Jeopardy! Math

Alex Trebek, the most pompous of all game-show hosts.


As a Jeopardy! fan, I enjoyed this article from 10+ years ago that presents some interesting scenarios about how one should bet in Final Jeopardy. I wonder if actual players on the show ever think that deeply and mathematically about their final bets. Not that the ideas in the article are super mathematically complicated, but still very interesting. Read on.

(And, the author of the article is named Matt Gaffney!)


Math for Jeopardy! Players
How contestants regularly blow their final bet.
By Matt Gaffney
Posted Thursday, July 20, 2000, at 3:00 AM ET

Why do so many Jeopardy! contestants blow it in the final round? Picture this scene, from the March 21, 2000 show: Going into "Final Jeopardy!" Andrew was in the lead with $8,000, Haley was in second with $5,700, and Dave was back in third with $2,700.

If you're Andrew in this situation, deciding on your bet is simple, assuming for argument's sake that the Final Jeopardy! category is neutral, i.e., one you know neither particularly well nor particularly poorly. Andrew's rational path is to wager the minimum he needs to put himself out of Haley's reach—that is, enough to give him twice her current score, plus $1. That's $3,401 in this case, which is precisely what Andrew wound up betting.

For Haley, betting is more complicated. Before I tell you how she should have bet, consider how she did bet. Like most contestants, she took a deep final-scene-of-Thelma-&-Louise breath, bet $5,600, got the final question wrong, and lost. Andrew got it right, won $11,401, and went back the next day. Dave, if anyone cares, bet the house, got Final Jeopardy! wrong, and wound up with nothing.

Here's what Haley should have bet: $299. Notice that the way she actually bet, the only way she could have won is if she'd gotten Final Jeopardy! right and Andrew had gotten it wrong. Obviously, if Andrew answers correctly, the game's over, no matter what Haley does.
By betting $299, Haley gives herself an extra chance. If Andrew gets it right, he still wins, as before. And, as before, if Haley gets it right and Andrew misses it, Haley wins. Here's the difference: If Haley bets $299 and they both miss Final Jeopardy! Haley wins. Her final total would be $5,401, while Andrew would be down at $4,599.

Why can't Haley bet more than $299? Because she has to guard against Dave, whose maximum score, if he bets everything and gets Final Jeopardy! right, would be $5,400. Note that, with correct wagering, Dave is a non-factor in this Final Jeopardy! equation. Even if he bet it all and got it right, he still wouldn't be able to overcome Haley, even if she answered incorrectly.
All this wouldn't have helped Haley in this case, since Andrew answered Final Jeopardy! correctly. But had he missed it, she would've won.

For the player in second place, this all boils down to betting an amount that still gives you the win if both you and the player in the lead miss Final Jeopardy! A wagering-savvy former Jeopardy! champ has labeled this "The Two-Thirds Rule," because the second-place player needs at least two-thirds of the leading player's score going into Final Jeopardy! to be able to pull this off. (Click here for more on the two-thirds rule.)



If the third player is close enough to worry about, as in the example above, you need to guard as much as possible against him. The following scenario from a recent show is a perfect illustration of this principle. Going into Final Jeopardy!, Melizza was in the lead with $7,500. Second was Miles with $7,300, and third was Judy with $5,800.

Again, the leader's bet is easy to calculate, and Melizza did in fact wager the correct amount: $7,101 (again, that gives Melizza twice Miles' score plus $1 if she gets it right). Miles should bet $4,301, while Judy should bet $2,800.

Why? To answer that, we'll only deal with scenarios in which Melizza gets it wrong — because if she bets correctly and gets the answer right, the game's over no matter what.
Miles' bet of $4,301 puts him out of Judy's striking range should he answer correctly, as he'd finish with $11,601, or two times Judy's score plus a dollar.

Judy's correct bet, $2,800, gives her that extra chance that Haley should have had in the first example. By betting $2,800, she can win if all three players miss the last question. With correct bets and all players missing Final Jeopardy!, Melizza winds up with $399, Miles finishes with $2,999, and we'll see Judy again tomorrow as our returning champion, as she's just finished with $3,000. (Naturally, if Judy gets it right and the others don't, she wins anyway.)

