Sunday, May 29, 2016
Saturday, May 28, 2016
Season 6 Episode 5
A very exciting episode, but nothing scored. Irene, therefore, holds a slight edge on the rest of the field.
Below we see the expected 95 percent confidence interval for each player’s score by season’s end— assuming that the corresponding player offered the correct odds on each event.
I will try to update quickly this week, regardless of the score. Look forward to seeing folks tomorrow!
Sunday, May 22, 2016
Season 6 Episode 4
So far, not much has happened. I mean, Jon Snow lives. That is not nothing. As we can see, Irene was least surprised by this outcome (having insisted on a 100 percent chance of Jon Snow’s revival.
Below we see the expected 95 percent confidence interval for each player’s score by season’s end— assuming that the corresponding player offered the correct odds on each event. Note that Mark seems to have the greater expectations for himself than anyone else’s own expectations. I am pretty sure this means only that Mark’s predictions are most out of line relative to the rest of the field.
Scoring... Conclusion
We have come up with a two-way bet in which both parties have incentive to reveal exactly their beliefs regarding the true probability of an event. Neither expects to lose, at worst breaking even— and then only when they agree regarding the true probability. It turns out that all these properties still hold when multiple players are involved. We simply have each player bet against every player. Let $p_i$ be the probability declared by player $i$ of $n$. Then let
$$
U=\frac{1}{n}\sum_{i=1}{\lg{p_i}}
$$
and
$$
V=\frac{1}{n}\sum_{i=1}{\lg{\!\left(1-p_i\right)}}
$$
We can set the payout to player $j$ to $\lg{p_j}-U$ if the event occurs, and $\lg{\!\left(1-p_j\right)}-V$ if it does not. Note that the total of all $n$ payouts is always zero.
And this is how the Game of Game of Thrones is scored. The score for the four-way Arya “Dead Pool” at the end works basically the same way, but with four possible outcomes rather than just two. The average score will always come to zero, but every player expects that they will do better than everyone else on any unresolved questions. But most importantly, the winner will be the player (ostensibly) least surprised by the events on the show.
A final detail on scoring. Since we in the GoGoT are not actually paying each other based on scores, the real objective is strictly not to maximize score, but merely to have the best score in the group. This might have made it worthwhile for someone to lie a little about what they thought the probabilities were. If you figured that out, then kudos to you.
And this is how the Game of Game of Thrones is scored. The score for the four-way Arya “Dead Pool” at the end works basically the same way, but with four possible outcomes rather than just two. The average score will always come to zero, but every player expects that they will do better than everyone else on any unresolved questions. But most importantly, the winner will be the player (ostensibly) least surprised by the events on the show.
A final detail on scoring. Since we in the GoGoT are not actually paying each other based on scores, the real objective is strictly not to maximize score, but merely to have the best score in the group. This might have made it worthwhile for someone to lie a little about what they thought the probabilities were. If you figured that out, then kudos to you.
Scoring, Part IV
In the previous post, we found that if we disagree to the odds of an event happening then it may be difficult to offer an appropriate bet.
Can we come to a suitable arrangement? Certainly. If I believe I am offering a fair bet with $H\approx 0.080793\ldots$, I should be willing to take the opposite side of my bet (at $p=1\%$) as payment. That is, I charge nothing up front. If the event comes to pass, I pay $\lg{0.49}-\lg{0.01}$ and otherwise pay $\lg{0.51}-\lg{0.99}$. On average I expect I will pay $$ 0.01\lg{\frac{0.49}{0.01}}+0.99\lg{\frac{0.51}{0.99}}\approx -0.891214\ldots $$ (That is, that I will win about 89 cents.) Yet you expect I will pay you $$ 0.49\lg{\frac{0.49}{0.01}}+0.51\lg{\frac{0.51}{0.99}}\approx 2.263172\ldots $$ So we both think that we will win on average.
