P2 = 1 (throwing 1 or 2 times will definitely not occur three times in a row, so the probability is 1).

P3 = 1-( 1/2) 3 = 7/8 (for three consecutive times)

P4 =1-3 * (1/2) 4 =13/16 (positive and negative, positive and negative anyway).

P _ n = P _(n- 1)-P _(n-4)* 1/ 16

If n- 1 time does not appear, the last n-th time must be positive and negative.

The general formula is difficult to find, so we need the knowledge of combinatorial mathematics.

The limit is 0.

The practical significance is that after countless attempts, small probability events will inevitably occur.