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.
- we start with 1000 coins
- we always bet that heads come up
- minimum bet is 10 coins
- maximum bet is 80 coins
- if tails comes up, we lose our bet
- if heads comes up, we win twice as much as we bet
- the coin is fair and so the probability of heads is 50%
- we start with a bet of 10
- if we lose we double our bet
- if we lose the maximum bet of 80 we start at a bet of 10 again
- 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.