The Monty Hall Problem

The Monty Hall Problem derives its name from the American TV show Let’s Make A Deal, originally hosted by Monty Hall.

The problem in its essential format is:

You are shown 3 doors. Behind one of them is a prize or reward. You are asked to choose which door you think has the prize. The host then opens one of the other 2 doors, revealing that it does not contain the prize. He then asks if you want to stick with your original choice or switch to the remaining closed door. 

You also know from the game rules that the host knows where the prize is and will not open a door containing it. Should you switch?

As a concrete example, suppose you choose Door 1. The host opens Door 2 and you see there is no prize behind it. He then asks: “Do you want to stick with Door 1 or change to Door 3?”

The concept behind this question predates Let’s Make A Deal. An equivalent puzzle involving prisoners appeared in Martin Gardner’s Mathematical Games column in the Oct 1959 issue of Scientific American, with discussion of the solution in the November issue.

The problem in the form I have described above appears in a letter to the editor by Steve Selvin in the Feb 1975 issue (Vol 29 (1)) of American Statistician, with a follow up discussion in the Aug 1975 issue (Vol 29 (3)).

However it really gained public attention as a result of Marilyn vos Savant’s solution in her Ask Marilyn column in the 9 Sep, 1990 issue of Parade magazine.

She explains that after the host opens another door, your probability of winning is 1/3 if you stick to your choice and 2/3 if you switch.

Remarkable is the number of bombastic letters, many signed PhD, all confidently telling her she’s wrong and the probabilities are both 1/2.

Their common reasoning is that opening Door 2 in my concrete example above changes the information about both Doors 1 & 3. All we know now is that the prize is behind one of them, so the probabilities are equal ie equal to 1/2.

Many people with mathematics qualifications give this answer, even initially a genuine giant of the field, Paul Erdos.

But this reasoning is wrong: it misunderstands the symmetry breaking in the information caused by the action of the host in opening the door. The solution is a fairly straightforward exercise in conditional probability and Marilyn is correct.

I’ll explain why.

First note that if we brought in another contestant who had not witnessed the previous events and asked them to choose between Doors 1 & 3, their probabilities would be {1/2, 1/2}. But that is because they do not know that Door 1 was chosen by the first contestant and the host, knowing the prize was not behind Door 2, chose that door to open.

This is a good example of probabilities being representations of information. The second contestant has validly (in terms of logic) assigned probabilities of {1/2, 1/2} to Doors 1 & 3, whereas the first contestant, if they are reasoning correctly, assigns probabilities of {1/3, 2/3}. The reason for the difference is the first contestant has information provided by the action of the host that is not available to the second contestant.

Both contestants’ probability assignments are logically correct in that they are correctly derived from the information each has. However, being different, they can’t both be correct in the sense of being true. Suppose the prize is actually behind Door 3, which the host knows. Then the host’s probabilities are {0, 1}.

So, 3 different sets of probabilities, each a valid representation of the available information.

Which one is actually true in the sense that it is a representation of reality?

The host’s: {0, 1}. That’s because the event has already occurred.

Another example is:

Suppose Person A is told that a 6 sided die will be rolled and if it shows a 1, they will win a prize. They calculate a probability of winning of 1/6.

Now suppose Person B is told that A was lied to and shown the die to be rolled: it is 12 sided. They calculate a probability of winning of 1/12. 

A and B’s probabilities will both be valid representations of the information they have. However only B’s probabilities are correct in the sense of being a true representation of reality, because the event of rolling the die has not yet occurred … unless we believe in determinism, in which case both A’s and B’s probabilities are validly assigned but false.

Solution to the Monty Hall Problem

As a general principle, suppose an event has some set S of possible outcomes. If there is a permutation of a subset T of S which does not change the information we have about the event, then the probabilities we assign to the outcomes in T are equal, since permuting any of them does not change our view.

In particular, if all we know is that an event has n possible outcomes, we assign a probability of 1/n to each. For example, we think of the chances of a fair coin as being {1/2, 1/2}.

So our familiar assignment of equal chances to outcomes where we know nothing more than the number of possibilities actually derives from a symmetry argument. It is also the probability distribution with maximum entropy, however that principle also derives in part from symmetry arguments.

Presented with 3 doors and no other information, we assign probabilities of {1/3, 1/3, 1/3} for the prize.

Because the classic Monty Hall initial situation has full symmetry ie is unaltered by any permutation of the labels {1, 2, 3} on the doors, we can assume without loss of generality that we choose Door 1.

Write R = k for the reward being behind Door k and D = k for the host opening Door k.

We have chosen Door 1. Suppose the host opens Door 2. The reasoning will be the same had the host opened Door 3, so for concreteness, we will do the calculation for Door 2, seeing that the answer will be the same had we done it for Door 3.

