Dice Probability Puzzle

https://www.omnicalculator.com/statistics/dice#how-to-calculate-dice-roll-probability

Looks like a GA to me,but I don't understand all I know about it.

Tx Solar, still looking through it, but it is not clear if the calculator can handle the specific puzzle. Maybe it is covered somewhere in the referenced links...

It is probably easier to solve the inverse, that no pair adds up to 10 or more, and then subtract from 1.

Hi Jorrie,

It's good to see you again. We haven't crossed paths lately - not even on the Physics Forums. All I can say is the probability of me being alive right now is over 50%, but I can't prove it.

-S

Hi S, good to hear from you as well!

Yes, even here at the southern tip of Africa the C-19 survival rate of the elderly is not good, so the wife and I stay in lockdown as much as we can.

A picture, if that helps...

Looks like the union of 3 triangular prisms + pyramid corner

Number = 3 prisms x 6 per layer x 3 layers + 5x3 + 3x2 +1x1

Combination cube seen from below for clarity.

It's a little easier to see by slicing the cube with the value on Die C. You always have the 6-5-4 triangle in the bottom corner above, the sums from A and B regardless of C. Adding C to the mix fills in combinations except for square on the top corner above. As C value gets smaller the excluded square in that layer gets larger.

The algorithm to generate the entire cube would be to populate all nodes, clear the lower right-hand squares for each value of C, and then set the 4-5-6 triangle nodes in the upper right corner for the combinations met by A and B alone. This algorithm is a multistep process (first subtractive, then additive), so I don't believe there is a formula that would produce the number of combinations of sums equal or greater than 10.

The combination cube by layers:

Layer 6, Die C = 6, Dice A and B = 4 to 6

Sum of 10 or more using Die C

Number = 6² - 3² = 27

Layer 5, Die C = 5, Dice A and B = 5 or 6

Sum of 10 or more using Die C

Number = 6² - 4² = 20, 4² combinations sum to less than 10

Layer 4, Die C = 4, Dice A and B = 6

Sum of 10 or more using Die C

Number = 6² - 5² = 11, 5² combinations sum to less than 10 using die C

Dice A and B = 5

Sum of 10 or more not using Die C

Number = 1 combination with A and B

Total = 11 + 1 = 12 combinations

Layers 1 - 3, Die C =1 - 3

No help from Die C, dice A and B are on their own...

3 layers x 6 per layer = 18 combinations from dice A and B.

Great Answer! The visualization is great, thanks Rixter. It gives the 77 combinations required.

Apart from my .js script that pops the 77 out, I have before done a laborious eyeball, pen and paper analysis from the table, to get the same 27+20+12+6+6+6 that your layers yield. Your combination cube is much superior.

Thanks, Jorrie. I think 4 dice is going to be more challenging. Maybe as the resident Relativity expert, you can visualize a 4D cube...?

here is the pattern for 2, 3, and 4 dice:

2 dice:

Self-explanatory - Only the 456 triangle

3 dice:

Algorithm:

Set all.

Clear cut-out square, 3x3 on level 6, 4x4 on level 5, 5x5 on level 4, 6x6 on level 3,...

Set the 456 triangle projected along edges terminating at vertex (6,6,6)

4 dice:

To visualize, we need a 6x6x6 cube for each value of the 4th die.

4th die = 6:

Set all;

Clear 3x3x3 cube at (1,1,1) vertex

Set 456 trianglar prism along edges terminating at vertex (6,6,6)

4th die = 5:

Set all;

Clear 4x4x4 cube at (1,1,1) vertex

Set 456 trianglar prism along edges terminating at vertex (6,6,6)

4th die = 4:

Set all;

Clear 5x5x5 cube at (1,1,1) vertex

Set 456 trianglar prism along edges terminating at vertex (6,6,6)

4th die = 1, 2, or 3: (4th die cannot contribute to pair with >=10)

Set all;

Clear 6x6x6 cube at (1,1,1) vertex (everything).

Set 456 triangular prism along edges terminating at vertex (6,6,6)

Note this is identical to 3 dice situation.

Conclusion: For number of dice N, you have the 456 hyper-prisms along the hyper-edges corresponding to the combinations between 2 dice >=10. In addition to this, you add all combinations but a square for 3 dice, all but a cube for 4 dice, etc.

In a standard 3-dice roll, what is the probability of any one or more pairs adding up to 10 or more?

In trying to find a formula for the probability for one or more pairs, "one or more pairs" is the troublemaker here. For example, in the 3 dice combination cube, the "456 prisms" and the combinations added by the 3rd die (the excluded squares) overlap. For example, we only count "555" as one combination when it is actually 3. Just an observation.

I'm still trying to wrap my brain around your 4-dice post.

In attempting to find a formula, I have identified the counting wedges in your 3-dice populated cube. I can see a pattern, but not quite a general one, i.e. one that would also work for 2 and 4 dice, for which we know the count.

I did not look from the angle of the non-counting wedges yet.

