I'm trying to do the maths on a coin toss game after 100 games but think I am failing. 

The rules are as follows.

1. we start with 1000 coins
2. we always bet that heads come up
3. minimum bet is 10 coins
4. maximum bet is 80 coins
5. if tails comes up, we lose our bet
6. if heads comes up, we win twice as much as we bet
7. the coin is fair and so the probability of heads is 50%
8. we start with a bet of 10
9. if we lose we double our bet
10. if we lose the maximum bet of 80 we start at a bet of 10 again
11. if we win we start at a bet of 10 again

So in this set up, 50% of the time we would get our 10 coins back plus 10 additional coins.

When we lose we need to lose four times in a row which I've figured out will be 6.25% which will cost 150 coins.

50% (10 coins) \* 50% (20 coins) \* 50% (40 coins) \* (80 coins) 50% = 6.25% (150 coins total)

Would I be right in thinking that this means that the chance of breaking even is 43.75%?

So after 100 games we would have the win and loss as below and end up with the balance after the wins and losses are applied.

Win: (50% win chance \* 100 games) \* 10 coins won = 500 coins won
Loss: (6.25% loss chance \* 100 games) \* 150 coins = 937.5 coins lost
Total: 1000 start + 500 win - 937.5 loss = 562.5 end balance

I am not confident that my maths is correct. I'd appreciate it if someone eyed it over and told me if I'm going horribly wrong.