We want the probability that R = 1, conditional on the host opening Door 2:

P(R = 1 | D = 2) = P(R = 1 ∧ D = 2) / P(D = 2)

We can calculate conditionals the other way around:

Prior to opening, P(D = 2 | R = 1) = 1 / 2, since the host could have opened Door 2 or 3.

But P(D = 2 | R = 2) = 0 and P(D = 2 | R = 3) = 1.

That’s because if R = 3, the host must necessarily open Door 2.

So P(D = 2) = ∑_k P(D = 2 | R = k) P(R = k) = 1/3 * 1/2 + 0 + 1/3 * 1 = 1/2.

Now from above, P(R = 1 ∧ D = 2) = P(D = 2 | R = 1) P(R = 1) = 1/2 * 1/3 = 1/6.

Put these together to get 

P(R = 1 | D = 2) = P(R = 1 ∧ D = 2) / P(D = 2) = 1/6 / 1/2 = 1/3.

That means P(R = 3 | D = 2) = 2/3, since R = 2 is impossible if D = 2.

So there is a 1/3 probability of the prize being behind Door 1 and a 2/3 probability of it being behind Door 3. That is, a 2/3 probability of winning if we switch.

Clearly the same reasoning carries through if the host opens Door 3 instead.

There is a 2/3 probability of winning if we switch and 1/3 if we stick to our first choice.

Why Does This Happen?

It’s because both our guess and the location of the prize influence which door is opened, but not symmetrically.

In terms of a causal graph:  Choice → Door ← Prize.

Door is what’s known as a collider. Information on the door D opened makes all probabilities conditional on D.

If you condition on a collider, you introduce a dependency between Choice and Prize, despite them being independent without the conditioning.

An easier to see example is:

Toss a pair of coins many times, say 100 or 1000, making sure all tosses are independent, including the pair in each trial. Record the result H or T of coin 1 in column 1 and coin 2 in column 2.

You should see the data in each column are statistically independent.

Now censor the data to remove all rows with T, T. 

The 2 columns are now clearly dependent: seeing a T in column 1 tells you column 2 has an H in that row and vice versa.

The causal graph:  Coin 1 → Retain Data ← Coin 2

Since both Coin 1 and Coin 2 causally influence Retain Data, it is a collider, so conditioning on it by only observing Retain Data = 1 introduces a dependence between the previously independent Coin 1 and Coin 2.

Simulation

We can easily simulate the Monty Hall game. It is trivial to extend it to n doors.

In fact, it was a simulation which convinced Paul Erdos.

For each simulation path:

1. Due to the discussion above, set the contestant’s choice H to Door 1: H = 1.
2. Generate a random sample from {1, 2, …, n}. That is the value of R, the location of the prize.
3. If R = 1, generate another random sample from {2, …, n}. 
   Else if R = k > 1, generate the random sample from {2, …, n} – {k}.
   That is the value D = d, the door opened by the host.
4. Generate a random sample W from {0, 1}. 
5. If W = 1, the contestant switches, simulated by generating a random sample H = r from {2, …, n} – {d}. 
   Note that if n = 3, we don’t need a random sample: if d = 2, H = 3 else d = 3 so H = 2.
6. If W = 0, record a 1 in the W = 0 column if R = 1 and a 0 otherwise.
   If W = 1, record a 1 in the W = 1 column if R = r (see Step 5) and a 0 otherwise.

If you run this simulation over many paths, you’ll see the proportion of 1s in the W = 0 column (the instances when the contest doesn’t switch and is correct) converges to P(0) = 1/n.

Harder to see is that the proportion of 1s in the W = 1 column (the instances when the contest switches and is correct) converges to P(1) = (n – 1) / n (n – 2).

For n = 3, P(1) = 2/3, so you should be able to see the proportion in your output is close to 0.667.

Generalization of the Problem

We can extend the problem to any number n of doors. After your guess, the host opens m of the other n – 1 doors.

If m = 1, it’s not hard to mirror the argument in the n = 3 case above.

Prior to opening, for any k > 1:

P(D = k | R = 1) = 1 / (n – 1).

P(D = k | R = r) = 1 / (n – 2) for k ≠ r.

Thus P(D = k) = 1 / (n – 1).

Also P(R = 1 ∧ D = k) = P(D = k | R = 1) P(R = 1) = 1 / n (n – 1).

If r > 1 & k ≠ r, P(R = r ∧ D = k) = P(D = k | R = r) P(R = r) = 1 / n (n – 2).

Put these together to get 

P(R = 1 | D = k) = P(R = 1 ∧ D = k) / P(D = k) = 1 / n.

P(R = r | D = k) = P(R = r ∧ D = k) / P(D = k) = (n – 1) / n (n – 2).

That is, after the host opens door k ≠ 1, if you stay with your original choice of Door 1, the probability of winning the prize is P(R = 1 | D = k) = 1 / n.

If you switch your choice to Door r, the probability of winning the prize is P(R = r | D = k) = (n – 1) / n (n – 2).