In fact, this works in general. Let us suppose I get to pick a $q_1$ so that I will pay you $-\lg{q_1}$ if the event happens and $-\lg{\!\left(1-q_1\right)}$ if it does not. Likewise, you will get to pick a $q_2$ so that you will have to pay me $-\lg{q_2}$ if the event happens and $-\lg{\!\left(1-q_2\right)}$ if not. Now, if I believe the actual probability is $p$, then I expect to win $$ w_1\!\left(q_1,q_2\right)=p\lg{\frac{q_1}{q_2}}+\left(1-p\right)\lg{\frac{1-q_1}{1-q_2}} $$ which is maximized when $$ 0=\frac{\partial w_1\!\left(q_1,q_2\right)}{\partial q_1}=\frac{p}{q_1}-\frac{1-p}{1-q_1} $$ regardless of your choice $q_2$. Thus, I always maximize my expected winnings by choosing $q_1=p$. When I do, my expected winnings are $$ w_1\!\left(p,q_2\right)=p\lg{\frac{p}{q_2}}+\left(1-p\right)\lg{\frac{1-p}{1-q_2}} $$ What is the least, then, I can win by picking $q_1=p$? My winnings are minimized when $$ 0=\frac{\partial w_1\!\left(p,q_2\right)}{\partial q_2}=-\frac{p}{q_2}+\frac{1-p}{1-q_2} $$ or when you choose $q_2=p$. This results in expected winnings $w_1\!\left(p,p\right)=0$. In other words, so long as I choose $q_1=p$ I maximize my expected winnings and I expect you to pay me no matter what $q_2$ you choose!
Of course, the structure of the bet is completely symmetric, so provided you choose $q_2$ to equal what you believe to be the correct probability, then you maximize your expected winnings and expect me to pay you no matter what $q_1$ I choose.
In the end, the winner will be the one of us least surprised by the outcome! Mathe-magical!
Can we come to a suitable arrangement? Certainly. If I believe I am offering a fair bet with $H\approx 0.080793\ldots$, I should be willing to take the opposite side of my bet (at $p=1\%$) as payment. That is, I charge nothing up front. If the event comes to pass, I pay $\lg{0.49}-\lg{0.01}$ and otherwise pay $\lg{0.51}-\lg{0.99}$. On average I expect I will pay $$ 0.01\lg{\frac{0.49}{0.01}}+0.99\lg{\frac{0.51}{0.99}}\approx -0.891214\ldots $$ (That is, that I will win about 89 cents.) Yet you expect I will pay you $$ 0.49\lg{\frac{0.49}{0.01}}+0.51\lg{\frac{0.51}{0.99}}\approx 2.263172\ldots $$ So we both think that we will win on average.
In fact, this works in general. Let us suppose I get to pick a $q_1$ so that I will pay you $-\lg{q_1}$ if the event happens and $-\lg{\!\left(1-q_1\right)}$ if it does not. Likewise, you will get to pick a $q_2$ so that you will have to pay me $-\lg{q_2}$ if the event happens and $-\lg{\!\left(1-q_2\right)}$ if not. Now, if I believe the actual probability is $p$, then I expect to win $$ w_1\!\left(q_1,q_2\right)=p\lg{\frac{q_1}{q_2}}+\left(1-p\right)\lg{\frac{1-q_1}{1-q_2}} $$ which is maximized when $$ 0=\frac{\partial w_1\!\left(q_1,q_2\right)}{\partial q_1}=\frac{p}{q_1}-\frac{1-p}{1-q_1} $$ regardless of your choice $q_2$. Thus, I always maximize my expected winnings by choosing $q_1=p$. When I do, my expected winnings are $$ w_1\!\left(p,q_2\right)=p\lg{\frac{p}{q_2}}+\left(1-p\right)\lg{\frac{1-p}{1-q_2}} $$ What is the least, then, I can win by picking $q_1=p$? My winnings are minimized when $$ 0=\frac{\partial w_1\!\left(p,q_2\right)}{\partial q_2}=-\frac{p}{q_2}+\frac{1-p}{1-q_2} $$ or when you choose $q_2=p$. This results in expected winnings $w_1\!\left(p,p\right)=0$. In other words, so long as I choose $q_1=p$ I maximize my expected winnings and I expect you to pay me no matter what $q_2$ you choose!
Of course, the structure of the bet is completely symmetric, so provided you choose $q_2$ to equal what you believe to be the correct probability, then you maximize your expected winnings and expect me to pay you no matter what $q_1$ I choose.
In the end, the winner will be the one of us least surprised by the outcome! Mathe-magical!
Scoring, Part III
In our previous part, we noted that a fair bet concerning an event with probability $p$ could be made with me paying you $\$S\!\left(p\right)$ up front and getting back $-\$\lg{p}$ if the event came to pass and $-\$\lg{\!\left(1-p\right)}$ if the event failed to come to pass.
