The best thing that could happen to the U.S.A. this year would be for Bernie Sanders to become the Democratic Party’s nominee for the presidency, and then for Sanders to win the general election in November. The Republican Party’s nominee will most certainly be Donald Trump. However, the possibility exists that due to means mostly foul that Bernie Sanders would not be the Democratic Party nominee. If this disappointment materializes in July, then you may find yourself harangued by rabidly passionate partisans telling you how to vote based on their preferences. So, in this essay I present a tool — Bayesian analysis — that can help you to clarify your own thoughts about how to proceed in a period of uncertainty, and to strengthen your convictions in a logical manner. This will be easy reading, stick with it.
The purpose of Bayesian analysis is to logically select the best course of action from a set of available options, despite uncertainties about the probabilities of the outcomes that may occur, and where the decision-making process takes into account your own personal preferences regarding those outcomes. You can easily learn the mechanics of basic Bayesian analysis by looking up articles on the Prisoner’s Paradox. I will proceed directly to our model “general election” problem:
Should “I” (a given Sandernista and/or Democratic Party voter) vote for Hillary Clinton, or not-vote for Hillary Clinton, if the general election is a Clinton versus Trump race?
A “not vote” means to instead vote for a third party candidate, or write in a candidate on your ballot, or abstain from voting for the presidency.
We identify two mutually exclusive actions: vote for Hillary Clinton (vote-H), and not-vote for Hillary Clinton (vote-not-H).
There are two mutually exclusive general (national) outcomes: Hillary wins (H-win) or Trump wins (T-win).
The probability of an H-win is designated by the letter p. The quantity p is an as yet uncertain number whose magnitude lies between zero (a certainty of a T-win) and one (a certainty of an H-win).
Because the probabilities of an H-win and T-win must add up to unity, the uncertain probability of a T-win is the quantity (1-p).
Since we have two actions (vote-H, vote-not-H) and two general outcomes (H-win, T-win) there are four possible specific (or personal) outcomes:
D1: vote-H, and H-win,
D2: vote-H, and T-win,
D3: vote-not-H, and H-win,
D4: vote-not-H, and T-win.
I, the voter, have very personal preferences, or desirabilities (D1, D2, D3, D4) regarding each of these outcomes.
For example, I might decide that voting H and having a T-win would rate on my personal desirability scale at -1000! You can use any numbers (positive or negative) you like for your personal subjective values D1, D2, D3 and D4 for the four objective outcomes.
Recall that specific probabilities for vote-H are: p (for an H-win), and (1-p) (for a T-win).
Recall that specific probabilities for vote-not-H are: p (H-win), and (1-p) (T-win).
My actions in a voting booth will not alter the actions of millions of other voters in their voting booths, so p is independent of what I (or any other single voter) does.
Recall that the desirabilities for the action vote-H are: D1 (H-win), and D2 (T-win).
Recall that the desirabilities for the action vote-not-H are: D3 (H-win), and D4 (T-win).
The expected value to me of any of the four outcomes is the quantity gotten by multiplying the probability of that specific outcome with the desirability I assigned to that outcome. So the four expectation values are:
For the action vote-H expectation values are: p*D1, and (1-p)*D2.
For the action vote-not-H expectation values are: p*D3, and (1-p)*D4.
The best action for me to take (in this model problem there is only a choice between two) is the one which has the highest utility value. The utility for an action is the sum of the expectation values of its consequences.
The utility value for vote-H is UH:
UH = p*D1 + (1-p)*D2.
The utility value for vote-not-H is Unot-H:
Unot-H = p*D3 + (1-p)*D4.
When UH is greater than Unot-H, vote Hillary (hang on, we’re still just talking math).
When UH is less than Unot-H, vote Not-Hillary.
I have taken the definitions and formulas described above, and worked out the general problem for any set of numbers D1, D2, D3, D4, and p (enough has been said that the mathematically inclined can easily duplicate this work). Now, I will lay out the algorithm for decision-making (picking an action) and show specific numerical examples, which you can use as templates to work out your own personal cases.
Define:
x = D2 – D4
y = D1 – D3
The extremes of x and y can be characterized as follows:
At large positive y (y >> 0): H-loyalty, happy to vote-H for an H-win.
At large negative y (y << 0): H-antipathy, unhappy to vote-H for an H-win.
At large positive x (x >> 0): H-guilt, at a failure to vote-H given a T-win.
At large negative x (x << 0): H-disgust, at a wasted H-vote with a T-win.
There are four possible classes of voters for this problem:
H-regardless
Bernie-or-Bust
Between guilt and disgust (over H)
Between anger and happiness (over H).
H-regardless voters are defined by:
x > 0, and y > 0,
and they will be most satisfied to vote-H regardless of any estimate they may make of p (the probability of an H-win). These are the “Hillary or bust” voters.
