3 ms·
Here’s a twist on the puzzle: What if there are L seats, where L >= N. (If L > N, not all the seats are assigned to people) What is the probability that the l
by Tinyyy 11y ago
Here’s a twist on the puzzle:
What if there are L seats, where L >= N. (If L > N, not all the seats are assigned to people)
What is the probability that the last person sits on his own seat?
- wrcwill 11y agoThe situation stops being a problem when someone either chooses seat 1 or seat N. Because: - If at any point someone chooses seat 1, then everyone else will get their chosen seat. -If at any point someone chooses the last seat, everyone but the last guy gets his seat. The last guy gets seat 1, because it has to be free (if it wasn't , then my first condition would have applied). So only seat 1 and N are possible, both having equal chance, thus 50%. You can notice that the two conditions that resolved the "seat being taken" problem were when someone sat in a seat that no one else would have a ticket for. What kinds of seats are those? The 1st seat, the last seat, and empty seats. This means that by adding L-N empty seats, the chances of the last guy having his seat is 1/(L-N + 1st seat + last seat) = 1/(L-N+2)
- Tinyyy 11y agoYep, except that the guy has his seat if someone chooses any of the unassigned seats, so probability should be (L-N+1)/(L-N+2).