Sound too theoretical? Consider what actually happened: All three players missed the final question. Melizza wound up with $399—which was good enough to win, as her opponents had wagered absurdly. Miles bet the insane sum of $7,000, which left him with $300. Judy bet the house and wound up with nothing. Had Miles bet correctly (or even close to correctly—six grand would still have won it for him), he would have been champ. Or, barring that, Judy could have won it herself.

You may be wondering: If the first person knows that the second-place person is going to have this back door, why doesn't he simply bet just enough to outstrip the second place player's more modest bet? Because there's always the possibility that the second-place person will bet it all. And in practice, the person in first place almost always bets $1 or $100 over what he needs to put himself out of reach. It's this tendency that the second- and third-place players should take advantage of.

There are also a few subtleties we're glossing over — for instance, Haley, Miles, or Judy would get a bigger payday if they bet everything and won. But the safer bet provides the extra chance of simply surviving and coming back the next day. And anyway, the basic premise is clear: If you're in second or third place going into Final Jeopardy!, don't just automatically bet it all. Your better chance may be backing into victory.

X is chillin'

Overheard during a free period I have...

I am sitting at my desk doing work in my room, next to another teacher who teaches in my room during my free period. He is giving a math test to his class. A student comes up to ask him a question. They are talking...

Teacher: "Well, before you do anything else, you have to get x by itself."
Student: "Ok..."

[uncomfortable pause]

Teacher: "You're not sure how to get x by itself?"
Student: "No."
Teacher: "Ok, well, what is x doing right now?"
[pause]
Student: "Chillin."
Teacher (laughing) "Ok, well, it's chillin' now with what kind of sign attached to it?"

... and the conversation continued.

However, in ten years of teaching math I'd never thought of a variable as chillin'. I guess I could see the student's point. X, indeed, at the moment wasn't doing anything - just chillin'.

The Torch to Bear (another math question)

Here's another fun math question, and I believe Side Bar may have posed this question some time ago in another context at a Margaret Street BBQ but anyway...

Four people have to get through a dark tunnel with a torch. A maximum of two people can go through at a time, with the torch always being needed to see where you are going (so the carrier of the torch must go each way). The four people have different rates of getting through the tunnel - 10 minutes, 5 minutes, 2 minutes, and 1 minute. What is the minimum number of minutes needed to get all 4 people through the tunnel?

Math Fact of the Day

A fellow teacher told me this today and it struck me as being interesting. Therefore, I wanted to share it.

Suppose you leave your house and walk a path to some other destination at 8 A.M. on Monday. You can walk at any speed you want, you can even stop your walk and stand still for a while, and you can even backtrack your path and then walk forward again. But, at some point on Monday, you reach your destination and sleep over.

Then, on Tuesday morning, you leave your destination at 8 A.M. and return home again. You walk with your speed being totally random and independent of your speed from Monday. Infact, for the sake of argument, you do something totally different (like, you sprint all the way home on Tuesday when on Monday you crawled, stood still, went back and forth, etc...)

So, here's the fact: There will be a location between your home and the destination that you passed both going to the destination and returning home at the exact same instant on both days.

Weird...

Math Questions VI - A Quick Quiz

Well, not really math, but logic.

Question 1.
The answer to question 2 is:
A. B
B. C
C. A


Question 2.
The first question with correct answer B is:
A. Question 3
B. Question 1
C. Question 2


Question 3.
The only answer you have not chosen yet is:
A. A
B. B
C. C

Birthday Paradox (Math Questions V)

To wrap up this string of math questions... the answers to Part II and III Jerry answered very well in the comments thread. The Monty Hall Paradox (Math Questions IV) is easily google-able... but the gist of it is that if you have a one in three chance of winning a prize (or, identically, a three-card monte winning card), and the host of the game shows you a non-winner of the remaining two options and offers to switch, you have a 2/3 chance of winning if you switch, and only a 1/3 chance of winning if you stay. That is because you will only choose correctly originally 1 out of 3 times, and if you have chosen incorrectly you should switch and will therefore win. This should happen two out of three times. Essentially the host is giving you the choice of "do you want to stay with the option you've chosen? Or, do you want BOTH of the other two?" to which one would obviously say, "Well, I'd rather have the other two." Very counterintuitive and a cool problem

The birthday paradox that Jerry referred to is not a question as much as an interesting fact: In a randomly chosen group of 23 people, the probability that two people have the same birthday is about 50%. So, if you are a teacher and have a class of 23 people and ask them all their birthdays, half the time two students will have the same birthday. More mathematical explanation that you probably care about is on the Wikipedia page, which is linked to above.

