Skip to content

The Three Prisoners Problem

First posed in Scientific American in 1959, the Three Prisoners Problem remains a classic of conditional probability. The problem, or a version of it, is simple to state. There are three prisoners on death row, Adam, Bob and Charlie. They are told that each of them has had their names entered into a hat and the lucky name to be randomly chosen will be pardoned as an act of clemency to celebrate the King’s birthday. The warden knows who has been pardoned, but none of the prisoners do.

Adam asks the warden to name one of the prisoners who will definitely NOT be pardoned. Either way, he agrees that his own fate should not be revealed. If Bob is to be spared, name Charlie as one of the men to be executed. If Charlie is to be spared, name Bob as one of the men to be executed. If it is he, Adam, who is to be pardoned, the warden should just flip a coin and name either Bob or Charlie as one of the men to be executed.

The warden agrees and names Charlie as one of the men going to the gallows.

Given this information, what is the probability that Adam is going to be pardoned, and what is the chance that Bob will instead be pardoned?

Adam reasons that his chance of being spared before the conversation with the warden was 1/3, as there are three prisoners, and only one of these will be pardoned by random lot. Now, though, he reasons that one of either he or Bob is to walk free, as he knows that Charlie is not the lucky one. So now Adam reasons that his chance of being pardoned has risen to 1/2.  But is he right?

 

Solution (Spoiler alert).

Before talking to the warden, Adam correctly concludes that his chance of evading the gallows is 1/3. It is either he, Bob or Charlie who will be released, and each has an equal chance, so each has a 1/3 chance of being pardoned.

When Adam asks the warden to name one of the OTHER men who will be executed, he is asking the warden not to name him either way, whether he is to be pardoned or not. The warden (as we are told in the question) selects which of the other men to name by flipping a coin. Now, Adam gains no new information about his fate. The information he does gain is about the fate of Bob and Charlie. By naming Charlie as the condemned man, the warden is ruling out the chance that Charlie is to be pardoned.

So Adam now knows the chance that Charlie will be spared has decreased from a 1/3 chance before the warden revealed this information to a zero chance after he reveals it.

But his own chance of being spared remains unchanged, because the warden was not able to reveal any new information relevant to his own fate. New information is a requirement for changing the probability that something will happen or not. So his probability of being pardoned remains at 1/3.

The new information he does have is that Charlie is not the lucky man, so the chance that Bob gets lucky is 2/3.

Put another way, how is it possible that Adam and Bob heard received the same information but their odds of surviving are so different? It is because, when the warden made his selection, he would never have declared that Adam was going to die. On the other hand, he might well have declared Bob to be the condemned man. In fact, there was a 50-50 chance he would have done so. Therefore, the fact that he didn’t name Bob provides valuable information as to the likelihood that Bob was pardoned while telling us nothing as to whether Adam was.

This is an example of the reality that belief updates must depend not merely on the facts observed but also on the method of establishing those facts.

In case there is still any doubt, imagine that there were 26 prisoners instead of 3. Adam asks the warden not to reveal his own fate but to name in random order 24 of the other prisoners who are to be executed. So what is the chance that Bob will be the lucky one of 26 before the warden reveals any names? It is 1/26, the same chance as each of the other prisoners. Every time, however, that the warden names a dead man walking, say Charlie or Daniel, that reduces their chances to zero and increases the chance of all those left except for Adam, who has expressly asked not be named, regardless of whether he is to be executed. So it means a lot to learn that the warden has eliminated everyone but Bob given that he had every opportunity to name Bob as one of those going to the gallows. It means nothing that he has not named Adam because he was expressly told not to, whatever his fate.

In a 26-man line-up, where the warden in random order names who are condemned, once everyone but Bob has been named for execution by the warden, Adam’s chance of surviving stays at 1/26. Bob’s chance of being pardoned rises to 25/26. This is despite the fact that there are only two remaining prisoners who have not been named for execution by the warden. Would you take 20/1 now that Adam will be spared? You might, if you were Bob, but you are not getting a good price, and you will not have long to spend it!

