#Coin flip(rare)

1 messages · Page 1 of 1 (latest)

nova swallow
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+$1 every round
At end of round, 1/2 chance of doubling the amount of money earned each round
1/8 chance of doubling odds to increase money earned

foggy rose
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oops + this:

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even without oops getting this on ante 4 would give liek 64 dollars per round by ante 8 ignoring the 1/8

nova swallow
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Fair

tight spade
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More like legendary

nova swallow
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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

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And with oops both effects can happen on a single round

tight spade
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That’s not bad actually

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Maybe kinda complex

nova swallow
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Can definitely be made less verbose

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Although considering wheel this thing is not getting above +4 dollars a round

foggy rose
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insanely inconsistent

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oh wait didnt see the last part

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isnt this still absolutely broken with oops, it doubles its money on 3/4 of rounds

nova swallow
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Saying something is really strong for only one joker shouldn't be considered insanely strong, it should be considered a good combo

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Especially when oops takes up a joker slot without increasing scoring

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And this only gives money

foggy rose
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those dont grow exponentailly but i see your point

weak cove
nova swallow
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Ok

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Maybe just make it add 2 half the time, the other half removes 2

weak cove
tardy rain
# weak cove 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.

weak cove
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Wouldn't it on average like not grow

tardy rain
# weak cove 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.