11-16-2016, 12:27 PM (This post was last modified: 11-16-2016, 12:28 PM by Power.

*Edit Reason: added source*) Here's a really simple straightforward coin toss strategy:

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

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

Quote:It really is as simple as "the bet is in your favor-take it." S=xS=x. You win 100(3100âˆ’1)100(3100âˆ’1) with probability 2âˆ’1002âˆ’100 and lose 100100 with almost certainty. This presumes somebody can pay you that much. The expected win is then 3100âˆ’12100â‹…100âˆ’100(1âˆ’12100)â‰ˆ4â‹…10193100âˆ’12100â‹…100âˆ’100(1âˆ’12100)â‰ˆ4â‹…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(xâˆ’S2+x+2S2)=1.5(x+S2)1.5(xâˆ’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