 

What’s in the box? Betting on the Bertrand’s Box paradox.

You are presented with three identical boxes. You are made aware that one of the boxes contains two gold coins, another contains two silver coins, and the third contains one gold coin and one silver coin. You do not know which box contains which.

Now, choose a box at random. Reach without looking under the cloth covering the coins and take out one of the coins. Now you can look. It is gold.

So you can be sure that the box you chose cannot be the box containing the two silver coins. It must be either the box containing two gold coins or the box containing one gold coin and one silver coin.

Withdrawing the gold coin from the box doesn’t provide you with the information to identify which of these two boxes it is. So the other coin must either be a gold coin or a silver coin.

Given what you now know, what is the probability the the other coin in the box is also gold, and what odds would you take to bet on it?

This is essentially the so-called ‘Bertrand’s Box’ paradox, first proposed by Joseph Bertrand in 1889 in his opus, ‘Calcul des probabilités’.

 

Spoiler alert (Solution)

After withdrawing the gold coin, there are only two boxes left. One is the box containing the two gold coins and the other is the box containing one gold and one silver coin. It seems intuitively clear that each of these boxes is equally likely to be the one you chose at random, and that therefore the chance it is the box with two gold coins is 1/2, and the chance that it is the box containing one gold and one silver coin is also 1/2. Therefore, the probability that the other coin is gold must be 1/2.

This sounds right, but it is in fact the wrong answer.

In fact, there are three equally likely scenarios that might have led to you choosing that shiny gold coin.

Let us separately label all the coins in the boxes to make this clear.

In the box containing two gold coins, there will be Gold Coin 1 and Gold Coin 2. These are both gold coins but they are distinct, different coins.

In the box containing the gold and silver coins, we have Gold Coin 3,which is a different coin to Gold Coin 1 and Gold Coin 2. There is also what we might label Silver Coin 3 in the box with Gold Coin 3. This silver coin is distinct and different to what we might label Silver Coin 1 and Silver Coin 2, which are in the box containing two silver coins, which was not selected.

So here are the equally likely scenarios when you withdrew a gold coin from the box.

a. You chose Gold Coin 1.

b. You chose Gold Coin 2.

c. You chose Gold Coin 3.

You do not know which of these gold coins you withdrew from the box.

If it was Gold Coin 1, the other coin in the box is also gold.

If it was Gold Coin 2, the other coin in the box is also gold.

If it was Gold Coin 3, the other coin in the box is silver.

Each of these possible scenarios is equally likely (i.e. each has a probability of being the true state of the world of 1/3), so the  probability that the other coin is gold is 2/3 and the probability that the other coin is silver is 1/3. So, if you are offered even money about the other coin being gold, the edge is very much with you.

Before withdrawing the gold coin, the chance that the box you had selected was that containing two gold coins was 1/3. By revealing the gold coin, however, you not only excluded the box containing two silver coins but also introduced the new  information that you could potentially have chosen a silver coin (if the selected box was that containing one gold and one silver coin) but in fact did not. That made it more likely (twice as likely) that the box you withdrew the gold coin from was that containing the two gold coins than the box containing one gold and one silver coin.

And that is the solution to the Bertrand’s Box paradox.

 

 

Election 2017: How well did the pollsters and pundits do?

When Theresa May announced on April 18 that she would call a snap general election, most commentators viewed the precise outcome of the vote as little more than a formality. The Conservatives were sailing more than 20% ahead of the Labour party in a number of opinion polls, and most expected them to be swept back into power with a hefty majority.
Even after a campaign blighted by manifesto problems and two terrorist attacks, the Conservatives were by election day still comfortably ahead in most polls and in the betting markets. According to the spread betting markets, they were heading for an overall majority north of 70 seats, while a number of forecasting methodologies projected that Jeremy Corbyn’s Labour could end up with fewer than 210.