The Monty Hall Paradox (Math Questions IV)

Since ChuckJerry provided a lucid, intelligent, well-thought answer to the last two math questions, I thought I'd provide another one. You can easily search for the answer on the web. Please don't. And if you've heard it already, let people guess first. Use your intuition and ponder the following scenario. This is one of my favorites.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others are goats. You pick a door, say No. 1, and the host, who knows what's behind the other doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to keep your choice (Door No. 1)? Or do you want to switch to Door No. 2?"

Should you keep Door #1? Should you switch to Door #2? Or does it not matter which option you take?

(Chuck, birthday paradox to follow.)

Math Questions II and III

First, the answer to the previous Math Question, as Pax and Evan answered, is 50%. Here's why:

The probability of the team of at least one run is 100% minus the probability of the team scoring no runs. There is a number of ways in which the team can score no runs. Let O be a strikeout, and H be a hit.

- The team can get three strikeouts (OOO). Probability = (1/2)(1/2)(1/2) = (1/8)
- The team can get three strikeouts and one hit in three ways:
HOOO, OHOO, OOHO. Probability of each: (1/2)(1/2)(1/2)(1/2) = 1/16. Probability of one hit and three strikeouts: 3 * (1/16) = (3/16).
- The team can get three strikeouts and two hits in six ways:
HHOOO, HOHOO, HOOHO, OHHOO, OHOHO, OOHHO. Probability of each: (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32. Probability of one hit and three strikeouts = 6 * 1/32 = 6/32

Probability of no hits: 1 - ( (1/8) + (3/16) + (6/32) ) = 1/2 = 50%.


Now, here's two more math questions:

Math Question #2: What's the most amount of money that a contestant can win on a single show of Jeopardy!? Assume that you can place the Daily Doubles wherever you want.

Math Question #3: Can you guarantee a tie in Tic-Tac-Toe for X? For O? How is it done?

Math Question of the Day

Suppose a baseball team has hitters who only gets singles or strikes out. Suppose further that when a batter gets a single, if there is already a man on first base he advances to second base, while any runner on second or third base scores a run. When a batter strikes out, any players on base stay at that base. Finally, suppoe that all the batters have equal hitting ability. If the probability of each batter getting a hit is 50%, what is the probability of the team scoring at least one run before having 3 outs?

7.11

I was waiting to get on a poker table at Bally's a couple of weeks back. I was donning my beautiful Rutgers sweatshirt. Here's a picture of it



R-U, rah rah, R-U, rah rah, HOO-rah, HOO-rah, Rutgers RAH!

So some guy also waiting for a table spots my sweatshirt and comes up to start up a conversation. Turns out, he taught math at Rutgers many years ago. We share our backgrounds, and he says, "Well, here's something to think about while you're playing..." He added a story that made the question more of a "riddle" than just a math question, but here is the core of it:

In a "seven-eleven" (7-11) store, a customer selected four items to buy. The check-out clerk says that he multiplied the costs of the items and obtained exactly 7.11, the very name of the store! The customer calmly tells the clerk that the costs of the items should be added, not multiplied. The clerk then informs the customer that the correct total is also $7.11. What are the exact costs of the 4 items?

I went to my 1-3 No-Limit table while he went to his 2-4 Limit table. While I was playing, during hands I wasn't in, the math geek in me got control of me and I started to try to work it out. Distracted from both the game and the problem, I put it aside intending to try more later. I got home and searched for the problem, intending to blog about it, and accidentally spotting the solution. Annoyed because I'll never know if I would've been able to figure it out, I post it here as a challenge.

You can find the solution if you google...