I recently read GameTek by Geoffrey Engelstein, which includes several interesting examples of how to improve your intuition when it comes to the design and the playing of games. Unfortunately, two of these examples are at best extremely misleading and at worst completely incorrect. However the way in which they are wrong is fascinating and prompted several hours of thought and discussion on my part, so I’d like to share them with you.

The Premises

Engelstein presents two similar examples in the book, which I will reproduce here. First, at the end of Chapter 6, “Feeling the Loss”, we have a presentation of a classic paradox in probability called The Two Child Paradox.

You’re talking to woman at a park and she say she has two children. Suddenly a boy runs up and grabs her hand and you ask if this is her son. She replies that he is.

What are the chances that her other child is a boy?

Most people say 50 per cent: the fact that one is a boy has no bearing on the gender of the other child. But this is not correct. There are four possible family arrangements that have two children: boy/boy, boy/girl, girl/boy and girl/girl. When I tell you that at least one is a boy, that eliminates one possible family: girl/girl. There are three combinations left: boy/boy, boy/girl, and girl/boy. Each is equally likely, but in only one out of three (and thus a ⅓ chance) is the other child a boy.

Then at the end of Chapter 14, “Tic-Tac-Toe and Entangled Pairs”, we have this other situation which might sound different but is actually intimately connected to the first.

Let’s say four people are playing Bridge. One of them says, ‘I have an Ace,’ and we know she is telling the truth. The chance that she’s holding more than one Ace is about 37 per cent. Later the same player says, again truthfully, ‘I have the Ace of Spades.’ Strangely, the chance that she has more than one Ace is now 56 per cent.

Take a moment to ponder these two examples. They probably feel unintuitive, but all you need to do at this point is understand the premise of each one. When you’re ready, proceed.

The Paradox

I don’t know about you, but I find the first example extremely unintuitive, if not totally wrong. In my view, the key point that Engelstein omits is that the children are distinguishable. Since you have met one of the children at the park, you can imagine labeling that as the “park” child and the other one as the “home” child. Going through the possible configurations we have:

\[\text{boy}_{\text{park}}/\text{boy}_{\text{home}},\text{boy}_{\text{park}}/\text{girl}_{\text{home}},\text{girl}_{\text{park}}/\text{boy}_{\text{home}},\text{girl}_{\text{park}}/\text{girl}_{\text{home}}\]

By discovering that the child in the park is a boy, we are not left with three possibilities as Engelstein argues, but only two, one of which involves two boys. So in my view the intuitive answer of 50% is actually correct. Engelstein’s summary of “at least one is a boy” clashes with his earlier statement that we learned that a specific child is a boy.

So why did Engelstein think it was 1 in 3? Is there even a paradox here? First we should restate the example to remove some ambiguitiy. Let’s imagine that you and I are playing a game with the following steps:

  1. I flip two fair coins and then hide the coins behind a screen. The random result is fixed but only I can see it
  2. I may reveal some information about the coins to you
  3. I reveal both coins and you win if both coins are Heads

Now consider how your chance of winning changes throughout the steps of the game. After step 1 your chance will always be the same: two fair coins were flipped, so there are four possibilities, so there is a 1 in 4 chance that both are Heads.

HH, HT, TH, TT

Suppose in step 2 that I look at the coins and tell you that at least one of them is Heads. What is your chance of winning? This is the situation that Engelstein was trying for earlier: this is equivalent to me saying that it is not the case that both coins are Tails; in other words, I have only eliminated one of the four possibilities. So by Engelstein’s argument, you have a 1 in 3 chance of winning.

Instead, suppose that in step 2 I pick up one coin from behind the screen and show you that it is Heads. In this case, we can apply my previous argument: the coins are distinguishable (we have the revealed coin and the hidden coin) thus there are only two remaining possibilities and so the chance of winning is 1 in 2.

\[\text{H}_{\text{revealed}}/\text{H}_{\text{hidden}},\text{H}_{\text{revealed}}/\text{T}_{\text{hidden}},\color{red}{\text{T}_{\text{revealed}}/\text{H}_{\text{hidden}},\text{T}_{\text{revealed}}/\text{T}_{\text{hidden}}}\]

Now the paradox: why is there a difference between these situations? In the first case, where at least one coin was Heads, certainly I could have also picked up a coin and showed you that it was Heads. Why does it matter if I actually show you or not? It’s as if, by introducing completely redundant information (physically showing you a Head rather than saying it), your chances of winning are magically increased!

Here is where the two examples from earlier connect. In the Bridge example, I first say “I have an Ace”, then by introducing completely redundant information and saying it’s specifically the Ace of Spades, the chances of having more than one Ace magically increase! Why does the redundant information help? Certainly when I say I have an Ace you can imagine me showing you that it is one of the specific Aces.

So what’s going on? Is Engelstein’s argument correct, or is mine? Or is something else going on entirely?

The Game

To help you understand how to resolve this, we need to put some money on the line. Now you have to pay to play, but there is a prize for winning:

  1. I flip two fair coins and then hide the coins behind a screen. The random result is fixed but only I can see it
  2. I may reveal some information about the coins to you and state both the prize for winning $W and the cost of entry $C
  3. If you pay, I reveal both coins and you win if both coins are Heads

The question is if the game is worth playing; based on the information I revealed, the prize money, and the cost of entry, do you expect to make money by playing? For example, if you can calculate that the probability of two Heads based on the information revealed is 50%, the prize is $24, and the cost of playing is $10, then

\[\text{expected winnings}=P(\text{win})\times($24-$10)-P(\text{lose})\times($10)=0.5\times $24-$10=$2\]

In other words, you expect to make $2 per play on average (assuming you have a big enough budget to play the game repeatedly and overcome some early bad luck). As you can see from this example, it is crucial that you are able to calculate the probability of winning in order to make a profit in this game. So imagine you find me at the carnival offering that exact situation: the prize is $24, the cost $10, and I’m showing you that a coin is Heads. Do you play?

