Need Help!

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Need Help!

Postby BadgerJelly on July 14th, 2020, 6:20 am 

Very quick math problem.

I’m making another board game and cannot figure out probability :(

If I roll 3 D6 what is the probability that two of them will add up to 10+?

(eg. 4,4,4 would only give total of 8, and 4,5,4 would only give total of 9, BUT 4,4,6 would give me the 10 I need.)

Thanks if you can throw me a formula for this and other dice probabilities involving rolling any number of dice.
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Re: Need Help!

Postby doogles on July 15th, 2020, 3:25 am 

I couldn't resist the challenge Badger.

My bush arithmetic may be a starter.

So you throw 3 dice and only 2 of them have to add up to ten.

This leaves you with a 4+6 or a 5+5 in any 2 of the 3 dice.

So the chances of throwing 4, 5 or 6 in the first die is 3 in 6 or 50%.

The chances of throwing a 6, 5 or 4 in the second die is also 50%, but the odds of them coming in the reverse order to match with the 4, 5 or 6 are 1 in 6 x 5 x 4 (= in 120)

So with 1 in 2 for getting a 4,5 or 6 with the first die, and the odds of getting a matching 6,5 or 4 with the second die of 1 in 120, the odds of getting a 10 with 2 dice only are 1 in 240.

Throwing a 3rd die is irrelevant as far as getting your sum of 10 from 2 dice, but in effect it gives you a second chance of matching a second die with the first. So it halves your odds of getting a corresponding match with your first die.

Thus the odds of throwing 3 dice and getting a sum of 10 with two of them becomes 1 in 120, according to my bush mathematics.

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Re: Need Help!

Postby charon on July 15th, 2020, 4:19 am 

I have to admit I'm fairly intrigued by this. I couldn't possibly begin to do the mathematics but - does this stuff actually work?

I mean, if I threw 3 dice 120 times I'm pretty sure that two of them would add up to 10 a lot more than once. I could be wrong but I'd be very surprised if they didn't.

Or am I misunderstanding this completely?
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Re: Need Help!

Postby BadgerJelly on July 15th, 2020, 4:57 am 

charon » July 15th, 2020, 4:19 pm wrote:I have to admit I'm fairly intrigued by this. I couldn't possibly begin to do the mathematics but - does this stuff actually work?

I mean, if I threw 3 dice 120 times I'm pretty sure that two of them would add up to 10 a lot more than once. I could be wrong but I'd be very surprised if they didn't.

Or am I misunderstanding this completely?


It’s certainly more likely than 1 in 120!

When I do twenty rolls I land between 25-50% success rate. It seems to be around 35-40% on average.

The odds for two dice is 1 in 6. Adding another dice would increase these odds.
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Re: Need Help!

Postby BurtJordaan on July 15th, 2020, 6:10 am 

BadgerJelly » 15 Jul 2020, 10:57 wrote:When I do twenty rolls I land between 25-50% success rate. It seems to be around 35-40% on average.

The odds for two dice is 1 in 6. Adding another dice would increase these odds.

Yip, the theoretical average value is 35.65%.

I also don't know the equation, but with only 3 dice, there are only 216 possible outcomes. A little program that runs through them all counted 77 outcomes with 10+ on any 2 dice. Better than 1 in 3 rolls.
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Re: Need Help!

Postby charon on July 15th, 2020, 9:26 am 

BadgerJelly » July 15th, 2020, 9:57 am wrote:
charon » July 15th, 2020, 4:19 pm wrote:I have to admit I'm fairly intrigued by this. I couldn't possibly begin to do the mathematics but - does this stuff actually work?

I mean, if I threw 3 dice 120 times I'm pretty sure that two of them would add up to 10 a lot more than once. I could be wrong but I'd be very surprised if they didn't.

Or am I misunderstanding this completely?


It’s certainly more likely than 1 in 120!

