>>>>>Take it to its simplest: if you toss a coin, it's 1:2 you'll get a heads.
>>>>>
>>>>>Next toss it's not 1:4 you'll get heads, then for a third 1:16 (1:8 or whatever) - it's still 1:2. By the end of 64 tosses, by that logic, like the old grains of sand on a chessboard conundrum, the chances of getting heads would be practically infinity:1.
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>>>>Yes, but if the question is whether tossing twice, you'll get two heads, then the odds are 1 in 4.
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>>>No. The odds of tossing twice and getting 2 heads is 50/50 because they are two independent events regardless of how you would like them not to be. The odds of tossing 2 coins at once and having them both come up heads is calculated differently and is 1 in 4.
>>>
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>>They are not independent events. You can't get two heads in two tosses unless you get one on the first toss. It's 50-50 on that first toss. If it winds up tails, your second toss is guaranteed to wind up not two heads in a row. If it winds up tails, your second toss has another 50-50 chance of being heads.
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>>What you are describing is the chance of the second toss being a head regardless of the result on the first toss.
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>>All I can add is to try it. Toss a coin and then toss another coin. Do that 100 times and count the number of times both wind up heads. It should be approximately 25, not 50.
>
>You've denied your own reasoning then. If I do it 100 times, shouldn't the odds against them both ending up heads be pretty astronomical rather than 1 in 4? See? even you realise somewhere down deep that these are separate events.

Toss the coin twice 100 times. Each set of two tosses is a separate trial and you have a total of 100 trials. Approximately 25 of the trials will result in two heads.