Assuming you found my argument from earlier convincing, your chances of winning are 1 in 2. Based on the previous calculation, you will make a profit! So you play and indeed you win! So you keep playing, following the exact same strategy: when I show you Heads you pay and, to keep things simple, when I don’t show you Heads you decline to pay and simply wait for me to flip the coins again. With this strategy you make a small but reliable profit ($2 per game on average); you’re going to bleed me dry if it takes all day!

Then something changes. Suddenly your strategy starts losing. Convinced it must simply be an unlucky streak, you play on, but eventually I completely drain your funds and you walk away with empty pockets. What happened?

Imagine you’re in my shoes and you need to run this game and you’ll quickly see what is missing from the picture: when do you decide to show the player a coin? For example, I could follow a procedure like this:

  1. Flip the coins and place one on the left and one on the right
  2. If the left coin is Heads, show them that coin
  3. Otherwise, show them nothing.

In this case, the 1 in 2 calculation is correct: whenever I show you a coin, the only remaining possibilities are

\[\text{H}_{\text{left}}/\text{H}_{\text{right}},\text{H}_{\text{left}}/\text{T}_{\text{right}}\]

So playing is always profitable.

But what if instead I follow this procedure:

  1. Flip the coins and place one on the left and one on the right
  2. If the left coin is Heads, show them that coin
  3. Otherwise, if the right coin is Heads, show them that coin
  4. Otherwise, show them nothing.

Now whenever I show you a coin, it could be

\[\text{H}_{\text{left}}/\text{H}_{\text{right}},\text{H}_{\text{left}}/\text{T}_{\text{right}},\text{T}_{\text{left}}/\text{H}_{\text{right}}\]

and you have no way to distinguish between them. The chance of winning has dropped to 1 in 3. Unfortunately, from your perspective, both procedures look nearly identical. The only difference would be the frequency of me showing nothing, but this is complicated by the fact that I can switch between these procedures at will. If I start with the first procedure I can sucker you in by making you think the game is rigged in your favor. When I switch strategies, your expected winnings flip to −$2 and hopefully for me your greed will cost you all of your profits and then some. I could even be sneakier and switch between these strategies randomly, slightly favoring the one that pays me more.

If you approach this game naively, you will always lose in the long run. Why? Because, like Engelstein, you will assume that the specific way in which you were given information reveals something additional about the nature of why that information was revealed; specifically, that me showing Heads versus saying “There is at least one Heads,” implies something about the choice I made. In general, there may be no connection between these things.

So this is the resolution of the paradox: it is not the case that revealing redundant information can change the chances of an event, rather it is possible to phrase a probability problem in an ambiguous way such that most people will make seemingly obvious (but unsupported) assumptions about it and thus come to incorrect conclusions.

The Resolution

Going back to the original Two Child Paradox. What were the unsupported assumptions we made in order to arrive at the paradox in the first place?

To arrive at the answer of 1 in 2 for the chances of the other child being a boy, we must assume that the gender of the child we met has no bearing on why we met them at the park. If that same child were a girl, we are assuming they would have still been brought to the park. This makes the gender irrelevant, since if they brought a boy, we have boy/girl or boy/boy (1 in 2) and if they brought a girl we have girl/boy or girl/girl (1 in 2). This seems like a reasonable assumption to me, however it is not stated in the problem.

To arrive at the answer of 1 in 3, we must assume some strange parental behavior, namely that only boys are brought to parks. In that case, the fact that we have met a boy at a park means that this could be a family with two boys, or boy/girl, or girl/boy since all of these equally likely scenarios would result in us meeting a boy. Again this seems like a strange assumption, but it is also not stated and thus cannot be excluded.

Back to Bridge. How do you arrive at the probabilities that Engelstein gave? If I say “I have an Ace” what is the probability that I have two or more Aces? In order to answer this we need to state our assumptions, namely, why did I say I have an Ace? If we assume that whenever I have any Aces I always say “I have an Ace” then indeed you arrive at about 37% chance of me having two or more. This seems like a reasonable assumption to make.

But what if I say that I have the Ace of Spades? How do we arrive at 56% chance of two or more Spades? That is the probability we get if we assume that I only tell you when I have the Ace of Spades; specifically, if I have other Aces and no Ace of Spades I say nothing. Similar to the Two Child Paradox this feels like a very strange assumption to make. If we’re interested in finding Aces, why wouldn’t I tell you about the others when I have them? This means that there would be situations where I have two or more Aces (none of them Spades) but I don’t reveal any information. A much more reasonable assumption here is if we assume that when I have any Aces I tell you the suit of one of them. It’s not hard to see that now I will tell you about an Ace (and its suit) whenever I have any Aces, in other words we’re back in the first situation. So now even though I am revealing additional information (the suit), the probability of having two or more Aces is still 37%, which is the intuitive result.

My Conclusion

If you only remember one thing from this, it is to be careful in your assumptions when turning word problems into math problems since they can completely change your answer. Engelstein mostly does a good job in his book of debunking common false assumptions such as the idea of “hot streaks” in gambling, but he ironically talks himself back into such bad assumptions with these two examples. Both answers he gives are unintuitive because they make unintuitive assumptions. If you make the more natural assumptions—in my opinion—you get the intuitive answers and no paradox appears.