Now if the host opens some set S of m > 1 doors:

P(D = S | R = 1) = 1 / choose(n - 1, m).

P(D = S | R = r) = 1 / choose(n - 2, m) for r ∉ {1, S}.

With a little algebra, P(D = S) = 1 / choose(n - 1, m).

Therefore

P(R = 1 | D = S) = P(R = 1 ∧ D = S) / P(D = S) = 1 / n.

If r ∉ {1, S}, P(R = r | D = S) = P(R = r ∧ D = S) / P(D = S) = (n – 1) / n (n – m – 1).

That is, after the host opens m doors, if you stay with your original choice of Door 1, the probability of winning the prize is still 1 / n.

If you switch your choice to one of the remaining n – m – 1 closed doors, the probability of winning the prize is = (n – 1) / n (n – m – 1) > 1 / n.

So no matter how many doors there are or how many doors the host opens, it’s always better to switch, even if your switch means having another guess from the multiple closed doors which remain.

Federal Election Betting Says ALP Majority Government Most Likely

I’ve done 2 simulations of tomorrow's (May 3) Australian federal election using the seat by seat betting odds from Sportsbet, TAB and Ladbrokes.

The simulation dates were 23 April and today, May 2.

The exact method of simulating the probability distribution of possible election results is described in detail in my post simulating the 2013 federal election.

In brief, any seat for which all 3 betting agencies have one candidate paying less than 1.10 is deemed safe and allocated to that candidate. As of today, there were 74 of these: 39 Labor, 33 Coalition and 2 independents: long time MPs Bob Katter in Kennedy (Far North QLD) and Andrew Wilkie in Clark (Hobart). There were 68 last week.

That's a historically low proportion of safe seats. It's driven by the large number of independents / Teals.

For the remainder of contested seats, I derive the probability of winning in each seat from an average of the implied probabilities from the 3 sets of bookmaker odds. I used 10 simulations of 100,000 runs each. In each run, a winner for each contested seat is decided by a correlated, random draw of each candidate, with the probabilities derived from the odds for that seat.

Each set of 100,000 runs gives an estimated probability distribution for the number of seats won by each party (and independents). The 10 sets of runs are to determine the stability of the distribution estimates. Runs are done across a range of correlation parameters, to simulate swings in general electoral mood.

In the current parliament, there are 151 federal seats, so 76 required to form a majority government. The Labor (ALP) government currently has 77, the Coalition (LNP) 53, with 19 on the cross benches, plus 2 vacancies. The vacant seats are Hinkler, centred on Bundaberg / Hervey Bay in QLD and Maribyrnong in northern Melbourne. 

Hinker is vacant because MP Keith Pitt (LNP) resigned to become ambassador to The Vatican. One less devout Catholic in parliament is a small improvement. Maribyrnong is vacant because MP Bill Shorten is retiring to a Vice-Chancellorship at the University of Canberra, where I'm sure his ego and penchant for slippery politics will benefit students and staff alike.

Interestingly, an electoral redistribution last year reduces the post-election parliament to 150 seats, still requiring 76 to form a majority government. NSW and Victoria have each lost 1 seat and WA gained one. That's a surprising outcome, since Victoria now loses the seat it gained in the 2018 redistribution, returning to 37.

The first simulation on April 23 gave a probability of 43% for an ALP majority, 55.5% for a hung parliament and 1.5% for an LNP majority. The most likely distribution of seats was 73 - 74 ALP, 62 - 63 LNP and 14 on the cross benches: 11 independents and 3 greens. That implies an 88% probability of a Labor government, either majority or minority.

With likely conservative independents in Calare and Cowper, plus independents in conservative electorates such as Indi, Monash, McPherson and possibly Bradfield, I set the bar at 71 seats for the LNP to form a minority government. That's because, conditional on the LNP gaining 71 seats, the mostly likely result is 68 for Labor, 9 independents and 2 Greens. That should be sufficient buffer over Labor to gain the support of 5 independents.

Mackellar and Warringah on Sydney's Northern Beaches are also conservative electorates. However Zali Steggall (Warringah) is essentially a Labor candidate and Sophie Scamps (Mackellar) has such a bee in her bonnet about failing to gain endorsement as the Liberal candidate for the 2022 election, she's just as likely to ignore her constituents and side with a Labor minority government. Who knows what Allegra Spender in Wentworth would do, let alone Bob Katter?

The simulation on election eve May 2 showed the Coalition's chances of even forming a minority government have significantly decreased. 

There were 39 safe ALP seats, 33 for the LNP and 2 independents: Bob Katter in Kennedy (Far North QLD) and Andrew Wilkie in Clark (Hobart). 

The additional 76 contested seats were simulated and returned the most likely scenario of 77 or 78 seats for Labor, 58 or 59 for the Coalition and 14 on the crossbenches: 11 independents and 3 greens. I'm counting Rebekha Sharkie (Centre Alliance) in Mayo (SA) as independent.