When I do twenty rolls I land between 25-50% success rate. It seems to be around 35-40% on average.

The odds for two dice is 1 in 6. Adding another dice would increase these odds.


So are we saying that the odds arrived at by the maths calculation aren't really true or realistic? Or are the calculations wrong? Or it doesn't really matter?
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Re: Need Help!

Postby BurtJordaan on July 15th, 2020, 10:26 am 

charon » 15 Jul 2020, 15:26 wrote:So are we saying that the odds arrived at by the maths calculation aren't really true or realistic? Or are the calculations wrong? Or it doesn't really matter?

The outcome of 77 out of 216 rolls is accurate, but all it says is that it will approach that ratio over a very large number of rolls. That's how probabilities work.

As far as I can recall, the probability of getting 77 or more after 216 rolls is 63.3 %, but I may be wrong.
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Re: Need Help!

Postby charon on July 15th, 2020, 10:36 am 

Thanks. Sounds fairly accurate, then.
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Re: Need Help!

Postby BadgerJelly on July 15th, 2020, 11:57 am 

A formula would still be nice.
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Re: Need Help!

Postby TheVat on July 15th, 2020, 3:37 pm 

(deleted, due to misreading OP)
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Re: Need Help!

Postby doogles on July 15th, 2020, 6:01 pm 

I made a gross error in my earlier bush maths. I actually calculated the odds of 6, 5 and 4 coming up in that sequence for throws of the second die. My figure was way out. It intrigued me so much I woke up in the middle of the night thinking about it.

Using TheVat's method, my P1 is also 0.5

But the my P2 is not 0.33. There are 36 (6x6) possible combinations of 2 digits when 2 dice are tossed eg 1,2 1,3 ... 4,1 4,2 etc. There can only be one 4,6 one 5,5 and one 6,4 according to my calculations. So P2 has to be 3/36 (= 0.08333, and not 0.33).

It's driving me silly. Although I think TheVat is on the right track, I can't see any error in my P2 here and would appreciate someone pointing out any such error before proceeding with the E value.

I'm off to work, so won't be able to respond for some hours.
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Re: Need Help!

Postby BadgerJelly on July 15th, 2020, 9:49 pm 

TheVat wrote:Hint: which die has which number (of the ten making pair) is immaterial. Order does not change value.
(and remember P1 already has eliminated all rolls of a 1,2, or 3, since they never yield a 10 pair, so those rolls are already out of the picture)

To get P2:

Combinations producing ten: 55, 64
Combinations not: 54, 44, 66, 65,

So: 2 combinations out of a total of 6 = .33


(if order matters, then we have:

55, 64, 46

54, 45, 44, 66, 65, 56. ....

But, as you see, the percent is unchanged, because you have
3 combinations out of a total of 9 = .33 = no different)


4+4 doesn’t equal 10. The odds of getting 10+ on two dive is 1 in 6. I’ve no idea how to crunch the number for what I want though.

Edit: just noticed what you meant. What makes this tricky to crunch is the issue of 4,4 requiring a 6 roll, 4,5 requiring a 6 or 5 roll and 6,1/2/3 requiring a 4,5 or 6 roll. It’s likely to be a VERY messy formula right?
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Re: Need Help!