The wedge on the left are combinations involving dice B & C. Die A is < 4 and can't contribute. Likewise, the wedge on the right involves dice A & C, where Die B is < 4. It's the same as the "456 case" for 2 dice. In the center, the bottom 3 layers of the wedge shown "6*(3+2+1)" would be the corresponding wedge involving dice A & B where C < 4. Since dice A, B, and C are interchangeable, you should have a symmetry in the combination cube.

Here's another way of looking at it. You have a wedge [6*(1+2+3)] along the top left edge of the combination cube - combinations involving dice B & C. The same size wedge runs along the top right edge, involving dice A & C. Likewise, there is a wedge running vertically of combinations A & B. Where these three wedges overlap, there is more than one combination. The total number of combinations should be 36 + 36 + 36 = 108. To get the "one or more" number, you need to subtract 1 for each point in the combination cube that is within the overlap of two of these wedges and 2 for each point that is in the intersection of all three wedges.

The overlap of the three wedges is within a 3x3x3 cube where all three dice have 4 or greater. The number of wedges overlapping for each point:

Top layer and overlap:

Middle layer and overlap:

Bottom layer and overlap:

There are 11 threes and 9 twos, or 11x2 + 9x1 = 31 extra combination due to overlap. 108 total combination - 31 extra = 77.

https://physics.info/curve-fitting/

SE, I don't quite know how to interpret your graph in terms of the dice problem. Could you enlighten?

Yes it's an attempt to visualize the possibilities as a graph, and then translate that to a mathematical formula that would hold true with any amount of dice...but the math is beyond my capabilities so far...I think the graph needs more work, it was just a first draft...It might be a blind alley, just exploring...Do you think it's possible?

Possibly, but for me your graph goes the wrong way, if the top scale is the number of dice. With 2 dice it is 1 in 6, with 3 dice it is slightly better than 1 in 3 (~35%) and with 4 dice it is slightly better than 1 in 2 (~53%).

Yeah I was using that omni calculator, it doesn't seem to be accurate....I will regroup and try again....

@SolarEagle: I have the actual count (and probability) for 2,3,4 and 5 dice and we know it must approach unity for a large number of dice. A curve fitter yielded this, with x the number of dice. I have guessed 0.995 for 10 dice.

For what it's worth, the lowest order reasonable fit equation is:

y = -0.280837 + 0.2504673*x - 0.01306944*x² + 0.00007993827*x^{^3}

I wrote: "@SolarEagle: I have the actual count (and probability) for 2,3,4 and 5 dice and we know it must approach unity for a large number of dice. A curve fitter yielded this, with x the number of dice. I have guessed 0.995 for 10 dice."

I think I have overestimated the 10 dice case considerably. After letting my script calculate the count for 6, 7 and 8 dice as well, it looks more like this, with ~0.92 for 10 dice.

y = -0.3463251 + 0.298422*x - 0.02251317*x^2 + 0.0005332211*x^3

It has no practical value, but is nevertheless intriguing..

Half dice? The dots all sit at integers.

"It has no practical value, but is nevertheless intriguing.."

Ha ha , yeah we get a lot of that here....

Your question poses "...adding up to 10 or more?" Your script answer "146 155 156 164..." is valid for 11 or more, but omits 136, 145, 226, 235, 244, 334

The question was perhaps not clearly enough stated, but the original was:

"In a standard 3-dice roll, what is the probability of any one or more pairs adding up to 10 or more?"

That rules out 136 etc.

Maybe a clearer formulation would have been:

"In a standard 3-dice roll, what is the probability of at least one of the three pairs adding up to 10 or more?"

A roll of 666 counts as only one pair, not 3 pairs.

as stated, you were looking for any series that was 10 or more, it appears that you missed the first iteration of each series: 145, 154, 163, 226...and so forth

Nope, read question carefully. No pair in the numbers that you have given adds up to 10 or more.

yes sir, missed that.

I have to admit that this math is beyond my present skill level, but I have some questions :

1. What is a " standard " roll, versus a nonstandard roll ?

2. In monopoly & craps you use 2 dice, what game of chance are 3 dice used ?

3. What would be the difference in outcome when using new dice rather than used dice ?

4. If a player rolled 3 dice three times and received an an average, how does the average change when the same player rolls 3 dice 676 times in relation to fatigue in the wrist and forearm ?

@tonyhemet:

1. "Standard 3-dice roll" was just shorthand for true dice with six sides.

2. No game that I know of uses 3 dice. It was just an exercise in probabilities and showing how such a simple-looking puzzle can be so difficult to solve - and looking for advise from the out-of-the box thinkers of CR4. Which I did get. :-)

3. As long as dice are true, age/usage should not matter.

4. Fatigue should not change the results. Depending on what average you are asking about, rolling 3 dice three times will give you an average that is quite far off the true probability count, while rolling 676 times will give an average quite close to that. There are 6³=216 possible combinations of 3 dice.

There is (or was) a game called Chuckaluck that uses three dice.