Jeopardy betting question..

Yesterday (4/13) I was watching Jeopardy when it came time for the final question. The leader had around $15k, second place had around $10k and third place had around $9.5k.

In my mind, optimal betting strategy for the third place contestant would be to bet $501.

The leader needs to bet $5k in order to guarantee that if he gets it right he would be back the next day. Second place should bet either $10k or $0, depending on how they feel about the topic. Betting 0 is a bet that the leader will get it wrong. Betting $10k is betting you will get it right and the leader will either bet only enough to tie or will get it wrong.

Given all that, third place can finish first only if both of the other players get the answer wrong and he gets it right. There is almost certainly no chance for him to finish in first if either of the other to get it right. So he is essentially play for second place. By betting $500, he will tie the leader and second place if they both get it wrong and will guarantee a second place if the present second place player or the leader get it wrong.

He bet all of it. I can’t figure out how that can possibly be considered a smart bet.

Well, what’s the point of living if you can’t feel alive?

I didn’t think you got anything for second place do you?
If you don’t it seems the only way you’ll get to keep anything is if you win.
In order to win, as you pointed out, you must get the question correct and both others must miss.
If I’m correct on those 2 points (and I conceed that is not a given) it seems logical to maximize your return.

If that’s true, why not bet all your money and maximize your winnings? The amount on the screen of the first place player is the actual amount they receive - those are actual dollars, not points. If he’s confident on the category, and thinks there’s no way to win unless both the others get it wrong, why hold back on the bet?

You are aware that 2nd & 3rd, while they don’t leave empty handed, do NOT receive the final amount they ended the game with, right? Whether you end with $0 or $15k, you get the same 3rd place prize.

ETA: zoid, you are correct. First place wins however much money they end the game with, as well as returning the next night. 2nd & 3rd win set consolation prizes that are not based on how much they had at the end. It’s completely different than “Wheel of Fortune”, where everyone leaves with what they ended the game with, and winning only gets you a chance at the final puzzle.

player 1 has two likely options. betting 5k+ to ensure the win, or 0 for loss aversion.

player 2 has 3 options. betting 10k for the win, $1 for the win if p1 loses, or 0 for loss aversion (tie p1 if he loses).

you’re right in assuming that player 3 can only win if he bets at least $501 (assuming p2 and p3 loses). However that wouldn’t be the optimal bet. betting $501 is already operating on the assumption that he will get it right. that means gaining $9.5k or gaining $501 both have the same probability of happening. However, if he’s loss averse, he would bet 0 dollars and be happy with what he has.

risk aversion, loss aversion, expected payoffs… are all topics of auction theory, which is a branch of game theory, which is big in behavioral finance, all of which fall under that dismal science - economics.

Nowadays, second place takes home $2000, third place takes home $1000.

It SHOULD be third place must perforam an act on second place of first places choosing.
I’m tellin’ ya that’s the ticket.
I need to get into television, my talents are wasted here.

Everyone’s talents are wasted here.

Not mine. I keep them hidden under this bushel.

So what happened?

I’m not sure I understand this. If p2 bets $1 and p3 bets $9k, both get it right and p1 gets it wrong, p3 would win.

As mentioned elsewhere, 2nd gets $3k and 3rd gets $2k. Nothing more and nothing less. He can gain $1k by moving into 2nd but loses nothing by losing by more. Betting it all and hoping the other two get it wrong means he could win an extra $9.5k, but that also eliminates any chance at 2nd place. Betting $501 gives him the same chance at first place (although he would win less) but also gives him two chances at 2nd place (if either of the other two get it wrong).

Everyone got it right and p1 won. It was an incredibly easy question. Something like, “At their founding in 1976, the logo for this company was Isaac Newton sitting under a tree.”

The bet calculator at J-Archive spits out almost exactly the strategies in the OP, except noting that the third-place player can bet anything between $501 and $9,499 and have the same result.

That being said, “Company Logos” sounds like an incredibly easy category and I would have been tempted to risk everything regardless.

If I ever go on Jeopardy, and I get a Final Jeopardy or Daily Double in a category I’m confident in (like, say, almost anything in science), then I’m going to bet it all. The way I see it, if it’s (say) a physics question, then I’ve got a nearly 100% chance of winning it, so it’s just a question of how much I win.

I once predicted the FJ question. The catagory was “The Commonwealth” and during the commercial break, I mulled over what kinds of questions one could ask three Americans and give them, a fighting chance, and guessed it would be “A: This is the largest by area. Q: What is Canada?”

I was right, though at least one contestant wrote “What is Australia?”

Strangely, I can’t find this at the Jeopardy Archive. I just hope it wasn’t a glorious dream.

I would bet 500 exactly, thus messing with the mind of #1 who bet that extra dollar to greedily take it all from #2.

The $9,499 bet doesn’t make sense to me. If all 3 get it wrong he would be in 2nd place with a $501 bet. With a $9,499 he would end up in third. Lets say each player has an 85% chance of getting it right, just to make up numbers.

His probability of winning is (0.0225)*(0.85) = 1.9%
Probability of one right answer from the other two: 25.5%
Probability of two right answers from the other two: 72.25%

0.019*$9,000 = $171
0.255 * $1,000 = $255

In other words, betting it all would, on average, get him an extra $171. Betting $501 would, on average, get him $255.

Of course, this assumes these odds are static and don’t take into account familiarity with the topic. It also doesn’t track the fact that I’m guessing the probability is not independent. If one contestant knows the answer it is probably a higher likelihood that other contestants will know it as well.

Clearly the contestant needed to engage in a lengthy game-theory analysis. I’m sure if he asked politely to stop tape for a few hours, the studio would be glad to oblige.

If he had only known he’d face that sort of decision he could have thought about the relatively small number of situations and planned a strategy beforehand.

My original analysis took me all of 10 seconds to figure out when I was watching the show. Of course I have the luxury of sitting in my living room and not under pressure of being on stage, but this kind of strategy can be worked out long before he shows up for the game.