Postby BurtJordaan on July 16th, 2020, 1:46 am 

Maybe the list of combinations that fit the problem will help someone with a statistical mind.
The following 77 outcomes yield 10 or above from any pair (algorithm shown at end of list):
1 4 6
1 5 5
1 5 6
1 6 4
1 6 5
1 6 6
2 4 6
2 5 5
2 5 6
2 6 4
2 6 5
2 6 6
3 4 6
3 5 5
3 5 6
3 6 4
3 6 5
3 6 6
4 1 6
4 2 6
4 3 6
4 4 6
4 5 5
4 5 6
4 6 1
4 6 2
4 6 3
4 6 4
4 6 5
4 6 6
5 1 5
5 1 6
5 2 5
5 2 6
5 3 5
5 3 6
5 4 5
5 4 6
5 5 1
5 5 2
5 5 3
5 5 4
5 5 5
5 5 6
5 6 1
5 6 2
5 6 3
5 6 4
5 6 5
5 6 6
6 1 4
6 1 5
6 1 6
6 2 4
6 2 5
6 2 6
6 3 4
6 3 5
6 3 6
6 4 1
6 4 2
6 4 3
6 4 4
6 4 5
6 4 6
6 5 1
6 5 2
6 5 3
6 5 4
6 5 5
6 5 6
6 6 1
6 6 2
6 6 3
6 6 4
6 6 5
6 6 6
count = 77

Here's the little algorithm that produced it:

number=10
count=0
for a = 1 to 6
_for b = 1 to 6
__for c = 1 to 6
___if((a+b >= number) or (a+c >= number) or (b+c >=number)) then
____output a,b,c
____increment count
__ end if
__next c
_next b
next a
output "count = " count

The underscores are just to make leading spaces show on this forum, to keep structure looking sensible.
Last edited by BurtJordaan on July 16th, 2020, 6:33 am, edited 4 times in total.
Reason: added algorithm
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Re: Need Help!

Postby doogles on July 16th, 2020, 6:41 am 

Thank you BurtJordaan for listing all 216 possibilities, but I'm still annoyed that I can't get my head around it to understand it clearly. I have been making the error of thinking that we needed the odds of 2 numbers adding up to 10 instead of 10-plus.
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Re: Need Help!

Postby BurtJordaan on July 16th, 2020, 7:01 am 

My pleasure. The algorithm only listed the 77 "good" ones, not all 216 of them.
Analytically it's a tough little problem - I still hope someone finds the general solution...
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Re: Need Help!

Postby charon on July 16th, 2020, 8:05 am 

the general solution


Oh, that's probably roll 'em and keep your fingers crossed :-)
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Re: Need Help!

Postby TheVat on July 16th, 2020, 9:02 am 

Sorry, I had not registered that you wanted a formula for getting ten PLUS. MY CALCULATIONS WERE BASED ON THE MISTAKEN IDEA YOU JUST WANTED IT FOR TWO DICE THAT SUMMED TO TEN. PLEASE DISREGARD MY POSTS, WHICH ADDRESS A SIMPLER PROBLEM.
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Re: Need Help!

Postby TheVat on July 16th, 2020, 8:30 pm 

Chances of any die coming up 456 = .5

Chances, for two dice, among those 456 rolls, of sums of ten plus = .6

So, with just two dice, chances = (.5 x .6) = .30

So, add a third die. Doubles chances overall. = .60

Am I nuts, or is that last bit about doubling chances intuitive?
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Re: Need Help!

Postby BurtJordaan on July 17th, 2020, 1:41 am 

TheVat » 17 Jul 2020, 02:30 wrote:Am I nuts, or is that last bit about doubling chances intuitive?

Can't say about trace of nut between either of us, but overnight my subconscious came up with what looks like a correct formula. If correct, it supports one nut. If false, it doubles the odds to at least two nuts. ;-)

My formula for this particular problem: probability = (62+52+42)/63 = 77/216 = 0.3565.

The argument is as follows: out of 63 = 216 rolls, any specified pair has a 1 in 6 chance to add up to 10+, hence 62 = 36 combinations.
There are 3 pairs in one throw, but they are not independent, in the sense that two or three pairs may qualify in the same throw, but counts as only 1 occurrence (as I understand Badger's game). So the second pair that you pick adds only 25 occurrences and the third pair only 16 extra occurrences.

I guess to generalize it to N dice, with 6N possible combinations, will need another subconscious occurrence, with unknown probability...
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Re: Need Help!

Postby doogles on July 18th, 2020, 2:47 am 

I've added a Table at the end of this post, TheVat, that sets out long-hand possibilities with just two dice.