Overall, the simulation gave a 59% probability of a majority Labor government and a 41% chance of a hung parliament, so in the course of little over a week, Labor has firmed from a likely minority government to a likely majority.

The LNP's chances of even getting to 71 seats and forming a minority government have slid to about 3 or 4%. The chances of them governing outright are negligible.

That's because the Coalition's recent campaign has been disastrous. Dutton's selling point is a strong Conservative who knows what he believes and you'll get what you see. Yet he has stumbled and vacillated over policies. Then he became desperate and tried throwing money at voters, losing credibility as a careful financial manager in an economy with far too much debt as it is.

His nuclear policy is too expensive and will take too long to build versus alternatives like battery farms. Although, if we're buying nuclear submarines, we should be able to produce our own fuel for them. It may be prudent to build at least a couple of reactors. But no-one in the Coalition has effectively articulated that idea.

We've seen very little from his potential front bench. People who might be naturally wary of the agenda of the ALP left have lost confidence in the Coalition as a viable alternative: competent and ready to govern.

This has certainly been reflected in movements in the betting odds in many seats.

Interestingly, the number on the cross benches is likely to decrease from 19 to about 14. Modelling suggests all of those seats will be picked up by the LNP. Labor will likely be in close to the same position they are now.

But increasing LNP representation to 58 or 59 from the current 53 will be nothing for Dutton and the Coalition to feel good about, if they even manage it. That's a poor result against an unpopular government full of very ordinary performers and a testament to the current lack of talent in the LNP ranks.

US Election Betting Has Trump Shading Harris

Betfair, Ladbrokes, Sportsbet and TAB are offering state by state betting on the November 5 US election.

I derived implied probabilities from the state by state odds from each agency and used them to simulate the election result. The method is discussed here. 

The odds are as at October 20. A lot may change between now and November 5, so I’ll run another simulation just before election day.

A simulation over 1,000,000 paths with a range of correlation parameters gives a 58% probability of a Trump victory. The variation in the 58% result was very small over the range of reasonable correlations. The most likely outcome in the 538 member Electoral College is 271 Republican and 267 Democrat, so whatever happens it is likely to be close. 

Looking at the odds, of the 50 states + Washington DC (3 delegates), only 13 states are in play at all. Those are Arizona (11), Florida (30), Georgia (16), Maine (4), Michigan (15), Minnesota (10), Nevada (6), New Hampshire (4), New Mexico (5), North Carolina (16), Pennsylvania (19), Virginia (13) and Wisconsin (10). The numbers in brackets are their Electoral College delegates. They are available here.

Of those, Florida is highly likely to go Republican and Maine, Minnesota, New Hampshire, New Mexico and Virginia are all paying between 1.05 and 1.15 for the Democrats. So that leaves 7 states which are highly uncertain. These are the 7 “battleground states” described in the media: Arizona, Georgia, Michigan, Nevada, North Carolina, Pennsylvania and Wisconsin.

States in play in 2020 such as Texas and Ohio are not in 2024, according to the betting. They appear to be firmly Republican.

Maine and Nebraska are the only states which do not use a winner take all system. However as Maine is usually 3 Democrat and 1 Republican and Nebraska 4 Republican and 1 Democrat, this usually balances out.

Any state where the odds of one party winning were at most 1.05 on all four sites was deemed to be already won and its Electoral College seats allocated by the model. That put the allocated standings at 190 Harris and 189 Trump. The 159 delegates from the other 13 states listed above were simulated from the odds.

Allocating the 6 low odds states and simulating just the 7 battleground states starts at 226 Harris and 219 Trump + the simulation of the remaining 93. The results obtained by this method show little variation from the results obtained by simulating the 13 states.

Interestingly, the head to head Trump (1.71) v Harris (2.40) odds imply a 58% probability of a Trump win. There is usually some arbitrage between head to head and state by state betting, but not in this market.

On a political note, if either party had a quality candidate, this election wouldn’t be close. If the Democrats had run anyone with the presence and electability of Bill Clinton or Barack Obama in 2020, they would likely have won by a much wider margin. But apparently the best they could put up was a senile, incompetent, corrupt old hack. Given the closeness of results in Arizona, Georgia, Pennsylvania and Wisconsin, the 306 – 232 margin was highly flattering to Biden.

Now, after 4 years of gross incompetence, manifest failure, lies and left wing ideology over pragmatism and substance, we have another close election. How?

Because the Republicans can’t find themselves a strong, credible, right-of-centre candidate. So Trump is back for another round of chaotic charlatanism. He might well win, because he can connect with people; at least enough of them to win against Kamala Harris, whose transparent insincerity connects with almost no-one.

Americans should rightly be asking what is so dysfunctional about their political system. In a nation of 340 million people, the best candidates they were offered were Trump and Biden, who was then replaced via backroom dealing by the possibly even more useless and self-serving Kamala Harris. How can America not do better?

