Heads up

heads

I recently posted a problem that had been shown to me by Benoit.Jottreau@smartodds.co.uk. Basically, you have a bunch of coins. You toss them and remove the ones that come up heads. You then repeat this process over and over till all the coins have been removed. The question was, if you start with respectively 10 or 100 coins, how many rounds of this game does it take on average till all the coins have been removed?

I’m really grateful to all of you who considered the problem and sent me a reply. The answers you sent me are summarised in the following graphs.

 

 

The graph on the left shows the counts of the guesses for the number of rounds needed when starting with 10 coins; the one on the right is the counts  but starting with 100 coins. The main features are as follows:

  • Starting with 10 coins, the most popular answer was 4 rounds; with 100 coins the most popular answer was either 7 or 8 rounds.
  • Almost everyone gave whole numbers as their answers. This wasn’t necessary. Even though the result of every experiment has to be a whole number, the average doesn’t. In a similar way, the average number of goals in a football match is around 2.5.
  • The shape of the distribution of answers for the two experiments is much the same: heavily skewed to the right. This makes sense given the nature of the experiment: we can be pretty sure a minimum number of rounds will be needed, but less sure about the maximum. This is reflected in your collective answers.
  • Obviously, with more coins, there’s more uncertainty about the answer, so the spread of values is much greater when starting with 100 coins.

Anyway, I thought the replies were great, and much better than I would have come up with myself if I’d just gone with intuition instead of solving the problem mathematically.

A few people also kindly sent me the logic they used to get to these answers. And it goes like this…

Each coin will come up heads or tails with equal probability. So, the average number of coins that survive each round is half the number of coins that enter that round. This is perfectly correct. So, for example, when starting with 10 coins, the average number of coins in the second round is 5. By the same logic, the average number of coins in the second round is 2.5. And the average number of coins in the third round is 1.25. And in the fourth round it’s 0.625. So, the first time the average number of coins goes below 1 is on the fourth round, and it’s therefore reasonable to assume 4 is the average number of rounds for all the coins to be removed.

Applying the same logic but starting with 100 coins, it takes 7 rounds for the average number of coins to fall below 1.

With a slight modification to the logic, to always round to whole numbers, you might get slightly different answers: say 5 and 8 instead of 4 and 7. And looking at the answers I received, I guess most respondents applied an argument of this type.

This approach is really great, since it shows a good understanding of the main features of the process: 50% of coins dropping out, on average, at each round of the game. And it leads to answers that are actually very informative: knowing that I need 4 rounds before the average number of coins drops below 1 is both useful and very precise in terms of explaining the typical behaviour of this process.

However… you don’t quite get the exact answers for the average number of rounds, which are 4.726 when you start with 10 coins, and 7.983 when you start with 100 coins. But where do these numbers come from, and why doesn’t the simple approach of dividing by 2 until you get below 1 work exactly?

Well, as I wrote above, starting with 10 coins you need 4 rounds before the average number of coins falls below 1. But this is a statement about the average number of coins. The question I actually asked was about the average number of rounds. Now, I don’t want to detract from the quality of the answers you gave me. The logic of successively dividing by 2 till you get below one coin is great, and as I wrote above, it will give an answer that is meaningful in its own right, and likely to be close to the true answer. But, strictly speaking, it’s focussing on the wrong aspect of the problem: the number of coins instead of the number of rounds.

The solution is quite technical. Not exactly rocket science, but still more intricate than is appropriate for this blog. But you might still find it interesting to see the strategy which leads to a solution.

So, start by considering just one of the coins. Its pattern of results (writing H for Heads and T for Tails) will be something like

  • T, T, H; or
  • T, T, T, H; or
  • H

That’s to say, a sequence of T’s followed by H (or just H if we get Heads on the first throw).

But we’ve seen something like this before. Remember the post Get out of jail? We kept rolling a dice until we got the first 6, and then stopped. Well, this is the same sort of experiment, but with a coin. We keep tossing the coin till we get the first Head. Because of the similarity between these experiments, we can apply the same technique to calculate the probabilities for the number of rounds needed to get the first Head for this coin. One round will have probability 1/2, two rounds 1/4, three rounds 1/8 and so on.

Now, looking at the experiment as a whole, we have 10 (or 100) coins, each behaving the same way. And we repeat the experiment until all of the coins have shown heads for the first time. What this means is that the total number of rounds needed is the maximum of the number of rounds for each of the individual coins. It turns out that this gives a simple method for deriving a formula that gives the probabilities of the number of rounds needed for all of the coins to be removed, based on the probabilities for a single coin already calculated above.

So, we now have a formula for the probabilities for the numbers of rounds needed. And there’s a standard formula for converting this formula into the  average. It’s not immediately obvious when you see it, but with a little algebraic simplification it turns out that you can get the answer in fairly simple mathematical form. Starting with n coins – we had n=10 and n=100 –  the average number of rounds needed turns out to be

1+\sum_{k=1}^n(-1)^{k-1}{n \choose k} (2^k-1)^{-1}

The \sum bit means do a sum, and the ~{n \choose k}~ term is the number of unique combinations of k objects chosen from n. But don’t worry at all about this detail; I’ve simply included the answer to show that there is a formula which gives the answer. 

With 10 coins you can plug n=10 into this expression to get the answer 4.726. With 100 coins there are some difficulties, since the calculation of ~{n \choose k}~  with n=100 is numerically unstable for many values of k. But accurate approximations to the solution are available, which don’t suffer the same numerical stability problems, and we get the answer 7.983.

So, in summary, with a fair bit of mathematics you can get exact answers to the problem I set. But much more importantly, with either good intuition or sensible reasoning you can get answers that are very similar. This latter skill is much more useful in Statistics generally, and it’s fantastic that the set of replies I received showed collective strength in this respect.

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