it shore looks like that but something is not logical
bet 1 | $2
bet 2 | $4
bet 3 | $8
bet 4 | $16
bet 5 | $32
bet 6 | $64
bet 7 | $128
bet 8 | $256
bet 9 | $512
bet 10 | $1024
On a 10 bets game like this you can win $1024 or you could lose your $1
So it does not make sense that each event is independent ... and last bet probability of getting heads is 50%
My instinct / gut is telling me that 50% chance for last bet is false ...
I think your question is different from what you're asking then. Are you asking: what is the probability I can win $1024 on the 10th round?
This is different from the coin question. If you're asking the chance of winning a coin flip back to back 10 times, then you calculate as such:
On each round you have a 1/2 chance of winning or doubling your initial bet. This is because it's a coin flip.
However, now you need to keep in mind that you need to win all of these in a row. As in, you need a coin flip to be heads on the first, second, third, etc. flips. This means out of 10 flips, all 10 must be heads to double your money.
In the case that we have 10 flips and each flip has a 1/2 chance of giving you a payout, we have (1/2) * (1/2) * (1/2) ... ten times, or (1/2) ^ 10 which is a
0.0009765625 chance of doubling your money 10 times in a row.
Your answer over here makes more sense. He is probably talking about back to back bets. I mean, to reach the 10th bet and to win the amount he will have to win the rest of the 9 bets (hence you either lose $1 or win $1024). So, in this case you are correct.
That's a pretty small chance. You might end up spending more than what you would win.