Pink Poogle Toy Forum

[dice]pzfczotlgs[/dice]
[dice]ybyhwcvcie[/dice]

Ooooh, cheating tiemz.

[dice]ellmoeqmek[/dice]
[dice]sjvzozmtxg[/dice]

[dice]ajkjrkuawn[/dice]
[dice]mdspjwfcva[/dice]

[dice]gomihpycgr[/dice]
[dice]epdybokiug[/dice]

[dice]acnvllbeuv[/dice]
[dice]gwyakoeqhl[/dice]

Haha, everyone's using the amounts I calculated... I'm so honored...

[dice]bvsveljwyr[/dice]
[dice]ljkardmnzq[/dice]

[dice]lqecaatlel[/dice]
[dice]xfoplrsokg[/dice]

[dice]ckffzikqyu[/dice]
[dice]rduatanrfk[/dice]

[dice]uoecqitdsv[/dice]
[dice]ebbantawsy[/dice]

[dice]bzvfpqxkvk[/dice]
[dice]xtwdgixvcn[/dice]

[dice]mpjpifboyw[/dice]
[dice]aakrmfddmz[/dice]

[dice]nryekxhbuw[/dice]
[dice]qqxafvkibd[/dice]

ETA: WOOOO!!!!!!!!!!!!!!!!!

Okay...new number: 575

OK, I know nothing about calculating, so I'm just making a wild guess:

[dice]xkfgzgrkls[/dice]
[dice]twvwjadfbm[/dice]

EDIT: Way too high

The best odds will occur at the average score for that sided-dice times the number of dice:
e.g.: 2 six-sided dice: 2*3.5=7

So to calculate the # of dice to roll, you can take the target and divide by the number of sides:
999/3 (ave. for 5-sided dice) = 333
575/3 = 191.66666 (so 192 5-sided dice)

The fewer the number of sides, the more weight there will be at the center. The extremes are 575d1, which will give you a 575 100% of the time, and 1d575, which will give you a 575 1/575 of the time.

I already rolled today, so I'll try for 575 tomorrow.

[dice]jlkwonvgka[/dice]
[dice]frwgiojubw[/dice]

oof did that wrong