In particular, an analysis of the favourite in each of the seats traded on the Betfair market gave the Tories 366 seats and Labour 208. The Predictwise betting aggregation site gave the Conservatives an 81% chance of securing an overall majority of seats, in line with the large sums of money trading on the Betfair exchange.

The PredictIt prediction market, meanwhile, estimated just a 15% chance that the Tories would secure 329 or fewer seats in the House of Commons (with 326 technically required for a majority), while the Oddschecker odds comparison site rated a “hung parliament” result an 11/2 chance (an implied probability of 15.4%). Only the Almanis crowd forecasting platform expressed any real doubt, putting the chance of a Conservative overall majority at a relatively paltry 62%.
In reality, the Conservative party lost more than a dozen seats net, ending up with 318 – eight short of a majority. Labour secured 262 seats, the Scottish National party 35, and the Liberal Democrats 12. Their projected vote shares are 42.4%, 40%, 3% and 7.9% respectively.
So did the opinion polls do any better than the betting markets? With the odd exception, no.

In their final published polls, ICM put the Tories on 46%, up 12% on Labour. ComRes predicted the Tories would score 44% with a 10-point lead. BMG Research was even further out, putting the Conservatives on 46% and a full 13% clear of Labour. YouGov put the Tories seven points clear of Labour (though their constituency-level model did a lot better), as did Opinium; Ipsos MORI and Panelbase had them eight points clear on 44%.

Other polls were at least in the ballpark. Kantar Public put the Tories 5% ahead of Labour, and SurveyMonkey (for the Sun) called the gap at 4%. Survation, the firm closest to the final result in their unpublished 2015 poll, this time put the Conservatives on 42% and Labour on 40%, very close to the actual result. Qriously (for Wired)was the only pollster to put Labour ahead, by three points.

According to the Chris Hanretty 2017 UK Parliamentary Election Forecast polling model, the Conservatives were heading for 366 seats, Labour 207, and the Liberal Democrats seven. Allowing for statistical uncertainty, the projection was of an “almost certain” overall majority for the Conservatives. The probability of a hung parliament was put at just 3%. All very bad misses.

Many others were wrong, too. The 2017 General Election Combined Forecast, which aggregates betting markets and polling models, forecast a Conservative majority of 66 seats. Other “expert” forecasts came from Britain Elects (Tories 356 seats, Labour 219 seats), Ashcroft (363, 217), Electoral Calculus (358, 218), Matt Singh (374, 207), Nigel Marriott (375, 202), Election Data (387, 186), Michael Thrasher (349, 215), Iain Dale (392, 163) and Andreas Murr and his colleagues (361, 236).
So what went wrong?

In the wake of the 2015 election, the Brexit referendum and Donald Trump’s victory, forecasters are getting used to fielding that question. But the answer isn’t that difficult: the problem is in quantifying the key factor in the common forecasting meltdown in advance. That factor is turnout, and notably relative turnout by different demographics.

In the Brexit referendum and 2016 US presidential election, turnout by poorer and less educated voters, especially outside urban areas, hit unprecedentedly high levels, as people who had never voted before (and may never vote again) came out in droves. In both cases, forecasters’ pre-vote turnout models had predicted that these voters wouldn’t show up in nearly the numbers they did.
In the 2017 election, it was turnout among the young in particular that rocketed. This time the factor was widely expected to matter, and indeed get-out-the-vote campaigns aimed at the young were based on it. But most polling models failed to properly account for it, and that meant their predictions were wrong.

Polling is a moving target, and the spoils go to those who are most adept at taking and changing aim. So will the lesson be learned for next time? Possibly. But next time, under-25s might not turn out in anything like the same numbers – or a different demographic altogether might surprise everyone. We might not have long to wait to find out.

 

References:

Leighton Vaughan Williams. Report card: How well did UK election forecasters perform this time?  Article in The Conversation. Link below:

https://theconversation.com/report-card-how-well-did-uk-election-forecasters-perform-this-time-79237

The Nash Equilibrium: Snappy Slides

Game Theory_YT1

Can Game Theory teach us how to play the game of life?