The mainstream media should be asking these questions. But they are not. Instead of seeking the answers the public really wants and needs, they have largely descended into party political activism.

ABC, MSNBC, CNN, The New York Times and The Washington Post are the media organs of the Democratic Party. Fox has filled the gap and become the organ of the Republicans.

Until the media start acting with their true purpose: honest investigation and careful analysis, future elections are likely to evince as little genuine hope as this one. But that will require replacing the current party political activists with proper journalists.

Interestingly, current owner Jeff Bezos has directed long time Democratic Party mouthpiece The Washington Post to not endorse a candidate in this election. His reason?

"What presidential endorsements actually do is create a perception of bias. A perception of non-independence. Ending them is a principled decision, and it’s the right one."

Here is the Post's own piece on the matter.

Unsurprisingly, left wing "commentators" have responded with the line that Jeff Bezos is the wrong person to own a newspaper like The Post, because he is too rich. Yes, if The Post had principled owners, it would have done the right thing and endorsed the Democratic candidate, no matter who that was. And where was this honest commentary published? That principled defender of independent journalism, The Guardian.

Update:

Thought I'd go with orange due to who's in the lead in the betting.

The same simulation on the latest betting odds as of the morning of Tues Nov 5 (in Australia) has Kamala Harris edging slightly closer, but with Trump still more likely to win. The Nov 5 result was a 56% probability of a Trump victory, versus a 58% probability 2 weeks ago.

The most frequent outcome in the simulation was a 281 - 257 victory for Trump.

What has changed?

Most notably, the polling in Iowa, which has put that state in play. It was firmly Republican 2 weeks ago. Now the Republicans are paying 1.20 - 1.25.

The betting in Nevada and North Carolina has moved significantly toward Trump. In Michigan, it has moved significantly toward Harris. Pennsylvania has moved a little more toward Trump. Georgia and Wisconsin have moved a little more toward Harris.

Florida (Republican) is now on the border of the no contest criterion, although removing it from the simulated states does not materially alter the prediction. The movement of the betting in Iowa was the dominant factor in changing the Trump win probability from 58% to 56%.

 

NSW Election Betting Says Most Likely Outcome Is Hung Parliament

Sportsbet and TAB are offering seat by seat betting on the NSW election this Saturday March 26. Interestingly, Ladbrokes have decided not to.

I derived implied probabilities from the seat by seat odds from Sportsbet and TAB and used them to simulate the election result. The method is discussed here. 

A simulation over 1,000,000 paths with a range of correlation parameters give a 70 - 75% probability of a hung parliament with a Labor minority government. The most likely outcome in the 93 seat NSW lower house is ALP 45, Coalition (LNP) 37, Greens 3 and 8 independents. 

With the support of 47 members required to govern, the Coalition has almost no chance. To form a minority government, they will likely need the support of all 8 independents, plus possibly the Greens, which will not happen. 

The Greens are highly likely to win Newtown, Balmain and Ballina, so let's assume they do. 

The almost certain independent wins are Lake Macquarie (left leaning Greg Piper), Sydney (left leaning Alex Greenwich) and Wagga Wagga (Joe McGirr - conservative but from a Labor family and not a big fan of the current government). Highly likely independent wins are Orange (Philip Donato), Barwon (Roy Butler) and Murray (Helen Dalton). All three are conservative; ex Shooters, Farmers and Fishers. 

Let's assume the above seats give us 6 independents, of which 3 will likely support the LNP. 

Kiama is an interesting seat. Holsworthy MP Melanie Gibbons is running as the LNP candidate, with sitting Liberal MP Gareth Ward running as an independent after being charged with sexual offences.

Then we have the challenges to normally LNP seats by the "Teals" in Manly, North Shore, Pittwater, Wakehurst and Wollondilly. It is possible that some of these 5 and possibly Kiama will elect independents. The simulation implies at least 1 or 2. 

The outcome is that if Labor get close to 47 seats, even significantly above 40, they can probably count on Greg Piper and Alex Greenwich, plus possibly Joe McGirr to form a government without having to try to deal with the Greens, who will almost certainly try to hold them to ransom with delusional demands; self-righteous, glorified uni students that they are. 

The probability that the Coalition get to 43 seats and can rely on the 3 ex Shooters plus possibly Gareth Ward or Joe McGirr or one of the Teals is very low: about 8%. 

Thus, the overwhelmingly most likely outcome is a Labor government, probably a minority one supported by the left leaning independents and possibly requiring a deal with the Greens, the latter option leading to chaos.

Post election:

Looks like the money was on the money. Betting odds aren't always a good predictor, but in this election they were: ALP 45 LNP 36 GRN 3 and 9 other independents. We were only one cross bencher out.

There were a few seats paying 1.10 to 1.20 which at one time looked like they might swing to Labor: Holsworthy, Ryde and Terrigal, but the Coalition just got over the line in all 3.

