I have been trying to work this out, no-one so far can help me.
I'm sure you can all see the relevance of this to roulette,
Q. With a fair coin, my opponent flips 50 times, & I call heads or tails.
It is fair odds, ie I get back 2xbet if I win, 0 if I lose.
The average outcome is 25 wins, 25 losses.
However, if I make the assumption that I will win at least 15 times, how do I proportion my bets to be ahead at the 15th win, whilst minimising bets?
Further, write an equation to illustrate this with e= expected wins, n= total flips.
I know e=15, n=50.
I think if c=expected wins at any point in the series, bet=1+(losses/c) works, but
it does not take into account n.
And unfortunately, when we are down to 1 expected win, we are betting all our losses +1.
So could anyone help with this please?
(0.486) to the 50th power=house edge (Euro)-Even Chances
0.486 to the 50th power/0.30 success rate(Letter e)=what you're looking for imo
OK, so 0.486 ^ 50 = 2.14 e-16
I'm not following you how this translates to how we proportion our bets, I'm afraid...?