#Coin flip(rare)
1 messages · Page 1 of 1 (latest)
oops + this:
even without oops getting this on ante 4 would give liek 64 dollars per round by ante 8 ignoring the 1/8
Fair
More like legendary
Alright, how about:
+$1 per round
At end of round, 1/2 chance to double it, 1/2 chance to half it(cannot go below 1)(both effects cannot happen on the same round)
If with oops: the chance to double it is doubled like normal, but to avoid bugging out, it goes to a 1/4 chance to half it
And with oops both effects can happen on a single round
Can definitely be made less verbose
Although considering wheel this thing is not getting above +4 dollars a round
insanely inconsistent
oh wait didnt see the last part
isnt this still absolutely broken with oops, it doubles its money on 3/4 of rounds
Sorry for not responding for a bit, but isn't bloodstone insanely strong with oops? Isn't photograph insanely broken with hanging Chad?
Saying something is really strong for only one joker shouldn't be considered insanely strong, it should be considered a good combo
Especially when oops takes up a joker slot without increasing scoring
And this only gives money
those dont grow exponentailly but i see your point
Exponential money = no
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Get th...
"only"
Without oops this is just bad
^
Random walk has an expected variance of sqrt(n) where n is the number of rounds. 2^(√n) is still exponential growth, but it might be slow enough that it's not completely, cryptid levels of broken. It would still easily outclass the rocket on average though.
If half the time it halves half it doubles
Wouldn't it on average like not grow
This joker will undergo a martingale process, i.e. a random walk where the expected value is its current value. Random walks of this nature (starting at 0) tend to stay in a range from -√n to √n after n steps. However, there is a bias. The joker can never dip under $1. Because there is a bias, the variance would be the same, but the mean must increase as a result. So from 0 to 2√n with mean √n. Hence 2^(√n) after n steps.