Federal Election Betting Says ALP With Slim Majority

I derived implied probabilities from the seat by seat odds from Sportsbet, Ladbrokes and the TAB for today's federal election and used them to simulate the election result. The method is discussed here.

The most likely outcome was ALP 79 seats, the Coalition (LNP) 63, 1 for the Greens (Melbourne) and 8 independents (including Bob Katter in Kennedy), giving the ALP a slim, but workable majority.

The simulation implies a 79% chance of a Labor majority, 20% chance of a hung parliament and < 0.5% chance of the Coalition government being returned.

Given that 2 of the cross bench MPs will be Andrew Wilkie and Adam Bandt, the ALP really only need 74 seats for government, as opposed to 76 to govern outright. When that is taken into account, Labor's chances of forming a government increase to about 90%.

Federal Election Betting Shows Coalition Firming But Labor Victory

I’ve done a series of 3 simulations of the May 18 Australian federal election using the betting odds. Sportsbet, Beteasy and Ladbrokes have all provided odds for every seat from mid April, just after the election was first called.

The simulation dates were 15 April, 25 April and today (election eve), May 17.

The exact method of simulating the probability distribution of possible election results is described in detail in my post simulating the 2013 federal election.

In brief, any seat for which all 3 betting agencies have one candidate paying 1.10 or less is deemed safe and allocated to that candidate. For the remainder of contested seats, I derive the probability of winning in each seat from an average of the implied probabilities from the 3 sets of bookmaker odds. I used 10 simulations of 100,000 runs each. In each run, a winner for each contested seat is decided by a random draw of each candidate, with the probabilities derived from the odds for that seat.

Each set of 100,000 runs gives an estimated probability distribution for the number of seats won by each party (and independents). The 10 sets of runs are to determine the stability of the distribution estimates.

In the current parliament, there are 150 federal seats, so 76 required to form a majority government. The Coalition (LNP) currently has 73, with Labor (ALP) 69 and 8 on the cross benches, although one of those is the Nationals’ Kevin Hogan from Page (northern NSW), who is there in protest at the ousting of Malcolm Turnbull.

However, with significant recent population changes in Victoria, South Australia and the ACT, an electoral redistribution last year gives the new parliament 151 seats, still requiring 76 to form a majority government. Victoria’s seat allocation has increased from 37 to 38, the ACT’s from 2 to 3 and South Australia’s is down from 11 to 10.

In the first simulation on April 15, there were 52 safe ALP seats, 20 for the Coalition and 1 independent: Andrew Wilkie in Clark in Tasmania. The imbalance was due to the bookmakers initially giving odds of 1.15 or 1.20 for many normally safe LNP seats in Queensland and Western Australia in particular.

The additional 78 contested seats were simulated and returned the most likely scenario of 91 or 92 seats for Labor, 52 or 53 for the Coalition and 7 on the crossbenches, most likely: Andrew Wilkie in Clark, Bob Katter in Kennedy (Nth Qld), Rebekha Sharkie in Mayo (Adelaide), any 2 or 3 from Rob Oakeshott in Cowper (centred on Port Macquarie), Kevin Mack in Farrer (western NSW), Helen Haines in Indi (country Victoria), Zali Steggall in Warringah (Sydney northern beaches), Kerryn Phelps in Wentworth (Sydney eastern suburbs), plus the Greens’ Adam Bandt in Melbourne and possibly Steph Hodgins-May in the neighbouring seat of MacNamara (inner city south and south east Melbourne).

The probability of an outright Labor majority was > 99%.

The simulation 10 days later on April 25 showed the Coalition closing slightly. The most likely scenario was 89 seats for Labor, 55 for the Coalition and the same 7 on the crossbenches. The probability of an outright Labor majority was still > 99%.

Over the past 3 weeks, the Coalition have been gaining some traction and opinion polls have been gradually shifting more toward them, though probably not sufficiently for a victory tomorrow. This has certainly been reflected in movements in the seat betting odds.

In today’s analysis (May 17), there were 59 safe ALP seats, 26 for the Coalition, 3 independents (Andrew Wilkie in Clark, Bob Katter in Kennedy, Rebekha Sharkie in Mayo) and 1 Green (Adam Bandt in Melbourne). The most likely scenario was 82 seats for Labor, 62 for the Coalition and 7 on the crossbenches.

The remaining 4 cross benchers are most likely Rob Oakeshott in Cowper, Kevin Mack in Farrer, Helen Haines in Indi and Zali Steggall in Warringah. Zali Steggall has come in to 1.40 and looks like removing Tony Abbott (if the betting can be trusted).

That ultimately may not be a bad thing for the Liberals, because short of an “unfortunate” accident, it’s pretty much the only way they will get the ignorant wrecker Tony Abbott out of parliament. They are almost certainly going to be in opposition for at least 3 and possibly 6 years. Zali Steggall can’t do much damage because Labor will have an absolute majority and will implement their agenda (which she largely supports) without her. The Liberals can then preselect a new candidate in Warringah who will win back the seat, as Dave Sharma now looks increasingly likely to do in Wentworth, removing the aberration of barely disguised left wing activist Kerryn Phelps representing one of the wealthiest electorates in Australia.