If there is a set of ‘game’ strategies with the property that no ‘player’ can benefit by changing their strategy while the other players keep their strategies unchanged, then that set of strategies and the corresponding payoffs constitute what is known as the ‘Nash equilibrium’.

This leads us to the classic ‘Prisoner’s Dilemma’ problem. In this scenario, two prisoners, linked to the same crime, are offered a discount on their prison terms for confessing if the other prisoner continues to deny it, in which case the other prisoner will receive a much stiffer sentence. However, they will both be better off if both deny the crime than if both confess to it. The problem each faces is that they can’t communicate and strike an enforceable deal. The box diagram below shows an example of the Prisoner’s Dilemma in action.

Prisoner 2 Confesses Prisoner 2 Denies
Prisoner 1 Confesses 2 years each Freedom for P1; 8 years for P2
Prisoner 1 Denies 8 years for P1; Freedom for P2 1  year each

The Nash Equilibrium is for both to confess, in which case they will both receive 2 years. But this is not the outcome they would have chosen if they could have agreed in advance to a mutually enforceable deal. In that case they would have chosen a scenario where both denied the crime and received 1 year each.

So a Nash equilibrium is a stable state that involves interacting participants in which none can gain by a change of strategy as long as the other participants remain unchanged. It is not necessarily the best outcome for the parties involved, but it is the outcome we would most likely predict.

The Prisoner’s Dilemma is a one-stage game, however. What happens in games with more than one round, where players can learn from the previous moves of the other players?

Take the case of a 2-round game. The payoff from the game will equal the sum of payoffs from both moves.

The game starts with two players, each of whom is given £100 to place into a pot. They can then secretly choose to honour the deal or to cheat on the deal, by means of giving an envelope to the host containing the card ‘Honour’ or ‘Cheat’.  If they both choose to ‘Honour’ the deal, an additional £100 is added to the pot, yielding each an additional £50. So they end up with £150 each. But if one honours the deal and the other cheats on the deal, the ‘Cheat’ wins the original pot (£200) and the ‘Honour’ player loses all the money in that round.  A third outcome is that both players choose to ‘Cheat’, in which case each keeps the original £100. So in this round, the dominant strategy for each player (assuming no further rounds) is to ‘Cheat’, as this yields a higher payoff if the opponent ‘Honours’ the deal (£200 instead of £150) and a higher payoff if the opponent ‘Cheats’ (£100 instead of zero). The negotiated, mutually enforceable outcome, on the other hand, would be to agree to both ‘Honour’ the deal and go away with £150.

But how does this change in a 2-round game.

Actually, it makes no difference. In this scenario, the next round is the final round, in which you may as well ‘Cheat’ as there are no future rounds to realise the benefit of any goodwill realised from honouring the deal. Your opponent knows this, so you can assume your opponent who wishes to maximise his total payoff, will be hostile on the second move. He will assume the same about you.

Since you will both ‘Cheat’ on the second and final move, why be friendly on the first move?

So the dominant strategy is to ‘Cheat’ on the first round.

What if there are three rounds? The same applies. You know that your opponent will ‘Cheat’ on the final round and therefore the penultimate round as well. So your dominant strategy is to ‘Cheat’ on the first round, the second round and the final round. The same goes for your opponent. And so on. In any finite, pre-determined number of rounds, the dominant strategy in any round is to ‘Cheat.’

But what if the game involves an indeterminate number of moves? Suppose that after each move, you roll two dice. If you get a double-six, the game ends. Any other combination of numbers, play another round. Keep playing until you get a double-six. Your score for the game is the sum of your payoffs.

This sort of game in fact mirrors many real-world situations. In real life, you often don’t know when the game will end.

What is the best strategy in repeated play? For the game outlined above, we shall denote ‘Honour the deal’ as a ‘Friendly’ move and ‘Cheat’ as a hostile move. But the notion of a Friendly or Hostile approach can adopt other guises in different games.