We may even loosen up the game somewhat. Suppose I offer $\$H$ as a bet, but you get to pick the $q$ which determines the payouts $-\$\lg{q}$ and $-\$\lg{\!\left(1-q\right)}$. What $q$ should you pick, knowing that the probability of the event is $p$? You may maximize your net payment by setting $$ 0=\frac{\partial}{\partial q}\left[H-S\!\left(q\right)\right]=\frac{\partial}{\partial q}\left[H+p\lg{q}+\left(1-p\right)\lg{\!\left(1-q\right)}\right]=\frac{p}{q}-\frac{1-p}{1-q} $$ or, conveniently enough, $q=p$. By choosing this way, you expect that I will pay you an amount $H-S\!\left(p\right)$. Obviously, it is hardly worth it to me to set $H>S\!\left(p\right)$. But if I set $H=S\!\left(p\right)$ then I expect to profit any time you choose $q\neq p$. However, this all assumes we know— or at least agree upon— $p$. Yet suppose I believe that the event is unlikely, say, $p=1\%$ and so I offer what I think is a fair bet with $H\approx 0.080793\ldots$ But if you are less certain, thinking $p=49\%$, then you expect to lose almost 92 cents on a bet which I believe fair. You may be enticed by an offer of at least $H\approx 0.999711\ldots$ but I will not make such an offer, because I expect that I will wind up paying you the same 92 cents.
In the next section, we make a deal.
We may even loosen up the game somewhat. Suppose I offer $\$H$ as a bet, but you get to pick the $q$ which determines the payouts $-\$\lg{q}$ and $-\$\lg{\!\left(1-q\right)}$. What $q$ should you pick, knowing that the probability of the event is $p$? You may maximize your net payment by setting $$ 0=\frac{\partial}{\partial q}\left[H-S\!\left(q\right)\right]=\frac{\partial}{\partial q}\left[H+p\lg{q}+\left(1-p\right)\lg{\!\left(1-q\right)}\right]=\frac{p}{q}-\frac{1-p}{1-q} $$ or, conveniently enough, $q=p$. By choosing this way, you expect that I will pay you an amount $H-S\!\left(p\right)$. Obviously, it is hardly worth it to me to set $H>S\!\left(p\right)$. But if I set $H=S\!\left(p\right)$ then I expect to profit any time you choose $q\neq p$. However, this all assumes we know— or at least agree upon— $p$. Yet suppose I believe that the event is unlikely, say, $p=1\%$ and so I offer what I think is a fair bet with $H\approx 0.080793\ldots$ But if you are less certain, thinking $p=49\%$, then you expect to lose almost 92 cents on a bet which I believe fair. You may be enticed by an offer of at least $H\approx 0.999711\ldots$ but I will not make such an offer, because I expect that I will wind up paying you the same 92 cents.
In the next section, we make a deal.
Saturday, May 21, 2016
Scoring, continued
In the previous post, we saw measure the average amount of surprise resulting from realizing one of $2^S$ equally-likely outcomes is $S$.
Each outcome therefore having probability $p=2^{-S}$, we may rewrite $$ S=-\lg{p} $$ where $\lg{2^x}=x$. Again, this is the average surprise. Sensibly, we may write that the surprise of any single outcome with probability $p_i$ is $$ s_i=-\lg{p_i} $$ This makes intuitive sense. If something happens which is absolutely certain ($p=1$) then we are not at all surprised ($s=-\lg{1}=0$). Likewise, if something is absolutely impossible, then we are infinitely surprised ($s=-\lg{0}=\infty$) if that impossible something comes to pass.
Now, let us use this to construct a bet based on how surprised we are. There is some event which may or may not happen (say, your favorite hockey team winning its next game.) Suppose that we agree that the probability of the event actually happening is $p$. Then with probability $p$ we will be surprised by an amount $-\lg{p}$ and with probability $1-p$ we will be surprised by an amount $-\lg{\!\left(1-p\right)}$. We expect, then, to be surprised by an amount $$ S=-p\lg{p}-\left(1-p\right)\lg{\!\left(1-p\right)} $$ Thus, we could construct a fair bet based on surprise. If you pay me $\$1$ per unit of surprise you actually experience, then to be fair I could pay you $\$S$ up front. If we played such a game over and over again with events of probability $p$ (my paying you $\$S$ each time, and you paying me $\$1$ per unit of surprise you experience) then on balance we should both break even.
In the next part, we will consider how to bet we do not actually know $p$.
Each outcome therefore having probability $p=2^{-S}$, we may rewrite $$ S=-\lg{p} $$ where $\lg{2^x}=x$. Again, this is the average surprise. Sensibly, we may write that the surprise of any single outcome with probability $p_i$ is $$ s_i=-\lg{p_i} $$ This makes intuitive sense. If something happens which is absolutely certain ($p=1$) then we are not at all surprised ($s=-\lg{1}=0$). Likewise, if something is absolutely impossible, then we are infinitely surprised ($s=-\lg{0}=\infty$) if that impossible something comes to pass.
