Ok, That was difficult to create that title. I hope the only way to calculate this isn't by mapping out all the possibilities!
I have a game where I have three fair coins. The goal is to get all three on heads and you have three tries.
If any of them land on heads then you don't need to flip them again.
For example if you get HHT on your first toss you can flip the T twice more to get a head.
If you get HTT you can try the TT twice more so your second toss could be HTT, HTH, HHT, or HHH. If you get HHH you'd obviously stop, but if you get any of the other outcomes you'd get one more try. I think this means that given you getting HTT you have a 4/9 chance of getting HHH in the end.
So I know that getting all three to land on heads on the first flip is 1/8, but I start losing the math after that.
Here's my guess after trying to map out all the possibilities and I came to 27/64 which is close to 42%. This doesn't seem right because I tend to get all three heads more often than not in reality!
Best Answer
Each coin toss is independent from other ones. So the probability of tossing a coin 3 times and getting at least one head is $1-(1/2)^3 = 7/8$. Hence, for this to happen to all three coins, the probability is $(7/8)^3 = 343/512$.