[Power]

Here's a really simple straightforward coin toss strategy:

It really is as simple as "the bet is in your favor-take it." S=xS=x. You win 100(3100−1)100(3100−1) with probability 2−1002−100 and lose 100100 with almost certainty. This presumes somebody can pay you that much. The expected win is then 3100−12100⋅100−100(1−12100)≈4⋅10193100−12100⋅100−100(1−12100)≈4⋅1019 To maybe make this less unbelievable, imagine a two round game. Clearly on the last throw, you want to bet all you have, increasing your expected fortune by 50%50%. On the first throw, then your expectation is 1.5(x−S2+x+2S2)=1.5(x+S2)1.5(x−S2+x+2S2)=1.5(x+S2) which (given the rules) is maximized when S=xS=x. Alternately, your result is the same if you interchange the two flips. Since you should be all on the last flip, you should on the first as well.

I think I'll have better luck without it lol :P

Source: http://math.stackexchange.com/questions/...l-strategy

[Silent Storm™]

I've flipped for heads 6 times in a row and haven't won one yet.. haha

[Arma]

I have won i think 9 tosses and lost 4 , won a total of 2000 ish coins and donated about 1200 coins

[Resize]

Shit is about to go down, i just won some coins hehe. Not following this strategi haha

Shit man i have made some BANK!

[Cyber]

We will be fixing the win probability soon. ;)

[Silent Storm™]

I think I'm 1 for like 8 right now haha, we should be able to see our W/L ratio.

[Resize]

We will be fixing the win probability soon. ;)

You should make a shop for the coins so we can buy some awards ::P

I have made som bank in the last 30 min

[Cyber]

[Resize]

[Arma]

If its heads and tails the win % should stay 50/50