Obviously the number of possibilities where 2 dice produce a total of 10+ is 6/36 (= 1/6 or 0.167 or 16.7%).
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Re: Need Help!

Postby doogles on July 18th, 2020, 3:08 am 

BurtJordaan appears to have the right answer in terms of a result. Certainly 6>2 + 5>2 +4>2 add up to 77.
A formula could look like
[d1>(N-1) + d2>(N-1) + d3>(N-1)]/ r>N
where d is one of each of the digits involved in the sum required, where r is the number of possibilities in a single roll, and where N is the number of dice being rolled.

But when applied to a double roll, it results in 6+5+4/36, which is not the longhand answer in the Table below.

I used the Table below to calculate the possibilities of adding the rolls of a third die to the combinations.

A third throw of 1s will account for 6 collections with 10+ (bottom right corner 6 cells) = 6
A third throw of 2s will account for 6 collections with 10+ (same 6 cells) = 6
A third throw of 3s will account for 6 collections with 10+ (same 6 cells) = 6
A third throw of 4s will combine with 12 (bottom row, plus first 5 rows in column 6, plus 5,5) =12
A third throw of 5s will combine with 20 of these cells (bottom 2 rows plus top 4 rows columns 5 and 6) =20
A third throw of 6s will combine with 27 of these cells (bottom 3 rows plus top 3 rows of columns 4,5 and 6) =27
This gives a total of 77 in 216 throws of 3 dice.

But the numbers 36, 25 and 16 do not come up anywhere. Could that sequence be just a coincidence BurtJordan?

I found a website a couple of days ago that dealt with a similar type of thing using 5 dice. But they could not get away from having to use a spread sheet to estimate the number of possibilities. Unfortunately, I've been unable to relocate that site.
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Re: Need Help!

Postby BurtJordaan on July 18th, 2020, 3:11 am 

Nice visualization! Now just make it a 6x6x6 cube.
Then start checking for the duplicates and discard them.
Then add another dimension... !@#$%& ;)
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Re: Need Help!

Postby BurtJordaan on July 18th, 2020, 3:18 am 

doogles » 18 Jul 2020, 09:08 wrote:[b]But the numbers 36, 25 and 16 do not come up anywhere. Could that sequence be just a coincidence BurtJordan?

Yup, I'm also starting to suspect that my suggested formula is a coincidence.
I tried to expand it for 4 dice, but it does not come up with the right answer.

My little brute force program, updated to 4 dice, kicks out a count of 676 out of 1296 possibilities, giving a probably of 52.16%.

Ps: here is the massive block of "unique" combinations:

