chance to freeze over N turns

[lilith]

well, it's nothing too exciting, but i'd never done it out before...

>>> f = lambda acc: lambda n: 1-pow(1-(acc*.1), n)
>>> for i in xrange(17):
...     # blizzard
...     print '{}: {}'.format(i, f(.9)(i))
...
0: 0.0
1: 0.09
2: 0.1719
3: 0.246429
4: 0.31425039
5: 0.3759678549
6: 0.432130747959
7: 0.483238980643
8: 0.529747472385
9: 0.57207019987
10: 0.610583881882
11: 0.645631332513
12: 0.677524512586
13: 0.706547306454
14: 0.732958048873
15: 0.756991824474
16: 0.778862560272
>>> for i in xrange(17):
...     # ice beam
...     print '{}: {}'.format(i, f(255/256.0)(i))
...
0: 0.0
1: 0.099609375
2: 0.189296722412
3: 0.270050369203
4: 0.342760195708
5: 0.408227441839
6: 0.467173536499
7: 0.520248047512
8: 0.568035839655
9: 0.611063519689
10: 0.649805239407
11: 0.684687920638
12: 0.716095959794
13: 0.744375463798
14: 0.769838064084
15: 0.79276435067
16: 0.813406964177
>>>

and while paralyzed:
>>> f = lambda acc: lambda n: 1-pow(1-(.75*acc*.1), n)
>>> for i in xrange(17):
...     # blizzard
...     print '{}: {}'.format(i, f(.9)(i))
...
0: 0.0
1: 0.0675
2: 0.13044375
3: 0.189138796875
4: 0.243871928086
5: 0.29491057294
6: 0.342504109267
7: 0.386885081891
8: 0.428270338864
9: 0.46686209099
10: 0.502848899848
11: 0.536406599109
12: 0.567699153669
13: 0.596879460796
14: 0.624090097192
15: 0.649464015632
16: 0.673125194577
>>> for i in xrange(17):
...     # ice beam
...     print '{}: {}'.format(i, f(255/256.0)(i))
...
0: 0.0
1: 0.07470703125
2: 0.143832921982
3: 0.207794622635
4: 0.266977934518
5: 0.321739836871
6: 0.372410640073
7: 0.419295977997
8: 0.462678651516
9: 0.502820334289
10: 0.539963151112
11: 0.574331138358
12: 0.606131595307
13: 0.635556334525
14: 0.66278283883
15: 0.687975331828
16: 0.711285768464
>>>