# Help me find the formula for this problem please (probably easy!) [closed]

This is kind of a gambling question but is solely about the numbers. You may have heard of the Martingale strategy, which to put in simple terms if you have unlimited money and you are allowed to bet an unlimited amount, you can never lose if you bet on a 1-1 (Evens) bet and whenever you lose you double the bet until you win, then you reset back to your starting bet.

But what if the bet odds are 1.66 every time instead? Is there a formula that can tell us exactly how much the previous bet would need to be multiplied by in the event it lost, for the next bet to mean you end up exactly 1 in profit once you do get a win.

I hope this makes sense, I'm not the best at explaining stuff (I wasn't even sure on good keywords to use, sorry). I look forward to seeing your replies.

• Not the one who downvoted, but it's probably because this question doesn't feel like it belongs on this forum (a bit subjective, yes). Also I'm no expert here, but I think you're trying to solve $1.66 \times (\text{amount you should bet}) = (\text{amount you lost}) + 1$ Jan 26 at 23:37

It doesn't matter what the odds are.

You will always eventually win and if the pay out is 1 to 1 if you double the bet it will work. If the odds are against you that just means it will take longer. So it's the payout the matters.

If you increase your bet by $$p$$ each time and you lose $$k$$ times your lose will be $$1+ p^2 + p^3+ .... + p^k = \frac {p^{k+1} -1}{p-1}$$. If you win one the $$k+1$$ bet and the payout is $$m$$ to $$1$$ then you will win $$m\times p^{k+1}$$.

So if you want to win $$m$$ then you need to bet enough so that $$m\times p^{k+1} - \frac {p^{k+1} - 1}{p-1} = m$$

So $$m(p^{k+1} -1)=\frac {p^{k+1} -1}{p-1}$$

$$m = \frac 1{p-1}$$

And $$p = \frac 1m +1$$

I guess you are asking if the payout is $$1\frac 23=\frac 53$$ to $$1$$. So you should increase your bet $$1\frac 35$$ each time.

If you win on the first bet you get $$1\frac 35$$. If you win on the second bet you lose $$1$$ but will win $$\frac 85\times 53=\frac 83$$ to net $$53$$. If you win on the third bet you lose $$1 + \frac 85=\frac {13}5$$ but you win $$(\frac 85)^2\times 53=\frac {64}{15}$$ to net $$\frac {64}{15}- \frac {13}5 = \frac {25}{15} = \frac 53$$ and so on....