1146 1155 1156 1164 1165 1166 1246 1255 1256 1264 1265 1266 1346 1355 1356 1364 1365 1366 1416 1426 1436 1446 1455 1456 1461 1462 1463 1464 1465 1466 1515 1516 1525 1526 1535 1536 1545 1546 1551 1552 1553 1554 1555 1556 1561 1562 1563 1564 1565 1566 1614 1615 1616 1624 1625 1626 1634 1635 1636 1641 1642 1643 1644 1645 1646 1651 1652 1653 1654 1655 1656 1661 1662 1663 1664 1665 1666 2146 2155 2156 2164 2165 2166 2246 2255 2256 2264 2265 2266 2346 2355 2356 2364 2365 2366 2416 2426 2436 2446 2455 2456 2461 2462 2463 2464 2465 2466 2515 2516 2525 2526 2535 2536 2545 2546 2551 2552 2553 2554 2555 2556 2561 2562 2563 2564 2565 2566 2614 2615 2616 2624 2625 2626 2634 2635 2636 2641 2642 2643 2644 2645 2646 2651 2652 2653 2654 2655 2656 2661 2662 2663 2664 2665 2666 3146 3155 3156 3164 3165 3166 3246 3255 3256 3264 3265 3266 3346 3355 3356 3364 3365 3366 3416 3426 3436 3446 3455 3456 3461 3462 3463 3464 3465 3466 3515 3516 3525 3526 3535 3536 3545 3546 3551 3552 3553 3554 3555 3556 3561 3562 3563 3564 3565 3566 3614 3615 3616 3624 3625 3626 3634 3635 3636 3641 3642 3643 3644 3645 3646 3651 3652 3653 3654 3655 3656 3661 3662 3663 3664 3665 3666 4116 4126 4136 4146 4155 4156 4161 4162 4163 4164 4165 4166 4216 4226 4236 4246 4255 4256 4261 4262 4263 4264 4265 4266 4316 4326 4336 4346 4355 4356 4361 4362 4363 4364 4365 4366 4416 4426 4436 4446 4455 4456 4461 4462 4463 4464 4465 4466 4515 4516 4525 4526 4535 4536 4545 4546 4551 4552 4553 4554 4555 4556 4561 4562 4563 4564 4565 4566 4611 4612 4613 4614 4615 4616 4621 4622 4623 4624 4625 4626 4631 4632 4633 4634 4635 4636 4641 4642 4643 4644 4645 4646 4651 4652 4653 4654 4655 4656 4661 4662 4663 4664 4665 4666 5115 5116 5125 5126 5135 5136 5145 5146 5151 5152 5153 5154 5155 5156 5161 5162 5163 5164 5165 5166 5215 5216 5225 5226 5235 5236 5245 5246 5251 5252 5253 5254 5255 5256 5261 5262 5263 5264 5265 5266 5315 5316 5325 5326 5335 5336 5345 5346 5351 5352 5353 5354 5355 5356 5361 5362 5363 5364 5365 5366 5415 5416 5425 5426 5435 5436 5445 5446 5451 5452 5453 5454 5455 5456 5461 5462 5463 5464 5465 5466 5511 5512 5513 5514 5515 5516 5521 5522 5523 5524 5525 5526 5531 5532 5533 5534 5535 5536 5541 5542 5543 5544 5545 5546 5551 5552 5553 5554 5555 5556 5561 5562 5563 5564 5565 5566 5611 5612 5613 5614 5615 5616 5621 5622 5623 5624 5625 5626 5631 5632 5633 5634 5635 5636 5641 5642 5643 5644 5645 5646 5651 5652 5653 5654 5655 5656 5661 5662 5663 5664 5665 5666 6114 6115 6116 6124 6125 6126 6134 6135 6136 6141 6142 6143 6144 6145 6146 6151 6152 6153 6154 6155 6156 6161 6162 6163 6164 6165 6166 6214 6215 6216 6224 6225 6226 6234 6235 6236 6241 6242 6243 6244 6245 6246 6251 6252 6253 6254 6255 6256 6261 6262 6263 6264 6265 6266 6314 6315 6316 6324 6325 6326 6334 6335 6336 6341 6342 6343 6344 6345 6346 6351 6352 6353 6354 6355 6356 6361 6362 6363 6364 6365 6366 6411 6412 6413 6414 6415 6416 6421 6422 6423 6424 6425 6426 6431 6432 6433 6434 6435 6436 6441 6442 6443 6444 6445 6446 6451 6452 6453 6454 6455 6456 6461 6462 6463 6464 6465 6466 6511 6512 6513 6514 6515 6516 6521 6522 6523 6524 6525 6526 6531 6532 6533 6534 6535 6536 6541 6542 6543 6544 6545 6546 6551 6552 6553 6554 6555 6556 6561 6562 6563 6564 6565 6566 6611 6612 6613 6614 6615 6616 6621 6622 6623 6624 6625 6626 6631 6632 6633 6634 6635 6636 6641 6642 6643 6644 6645 6646 6651 6652 6653 6654 6655 6656 6661 6662 6663 6664 6665 6666

count = 676 out of 1296, giving probably of 52.16%
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Re: Need Help!