Now, let us use this to construct a bet based on how surprised we are. There is some event which may or may not happen (say, your favorite hockey team winning its next game.) Suppose that we agree that the probability of the event actually happening is $p$. Then with probability $p$ we will be surprised by an amount $-\lg{p}$ and with probability $1-p$ we will be surprised by an amount $-\lg{\!\left(1-p\right)}$. We expect, then, to be surprised by an amount $$ S=-p\lg{p}-\left(1-p\right)\lg{\!\left(1-p\right)} $$ Thus, we could construct a fair bet based on surprise. If you pay me $\$1$ per unit of surprise you actually experience, then to be fair I could pay you $\$S$ up front. If we played such a game over and over again with events of probability $p$ (my paying you $\$S$ each time, and you paying me $\$1$ per unit of surprise you experience) then on balance we should both break even.
In the next part, we will consider how to bet we do not actually know $p$.
Introduction to Scoring
Let us agree that we have a fair coin and a fair way to flip that coin. Neither of us, then, should be very surprised should a flip come up heads. Nor should either of us be very surprised should a flip come up tails.
Suppose I am surprised by an amount $s_H$ if the outcome is heads, and an amount $s_T$ if the outcome is tails. Then on average I will be surprised by an amount $$ S = \frac{1}{2}\times s_H+\frac{1}{2}\times s_T $$ Very reasonably, we should each be equally surprised regardless of the outcome, so $S=s_H=s_T$. Surprise is a weird thing to quantify, so let us define the surprise caused by a fair coin toss to be 1.
The surprise of two coin tosses, then should be 2, right? There are four equally likely outcomes: heads followed by heads, heads followed by tails, tails-tails, and tails-heads, so $$ 2=S=\frac{1}{4}\times s_{HH}+\frac{1}{4}\times s_{HT}+\frac{1}{4}\times s_{TT}+\frac{1}{4}\times s_{TH} $$ and therefore equal surprise from each: $$ S=s_{HH}=s_{HT}=s_{TT}=s_{TH} $$ With three tosses, $$ 3=S=8\times\frac{1}{8}\times s_{ttt} $$ and so on. The important takeaway here is that $S$ units of surprise result from realizing one of $2^S$ equally likely outcomes.
Suppose I am surprised by an amount $s_H$ if the outcome is heads, and an amount $s_T$ if the outcome is tails. Then on average I will be surprised by an amount $$ S = \frac{1}{2}\times s_H+\frac{1}{2}\times s_T $$ Very reasonably, we should each be equally surprised regardless of the outcome, so $S=s_H=s_T$. Surprise is a weird thing to quantify, so let us define the surprise caused by a fair coin toss to be 1.
The surprise of two coin tosses, then should be 2, right? There are four equally likely outcomes: heads followed by heads, heads followed by tails, tails-tails, and tails-heads, so $$ 2=S=\frac{1}{4}\times s_{HH}+\frac{1}{4}\times s_{HT}+\frac{1}{4}\times s_{TT}+\frac{1}{4}\times s_{TH} $$ and therefore equal surprise from each: $$ S=s_{HH}=s_{HT}=s_{TT}=s_{TH} $$ With three tosses, $$ 3=S=8\times\frac{1}{8}\times s_{ttt} $$ and so on. The important takeaway here is that $S$ units of surprise result from realizing one of $2^S$ equally likely outcomes.
Friday, May 20, 2016
Welcome
This is the Game of Game of Thrones blog. Here, I will be reporting weekly results from the game as well as offering some background on how the scoring system works.
That is, I am going to muck up a masterful story with math.
I will try not to spoil anything without a sign. If I title a post with an episode number, for example, I expect the reader to be caught up through that episode. This will surely include all scoring updates. I will include updates every week so that even if nothing game-worthy happens, nobody will be spoiled by a lack of an update.
In the next post, I will begin to explain the idea behind the scoring.
That is, I am going to muck up a masterful story with math.
I will try not to spoil anything without a sign. If I title a post with an episode number, for example, I expect the reader to be caught up through that episode. This will surely include all scoring updates. I will include updates every week so that even if nothing game-worthy happens, nobody will be spoiled by a lack of an update.
In the next post, I will begin to explain the idea behind the scoring.
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