There are seven proposed strategies here.

  1. Always Friendly. Be friendly every time
  2. Always Hostile. Be hostile every time
  3. Retaliate. Be Friendly as long as your opponent is Friendly but if your opponent is ever Hostile, you be Hostile from that point on.
  4. Tit for tat. Be Friendly on the first move. Thereafter, do whatever your opponent did on the previous move.
  5. Random. On each move, toss a coin. If Heads, be Friendly. If tails, be Hostile.
  6. Alternate. Be Friendly on even-numbered moves, and Hostile on odd-numbered moves, or vice-versa.
  7. Fraction. Be Friendly on the first move. Thereafter, be Friendly if the fraction of times your opponent has been Friendly until that point is less than a half. Be Hostile if it is less than or equal to a half.

Which of these is the dominant strategy in this game of iterated play? Actually, there is no dominant strategy in an iterated game, but which strategy actually wins if every strategy plays every other strategy.

‘Always Hostile’ does best against ‘Always Friendly’ because every time you are Friendly against an ‘Always Hostile’, you are punished with the ‘sucker’ payoff.

‘Always Friendly’ does best against Retaliation, because the extra payoff you get from a Hostile move is eventually negated by the Retaliation.

Thus even the choice of whether to be Friendly or Hostile on the first move depends on the opponent’s strategy.

For every two distinct strategies, A and B, there is a strategy C against which A does better than B, and a strategy D against which B does better than A.

So which strategy wins when every strategy plays every other strategy in a tournament? This has been computer simulated many times. And the winner is Tit for Tat.

It’s true that Tit for Tat can never get a higher score than a particular opponent, but it wins tournaments where each strategy plays every other strategy. In particular, it does well against Friendly strategies, while it is not exploited by Hostile strategies. So you can trust Tit for Tat. It won’t take advantage of another strategy. Tit for Tat and its opponents both do best when both are Friendly. Look at this way. There are two reasons for a player to be unilaterally hostile, i.e. to take advantage of an opponent or to avoid being taken advantage of by an opponent. Tit for Tat eliminates the reasons for being Hostile.

What accounts for Tit for Tat’s success, therefore, is its combination of being nice, retaliatory, forgiving and clear.

In other words, success in an evolutionary ‘game’ is correlated with the following characteristics:

Be willing to be nice: cooperate, never be the first to defect.

Don’t be played for a sucker: return defection for defection, cooperation for cooperation.

Don’t be envious: focus on how well you are doing, as opposed to ensuring you are doing better than everyone else.

Be forgiving if someone is willing to change their ways and co-operate with you. Don’t bear grudges for old actions.

Don’t be too clever or too tricky. Clarity is essential for others to cooperate with you.

As Robert Axelrod, who pioneered this area of game theory in his book, ‘The Evolution of Cooperation’: Tit for Tat’s “niceness prevents it from getting into unnecessary trouble. Its retaliation discourages the other side from persisting whenever defection is tried. Its forgiveness helps restore mutual cooperation. And its clarity makes it intelligible to the other player, thereby eliciting long-term cooperation.”

How about the bigger picture? Can Tit for Tat perhaps teach us a lesson in how to play the game of life? Yes, in my view it probably can.

 

Further Reading and Links

Axelrod, Robert (1984), The Evolution of Cooperation, Basic Books

Axelrod, Robert (2006), The Evolution of Cooperation (Revised ed.), Perseus Books Group

Axelrod, R. and Hamilton, W.D. (1981), The Evolution of Cooperation, Science, 211, 1390-96. http://www-personal.umich.edu/~axe/research/Axelrod%20and%20Hamilton%20EC%201981.pdf

https://en.wikipedia.org/wiki/The_Evolution_of_Cooperation

What should Messi do? Applying Game Theory to Penalty Kicks.

The El Clasico game between Real Madrid and Barcelona is in the 23rd minute at the Santiago Bernabeu when Lionel Messi is brought down in the penalty box and rewarded with a spot kick against the custodian of the Los Blancos net, Keylor Navas.

Messi knows from the team statistician that if he aims straight and the goalkeeper stands still, his chance of scoring is just 30%. But if he aims straight and Navas dives to one corner, his chance of converting the penalty rises to 90%.

On the other hand, if Messi aims at a corner and the goalkeeper stands still, his chance of scoring is a solid 80%, while it falls to 50% if the goalkeeper dives to a corner.

We are here simplifying the choices to two distinct options, for the sake of simplicity and clarity.

Navas also knows from his team statistician that if he dives to one corner and Messi aims straight, his chance of saving is just 10%. But if he stands still and Messi aims at one corner, his chance of saving the penalty rises to 50%.

On the other hand, if Navas stands still and Messi aims at a corner, his chance of making the save is just 20%, while it rises to 70% if Messi aims straight.

So this is the payoff matrix, so to speak, facing Messi as he weighs up his decision.

Goalkeeper – Stands Still Goalkeeper – dive to one corner
Lionel Messi – Aims straight 30% 90%
Lionel Messi – Aims at corner 80% 50%

 

So what should he do? Aim straight or to a corner. And what should Navas do? Stand still or dive?

Here is the payoff matrix facing Navas.

Messi – Aims straight Messi – Aims at a corner
Navas – Stands still 70% 20%
Navas – Dives to one corner 10% 50%

 

Game theory can help here.

Neither player has what is called a dominant strategy in game-theoretic terms, i.e. a strategy that is better than the other, no matter what the opponent does. The optimal strategy will depend on what the opponent’s strategy is.

In such a situation, game theory indicates that both players should mix their strategies, in Messi’s case aiming for the corner with a two-thirds chance, while the goalkeeper should dive with a 5/9 chance.

These figures are derived by finding the ratio where the chance of scoring (or saving) is the same, whichever of the two tactics the other player uses.

 The Proof

Suppose the goalkeeper opts to stand still, then Messi’s chance (if he aims for the corner 2/3 of the time) = 1/3 x 30% + 2/3 x 80% = 10% + 53.3% = 63.3%

If the goalkeeper opts to dive, Messi’s chance = 1/3 x 90% + 2/3 x 50% = 30% + 33.3% = 63.3%

Adopting this mixed strategy (aim for the corner 2/3 of the time and shoot straight 1/3 of the time), the chance of scoring is therefore the same. This is the ideal mixed strategy, according to standard game theory.

From the point of view of Navas, on the other hand, if Messi aims straight, his  chance of saving the penalty kick (if he dives 5/9 of the time) = 5/9 x 10% + 4/9 x 70% = 5.6% + 31.1% = 36.7%

If Messi opts to aim for the corner, Navas’ chance = 5/9 x 50% + 4/9 x 20% = 27.8% + 8.9% = 36.7%

Adopting this mixed strategy (dive for the corner 5/9 of the time and stand still 4/9 of the time), the chance of scoring is therefore the same. This is the ideal mixed strategy, according to standard game theory.

The chances of Messi scoring and Navas making the save in each case add up to 100%, which cross-checks the calculations.

Of course, if the striker or the goalkeeper gives away real new information about what he will do, then each of them can adjust tactics and increase their chance of scoring or saving.

To properly operationalise a mixed strategy requires one extra element, and that is the ability to truly randomise the choices, so that Messi actually does have exactly a 2/3 chance of aiming for the corner, and Navas actually does have a 5/9 chance of diving for the corner. There are different ways of achieving this. One method of achieving a 2/3 ratio is  to roll a die and go for the corner if it comes up 1, 2, 3 or 4, and aim straight if it comes up 5 or 6. Or perhaps not! But you get the idea.

For the record, Messi aimed at the left corner, Navas guessed correctly and got an outstretched hand to it, pushing it back into play. Leo stepped forward deftly to score the rebound. Cristiano Ronaldo equalised from the spot eight minutes later. And that’s how it ended at the Bernabeu. Real Madrid 1 Barcelona 1. Honours even in El Clasico.

