If a coin is tossed 4 times and we get Head all the time then what is probability of getting exactly 3 heads?

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How to find Probability in this question? It gives different answer based on context.

Scenario 1:

Answer = 1

because we are getting Heads always. So probability of getting any number of Head such as be it 1, 2 , 3 or 4 it will be equal to 1.

Scenario 2 :

Answer = 0

because None of the times we got exactly 3 Head , so it should be 0.

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I would say that this is conditional probability, although the wording is a bit strange. Usually we have wording like "given that" or "conditioned on" or "on the condition that". Your problem says: "and we get heads all the time" but to me this is acceptable way to state a condition.

If we assume that this is conditional probability then the answer is indeed 0.

$$ P(exactly\ three\ heads\ |\ all\ heads)=\frac{P((HHHT\ or\ HHTH\ or\ HTHH\ or\ THHH)\cap\ HHHH)}{P(HHHH)}=\frac{0}{P(HHHH)}=0 $$