Dynasty, in Theory: What Are Odds?

There's a lot of strong dynasty analysis out there, especially when compared to five or ten years ago. But most of it is so dang practical-- Player X is undervalued, Player Y's workload is troubling, the market at this position is irrational, and take this specific action to win your league. Dynasty, in Theory is meant as a corrective, offering insights and takeaways into the strategic and structural nature of the game that might not lead to an immediate benefit but which should help us become better players over time.

The Most Dangerous Game

I am hanging out with five friends (who all just happen to be skilled logicians) when I pull out a deck of cards and suggest we play a game. I deal one card face-down in front of me, turn to my first friend, and ask, "What are the odds the card in front of me is a seven?"

My friend quickly responds, "One in thirteen". I then turn to my second friend, deal eight cards to her, and ask, "What are the odds the card in front of me is a seven?" She glances at the cards I gave her, thinks for a moment, and then says, "One in eleven".

I deal eight more cards to the next friend in line and repeat my question: "What are the odds the card in front of me is a seven?" My friend glances, thinks, and then says "one in nine". I deal eight more cards to my fourth friend and ask again. He looks at them for a long time with a very puzzled look on his face before answering, "One in thirteen, I guess?"

While everyone looks at each other in bewilderment, I place eight cards face-up in the middle of the table: an Ace of Spades, a Nine of Hearts, a King of Diamonds, a King of Clubs, an Ace of Hearts, a Queen of Hearts, a Ten of Clubs, and a Nine of Clubs. I then look at my last friend and ask one last time: "What are the odds the card in front of me is a seven?" My friend thinks about the previous estimates for a while before suddenly responding with "zero percent". Everyone else immediately realizes what happened and groans.

What Happened?

My first friend-- tasked with estimating my odds of having a seven with no other information to go on-- resorted to basic math. A standard deck has 52 cards, 4 of them are sevens, so there was a 4-in-52 chance a card I selected randomly would be a seven, which simplifies to 1-in-13.

My second friend was able to peek at eight cards. None of those cards were a seven, which meant there were 44 cards that were still unknown and 4 of them were sevens, giving me a 1-in-11 chance of having a seven. Since everyone at the table was a skilled logician, as soon as my second friend gave this estimate, everyone knew that none of her cards was a seven.

My third friend received eight more cards, none of which were a seven. Between the eight he saw and the prior eight my second friend had seen, that was 16 non-sevens removed from the deck, which left 4 sevens in 36 cards; therefore, friend #3 set my odds at one in nine.

After my fourth friend looked at his cards, there would only be 28 cards still unaccounted for, so my odds would be 4 in 28, 3 in 28, 2 in 28, 1 in 28, or 0 in 28, depending on how many sevens friend #4 saw in his pile; instead, he answered one in thirteen, which didn't match any of those expectations.

Why? Because in his eight cards, he had two copies of the Jack of Spades and two copies of the Ace of Diamonds, which immediately let him know I wasn't using a standard deck, and since he didn't know the distribution of cards, he couldn't calculate the odds. Instead, he retreated back to the base rate-- one out of every thirteen playing cards is a seven.

It took a while for my fifth friend to work out why they had gone back to the one-in-thirteen estimate, but eventually, he realized that friend four must have discovered we were working with a non-standard deck. Friend five then noticed that the cards on the table were all nine or higher and realized that I had pulled out a Pinochle deck, which consists of two copies of every 9, 10, Jack, Queen, King, and Ace (for a total of 48 cards).

Since there are no sevens in a Pinochle deck, my last friend estimated my odds of having a seven at zero. Everyone at the table noticed the estimate fell to zero even though I still hadn't revealed any sevens, thought back to the cards they had already seen, and likewise quickly realized what I had done.

So Who Was Right?

Our first friend gave a confident 1-in-13 estimate. Our second friend gave a confident 1-in-11. Our third friend gave a confident 1-in-9. Our fourth friend gave an unconfident 1-in-13, and our fifth friend gave a 0%. Which of these estimates was most correct?

Before we answer, a twist. I flip up the card in front of me to reveal that it is, in fact, a seven. The game was rigged. I had set up the deck ahead of time specifically to challenge their assumptions. (Had these friends known me a bit better, they might have seen this coming.)