Postby BurtJordaan on July 18th, 2020, 5:03 am 

doogles » 18 Jul 2020, 09:08 wrote:A third throw of 1s will account for 6 collections with 10+ (bottom right corner 6 cells) = 6
A third throw of 2s will account for 6 collections with 10+ (same 6 cells) = 6
A third throw of 3s will account for 6 collections with 10+ (same 6 cells) = 6

I don't quite follow the scheme, because a third row of 2s or 3s should add nothing to the count for the 1s (which I agree is 6), because they are duplicates. Perhaps I misread you?
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Re: Need Help!

Postby doogles on July 18th, 2020, 6:46 am 

Sorry for not making it clear.

The Table that I uploaded contained the 36 possibilities for a roll of two dice.

When we throw a third dice separately instead of at the same time, it will come with a 1 or 2 or 3 or 4 or 5 or 6.

If it's a 1, and if it's used as if it were the the third die in a triple throw, it will count as a 10+ with the 4,6, 5,6, 6,6 5,4 5,5 and 5,6 possibilities in the bottom right hand corner of the Table. Six possibilities exist for a triple throw of 4,6,1, 5,6,1, 6,6,1, 5,4,1 5,5,1 and 5,6,1. The 10+ score is already there even though a dead '1' has been added to each of them. But they count as 6 of the 77 possibles.

It's the same with the third die rolls of 2s and 3s.

The 4s, 5s and 6s add to some of the possibilities of the 2-dice throw, but they also are 'dead' weight when added to some of the existing possibilities that already score 10+ in the 2-dice Table. Each of these accounts for 12, 20 and 27 possibilities respectively.

Looking at your formidable list of figures for a 4-dice throw, I think I'll leave it at
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Re: Need Help!

Postby BurtJordaan on July 18th, 2020, 11:05 am 

Ok thanks, I follow now. One feels there must be a combination/permutation formula, but it may need a real statistician to figure that out.
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Re: Need Help!

Postby BurtJordaan on July 20th, 2020, 2:03 am 



An engineer-friend came up with this combination cube for 3 dice:

Combination Cube.png

Each horizontal layer represents one of the cases that you have calculated before.

He reckons there probably exists no analytical formula for this problem, but I will still ponder that a bit.
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Re: Need Help!

Postby doogles on July 20th, 2020, 4:58 pm 

Thanks BurtJ.

I can see where it would be a good visualisation of the 216 possible outcomes of a 3-dice throw, if he numbered Die C from 1 to 6. It demonstrates 6>3 beautifully.

But I'm having trouble with the zeros as a tool to indicate those 77 possibilities where 2 of the 3 dice have numbers that add up to 10 (in his case) or 10+. I may be missing something, but I can't see the 77 possibilities at a glance.

It seems to me that a spread sheet has to be used first to do preliminary transformations before an analytical formula can be applied.

You have cleverly done this yourself with the three and 4-dice outputs.
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Re: Need Help!

Postby doogles on July 20th, 2020, 5:56 pm 

Whoops! I just went for a walk and my subconscious was still crunching numbers.

I just realised that the '0s' weren't zeros but were actually hits on the relevant numbers.

Brilliant!

He has hit all 77 of them.
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Re: Need Help!

Postby BurtJordaan on July 21st, 2020, 1:06 am 

doogles » 20 Jul 2020, 23:56 wrote:Whoops! I just went for a walk and my subconscious was still crunching numbers.

I just realised that the '0s' weren't zeros but were actually hits on the relevant numbers.

Brilliant!

He has hit all 77 of them.

The wonders of a walk! My little Chihuahua pesters me for a walk at least once a day and try as I might, I can't switch off from what I consider as a solvable problem.

In attempting to find a formula, I have identified the counting wedges in his 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.

Combination Cube Wedges.png
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