 

Appendix

Messi’s strategy

x = chance that Messi should aim at corner

y = chance that Messi should aim straight

So,

80x + 30y (if Navas stands still) = 50x + 90y (if Navas dives)

x + y = 1

So,

30x = 60y

30x = 60 (1-x)

90x = 60

x = 2/3

y=1/3

 

Navas’ strategy

x = chance that Navas should dive to corner

y  = chance that Navas should stand still

So,

10x + 70y (if Messi aims straight) = 50x + 20y (if Messi aims at corner)

x+y = 1

So,

10x + 70y = 50x + 20y

40x = 50y

40x = 50(1-x)

90x = 50

x = 5/9

y = 4/9

 

 

Can the Super Bowl Predict the Stock Market?

“Few prediction schemes have been more accurate, and at the same time more perplexing, than the Super Bowl Stock Market Predictor, which asserts that the league affiliation of the Super Bowl winner predicts stock market direction. In this study, the authors examine the record and statistical significance of this anomaly and demonstrate that an investor would have clearly outperformed the market by reacting to Super Bowl game outcomes.” Thus read the abstract to a paper published in 1990 by Thomas Krueger and William Kennedy in the very well regarded Journal of Finance.

“If the Super Bowl is won by a team from the old National Football League (now the NFC, or National Football Conference),” they wrote, “then the stock market is very likely to finish the year higher than it began. On the other hand, if the game is won by a team from the old American Football League (now the AFC, or American Football Conference), the market will finish lower than it began.”

It is important to note, though, that some AFC teams count as NFL wins because they originated in the old NFL, i.e. Pittsburgh Steelers, Baltimore Ravens (formerly Cleveland Browns, Baltimore/Indianapolis Colts).

Over the 22-year history of the Super Bowl to the date of submission of their study in 1988, they documented a 91% accuracy rate for their predictor.

What happened in 1989? The NFC team, San Francisco 49ers, beat the AFCs Cincinnati Bengals the stock market rose 27%.

Further confirmation of an idea first proposed by New York Times sportswriter Leonard Koppett, published as The Super Bowl Predictor by investment advisor Robert H. Stovall in the January 1988 issue of Financial World.

So what happened in 1990?  Well, the NFC’s San Francisco 49ers won a second consecutive victory, beating the AFC’s Denver Broncos, by 55 points to 10. But the stock market fell in 1990, by 4.3%.

But then the Super Bowl Predictor returned to form, correctly predicting the direction of the stock market in 1991, 1992, 1993, 1994, 1995, 1996, 1997. Since the launch of the Super Bowl that made for 28 correct predictions out of 31 (a success rate of 90.3%).

Since then, the Super Bowl Predictor has had a much more chequered record. predicted correctly only about half the time since 1997. In 2009, Robert Stovell, a strategist for Wood Asset Management in Sarasota, Florida, and an early champion of the Stock Market Indicator wrote: Nothing seems to be working anymore {in the stock market]. Used to be, I was only happy when it was over 90% (accurate), and when it was still above 80% I was pleased. But certainly 79% is still far above a failing grade. (quoted on January 12, 2009, in MarketBeat (WSJ.coms inside look at the markets).

Since then, the Predictor has called it right five times (2010, 2011, 2012, 2014 and 2015, and wrong twice (2013 and 2016). As of January, 2017, the indicator been right a total of 40 times out of 50, as measured by the S&P 500 index. This year the AFCs New England Patriots stormed from 25 points behind at one point in the game to beat the Atlanta Falcons by 34 points to 28 in overtime. That should presage a bad year for the stock markets in 2017. We shall see.

So is the Super Bowl Indicator a real forecasting tool, or is it simply descriptive of wat has happened rather than containing any predictive value?

You decide!

 

Further Reading

Krueger, T.M. and Kennedy, W.F. (1990), An Examination of the Super Bowl Stock Market Predictor, Journal of Finance, 1990, 45 (2), 691-697.