How to cheat on a poll without actually lying

As you can read on various blogs (I did on Mark Chu-Carroll’s), there is some fuss over Washington State Republican caucus. Apparently, the GOP stopped counting votes on 87%, declaring McCain a winner before all votes got counted, and said the final results would be determined sometime next week. Now I won’t get into the gory details of the Republican party’s election methods (and gory they are, what a mess!). Some people, however, claim none of it matters. After all, was this really cheating? Isn’t the GOP right? They’ve counted 87% of the caucus. McCain has an edge, so what’s the chance that the total caucus will be different?

Well, today we shall answer that first question. Bottom line: it does matter, it was cheating, and pollsters do cheat like that all the time.

This is election lesson no. 1:

How to cheat on a poll without actually lying

What if Huckabee did have an edge? Suppose, hypothetically, that he had 55% of the votes, and that the other 45% were McCain’s. The chairman, who is a McCain supporter, supervises the votes counting. There is a very simple thing which he can do if he wants McCain to win – he can wait for a McCain to gain a temporary advantage, then stop the count, and declare McCain the winner. If you like riddles, you should try this question out, and then skip to the end of the post: what are the chairman’s chances of success?

Here comes a very simple and surprising bit of math. We’ll divide the problem. There are two ways in which McCain can have a temporary advantage. One happens if the first few votes counted are McCain’s, so he gains an early lead (which will be closed later on, when more voted are counted). What are the chances for that to happen? Well, McCain gets an early edge if the very first vote counted is his. As he has 45% of the votes, that’s his chance for having an early lead.

But there is another thing that can happen: it might be that the first votes are Huckabee’s, which has the early edge, and that as more votes get counted, McCain’s votes accumulate beyond Huckabee’s. Of course, when even more votes get counted, McCain’s advantage will dwindle and disappear. But the chairman can stop the vote counting before Huckabee gains his lead back. What are the chances for that?

Let’s write an M for a McCain vote, H for a Huckabee vote. Write down the sequence of votes as the get counted. Look at this sample of votes:


McCain has an early lead of 3 votes. Only on the 8th vote does Huckabee close the gap. If I were the chairman, I would have stopped by the 7th vote counted. Now let’s take the first 7 votes, and flip them. We get:


In this sequence, Huckabee starts off with an advantage, and McCain still catches up and gets a temporary lead, which is again closed by the 8th vote. Now comes the really neat bit:

Every sequence in which McCain has an early lead is the flip of a sequence in which McCain gets a temporary lead later on. For every sequence of the first type, take the part where McCain still has the advantage, flip it, and you have a sequence of the second type.

To conclude:

There are two types of voting sequences which would allow the chairman to succeed. As we have just seen, both types are commensurable. Thus, if McCain had 45% of the votes, then 45% of the voting sequences are of the first type, and another 45% are of the second type. Put together they give a 90% chance that he will have a lead somewhere throughout the count. This is a chance you can count on. In general, the minority candidate has some fraction of the votes, which we denote β. This fraction is exactly the chance to have a voting sequence of either type. So, together, there’s a 2β chance for both types, and 2β is the chairman’s chance.

If it still looks fishy, look at it this way: if Huckabee had a majority of the votes, then every sequence of votes ends with a Huckabee victory. Now look at all subsequences. Most of them will give you a Huckabee victory – which is the reason polls actually give you information. If you decide to count just 10% of the votes, most chances are you’ll observe a Huckabee victory. But some subsequences will give you a McCain victory. By cherry picking the right subsequences, you may be able to make the minority victorious.

As it happens, pollsters know this fact. Let’s get on to polls. Suppose you want a poll to prove that most people like your product/candidate/whatever better than they like mine. You call a pollster and ask him to prove you’re better. If you do have a majority on your side, then all your pollster needs to do is count enough people, and he can show it. But what if most people prefer your nemesis? As we have just shown (and assuming that the margin is not too high), the pollster needs to count votes until he sees a lead to your side. He goes on counting until your advantage starts to dwindle. He stops counting at that point, and publishes the results, showing that most people prefer you.

I had the pleasure of learning games theory with Yisrael Auman, Nobel laureate in Economics. From him I heard the story about some university officials, who asked a polling company to check something for them. About a week after they invited the pollsters, they got a call from the company CEO.

“You have forgotten”, he told them, “to give us a vital piece of data, without which it will be very hard to do the poll successfully”.

“What is it that you need to know”? they asked.

“The result you want to get”, the CEO answered.

4 thoughts on “How to cheat on a poll without actually lying

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