$$E(\text{three 6 sided die}) = E(X_1 + X_2 + X_3) = 3 \cdot E(\text{one 6 sided die}) = 3 \cdot \frac{7}{2} = 10.5$$
$$Var(\text{three 6 sided die}) = Var(X_1 + X_2 + X_3) = 3 \cdot \frac{n^2 - 1}{12} = 3 \cdot \frac{(6)^2 - 1}{12} = \frac{35}{4}$$

Thus, both of the distributions have the same expected value/mean. Therefore, perhaps we would rather be the person who chooses the lower variance side, which is the three 6-sided die. Does that seem reasonable?

Also, what does the 'is the game fair' part mean? Is that basically asking if the mean and variance are the same? If so, then based on the calculations above, I would say the game isn't fair...

Any help would be greatly appreciated.

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It kind of depends on the number to win.If the number is 10 or 11, then
- the 1D20 has a 1-in-20 chance of winning
- the 3D6 have a 27-in-216 (1-in-12.5) chance of winning .

If the number is 3 or 18, then
- the 1D20 has a 1-in-20 chance of winning
- the 3D6 have a 1-in-216 chance of winning.

Heck, if the number is 1, the 3D6 has a zero chance of winning.

Or am I misunderstanding the rules?

DaveC426913 said:
It kind of depends on the number to win.

If the number is 3 (or 18), then
- the 1D20 has a 1:20 chance of winning
- the 3D6 have a 1:216 chance of winning.

Heck, if the number is 1, the 3D6 has a zero chance of winning.

Or am I mis-understanding the rules?
Hi @DaveC426913 , thanks for the reply! Sorry, I think there was a word missing from my original question. Basically, the person with the three 6-sided die adds up their 3 numbers and compares the sum to the 1 number from the person who rolled the 20-sided die. Then whichever number is higher wins the game.

Apologies - I hope this is more clear

Master1022 said:
whichever number is higher wins the game.
Ah.

There's a super-duper easy way to answer this question, but your teach might not like it.

Master1022
DaveC426913 said:
Ah.

There's a super-duper easy way to answer this question, but your teach might not like it.
Does it have something to do with the fact that the 20-sided die has two cases which cannot be beaten by the other player (19 and 20)? Or should I continue thinking along the mean and variance line?

For both dice sets, the probability curve is symmetrical.

You could just as easily make the rule "whoever gets the lowest score wins" and you have the exact same problem with the exact same solution - just mirrored.

I could produce a graph of the probabilities and show them to you horizontally flipped or unflipped, and you would not be able to tell me which rule is applied.

Therefore, neither dice set can have an advantage.

Master1022 and jbriggs444
DaveC426913 said:
For both dice sets, the probability curve is symmetrical.

You could just as easily make the rule "whoever gets the lowest score wins" and you have the exact same problem with the exact same solution - just mirrored.

I could produce a graph of the probabilities and show them to you horizontally flipped or unflipped, and you would not be able to tell me which rule is applied.

Therefore, neither dice set can have an advantage.
Okay thank you, I agree with regards to the symmetry. However, would the variance come into consideration if we had to choose one set with which to play the game?

Master1022 said:
Okay thank you, I agree with regards to the symmetry. However, would the variance come into consideration if we had to choose one set with which to play the game?
Both set of dice average 10.5. If the goal is to get the highest sum, then they are equal.

Master1022
Master1022 said:
However, would the variance come into consideration if we had to choose one set with which to play the game?
Not really. It would only indicate if you lost (won) a roll by a lot or a little. If the expected values are equal, then the game is fair.

Master1022 and phyzguy
Note that my "solution" is off-topic for the Mathematics Homework forum (though it might be on-topic in a Mathematics or Logic forum).

Master1022 said:
Okay thank you, I agree with regards to the symmetry. However, would the variance come into consideration if we had to choose one set with which to play the game?
Symmetry, not variance, is the key.

Master1022
DaveC426913 said:
I'm not sure the student is supposed to crunch the numbers here, in terms of probability calculations. If anything, I'd do a computer simulation to check the symmetry argument.

PeroK said:
I'm not sure the student is supposed to crunch the numbers here, in terms of probability calculations.
I guess. My solution bypass any math completely. Not sure if an elegant, intuitive proof is allowed.
PeroK said:
If anything, I'd do a computer simulation to check the symmetry argument.
You mean check that a D20 has a symmetrical distribution, and that 3D6 has a symmetrical distribution?
That's too trivial to warrant a simulation is it not? It's self-evident.

DaveC426913 said:
I guess. My solution bypass any math completely. Not sure if an elegant, intuitive proof is allowed.

You mean check that a D20 has a symmetrical distribution, and that 3D6 has a symmetrical distribution?
That's too trivial to warrant a simulation is it not? It's self-evident.
Your solution is probably what is expected.

There's no harm in running a simulation just to confirm. Even just writing the code may give an insight into why the symmetry argument works.

jedishrfu
When in doubt, write a program to figure it out.

Master1022, Dale and berkeman
jedishrfu said:
When in doubt, write a program to figure it out.
Or use Excel...

jedishrfu
berkeman said:
Or use Excel...
Gloria in excel-sis deo!

SammyS, jedishrfu and berkeman
If you turn it into a three person game, variance becomes a useful heuristic and the d20 has an advantage.

