The extra credit for the car catch needs a little clarification of the definition of average. This is a highly skewed result set where some outlying cases can go on indefinitely. So the standard statistical definition of average isn't really very helpful.
I assume that the instant the pursuer reaches the intersection we've been waiting at (for a red light), that counts as "caught"? (Specifically if the pursuer is about to get a red light - rather than assuming we pulled away first.)
Oct 20, 2023·edited Oct 20, 2023Liked by Zach Wissner-Gross
So to be clear, as phrased, the answer to last week's "Making the Rounds" is 50%. In the long run, half the flips following heads will be heads, and half the flips following heads will be tails. The riddle seems to be anticipating an answer less than half, but it's phrased incorrectly. If you flip every possible sequence of 5 flips, 64 will follow a heads, and 32 of those will be heads. Each sequence of 5 is equally likely, so in the long run, you'll flip them all equally often, say N times, getting 32N heads-after-heads and 32N tails-after-heads.
EDIT: rereading the tweet, it could be interpreted either way. "Average proportion" could mean the average of the proportions measured in each trial, or it could mean the expected proportion among all trials. I read it the second way, but she might have meant it to be the first way. I'm not changing the rest of this post, though. I still find it ambiguous.
Obviously every flip is independent, and therefore, after a heads, the next flip has exactly a 50% chance of being heads. However, there is another way of counting that gives a different result. Suppose I take the proportion of heads out of all flips following tails each time. So for instance, HHTHT has a proportion of 2/3, since out of three flips following heads, two were themselves heads. If I then consider the (unweighted) average of those individual proportions, that will be less than half. The reason is that some outcomes with high proportions should also have high weight, because they contain many heads in a row, but they are not being given that weight. Calculated this way, after flipping each sequence once, the average proportion of heads-after-tails out of flips-after-tails is 49/120. This comes from only considering the 30 flips with at least one heads in a non-final flip (i.e. excluding TTTTT and TTTTH, which have undefined proportion 0/0), taking the proportion for each, and then averaging those all together. So for instance, HHHHH counts only once as a proportion of 1.0, just as much as TTTHT counts with its 0, so they average to 0.5, even though out of a total of 5 flips-following-heads, 4 were heads. And taking the average of all 30 in this sense gives 11.25/30 = 49/120 = 40.833%
The extra credit for the car catch needs a little clarification of the definition of average. This is a highly skewed result set where some outlying cases can go on indefinitely. So the standard statistical definition of average isn't really very helpful.
Distributions with long tails can still converge to finite results (e.g., 1/x^2 from x = 1 to x = inf). Perhaps this is such a case?
For "making the rounds", should we assume that the water level in the tilted tank is at the corner?
Yes, that's an important point!
I assume that the instant the pursuer reaches the intersection we've been waiting at (for a red light), that counts as "caught"? (Specifically if the pursuer is about to get a red light - rather than assuming we pulled away first.)
Correct!
So to be clear, as phrased, the answer to last week's "Making the Rounds" is 50%. In the long run, half the flips following heads will be heads, and half the flips following heads will be tails. The riddle seems to be anticipating an answer less than half, but it's phrased incorrectly. If you flip every possible sequence of 5 flips, 64 will follow a heads, and 32 of those will be heads. Each sequence of 5 is equally likely, so in the long run, you'll flip them all equally often, say N times, getting 32N heads-after-heads and 32N tails-after-heads.
EDIT: rereading the tweet, it could be interpreted either way. "Average proportion" could mean the average of the proportions measured in each trial, or it could mean the expected proportion among all trials. I read it the second way, but she might have meant it to be the first way. I'm not changing the rest of this post, though. I still find it ambiguous.
Obviously every flip is independent, and therefore, after a heads, the next flip has exactly a 50% chance of being heads. However, there is another way of counting that gives a different result. Suppose I take the proportion of heads out of all flips following tails each time. So for instance, HHTHT has a proportion of 2/3, since out of three flips following heads, two were themselves heads. If I then consider the (unweighted) average of those individual proportions, that will be less than half. The reason is that some outcomes with high proportions should also have high weight, because they contain many heads in a row, but they are not being given that weight. Calculated this way, after flipping each sequence once, the average proportion of heads-after-tails out of flips-after-tails is 49/120. This comes from only considering the 30 flips with at least one heads in a non-final flip (i.e. excluding TTTTT and TTTTH, which have undefined proportion 0/0), taking the proportion for each, and then averaging those all together. So for instance, HHHHH counts only once as a proportion of 1.0, just as much as TTTHT counts with its 0, so they average to 0.5, even though out of a total of 5 flips-following-heads, 4 were heads. And taking the average of all 30 in this sense gives 11.25/30 = 49/120 = 40.833%
I ran a more carefully worded version of this "hot hand" problem several years ago, back over at FiveThirtyEight: https://fivethirtyeight.com/features/can-you-feed-the-hot-hand/