Now, knowing this, which of these estimates was the most correct?

One could say that none of the estimates were right because they all suggested my card was most likely not a seven, but my card was in fact a seven. One could say that all of the estimates were right because they were all the best estimate possible based on the information available at the time.

One could say that "one in seven" was the most correct estimate because it was the highest value, which meant that it was the closest to the reality. One could say that "one in thirteen, I guess?" was the most correct estimate because it was the least confident, the one that best recognized that it didn't fully understand the rules of the game. You could say all of the estimates were incorrect because they forgot to account for the most important variable-- me.

I would suggest that odds for one-off events are technically never "correct"-- in that "correct" is not a word that has any meaning in this context. It's like asking how fast is the color blue, or how wide is winter. Because odds cannot be "correct", they must be evaluated against a different standard.

I want you to notice something else, though: all of these estimates were purportedly about my card, but none of them had anything to do with my card. My card stayed the same the entire time, but the estimate kept changing based on other, completely different cards. We'll come back to that in a minute.

A Less Dangerous Game

You go on the show Let's Make a Deal and the host, Monty Hall, shows you three doors. He tells you behind one of the doors is a car and behind the other two are goats, then asks you to choose which door you would like. You choose Door #1.

Monty Hall then opens one of the other two doors to reveal that there was a goat behind it. He then asks if you would like to switch doors or stick with the one you already picked. What are your odds of winning the car if you stay put? What are your odds of winning if you switch?

This is known as the Monty Hall problem; it is one of the most famous logic puzzles of all time, in no small part because of how mad it tends to make people the first time they hear it-- because the answer is you should switch. If you picked the car originally, switching will move you to a goat. If you picked a goat originally, switching will move you to the car. There's a 1/3 chance you picked the car originally, which represents your odds of winning if you stay. There was a 2/3s chance you picked the goat, which represents your odds of winning if you switch.

(If this seems crazy to you, let me remind you that this puzzle is famously unintuitive. But many find it helpful to extend the problem: imagine there were a million doors, one of which hid a car. You choose a door, and then Monty Hall opens 999,998 others, revealing goats behind all of them. In this instance, would you switch? Either you got extraordinarily lucky and managed to land on the car with your first pick, or else you didn't get extraordinarily lucky, you picked a goat, and the car is behind that last remaining door. Your odds of winning if you stay are 1-in-1,000,000. Your odds of winning if you switch are 999,999-in-1,000,000.)

But just like my five friends made assumptions about the game I was playing, these estimated odds for the Monty Hall problem also hinge on some pretty big assumptions: first, that Monty Hall knows where the goats are and always deliberately reveals one when he opens the door.

If Monty Hall is just opening doors at random, your odds of winning the car are 50/50 whether you stay or switch. To illustrate, there are six possibilities: you pick the Car and he opens Goat #1, you pick the Car and he opens Goat #2, you pick Goat #1 and he opens the Car, you pick Goat #1 and he opens Goat #2, you pick Goat #2 and he opens the Car, you pick Goat #2 and he opens Goat #1.

When Monty Hall opens a door to reveal a goat, we can remove two of those possible situations (the ones where he reveals the car), leaving four. In two of those four, we are currently on the car. In the other two, we are currently on a goat. Therefore, our odds of success are 50% whether we stay or switch.

But the second assumption is especially pernicious: the problem relies on the assumption that Monty Hall must always open a door and offer a switch. Think back to my card game above; if I were Monty Hall and I were looking to save some money on prizes, I could only open a door and offer a switch when the contestant was already on the car. If the contestant declines the switch, that's no big deal, they were going to win anyway. But if they accept the switch, I just saved the show's prize budget the cost of a car. In this "Evil Monty Hall" hypothetical, your odds of winning a car if you stay are 100%, while your odds if you switch are 0%.

What Am I Getting At?

If someone gives you the odds of Event X, those odds are not a measure of Event X. The odds my friends assigned to my card were not a reflection of my card, they were a reflection of all the other cards my friends had seen. The estimated odds of winning a car when you take Monty Hall's switch say far less about which door has a car and far more about what you think of Monty's motivations. 

Ultimately, odds are a measure of ignorance.