Example #1 (H-regardless):
D1 = 10 (H-win is good),
D2 = 0 (T-win is not good),
D3 = 0 (I let the H-team down, but at least they didn’t lose),
D4 = -10 (shame! I didn’t support the H-team and result is a T-win).
Thus x = 10, and y = 10.
Analysis indicates they should vote-H regardless of any estimate of p, if they are to be most satisfied.
Bernie-or-Bust voters are defined by:
x < 0, and y < 0,
and they will be most satisfied to vote-not-H regardless of any estimate they may make of p (the probability of an H-win).
Example #2 (Bernie-or-Bust):
D1 = -5 (I hate the idea of vote-H to stop a T-win),
D2 = -10 (damn! I did a vote-H and still got the T-win),
D3 = 5 (an H-win is better than a T-win, and, yea!, I didn’t have to vote-H!),
D4 = 0 (if T-win was destined at least I didn’t waste my vote on the loser H-team).
Thus x = -10, and y = -10.
Analysis indicates they should vote-not-H regardless of any estimate of p, if they are to be most satisfied.
The other two classes require the calculation of the critical probability, q, defined as:
q = -x/(y-x),
which is equivalent to
q = x/(x-y).
Between guilt and disgust voters are defined by:
x > 0, and y < 0,
and they will be most satisfied to:
vote-H if p < q,
vote-not-H if p > q.
Example #3 (Between guilt and disgust):
D1 = -5 (H-win is better than a T-win, but I didn’t want to vote-H),
D2 = 0 (no guilt for the T-win, I did a vote-H),
D3 = 5 (H-win, which I didn’t have to vote for),
D4 = -10 (guilt over the T-win since I didn’t do a vote-H).
Thus x = 10, and y = -10, and q = 50%.
These people will be most satisfied if they:
vote-H if p < q = 50%
vote-not-H if p > q = 50%.
In the above example the voter only feels safe to vote their preference of vote-not-H (and avoid feeling guilt if the result is a T-win) if H is more than q = 50% likely to win the election.
Between anger and happiness voters are defined by:
x < 0, and y > 0,
and they will be most satisfied to:
vote-H if p > q,
vote-not-H if p < q.
Example #4 (Between anger and happiness):
D1 = 10 (happy to vote for an H-win),
D2 = -10 (unhappy with a T-win since I did a vote-H),
D3 = 0 (I didn’t vote-H, it doesn’t much matter),
D4 = 0 (I didn’t vote-H, it doesn’t much matter).
Thus x = -10, and y = 10, and q = 50%.
These people will be most satisfied if they:
vote-H if p > q = 50%
vote-not-H if p < q = 50%.
In the above example the voter only feels satisfied voting for the H-team if it is a sure winner, so they should vote-H only if they estimate that the probability of H-team success, p, is greater than the q (critical probability based on desirabilities) for this case, which is 50%.
Two more examples follow.
Example #5 (Between disgust and guilt, with a lot of guilt-fear):
D1 = 100 (H-win, okay I guess),
D2 = -100 (sad if a T-win, but no guilt as I did a vote-H),
D3 = 200 (I’d rather vote Bernie or Jill Stein if H-win is destined),
D4 = -1100 (lots of guilt over my vote-not-H with a T-win).
Thus x = 1000, and y = -100, q = 0.90909.
This guilt-fearing voter should only vote their not-H preference if they believe an H-win is over 91% likely! Specifically:
vote-H if p < q = 90.909%
vote-not-H if p > q = 90.909%
It would be so much better to jettison the guilt.
Example #6 (Between anger and happiness, with a lot of anger):
D1 = 10 (guess I had to vote-H to prevent a T-win),
D2 = -1100 (damn!, I vote-H and get a T-win),
D3 = 0 (H-win, and I didn’t have to use up my vote for it),
D4 = -100 (T-win anyway, glad mine wasn’t a loser vote-H).
Thus x = -1000, and y = 10, q = 0.99009.
This person is angry about the idea of “having to” vote-H to prevent a T-win, and then that vote-H being for a loser. They should:
vote-H if p > q = 99.009%
vote-not-H if p < q = 99.9909%
In the above example the voter only feels satisfied voting for the H-team if it is a sure winner. If T-team is destined to win, then they want to use their vote elsewhere instead of on vote-H.
Why don’t you try making up some examples by choosing D1, D2, D3, D4 and p? The value in actually working out numerical examples based on your own preferences (desirabilities) is that it helps to clarify your mind about all the possible choices and outcomes you may be faced with. That can improve your self-confidence and sense of calmness about the whole electoral spectacle. Also, it may give you ideas about other types of choices to play Bayesian games with. Enjoy.