The Greens have gone out to around 3.00 in MacNamara, but come in to 2.50 in the once apparently safe-ish neighbouring Liberal seat of Higgins, so they could also still pick up a second lower house seat.

The LNP gains from the 2 previous simulations were largely firming seats in Queensland and Western Australia, although Banks, Lindsay and Page in NSW have flipped from likely ALP gains to LNP holds.

The estimated probability of an outright Labor majority has reduced to 94%. However, given the left leaning Andrew Wilkie and Adam Bandt, the possibility of Zali Steggall and maybe even a 2^nd Green, Labor only really needs to get to 73 seats to form a government. The simulation estimates the chance to be 98.5%, with 74 seats at 97%.

With these odds, Labor should be paying about 1.03 as the sworn in government. But the 3 sites have them at 1.14, 1.14 and 1.15. Very generous odds, I suggest.

The Hawke Factor:

Will Bob Hawke’s death yesterday help swing the result to Labor?

Yes. A simple reason: brand recognition.

He was genuinely one of Australia’s best and most popular prime ministers. He had genuine vision and made important, lasting economic reforms, without the conflict. So it’s not unreasonable for there to be extensive coverage of his contribution to public life and the nation.

But his death couldn’t have come at a worse time for the Coalition campaign.

Hawke represented Labor ideals. Hawke was good. Therefore Labor is good. That will be the association in many people’s minds from all the coverage. It’s largely subconscious. It will almost certainly sway some undecided voters.

And of course the media arm of the ALP: the ABC, SMH and Age have been going out of their way to use their tributes as ads for their Labor masters.

For the Coalition, the one positive is the number of people who have already cast their vote.

Of course, that also increases the uncertainty in any simulation using current betting odds.

Election Betting Shows Coalition Minority Government Most Likely in NSW

I’ve done a belated simulation of the NSW election from the betting odds. I only used Sportsbet because by the time I got around to entering the data, Beteasy and Ladbrokes had removed their displayed odds and suspended betting.

There are 93 seats in NSW, so 47 required to form a majority government. The Coalition currently has 52, Labor (ALP) 34 and 7 on the cross benches.

46 safe seats were identified: 20 Coalition and 26 Labor. Normally there would be several more safe Coalition seats, however they are being challenged by the Shooters, Fishers and Farmers Party (SFF) in Barwon, Murray and Orange, and by conservative independents in Dubbo, North Shore and Wollondilly.

I ran a simulation for the 47 contested seats, deriving the probability of winning in each seat from the Sportsbet odds. A detailed explanation is given in my post simulating the 2013 federal election.

In brief, I used 5 simulations of 100,000 runs each. In each run, a winner for each contested seat is decided by a random draw of each candidate, with the probabilities derived from the Sportsbet odds for that seat.

25 of the 47 contested seats have the Coalition paying less than 1.40. In 17 of those, the Coalition are paying 1.20 or less. Comparatively, 9 of the 47 contested seats have the ALP paying less than 1.40. In 5 of those, Labor are paying 1.20 or less.

Thus, most contested seats are leaning toward the Coalition, so we should expect the simulation to show more contested seats going to the Coalition than Labor. Unless the betting money has completely misunderstood the mood of the electorate, which it did in Brexit and the 2016 US election, betting is pointing to the Coalition having stronger prospects of forming a (probably minority) government than Labor.

The simulation results show the Coalition with a 39% chance of forming a majority government, Labor a 1% chance and a 60% chance of a hung parliament. There was very little variation over the 5 simulations.

In terms of seats, the most likely outcomes are 45 or 46 for the Coalition, 40 for Labor, 2 for the Greens (Balmain and Newtown), 2 for the Shooters (Barwon and Orange) and 3 or 4 for independents (Lake Macquarie, Sydney, Wagga, with an outside chance of the 4^th being Dubbo or North Shore). There is also a small possibility of an 8^th on the cross benches being a 3^rd Green: either in Lismore or Ballina, or a 3^rd Shooter in Murray.

What does this mean to form government?

The Coalition, with 45 or 46 seats only need to cut a deal with the Shooters and/or the independent in Wagga. These are all conservative electorates, so it shouldn’t be too difficult. Although most will remember the 2010 federal election, when the perfidious Tony Windsor and Rob Oakeshott betrayed their conservative electorates to install a minority Labor government.

Anything less than 44 seats for the Coalition will make a minority government difficult, since at least 4 of the cross bench MPs will be left leaning. The simulation gives a 31% chance of this happening.

Update:

The Coalition achieved a slender outright majority with 48 seats to Labor's 36 and 9 on the cross benches. Of those 9, 3 are Shooters and 1 is the conservative independent Joe McGirr in Wagga. Thus, even if they lose a couple of by elections, they will still be able to maintain government.

