Saturday, February 6, 2010

Who are three people who have never been in my kitchen?

I've run into two separate discussions based on this article about Jeopardy! and baseball questions (er, answers). The relevant quote:
Let's say you're on 'Jeopardy!' and you're absolutely routing your two opponents. You have $40,000 going into the final round, while one of your opponents has, let's say, $15,000. You're guaranteed to move onto the next day, but the final category comes up and it has something to do with baseball, which is your favorite sport. How much — if anything — do you risk?
The discussions of this article go along two paths. One is to compare how many of the ten Final Jeopardy! answers everyone got right (most participants have gotten at least seven of the ten - both sites have a fair number of people who are very knowledgeable about baseball). The other is to loudly decry the article's premise that, since you should be fairly confident that you will respond correctly (the article, on the Yahoo! Sports site, assumes you're a baseball fan), you should bet your entire $40,000 haul, rather than the $10,000 that will guarantee a win. Since the Preschool is all about the math these days, let's look at this in a little more detail. Specifically, let's look at the following situation: you have $40,000, the second-place competitor has $15,000, and the Final Jeopardy! category is something squarely in your wheelhouse. How much do you bet?

This (like the Scat decisions I'm supposedly writing about) is a problem of expected value. Suppose that you estimate the probability you will respond correctly as p; for example, if p is 0.8, it means you think you're 80% likely to be correct. If you respond correctly (as you will do 100p% of the time), you gain whatever you've bet; call this number b. If you respond incorrectly (which will happen 100(1-p)% of the time), you will lose b dollars. Your expectation for this decision is then 40,000 + bp - b(1-p), which simplifies to 40,000 - b + 2bp. If your only goal is to maximize your expectation for this decision, then the answer is simple: if p is at least 0.5, you should bet everything; if it's less than 0.5, you should bet nothing.

However, this is not your only goal, because if you win the game, you get to come back tomorrow, and try to win more money. Thus, we must also factor in your expectation of future winnings. (Incidentally, the primary argument I'm seeing against betting everything is that "you're risking a lot in future winnings.") Let's assume, for the sake of argument, that the second-place competitor will bet his entire $15,000, and will respond correctly. Thus, if you bet $10,000 (or less), you will win (or tie), and come back tomorrow, regardless of your response. If you bet more than $10,000, and respond incorrectly, you will lose.

At this time, I should point out that we could make this a more complicated model, estimating the probabilities that the other competitors will respond correctly. Also, you could bet something other than $10,000 or $40,000. Even if you factor in the chance of the other competitors missing, betting something like $20,000 seems a doubly bad idea; you leave $20,000 "on the table" if you are right, and risk losing if you are wrong. You could bet slightly less than $40,000, thus improving the probability that you will win if everyone misses the final answer. In fact, I think $39,998 might be the best bet if you're "going for broke." That said, complicating the model takes time, and might be better suited to a paper than a blog post—any undergrads looking for a thesis, this might be a starting point! Once we decide to stick with a simple model, betting anything other than $40,000 or $10,000 makes the math work out less nicely, and a couple of dollars here and there isn't going to make much difference, nor will the tiny probability that you'll win with $2. Let's assume, then, that this is a binary decision: bet $10,000 and always win, or bet $40,000 and lose if you're wrong.

Assume that, if you win, your expected earnings from future shows is x (we could take some time and figure out what x might be, but let's wait to decide whether or not to deal with that). Now, we can look at two different expectations: one for playing conservatively, and one for betting everything. If you play conservatively, and bet $10,000, you will always win, thus adding x to your expectation. Using the formula above, your expectation is now 30,000 + 20,000p + x. If you bet your entire $40,000, you will lose everything, including future winnings, if you respond incorrectly. However, when you respond correctly, you double your money, and you still get to come back tomorrow. Thus, you add px to your expectation, which is now 80,000p + px.

An aside: I think this is what a lot of the folks making the "future winnings" argument miss—you aren't giving up your future winnings, just a portion of them. If you can win more today than you're likely to get on future days, there's a strong incentive to go for it.

Anyway, let's see what that incentive is. The difference between playing recklessly and playing safe is 60,000p - (30,000 + (1-p)x). Economists refer to this as an opportunity cost - by betting only $10,000, you are forfeiting whatever you could have earned by betting everything. If this cost is positive, then, economically, you are making a mistake by only betting $10,000; if it is negative, then playing it safe is the right call. Your decision is based on whether 60,000p is larger than (30,000 + (1-p)x).

Let's look at an example: Suppose you think you're 80% likely to get the Final Jeopardy! question correct. You are then comparing 48,000 to 30,000 + .2x. If x is less than 90,000, then 48,000 will be larger. In other words, you should bet everything unless you think you'll make at least $90,000 from future shows; given that $50,000 is a pretty hefty one-day total, this would require a fair amount of confidence on your part!

Does this mean Jeopardy! players should be wagering more recklessly? Well, perhaps. The example above is pretty solidly in favor of betting it all, but the big issue is estimating that probabilty. I managed to get 8 of the ten questions posed in the linked article (nine, actually, but I had to think about one for more than 30 seconds, so I'd have been hosed in Final Jeopardy), but that doesn't mean I'd be 80% confident I could answer a question based only the fact that it would be about baseball. Also, the necessary value for x in order to play conservatively goes down very quickly; if p = 0.7, then you're comparing $42,000 to 30,000 + .3x, and x only needs to be $40,000 to break even; considering that you've got $40,000 right now, this is getting to the realm of possibility. If we knew x (and we can probably come up with a statistical value, given the probability that the previous day's champion wins again, and the mean winning total), we could come up with a "break-even" value for p. I might return to this, but for now, I think the risk-averse behavior is better, unless you are very very sure of yourself.

Wednesday, February 3, 2010

Another Post Just To Prove I Still Post

The latest post on Scat strategy has been half-finished for over a week, as I have been working on presentations of old and new research. Also, I'm still trying to find a new trivia night; I may give up soon and just start posting my own stuff. For now, here's a fun definition:

Let Ak be the set {1, 2, ..., k}. A graph G has an optimal t-tone coloring if each vertex can be assigned a t-element subset of Ak in such a way that if vertices u and v are distance d apart (in other words, the shortest path in G connecting these vertices contains d edges), then their subsets have at most d-1 elements in common.

For example, vertices adjacent to each other can have no shared elements in their subsets, while if the distance between two vertices is greater than t, they may be assigned identical subsets.