We finally have the alternate commander for the Mutant Menace precon, and of course, it's the Master himself from Fallout 2.
Technically, The Master was from Fallout 1, but you can encounter "remnants of the Master's army" in Fallout 2. From FONV, Lily, Tabitha, Anderson, Keene, Marcus, and Neal all served under The Master...but only Marcus and Neal are not insane.
What's that, Rhonda? Nightkin are not crazy, you say? Okay.
I see luck bobble head has the 4 magic words to become a instantiously “Get rid of it” thing for opponents and that’s “you win the game”
Is it getting a 6 seven total times or rolling seven 6's in one roll? Which ever one this will definitely be removed for similar reasons people remove halo fountain even when the owner forgets about ability three or claimed for just the draw.
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It is a very narrow condition. You would have to roll 7 or more dice (that's a lot of bobbleheads to have) and then out of those dice there have to be exactly 7 results of a 6. At 7 dice, you are looking at (1/6)^7 = 1/279936. For n > 7, you are looking at [n C 7] * (1/6)^7 * (5/6)^(n - 7). If n = 10, that is 120 * (1/279936) * (5/6)^3 = 0.000248. It is exceedingly unlikely that anyone would ever win the game this way...but it sure would be cool and would make a great story.
I was going to think about Dynavolt Tower but then I reread Dr. Madison Li and I thought "okay, both".
Aren't all Yao Guai enraged?
Technically, The Master was from Fallout 1, but you can encounter "remnants of the Master's army" in Fallout 2. From FONV, Lily, Tabitha, Anderson, Keene, Marcus, and Neal all served under The Master...but only Marcus and Neal are not insane.
What's that, Rhonda? Nightkin are not crazy, you say? Okay.
Harold and Bob, First Numens doing their best Lotus Field impression.
The White Glove Gourmand is curious as to your opinion of your Brahmin Welllington. Philippe skillfully prepared it just for you.
It is for Veronica--it counts as that "fancy dress" she always wanted.
It is a very narrow condition. You would have to roll 7 or more dice (that's a lot of bobbleheads to have) and then out of those dice there have to be exactly 7 results of a 6. At 7 dice, you are looking at (1/6)^7 = 1/279936. For n > 7, you are looking at [n C 7] * (1/6)^7 * (5/6)^(n - 7). If n = 10, that is 120 * (1/279936) * (5/6)^3 = 0.000248. It is exceedingly unlikely that anyone would ever win the game this way...but it sure would be cool and would make a great story.