Independents Strengthening In Election Eve Betting

As promised, I have run a final simulation of the Federal election, using the individual lower house electorate betting odds as of this Friday night (1/7).

The most probable seat distribution is now almost evenly split between 79 - 65 - 6 and 78 - 65 - 7. Labor’s chances of forming a majority government is now less than 0.1%, but the chance of a hung parliament is still high: 25% ± 2.5%. Although several seats have firmed in either the Coalition or Labor’s favour, the extra uncertainty around the independents is offsetting this clarity.

Independent (ALP stooge) Rob Oakeshott now looks to be a real problem for the Coalition in the NSW electorate of Cowper. With Cowper now at almost even money, as well as Mayo and Batman, there is a real chance of 7 on the cross benches.

There are 14 lower house seats in which 2 candidates are both paying less than $3 on all sites and thus could be considered genuinely close contests: Cowper, Greenway, Macarthur and Macquarie in NSW, Batman and Chisolm in VIC, Brisbane, Capricornia and Petrie in QLD, Hindmarsh and Mayo in SA, Cowan in WA and Braddon in TAS. Of these, Cowper, Macarthur, Macquarie, Batman, Mayo and Cowan are very close.

This is a slight change from Wednesday night’s run. Greenway has tightened, but Gilmore now looks safer for the Liberals. Brisbane is now close as the ALP have improved and Forde now looks safe for the Liberals, as do Burt and Lyons for the ALP.

PS: Don’t forget to vote for minor parties in the senate before you choose a major (if you choose a major at all). There are plenty of liberal choices: Liberal Democrats, Voluntary Euthanasia, Drug Law Reform, HEMP . It’s hard to tell what the Secular Party and the Science Party / Cyclists Alliance are all about, but I voted for them anyway.

Latest Election Betting Implies 6 On Cross Benches, But Still Coalition Majority

Last Tuesday night (21/6) I ran a simulation of the looming Federal election using probabilities implied by the individual electorate betting odds on Sportsbet, Luxbet and Crownbet. The odds implied Coalition 79 seats, ALP 66 seats and 5 on the cross benches as the most probable result.

I just ran another, using the betting odds as of this Wed night (29/6).

The most probable seat distribution has changed to 79 - 65 - 6, with an extra cross bench seat for either the Greens or Nick Xenophon. As the odds in some marginal seats have firmed for the favourites, Labor’s chances of forming a majority government have fallen from about 1% to less than 0.25%. The chance of a hung parliament is still high: 25% ± 2.5%.

Nick Xenophon’s odds in the conservative Adelaide electorate of Mayo have significantly shortened, from about $3.00 to $1.50 - $1.60. He has also come in to between $3 and $4 in Barker, Boothby and Grey. All of these are normally conservative electorates, so the swing toward Xenophon must be a concern to the Coalition. The betting now puts Xenophon at a 35% chance of winning multiple lower house seats in SA.

With the Greens shortening from $4 to even money in the north Melbourne seat of Batman, they now have a 50% chance of winning 2 lower house seats. Moreover, Batman borders Adam Bandt’s seat of Melbourne, so a win here would give the Greens the makings of a contiguous heartland in inner city Melbourne.

The perfidious Tony Windsor and Rob Oakeshott are still chances in New England and Cowper (God help us).

Bob Katter, Andrew Wilkie, Cathy McGowan, plus 2 Greens and a Nick Xenophon Team or 1 Green and 2 Nick Xenophon Team ... not an easy cross bench for Malcolm Turnbull to deal with. Wilkie and the Greens will be obstructionist and Katter, McGowan and Xenophon will hold the government to ransom.

There are 14 lower house seats in which 2 candidates are both paying less than $3 on all 3 sites and thus could be considered genuinely close contests: Cowper, Gilmore, Macarthur and Macquarie in NSW, Batman in VIC, Capricornia, Forde and Petrie in QLD, Hindmarsh and Mayo in SA, Burt and Cowan in WA and Braddon and Lyons in TAS. Of these, Macarthur, Batman, Capricornia, Mayo and Cowan are very close.

I will look at the betting odds again this Friday night.

The Best Brexit Result: A Very Narrow Bremain

With voting about to get underway in the Brexit referendum, what is the best outcome for Britain?

I won't link to the betting sites, because the odds will change so rapidly, so there is not much point in deriving probabilities from them. Last estimate was 1.25 for Remain and 4.00 for Brexit. That translates to an implied 81% probability for Britain to remain in the EU. 

Better for Britain is a very close Bremain, say 51 - 49 or 52 - 48. This would force the Eurocrats to really put a dollar value on Britain remaining part of the EU, without the chaos of it actually exiting.

The salient problem driving the Brexit vote is the technocratic socialism which Britain has always hated. To imagine anything else in the EU is anathema to the EU MPs in Brussels and that is the nub of the problem