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Since the coins are fair, they have equal chances of getting heads or tails. Therefore, after each flip, about half of the people sit down and half still standing.
When there are 16 people, there is probably a person who gets 4 heads in a row. This is because log2(16)=4. The formula for the number of heads that the last one standing has flipped with x people at start can be estimated by log2(x).
Back to the starting problem, since there are 250 people, the last one standing is expected to flip about log2(250) heads (about 8). Also, the chances of getting 6 heads in a row is 1/64, so the chance of nobody getting at least 6 heads in a row is (63/64)^250, which is about 2%. So I would expect someone to get 6 heads in a row, but if it doesn’t happen, just blame your luck (I once got nobody standing after 2 flips with 16 people).
Also, for the first related problem, since the probability is 1 in 14 million, 14*2=28 million tickets are supposed to be sold each week (assume that everyone chooses their number randomly).
For the second related problem, the correct probability should be 1 in 133225 (1/365^2), ignoring leap years. The incorrect answer is because of forgetting the event can happen on any date. And since there are more than 1 million families, there should be about 10 (about 1 million/100 thousand) or more families with three children sharing a birthday. Maybe coincidence is not that rare!
For the third related problem, the probability of getting 10 heads in a row is 1/2^10=1/1024. If we assume each flip takes 2 seconds and he immediately restarts when he gets tails, it should take him about (512+256*2+128*3+...+4*8+2*9+10)*2=4054 seconds, which is about 68 minutes, to get 10 heads in a row. Therefore, it should take him about an hour to film this unlikely event.