Say I'm in an office pool pick 'em contest. My 10 buddies and I pick NFL winners each week, and the guy with the best record at the end of the season wins. My office mates aren't particularly good at handicapping football games, so I figure that if I pick the consensus favorite in every game (the team favored by the spread), I'll have a great chance to come out on top over the long haul.

My office buddies have access to point spreads too. They tend to look at the spread (or at least look at each team's respective record, which is just as accurate) and then pick a couple upsets each week. Over the past five years, the spread identifies winners correctly 66.2% of the time. So, normally, their upsets would be correct on average 33.8% of the time (100%-66.2%), but they won't be picking upsets in lopsided match-ups. (Although we can't always assume rationality, we will assume sanity.) So in their 34 chosen upsets (2 per week), my buddies will be right 45% of the time on average. I would have a 10% accuracy advantage in those games.

After doing the math, each of my buddies would average a 63.3% accuracy rate (66.2% * 232 games + 45% * 34 games). And I'd average 66.2% accuracy. Man, I can't wait to collect my winnings!

But wait. Because of luck, some would be slightly more accurate, and some would be less accurate. In fact, the only thing that really matters is how well each of them do on the 34 games they deviate from picking the published favorite. In the other 232 games, we'd have identical picks. Of the 34 games in question, each game that one of my buddies gets right, I must have been been wrong. One of my 10 friends needs to be correct greater than 50% of the time in his 34 games to beat me.

The mathematical bottom line is, "How often is someone correct at least in 18 out of 34 trials with a 0.45 probability of being correct in any given trial?" The binomial distribution gives us the answer--it's 22.4% of the time. That's pretty good, right? I have a 77.6% chance of beating any one of my opponents. The only problem is that there are 10 of them.

The chance I would beat all 10 of my buddies is the conjunctive probability of beating one of them. It's 77.6% * 77.6%... and so on, for however many opponents I have. In this case, it's:

0.776 ^{10} = 0.079

In other words, my chances of winning the office pool are just 7.9%--significantly less than a fair chance of 1 in 11. That's why just picking the favorites is a bad strategy. I'd actually be better off choosing the less accurate strategy of my buddies. At least then I'd have fair chance at 1 in 11.

I realize that it is counter-intuitive that a strategy that is less accurate overall is better than a more accurate strategy. But in a contest against several opponents, the more risky strategy--with a greater deviation of outcomes--may be best.

Note: Phil Birmbaum points out that the odds of the opponents are not independent of one another, and therefore the simple compound probability I calculated here is far too low. If one opponent happens to beat you, then the other opponents may be more likely to beat you as well, and vice versa. In the end, picking all favorites may be the better play. See his comments for an explanation.

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## The Office Pool 1

By
Brian Burke

published on 3/04/2008

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I don't think this is quite right ... you can't multiply the 10 probabilities together like that, because they're not independent. Suppose you beat the first opponent. Then the chance that you did very well on your favorites is higher, and so the chance you also beat the second, third, fourth ... opponent is more than 77.6%.

(Look at it this way: suppose you run the same calculation for the first opponent, who picks underdogs. His chance of beating you is 23.4%. His chance of beating the others is 50%. Multiply nine 50%s and one 23.4%, and you obviously get a lot less than 7.9%. So that can't be right.)

I think the best strategy is to pick all favorites. The disadvantage is not that it reduces your odds of winning, but that it increases the chance of a tie and a split pot. You might be better off taking all favorites except one underdog, and hoping nobody matches you.

The best strategy, assuming all your opponents also use it, is probably to take all favorites with probability X, and one underdog with probability (1-X). It's not that hard to calculate X, but I don't feel like it. :)

As long as the opponents aren't colluding to pick the same upsets, and there are enough upsets to pick, wouldn't it be independent? I guess not; your calculation based on the 'first opponent' proves that.

I suppose if there are only a handful of reasonable upsets to pick, that would make it non-independent. (I actually just went and calculated--there are 58.4 games per year with spreads <4 pts.) I guess that's not enough.

Would it really matter how well the favorites do, as long as the opponents are basing their picks on the "favorites except a couple upsets" method? In other words, the record of the favorites becomes a floating zero-point, a baseline from which the other contestants' records deviate. Each contestant has a 23.4% chance of beating that baseline, whatever it is. (But not independently as you pointed out.)

I suppose if the favorites had an unusually strong year (say 70%) then we'd have to slightly reduce the 45% upset accuracy rate I assumed (and that was just an assumption). But then again, there is equal chance of an unusually weak year for favorites. For example, in 2006 the Vegas favorites won only 59% of the time.

Yeah, the records of the underdog pickers (assuming they're all different games) are independent, but the *probability of beating the favorite picks* is not independent.

If the first underdog-picker beats the favorite-picker, it's likely the favorite-picker picked badly, and so more likely the second underdog-picker will beat him too, and the third, and so on.

I think the problem is that the effect of the favorites' record is not symmetrical. The baseline is 55%. If the favorites win only 50%, every picker has an equal chance (since there's no such thing as a favorite any more), and the favorite-picker has a 1/11 (9.1%) chance of winning the pool.

But if the favorites win 60%, then it's going to be very hard for a bunch of 40% underdogs to beat the favorites in 34 games. I won't do the math, but perhaps the probability of the favorite-picker winning will now be 40%.

Extrapolating linearly (which we shouldn't do), you can average the 50% and 60% numbers -- 9.1% and 40% (or whatever the correct number is) -- to estimate the chance of the favorite-picker winning at 55% per game. This naive estimate comes out to about 25%, which I bet is pretty close. Probably easy to do a simulation, maybe I'll try it.