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From a Bayesian perspective, or from a betting one, it doesn't make sense to put probabilities in a confidence interval. You might be uncertain about the world,

From a Bayesian perspective, or from a betting one, it doesn't make sense to put probabilities in a confidence interval. You might be uncertain about the world, but you can be certain about how much uncertainty you have, since it's a property of your own mind.

fragsworth 8y ago
How do we quantify the difference between coin tosses, which we are very certain is 50% likely to end up heads, from political elections, where we only have a few examples to go off of?
- edbaskerville 8y ago
  In terms of predicting what's going to happen next, it really isn't that different if your model's any good. Looking backwards, if the 2016 election had been determined by two coin tosses both coming up heads, would we be surprised? And yet 538's polling model indicated that the probability was even better than two coins coming up heads. They probably should have just written that right there: "71%, better odds than two coins coming up heads." But: in order to understand if the model is any good, you need to try to build a good model, describe your model, and describe what you know about the underlying processes. With words, visuals, and math. Which is pretty much what 538 does better than most.
- OscarCunningham 8y ago
  Good question. The difference is how the probability changes in response to new information. Learning more about the coin won't change the probability of heads from 50%, but doing more exhaustive polls would have likely improved our prediction about the election. However I don't think there's as much of a difference as you think. If we learnt more about how the coin was going to be flipped then that would certainly improve our estimate of the outcome. If we found out the exact way it would be flipped we could calculate the outcome exactly. So we can't really compare the two situations quantitatively, since there's no way to match up like-for-like the information we could receive. But we can say, for each possible piece of information we could receive, how much we expect it to change our probability.
edbaskerville 8y ago
This is worth emphasizing. If you have a model with a parameter that represents the probability of something happening, and then you have uncertainty in that parameter, your real probability needs to integrate over that uncertainty. But the final probability is still just a single number. But I suspect the GP's reasoning was getting interference from the valid point that poll numbers, which are not probabilities, are meaningless without confidence intervals or some other measure of uncertainty. That uncertainty will then feed, via some model, into a probability estimate of a candidate winning. And that's just a single number.
- 3pt14159 8y ago
  Hmm. Doesn't it depend on whether re-rolls are allowed as to how confident we can be? For example, I can imagine a well engineered coin that has a 0.5 probability of heads with a 0.005 confidence (ie, we suspect that the true weighting between heads and tails is likely between 0.495 and 0.505 19 times out of 20) and I can contrast it with a hastily made coin that can still have a 0.5 expected probability with a 0.1 confidence (ie, we expect the true probability to be between 0.4 to 0.6, 19 times out of 20). Coin A and Coin B have dramatically different impacts on our decision making. For example, selling insurance against 5 identical flips in a row is a much more expensive proposition for Coin B than Coin A. I feel like I really should know this given my data science background, but sometimes the basics slip away.
  - OscarCunningham 8y ago
    Right. Both coins have a probability of 0.5 on the next flip, but when you do multiple flips the hastily made coin has a higher chance of a result with lots of heads or lots of tails. We could do the same thing for the 2016 election, but we would have to specify exactly what we meant by a "repeat". Do we just let Hillary run for the 2020 election and see what happens? Or do we put back every atom to the exact position it had in 2014, so that the only divergences between the two elections are caused by quantum randomness? Or something in between, like looking at all elections where a demagogue outsider runs against an established insider? Really it doesn't matter what definition of "repeat" we choose. Since the repeat won't actually happen, we can't be called to bet on it, so knowing the probability in that case isn't too useful. Whereas a coin actually can be flipped multiple times.
TTPrograms 8y ago
It might be of some general interest to note that there exist such objects - distributions over distributions, or metaprobabilities. They have practical use - for example, if you are playing a game with uncertainty, you might observe something in the next step that changes your belief, or distribution over world state, b = P(s). If you have some expected distribution of next observations then you can talk about the probability of having some belief in the next state. Roughly P(b'|b) = P(P(s')|P(s)). You can collapse this into simple probabilities if you just care about your new observations, P(obs'|b) = \sum_s P(obs'|s)P(s|b). I could imagine some frameworks where confidence intervals in this way would be useful - ex. I have 3 fairly different world models that I think are equally likely, each has a P(election), what's the P(election|world) and get confidence intervals across world models rather than just summing over them to P(election). But I agree that for most common approaches that simple probabilities are most useful and clear.
- OscarCunningham 8y ago
  Something similar certainly is useful in the case of flipping a coin. Suppose we have two coins, one of which we know is fair and one of which we know is biased but we don't know which way. Then our knowledge about the bias of the first coin would be described by a distribution with a sharp peak around 0.5, but the distribution for the second coin would have two lumps either side of 0.5. Then the probability of heads on the next flip would be 0.5 for either coin. But if someone asked us to bet on the probability of heads on the next two flips, then the second coin would have a higher probability than the first. The different distributions for the two coins don't make a difference for one flip, but do for multiple flips. But there's no meaningful way to repeat an election, so I don't think similar distributions or confidence intervals are useful in that case. EDIT: We can imagine rerolling an election, but since we can't actually do it we don't have to bet on it so the information wouldn't be very useful to us.
AlexCoventry 8y ago
> From a Bayesian perspective, or from a betting one, it doesn't make sense to put probabilities in a confidence interval "Bayesian perspective" covers a lot of territory, and your assertion depends on the situation and the modeling objective. If a probability is a model parameter (for instance, frequency of heads for a particular coin), then summarizing the posterior distribution on that parameter with a confidence interval can be a sensible thing to do. Or do you mean as opposed to a credibility interval?