PeroK
jbriggs444 said:
If you turn it into a three person game, variance becomes a useful heuristic and the d20 has an advantage.
That depends on what the third player has, I imagine!

jbriggs444
PeroK said:
That depends on what the third player has, I imagine!
Indeed. We might constrain the third player to a set of dice with an expectation of 10.5. Or to a choice between 3d6 and 1d20.

jbriggs444 said:
Indeed. We might constrain the third player to a set of dice with an expectation of 10.5. Or to a choice between 3d6 and 1d20.
I see what you mean. The D20 player can't be worse, it seems. If the third player has a two-sided die showing 0 and 21, then the D20 and the 3xD6 are still equal.

PeroK said:
I see what you mean. The D20 player can't be worse, it seems. If the third player has a two-sided die showing 0 and 21, then the D20 and the 3xD6 are still equal.
I think that against an intentionally sub-optimal and asymmetric third player with something like a 5, 5, 5, 6, 21, 21 that you can get an advantage with a symmetric low variance (3d6) over a symmetric high variance (1d20).

jbriggs444 said:
I think that against an intentionally sub-optimal and asymmetric third player with something like a 5, 5, 5, 6, 21, 21 that you can get an advantage with a symmetric low variance (3d6) over a symmetric high variance (1d20).
I did a Python simulation and there seems to be a slight edge to D20 there. For simplicity I didn't count draws, only when one player won.

I can't immediately find any third option that gives 3D6 an advantage over D20.

PS I assigned the third player a fixed score of everything from 1-20 and there is no score that creates an advantage for 3D6. The higher the number, the greater the advantage to D20.

Therefore, there is no combination of numbers for the third player that gives 3D6 an advantage.

jbriggs444
PeroK said:
There's no harm in running a simulation just to confirm.
Here's a Python simulation with two players as in the original problem.
Python:
from random import choice

SixDie = [1, 2, 3, 4, 5, 6]
TwentyDie = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
SDTotal = 0
TDTotal = 0

NTrials = 1000000
for i in range(0, NTrials+1):
SDScore = choice(SixDie) +  choice(SixDie) +  choice(SixDie)
SDTotal += SDScore
TDScore =  choice(TwentyDie)
TDTotal += TDScore

print("Six dice average: ", {SDTotal / NTrials})
print("Twenty die average:  ", {TDTotal / NTrials})
Output with 1,000,000 trials is
Code:
Six dice average:  {10.501267}
Twenty die average:   {10.501052}
Running time in Python is about 8 seconds.

The game is who wins the most; not the highest average. The average scores are clearly equal.

Anyway, if a third player is involved, then in general the 20-sided die is better than the three 6-sided dice - or, at least, no worse.

PeroK said:
I did a Python simulation and there seems to be a slight edge to D20 there. For simplicity I didn't count draws, only when one player won.
I had forgotten that the aim was to see which was the winner. I edited my Python code to count wins, and saw a slight advantage for 1D20, likely for the reason already given, that for a 20-sided die, there are possible scores of 19 and 20 that can't be matched by three 6-sided dice.

If we modify the game with four 5-sided dice (no such Platonic solid exists, but never mind) vs. the 20-sided die, the four 5-sided dice have a clear advantage. Each has an expected value of 3, so the expected value of four of them would be 12, which is higher than the 10.5 expected value for the 20-sided die.

Mark44 said:
I had forgotten that the aim was to see which was the winner. I edited my Python code to count wins, and saw a slight advantage for 1D20, likely for the reason already given, that for a 20-sided die, there are possible scores of 19 and 20 that can't be matched by three 6-sided dice.
By a symmetry argument the game is equal. The automatically winning scores of 19 and 20 are counterbalanced by the losing scores of 1 and 2. Likewise scores of 18 and 3 counterbalance etc.

Master1022 and pbuk
PeroK said:
The game is who wins the most; not the highest average. The average scores are clearly equal.

Anyway, if a third player is involved, then in general the 20-sided die is better than the three 6-sided dice - or, at least, no worse.
Can you specify how the third player is involved? I am confused by this.

Master1022
FactChecker said:
Can you specify how the third player is involved? I am confused by this.
You have a third player with a third way of scoring. It could be anything. Absolutely anything. Then, excluding draws between two or more players, you count how often each player wins outright.

The hypothesis is that the 20-sided die is never worse than the three 6-sided dice. And, in general, is almost always better.

PeroK said:
You have a third player with a third way of scoring. It could be anything. Absolutely anything. Then, excluding draws between two or more players, you count how often each player wins outright.

The hypothesis is that the 20-sided die is never worse than the three 6-sided dice. And, in general, is almost always better.
Is the third player rolling another 20-sided die, or another set of three 6-sided die, or doesn't it matter?
Before I think hard about it, I want to know what I am thinking hard about. :-)

FactChecker said:
Is the third player rolling another 20-sided die, or another set of three 6-sided die, or doesn't it matter?
It could be those or anything at all. They might get a fixed score of 15 every time; or scores of 5 or 10 with equal probability; or, absolutely anything.

PeroK said:
It could be those or anything at all. They might get a fixed score of 15 every time; or scores of 5 or 10 with equal probability; or, absolutely anything.
So, as long as the winner has to beat two others, it is an advantage to have a larger variance than at least one other (and not less than others). Is that what you are saying? I'll buy that.
(That is, given equal means.)

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PeroK
Mark44 said:
NTrials = 1000000
for i in range(0, NTrials+1):
That loop will execute 1,000,001 times.