D&D 5E Point Buy vs Rolling for Stats

Tony Vargas said:

We passed 'starting point' many, many pages back.

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Truer words haven't been said.

I wouldn't consider 'try rolling 4d6 in order instead of arranging,' or 'using a different total for point buy' to be "pure house rules" they're just fairly minor variations that showcase the strengths of there respective methods, or illustrate how a supposed 'weakness' is easily addressed.

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To me, going beyond the rules is a house rule. It may not be pure in the sense that it's 100% house rule, but it's still not conducive to having a discussion. The opposite, really.

Maxperson said:

Sure you are. By picking the array, you have picked 15, 14, 13, 12, 10, 8.

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You didn't decide what numbers make up the standard array. The designers of D&D did that, not you.

And yet you can't just pick any other set of numbers.

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Not if you're taking the standard array, no. That isn't how the standard array works.

You are picking the numbers you end up with. That there is a more variety makes it more realistic than arrays, but less realistic than rolling.

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Variety adds to realism. I'll give you that.

Hriston said:

You didn't decide what numbers make up the standard array. The designers of D&D did that, not you.

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Just because I pick the number 5, doesn't mean you also can't pick the number 5. You have picked every number in that array. That they were the only ones to pick from, or that someone else also picked them changes nothing.

Hriston said:

And to be fair, in point-buy your choice is constrained to one of 65 possible combinations.

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Is it really that few? How did you come up with that? Did you just enumerate them? I can sort of see how to create a generalized recursive formula to calculate it, but a closed form solution is not immediately apparent to me. It's a weird kind of bin-packing problem, though, so maybe there isn't one.

I'm not really doubting you, I'm just slightly surprised.

Harzel said:

Is it really that few? How did you come up with that? Did you just enumerate them? I can sort of see how to create a generalized recursive formula to calculate it, but a closed form solution is not immediately apparent to me. It's a weird kind of bin-packing problem, though, so maybe there isn't one.

I'm not really doubting you, I'm just slightly surprised.

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Yes

13, 13, 13, 12, 12, 12
13, 13, 13, 13, 12, 11
13, 13, 13, 13, 13, 10
14, 12, 12, 12, 12, 12
14, 13, 12, 12, 12, 11
14, 13, 13, 12, 11, 11
14, 13, 13, 12, 12, 10
14, 13, 13, 13, 11, 10
14, 13, 13, 13, 12, 9
14, 13, 13, 13, 13, 8
14, 14, 12, 11, 11, 11
14, 14, 12, 12, 11, 10
14, 14, 12, 12, 12, 9
14, 14, 13, 11, 11, 10
14, 14, 13, 12, 10, 10
14, 14, 13, 12, 11, 9
14, 14, 13, 12, 12, 8
14, 14, 13, 13, 10, 9
14, 14, 13, 13, 11, 8
14, 14, 14, 10, 10, 10
14, 14, 14, 11, 10, 9
14, 14, 14, 11, 11, 8
14, 14, 14, 12, 9, 9
14, 14, 14, 12, 10, 8
14, 14, 14, 13, 9, 8
15, 12, 12, 12, 11, 11
15, 12, 12, 12, 12, 10
15, 13, 12, 11, 11, 11
15, 13, 12, 12, 11, 10
15, 13, 12, 12, 12, 9
15, 13, 13, 11, 11, 10
15, 13, 13, 12, 10, 10
15, 13, 13, 12, 11, 9
15, 13, 13, 12, 12, 8
15, 13, 13, 13, 10, 9
15, 13, 13, 13, 11, 8
15, 14, 11, 11, 11, 10
15, 14, 12, 11, 10, 10
15, 14, 12, 11, 11, 9
15, 14, 12, 12, 10, 9
15, 14, 12, 12, 11, 8
15, 14, 13, 10, 10, 10
15, 14, 13, 11, 10, 9
15, 14, 13, 11, 11, 8
15, 14, 13, 12, 9, 9
15, 14, 13, 12, 10, 8
15, 14, 13, 13, 9, 8
15, 14, 14, 10, 9, 9
15, 14, 14, 10, 10, 8
15, 14, 14, 11, 9, 8
15, 14, 14, 12, 8, 8
15, 15, 11, 10, 10, 10
15, 15, 11, 11, 10, 9
15, 15, 11, 11, 11, 8
15, 15, 12, 10, 10, 9
15, 15, 12, 11, 9, 9
15, 15, 12, 11, 10, 8
15, 15, 12, 12, 9, 8
15, 15, 13, 10, 9, 9
15, 15, 13, 10, 10, 8
15, 15, 13, 11, 9, 8
15, 15, 13, 12, 8, 8
15, 15, 14, 9, 9, 8
15, 15, 14, 10, 8, 8
15, 15, 15, 8, 8, 8

Here's the table I made to figure it out. I don't think I missed any. My method was to start with the highest numbers I could and work my way down. I was surprised it was only that many too.





edit: Ninja'd by Oofta

Hriston said:

Here's the table I made to figure it out. I don't think I missed any. My method was to start with the highest numbers I could and work my way down. I was surprised it was only that many too.


edit: Ninja'd by Oofta

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Yours is much prettier.

Maxperson said:

Just because I pick the number 5, doesn't mean you also can't pick the number 5. You have picked every number in that array. That they were the only ones to pick from, or that someone else also picked them changes nothing.

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Look, I'm not going to buy that you independently came up with the standard array all on your own.

Hriston said:

Look, I'm not going to buy that you independently came up with the standard array all on your own.

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Good, because I don't need to sell that. If I pick the standard array, I have picked those numbers. They are impossible to place on my sheet without picking them.

Harzel said:

Is it really that few? .

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65 is 64 more arrays to choose from before arranging than standard array or random give you.