Leaderboard

Suppose the leading bullet has 1/k chance of being the fastest. Then we will not get complete annihilation. So far, so good. But *still* we could get another collision, between other bullets. We could and very often we would, but we do not care. We ONLY consider the complementary case where the leading bullet is NOT the fastest. And that happens with the probability (1-1/k). So while it may seem true to say the probability of *at least* one future collision is 1-1/k, we are still measuring only a *sufficient* but not a *necessary* one. Look at it this way: - If the LB is the fastest, we quit the game. p=1/k - Necessary condition to continue the game: LB may not be the fastest. p=1-1/k. There SURE will be a collision. We have k-2 bullets Example for k=8: We do not care what happens if LB is the fastest because there will be no annihilation. EXIT In 7/8 cases, LB is not the fastest, so we are sure there will be a collision. It does not matter whether 3 collides with 4, 6 with 7 or 7 with 8. Just two bullets are gone. For future convinience, we renumber the bullets. Now, we have k=6 We do not care what happens if LB is the fastest... EXIT In 5/6 cases, LB is not the fastest, so we are sure there will be a collision. Again, it does not matter who collides with whom. Again, we renumber. We got here with the p that the LB was not the fastest when we had 8 bullets AND with the p that the LB was not the fastest when we had 6 bullets: 7/8 * 5/6 Now, we have k=4 We do not care what happens if LB is the fastest... EXIT In 3/4 cases, LB is not the fastest, so we are sure there will be a collision. Again, it does not matter who collides with whom. Again we renumber. We got here with the p that the LB was not the fastest when we had 8 bullets AND with the p that the LB was not the fastest when we had 6 bullets AND with the p that the LB was not the fastest when we had 4 bullets: 7/8 * 5/6 * 3/4 Now, we have 2 bullets We do not care what happens if LB is the fastest... EXIT In 1/2 cases, LB is not the fastest, so we are sure there will be a collision. To get here: 7/8 * 5/6 * 3/4 * 1/2 Think of it this way. For an initial ensemble of n bullets, there are n! permutations of their speeds, Oh, not again!!! It took me a long time to get freed of the relative speeds. The trouble is that it works well for k=2, but it gets very complicated for higher k. Take 4 bullets with v1>v2>v3>v4 when the distances are equal: 9 8 2 1 at first, 8 hits 2 9 8 6 1 at first, 6 hits 1 To know what happens, you have to consider not only the speed hierarchy, but the speed differences. There remain two bullets, it can be managed. However, imagine 6 bullets. After the first collision, the distances are not equal so the speed differences do not help. Gets really, really complicated.