Re-roll lowest X dice and keep the better result of the re-rolls

I've been racking my brain over the dice reference but I can't quite figure this one out. My character has an ability that lets him roll 5d8, but then also re-roll the lowest 2 dice and keep the higher result of each. At first I thought I could just do /r 7d8kh5 but that is not really the same. For example, let's say I roll 1-2-2-3-4; I re-roll the 1 and I get another 1, and I re-roll the 2 and get a 5, which gives me 1-2-3-4-5. If it was roll 7d8 keep highest 5, it would have been 2-2-3-4-5, so the two are not the same. Does that make sense? Is this possible to accomplish?

I feel pretty safe saying that isn't possible in the standard dice syntax. Brian might pop on and tell you how you could design a Rollable Table with the right statistical distribution to simulate it, but beyond that, you'd need to implement the logic with the API (Mentor Level Feature).

Not possible even with a rollable table I think. It would have to be pure API.

The Aaron said: Brian might pop on and tell you how you could design a Rollable Table with the right statistical distribution to simulate it That would be a lot of tables.

Yeah, I suppose that was partially tongue-in-cheek. You could get the right distribution while having it provide the result of the highest 5 dice, but you'd have to have an entry for each possibly set of 5 six sided dice (1-1-1-1-1, 1-1-1-1-2, 1-1-1-1-3, ...), with fairly non-obvious weights to anyone that isn't Brian. =D Edit: incidentally, getting 1-1-1-1-1 would be 1 in 10,077,696 134,217,728... effectively the chance of rolling 9 1s.

Ah, bummer. I guess I'll just have to eyeball it. Thanks for the answers!

You could get close with 5d8kh3+2d8, but you'd still have to be aware of the times when the reroll is lower than the initial roll (which should be fairly rare, given that you're rerolling the lowest rolls). Or if you want to do a rollable table, here are the odds of each roll: 5: 1 / 2097152 (4.76837158203e-05%) 6: 7 / 2097152 (0.000333786010742%) 7: 28 / 2097152 (0.00133514404297%) 8: 84 / 2097152 (0.00400543212891%) 9: 215 / 2097152 (0.0102519989014%) 10: 490 / 2097152 (0.023365020752%) 11: 1016 / 2097152 (0.0484466552734%) 12: 1948 / 2097152 (0.092887878418%) 13: 3498 / 2097152 (0.166797637939%) 14: 5943 / 2097152 (0.28338432312%) 15: 9617 / 2097152 (0.458574295044%) 16: 14874 / 2097152 (0.709247589111%) 17: 22051 / 2097152 (1.05147361755%) 18: 31440 / 2097152 (1.49917602539%) 19: 43229 / 2097152 (2.0613193512%) 20: 57399 / 2097152 (2.73699760437%) 21: 73680 / 2097152 (3.51333618164%) 22: 91504 / 2097152 (4.36325073242%) 23: 109998 / 2097152 (5.2451133728%) 24: 128018 / 2097152 (6.10437393188%) 25: 144220 / 2097152 (6.87694549561%) 26: 157178 / 2097152 (7.49483108521%) 27: 165548 / 2097152 (7.89394378662%) 28: 168233 / 2097152 (8.0219745636%) 29: 164608 / 2097152 (7.84912109375%) 30: 154683 / 2097152 (7.37586021423%) 31: 139113 / 2097152 (6.63342475891%) 32: 119168 / 2097152 (5.68237304688%) 33: 96622 / 2097152 (4.60729598999%) 34: 73576 / 2097152 (3.5083770752%) 35: 52059 / 2097152 (2.48236656189%) 36: 33704 / 2097152 (1.60713195801%) 37: 19431 / 2097152 (0.926542282104%) 38: 9525 / 2097152 (0.454187393188%) 39: 3610 / 2097152 (0.172138214111%) 40: 834 / 2097152 (0.0397682189941%)

manveti said: 5: 1 / 2097152 (4.76837158203e-05%) 5: 1 / 134,217,728 (7.4505806e-9) 5 (1-1-1-1-1) can only happen by rolling 7 1s initially, then re-rolling the lowest 2 dice and having them come up as 1s again, for a total of 9 1s rolled. That is the only way it can happen, so it is 1 chance in (8 raised to the 9th power): 134,217,728

I thought the initial post said start with 5d8, then reroll the two lowest. That means there are 7 dice being rolled in total.

7d8kh5, but reroll the lowest 2 dice, for 9 rolls.

No, it's 5d8 and re-roll 2 lowest, manveti is correct. I first thought I could approximate it with 7d8kh5 but I was wrong.

Ah! Well that's loads simpler... :)