fragsworth 8y ago

How do we quantify the difference between coin tosses, which we are very certain is 50% likely to end up heads, from political elections, where we only have a few examples to go off of?

edbaskerville 8y ago
In terms of predicting what's going to happen next, it really isn't that different if your model's any good. Looking backwards, if the 2016 election had been determined by two coin tosses both coming up heads, would we be surprised? And yet 538's polling model indicated that the probability was even better than two coins coming up heads. They probably should have just written that right there: "71%, better odds than two coins coming up heads." But: in order to understand if the model is any good, you need to try to build a good model, describe your model, and describe what you know about the underlying processes. With words, visuals, and math. Which is pretty much what 538 does better than most.
OscarCunningham 8y ago
Good question. The difference is how the probability changes in response to new information. Learning more about the coin won't change the probability of heads from 50%, but doing more exhaustive polls would have likely improved our prediction about the election. However I don't think there's as much of a difference as you think. If we learnt more about how the coin was going to be flipped then that would certainly improve our estimate of the outcome. If we found out the exact way it would be flipped we could calculate the outcome exactly. So we can't really compare the two situations quantitatively, since there's no way to match up like-for-like the information we could receive. But we can say, for each possible piece of information we could receive, how much we expect it to change our probability.

edbaskerville 8y ago

This is worth emphasizing. If you have a model with a parameter that represents the probability of something happening, and then you have uncertainty in that parameter, your real probability needs to integrate over that uncertainty. But the final probability is still just a single number. But I suspect the GP's reasoning was getting interference from the valid point that poll numbers, which are not probabilities, are meaningless without confidence intervals or some other measure of uncertainty. That uncertainty will then feed, via some model, into a probability estimate of a candidate winning. And that's just a single number.

3pt14159 8y ago
Hmm. Doesn't it depend on whether re-rolls are allowed as to how confident we can be? For example, I can imagine a well engineered coin that has a 0.5 probability of heads with a 0.005 confidence (ie, we suspect that the true weighting between heads and tails is likely between 0.495 and 0.505 19 times out of 20) and I can contrast it with a hastily made coin that can still have a 0.5 expected probability with a 0.1 confidence (ie, we expect the true probability to be between 0.4 to 0.6, 19 times out of 20). Coin A and Coin B have dramatically different impacts on our decision making. For example, selling insurance against 5 identical flips in a row is a much more expensive proposition for Coin B than Coin A. I feel like I really should know this given my data science background, but sometimes the basics slip away.
- OscarCunningham 8y ago
  Right. Both coins have a probability of 0.5 on the next flip, but when you do multiple flips the hastily made coin has a higher chance of a result with lots of heads or lots of tails. We could do the same thing for the 2016 election, but we would have to specify exactly what we meant by a "repeat". Do we just let Hillary run for the 2020 election and see what happens? Or do we put back every atom to the exact position it had in 2014, so that the only divergences between the two elections are caused by quantum randomness? Or something in between, like looking at all elections where a demagogue outsider runs against an established insider? Really it doesn't matter what definition of "repeat" we choose. Since the repeat won't actually happen, we can't be called to bet on it, so knowing the probability in that case isn't too useful. Whereas a coin actually can be flipped multiple times.

TTPrograms 8y ago

It might be of some general interest to note that there exist such objects - distributions over distributions, or metaprobabilities. They have practical use - for example, if you are playing a game with uncertainty, you might observe something in the next step that changes your belief, or distribution over world state, b = P(s). If you have some expected distribution of next observations then you can talk about the probability of having some belief in the next state. Roughly P(b'|b) = P(P(s')|P(s)). You can collapse this into simple probabilities if you just care about your new observations, P(obs'|b) = \sum_s P(obs'|s)P(s|b). I could imagine some frameworks where confidence intervals in this way would be useful - ex. I have 3 fairly different world models that I think are equally likely, each has a P(election), what's the P(election|world) and get confidence intervals across world models rather than just summing over them to P(election). But I agree that for most common approaches that simple probabilities are most useful and clear.

OscarCunningham 8y ago
Something similar certainly is useful in the case of flipping a coin. Suppose we have two coins, one of which we know is fair and one of which we know is biased but we don't know which way. Then our knowledge about the bias of the first coin would be described by a distribution with a sharp peak around 0.5, but the distribution for the second coin would have two lumps either side of 0.5. Then the probability of heads on the next flip would be 0.5 for either coin. But if someone asked us to bet on the probability of heads on the next two flips, then the second coin would have a higher probability than the first. The different distributions for the two coins don't make a difference for one flip, but do for multiple flips. But there's no meaningful way to repeat an election, so I don't think similar distributions or confidence intervals are useful in that case. EDIT: We can imagine rerolling an election, but since we can't actually do it we don't have to bet on it so the information wouldn't be very useful to us.

AlexCoventry 8y ago

> From a Bayesian perspective, or from a betting one, it doesn't make sense to put probabilities in a confidence interval "Bayesian perspective" covers a lot of territory, and your assertion depends on the situation and the modeling objective. If a probability is a model parameter (for instance, frequency of heads for a particular coin), then summarizing the posterior distribution on that parameter with a confidence interval can be a sensible thing to do. Or do you mean as opposed to a credibility interval?