# Another Commercial… What Are the Odds?

This year’s NCAA tournament is being spread over four tv networks, so that every game is shown in its entirety. Previously, you got whatever game was deemed to hold the most interest for your region of the country, plus occasional looks in at games elsewhere, that usually managed to miss the most exciting bits. For a hoops junkie like me, this was fantastic, with the only problem being how to manage to overload. I eventually settled on switching channels every time whatever game I was watching went to a commercial, so I got to see a bit of everything.

I was struck, though, but just how often there were multiple networks showing commercials. On one or two occasions, they managed to have all four channles showing ads rather than hoops.

So, what are the odds of that happening?

Well, first we need an estimate of how likely it is for a game to be in a break. So how often does a basketball game stop?

A typical basketball broadcast takes a bit more than two hours– let’s call it 140 minutes. Of that time, 20 minutes (give or take) is the halftime show (which is the moral equivalent of a long commercial break), and there are four guaranteed “media timeouts” in each half, running something like a minute and a half each. Each team also gets a number of timeouts, which are a mix of “full” and “30-second.” They don’t always use all of these over the course of the game, though it’s usually close, so let’s call that an additional 5 minutes of stoppage time for each team.

The total time spent in breaks, then, is 20+8(1.5)+5+5 minutes, or 42 minutes, out of 140, a nice, round 30% of the time. The chances of all four networks being in a timeout, then, are:

P(all 4) = 0.3*0.3*0.3*0.3 = 0.34 = 0.0081

So, there’s a bit less than a 1% chance of all four networks being in commercial at the same time. What are the odds of me catching one of those instances during yesterday’s channel-flipping?

There are two ways to determine this: one is to estimate the number of channel-flips during the afternoon. There are nine guaranteed stoppages per game, plus another 5-10 for coach-called timeouts, so you’re looking at a maximum of about 70 flips per game, times four games in a given block, times two game blocks, for a whopping 560 channel-flips. That’s an extreme upper limit, which you can use to estimate the maximum probability that I would encounter all four stations during a break.

The simpler and more elegant calculation is that the probability is 100%. Because that’s just the way my luck tends to run– if something annoying can happen, it will happen, odds be damned.

## 7 thoughts on “Another Commercial… What Are the Odds?”

1. This assumes that the commercial breaks are randomly distributed, and they aren’t. Timeouts are less likely at the very start of a half, for example (how often does anyone call a TO in the first minute?), and probably more likely as the half draws to a close, since they are used to set up plays. So there’s some correlation which increases the odds of a four-station commercial nexus.

2. Yes, but the start times for the games are staggered, so there shouldn’t be multiple games in an end-of-half situation at the same time. For the purposes of this kind of crude approximation, I think it’s probably a wash.

3. There’s also the possibility that they actually plan the officials time-outs all at around the same time, so as to force people to watch the commercials. So, again, it’s probably not a totally random distribution of breaks.

Certainly I’d believe this for stations in the same network family.

4. The official time-outs come at specified times in the game (first whistle after the clock passes 16, 12, 8, and 4 minutes in each half), so they can’t do too much to fiddle with the scheduling. Unless there’s some Buffalo Wild Wings foolishness going on with the officials.

5. Eric Lund says:

What you calculated was the probability that all four networks would be in a commercial break during any given minute. But there are, as you said, 140 minutes in a game, so the cumulative probability is not P but 1 – (1 – P)^140. If I have punched the numbers in my calculator correctly, that works out to 68%.

6. Eric Lund says:

And on further reflection, my post #5 underestimates the probability, since it is known a priori that at least one of the networks (the one you were watching just before it went to a commercial break) is in a commercial break. So P is not 0.0081 but 0.027. And there were two sets of games, so double the number in the exponent.

IOW, the probability approaches 100%.

7. CCPhysicist says: