Final Jeopardy wagering theory

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dmleach
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Final Jeopardy wagering theory

Post by dmleach »

In a fit of optimism, I'm trying to tackle the big "Jeopardy!" skill I lack: Final Jeopardy wagering. I'm getting better at it with practice, but there's still some scenarios where the Archive's calculator gives a different answer than what I produce. Could I enlist your help in understanding some of these disagreements?

Let's start with these scores going into FJ: (A) 10600 vs (B) 8000 vs (C) 4800. I came up with wagers of (A) 5401, (B) 2800, and (C) 3201. The archive's calculator says that I'm right, except (B) should instead bet 2601.

When I come up with something other than the calculator's result, I map out all the possibilities to understand why my bets aren't optimal. In this case, the two possible (B) bets don't change the outcomes: (A) wins in four scenarios, finishes second in two, and loses in two; (B) wins three, places four, and loses one; (C) wins one, places two, and loses five. The difference is that, with my bet of 2800, (B) has the potential to walk away with a couple of hundred extra bucks while not taking on any additional risk.

Now, obviously, in this case we're not talking a huge numerical difference. My point in asking is to help learn the theory behind the numbers. Am I thinking about the problem in the right way?

Thanks in advance!

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Woof
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Re: Final Jeopardy wagering theory

Post by Woof »

dmleach wrote:In a fit of optimism, I'm trying to tackle the big "Jeopardy!" skill I lack: Final Jeopardy wagering. I'm getting better at it with practice, but there's still some scenarios where the Archive's calculator gives a different answer than what I produce. Could I enlist your help in understanding some of these disagreements?

Let's start with these scores going into FJ: (A) 10600 vs (B) 8000 vs (C) 4800. I came up with wagers of (A) 5401, (B) 2800, and (C) 3201. The archive's calculator says that I'm right, except (B) should instead bet 2601.

When I come up with something other than the calculator's result, I map out all the possibilities to understand why my bets aren't optimal. In this case, the two possible (B) bets don't change the outcomes: (A) wins in four scenarios, finishes second in two, and loses in two; (B) wins three, places four, and loses one; (C) wins one, places two, and loses five. The difference is that, with my bet of 2800, (B) has the potential to walk away with a couple of hundred extra bucks while not taking on any additional risk.

Now, obviously, in this case we're not talking a huge numerical difference. My point in asking is to help learn the theory behind the numbers. Am I thinking about the problem in the right way?
You've got the right of it. Basically, player B's constraints are to bet enough to shut out an all-in bet by C (>1600), enough to overtake a zero bet by A (>2600) but no more than 2800 to win in the case that A gets FJ wrong. The calculator gives a value at the low end of the acceptable range (2601) whereas you've selected the highest value. You say that it gives you extra bucks, but that's only true if you get it right; otherwise, you're out 200 more. Pick your poison.

dmleach
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Re: Final Jeopardy wagering theory

Post by dmleach »

Woof wrote:You've got the right of it. Basically, player B's constraints are to bet enough to shut out an all-in bet by C (>1600), enough to overtake a zero bet by A (>2600) but no more than 2800 to win in the case that A gets FJ wrong. The calculator gives a value at the low end of the acceptable range (2601) whereas you've selected the highest value.
Okay, good! Usually the calculator presents such choices with Venusian and Martian bets, so when it shows only one number I tend to read that as the One True Bet. I'm glad to see that the way I'm evaluating the possible outcomes isn't way out in left field, too.

Now for a more confusing one: (A) 14000 vs (B) 11000 vs (C) 6200. My wagers are (A) 8001, (B) 4799, and (C) 4801. The calculator calls for (A) 8001, (B) 3001, and (C) 200.

In this case, it seems as though the calculator is just wrong. If C only wagers 200, that's conceding any possibility of winning the game. At the very least C should risk 1800, since that will win the game if A and B give wrong answers. Accepting that, then really C should bet the farm (minus a few bucks) to maximize potential winnings.

If I'm right about C, then the calculator's bet of 3001 by B carries the same risk as my 4799 bet, but with less potential payoff. This may just be a repeat of the scenario from post #1 where we're only being shown the lower end of the range.

Who's right this time?

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Re: Final Jeopardy wagering theory

Post by Bob78164 »

dmleach wrote:
Woof wrote:You've got the right of it. Basically, player B's constraints are to bet enough to shut out an all-in bet by C (>1600), enough to overtake a zero bet by A (>2600) but no more than 2800 to win in the case that A gets FJ wrong. The calculator gives a value at the low end of the acceptable range (2601) whereas you've selected the highest value.
Okay, good! Usually the calculator presents such choices with Venusian and Martian bets, so when it shows only one number I tend to read that as the One True Bet. I'm glad to see that the way I'm evaluating the possible outcomes isn't way out in left field, too.

Now for a more confusing one: (A) 14000 vs (B) 11000 vs (C) 6200. My wagers are (A) 8001, (B) 4799, and (C) 4801. The calculator calls for (A) 8001, (B) 3001, and (C) 200.

In this case, it seems as though the calculator is just wrong. If C only wagers 200, that's conceding any possibility of winning the game. At the very least C should risk 1800, since that will win the game if A and B give wrong answers. Accepting that, then really C should bet the farm (minus a few bucks) to maximize potential winnings.

If I'm right about C, then the calculator's bet of 3001 by B carries the same risk as my 4799 bet, but with less potential payoff. This may just be a repeat of the scenario from post #1 where we're only being shown the lower end of the range.

Who's right this time?
Neither one of you. The calculator is recommending that C wager only 200 so that C doesn't fall behind A if both give an incorrect response. But as you correctly point out. if that is C's wager, B can wager enough to pass A without falling behind C's score even if C gives a correct response. C should therefore wager more than 200.

But once C has made that determination, he or she may as well go all-in. That at least has the benefit of forcing B to give a correct response, and picks up on occasion when B mistakenly underwagers. I believe that historically that happens significantly more often that A materially overwagering. --Bob

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jpahk
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Re: Final Jeopardy wagering theory

Post by jpahk »

here's how i view the situation (from scratch). true 3-player scenarios are complicated.

A should wager at least $8k to shut out B.

B and C should both probably work off the premise that A will wager $8k+, and hopefully miss. then A goes down to at most $6k. so B should not wager more than $5k, staying ahead of A on the double-stumper.

B wants to stay ahead of C's score on a double-get, so should bet at least $1400. given that, C needs to wager at least $3400 to win on a single-get, so can't afford to make the small ($200) bet that keeps her ahead of A. and since she is crushed by B, there is no real reason for her not to bet it all; with any reasonable wager by all 3 players, C is going to be in 3rd place on a miss (since neither A nor B will come down much below $6k), so she might as well maximize the payday on a single-get.

so B's reasonable wagers are $1400 (venusian) to $5k (martian). he is close enough to afford the $3k wager to guard against A betting $0 (the safeguard that i was castigated for not making against franny in my oct 6 game) without reducing his winning chances.

i think the only thing unresolved is what should A's martian wager be. and ... i don't know. A cannot really predict with any confidence what B or C will wager, so it is pretty tough to say for sure. if i were A, i know i would not expect to stay ahead of B on a triple stumper, but i would have some hopes of staying ahead of C, since i don't expect most players to work out that C should go all-in (although many players go all-in, or nearly so, anyway for no discernible reason). that said, if you're not going to win on a triple-stumper, the $1000 difference between 2nd and 3rd is not much of a factor compared to the extra $6k you might win if you get it right. so i think A's martian wager is an all-in $14k, but practically speaking, something like $13k might well win you the game on a triple-stumper if B makes an irrationally large wager, which we've all seen happen.

i've never seen somebody bet it all from a (solo) FJ lead, but i would love to see it some day if A has sound reason to believe that B will wager rationally.

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Re: Final Jeopardy wagering theory

Post by KellyJ! »

In the (A) 14,000 vs (B) 11,000 vs (C) 6,200 situation C has a dilemma: Whether or not B will make a rational wager. If so, his/her only realistic hope of a win is a sole get (which calls for the big bet you described). The wagering calculator in this case assumes that B won't wager as recommended, and that $200 wager is to allow for a TS win in such a case. Back on the old board (in January 2009 to be exact) I coined a name for this situation: Kohlstedt's Dilemma (based on the position Matt Kohlstedt was in this game, similar to how someone else coined Stratton's Dilemma from the scenario in this game).
Last edited by KellyJ! on Mon Oct 31, 2011 6:21 pm, edited 2 times in total.

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jpahk
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Re: Final Jeopardy wagering theory

Post by jpahk »

KellyJ! wrote:In the (A) 14,000 vs (B) 11,000 vs (C) 6,200 situation C has a dilemma: Whether or not B will make a rational wager. If so, his/her only realistic hope of a win is a sole get (which calls for the big bet you described). The wagering calculator in this case assumes that B won't wager as recommended, and that $200 wager is to allow for a TS win in such a case. Back on the old board (in January 2009 to be exact) I coined a name for this situation: Kohlstedt's Dilemma (based on the position Matt Kohlstedt was in this game, similar to how someone else coined Stratton's Dilemma from the scenario in this game).
ah, that makes sense (although i think you mean this game).

it is really hard to predict whether other people will make rational wagers, but for me, i know that i would feel more than a little weird giving up on a legitimate chance to win on a single-get in order to improve my chances of winning on a (TS + irrational opponent) parlay. maybe that just means i should do more homework. probably something like 50% of second-place contestants significantly overwager in FJ, right?

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Re: Final Jeopardy wagering theory

Post by KellyJ! »

jpahk wrote:
KellyJ! wrote:In the (A) 14,000 vs (B) 11,000 vs (C) 6,200 situation C has a dilemma: Whether or not B will make a rational wager. If so, his/her only realistic hope of a win is a sole get (which calls for the big bet you described). The wagering calculator in this case assumes that B won't wager as recommended, and that $200 wager is to allow for a TS win in such a case. Back on the old board (in January 2009 to be exact) I coined a name for this situation: Kohlstedt's Dilemma (based on the position Matt Kohlstedt was in this game, similar to how someone else coined Stratton's Dilemma from the scenario in this game).
ah, that makes sense (although i think you mean this game).
Sorry, I copied the wrong game's URL (I forgot exactly which game it was and I was searching through his games to find the right one, and I had the wrong one showing when I copied). I edited the URL to correct this mistake.

ETA: Here's the conditions for Kohlstedt's Dilemma:
B has at least 3/4 of A's score
B is crushing C
C has less than half of A's score (when C has more than half he/she can bet enough to go over where B would fall if he/she covers C and misses without falling behind A on a TS)

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Re: Final Jeopardy wagering theory

Post by seaborgium »

The thing I find funny about Kohlstedt's Dilemma is that Jerry Pyzansky was in it in Matt's second game ($5,600 to Matt's $10,000 to the leader's $11,800), and if Jerry had made a wager comparable to Matt's game 4 wager, Matt would have been a one-and-done champ.

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Re: Final Jeopardy wagering theory

Post by GMB »

Hello.

I recently joined this forum, because I have become interested in Final Jeopardy wagering.

I created optimum wagers for Final Jeopardy, before I knew the wagering calculator did it.

Here is a scenario that the wagering calculator does not recognize:
A scenario where the third place contestant (Player C) has exactly two-thirds of Player B's pre FJ score, and Player B has more than three-quarters of Player A's score.Here is an example:
Player A - $16,000
Player B - $15,000
Player C - $10,000

I plugged these scores in the wagering calculator, and it said Player A should bet $14,001, Player B should bet either at least $5,001 (to shut out Player C) or bet up to $4,999 (to win on a triple stumper.) Notice the two dollar difference between Player B's two rational wagering options. The calculator does not take the two-thirds (exactly) rule into consideration between the second and third place contestants that it does with first and second place.

Here is what I think each contestant should wager
Player A - $14,001 (Same as the wagering calculator)
Player B - $5,000 to cover Player C's doubled score and tie him. If you were to cover Player C's doubled score by $1, and C bet $0, you would lose to him $1 dollar on a triple stumper, which you can't afford.
Player C - You basically have two options: wager all $10,000 or nothing. A $0 wager is preferable, because even if Player C bets correctly (doesn't overwager by $1) dollar, and you bet all your money you have to get FJ right, and Player A has to get it wrong in order for you to win. If Player A gets Final Jeopardy wrong, it is likely that you will get it wrong too, so a win on a triple stumper is probably the most likely outcome for you.

I basically used two-thirds (exactly) rule that the wagering calculator suggest between Players A and B and applied it to Players B and C.

What does everyone else think?

Here is a FJ scenario I was wondering about:
When the first place person's score is equal to the sum of the other two players score (A = B +C) AND third has exactly 1/2 of second, and second has exactly 2/3 of first. (Also the scores are evenly spaced.)

Example scores:
Player A - $12,000
Player B - $8,000
Player C - $4,000

Player A should go for tie and wager $4,000 (so he won't lose by $1 to EITHER of the other players) and Player C wager all her money, no question. But should Player B wager all or nothing? What do you guys think?

Gabriel

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Re: Final Jeopardy wagering theory

Post by RobotJepLady »

GMB wrote:Here is a FJ scenario I was wondering about:
When the first place person's score is equal to the sum of the other two players score (A = B +C) AND third has exactly 1/2 of second, and second has exactly 2/3 of first. (Also the scores are evenly spaced.)

Example scores:
Player A - $12,000
Player B - $8,000
Player C - $4,000

Player A should go for tie and wager $4,000 (so he won't lose by $1 to EITHER of the other players) and Player C wager all her money, no question. But should Player B wager all or nothing? What do you guys think?

Gabriel
I think it's entirely dependent on B's interpretation of the events around her. If she is confident about the category for herself, she should bet it all. If she thinks A is also good in that category, she can flip a mental coin about betting 0 to get (or tie for) second or betting it all in case A flubs it, or if she really wants 2nd place if she thinks the ship has sailed on finishing first. If she thinks no one will do well in that category, she should bet 0.

Then there are the emotional factors: if B is the returning champion, and got that way thanks to a risky bet in the previous game's FJ, that will probably influence her choice. Similarly, if she was rewarded for risky betting on Daily Doubles, or if she saw other players rewarded for risky bets on DDs that she also knew, she might be more inclined to go all in. If she got burned on the DDs, she might prefer to play it safe.

If she knows, from playing at home, that she gets FJs right most of the time even in categories that she doesn't thing of as her strong suits, she might want to bet it all. I think there's probably a fancy name for that logical fallacy, and since I'm talking about the emotional side of it, those kinds of things will influence her.

Also, if she cares most about winning and getting to play in the next game, she probably won't be overbothered about making sure she gets second rather than third. While the extra $1000 is nothing to sneeze at, second goes home just as much as third does.

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Re: Final Jeopardy wagering theory

Post by Rex Kramer »

GMB wrote:Here is a scenario that the wagering calculator does not recognize:
A scenario where the third place contestant (Player C) has exactly two-thirds of Player B's pre FJ score, and Player B has more than three-quarters of Player A's score.Here is an example:
Player A - $16,000
Player B - $15,000
Player C - $10,000

I plugged these scores in the wagering calculator, and it said Player A should bet $14,001, Player B should bet either at least $5,001 (to shut out Player C) or bet up to $4,999 (to win on a triple stumper.) Notice the two dollar difference between Player B's two rational wagering options. The calculator does not take the two-thirds (exactly) rule into consideration between the second and third place contestants that it does with first and second place.

Here is what I think each contestant should wager
Player A - $14,001 (Same as the wagering calculator)
Player B - $5,000 to cover Player C's doubled score and tie him. If you were to cover Player C's doubled score by $1, and C bet $0, you would lose to him $1 dollar on a triple stumper, which you can't afford.
Player C - You basically have two options: wager all $10,000 or nothing. A $0 wager is preferable, because even if Player C bets correctly (doesn't overwager by $1) dollar, and you bet all your money you have to get FJ right, and Player A has to get it wrong in order for you to win. If Player A gets Final Jeopardy wrong, it is likely that you will get it wrong too, so a win on a triple stumper is probably the most likely outcome for you.

I basically used two-thirds (exactly) rule that the wagering calculator suggest between Players A and B and applied it to Players B and C.

What does everyone else think?
Welcome, Gabriel. I think your analysis for B and C is correct. However, your analysis also points out that the advantage to B of betting just $5000 is so strong that, if I were player A, I would be much less inclined to bet to cover than under ordinary circumstances. It makes so much sense for B (and possibly C) to aim for $20,000 that I'd advise player A to bet just $4,001.

Rex

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Re: Final Jeopardy wagering theory

Post by GMB »

Thanks for your responses.

Another scenario where I disagree with the wagering calculator is when Player B (second place contestant) has between 3/4ths and 4/5ths of Player A's pre FJ score, but more than twice of Player C's score. (Note

Here is an example:
Player A - $19,000
Player B - $15,000
Player C - 7,000

I plugged these scores in the wagering calculator, and it said for B to wager $4,001. Because Player B has more than twice Player C's score, shouldn't she bet up to $999 instead, to keep Player C locked out? If B wagered $4,001, and C was the only contestant to respond to Final Jeopardy right, B would lose because she overwagered. I know the chances of C being the only contestant to answer FJ correctly are pretty slim, but they still seem more likely than a $0 wager from Player A (which is where the wager of $4,000 came from.) The wager of $4,001 for Player B makes sense to me if C had less than $5,500.

Does anyone know why the wagering calculator suggested this wager for Player B? Whenever B has at least 4/5ths of Player A, and twice of Player A the wagering calculator does suggest that B should wager to keep C locked out. (When I changed Player A's score in the wagering calculator to $17,000, it did suggest that Player B wager $999.)


Gabriel

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Re: Final Jeopardy wagering theory

Post by whoisalexjacob »

Rex Kramer wrote:
GMB wrote:Here is a scenario that the wagering calculator does not recognize:
A scenario where the third place contestant (Player C) has exactly two-thirds of Player B's pre FJ score, and Player B has more than three-quarters of Player A's score.Here is an example:
Player A - $16,000
Player B - $15,000
Player C - $10,000

I plugged these scores in the wagering calculator, and it said Player A should bet $14,001, Player B should bet either at least $5,001 (to shut out Player C) or bet up to $4,999 (to win on a triple stumper.) Notice the two dollar difference between Player B's two rational wagering options. The calculator does not take the two-thirds (exactly) rule into consideration between the second and third place contestants that it does with first and second place.

Here is what I think each contestant should wager
Player A - $14,001 (Same as the wagering calculator)
Player B - $5,000 to cover Player C's doubled score and tie him. If you were to cover Player C's doubled score by $1, and C bet $0, you would lose to him $1 dollar on a triple stumper, which you can't afford.
Player C - You basically have two options: wager all $10,000 or nothing. A $0 wager is preferable, because even if Player C bets correctly (doesn't overwager by $1) dollar, and you bet all your money you have to get FJ right, and Player A has to get it wrong in order for you to win. If Player A gets Final Jeopardy wrong, it is likely that you will get it wrong too, so a win on a triple stumper is probably the most likely outcome for you.

I basically used two-thirds (exactly) rule that the wagering calculator suggest between Players A and B and applied it to Players B and C.

What does everyone else think?
Welcome, Gabriel. I think your analysis for B and C is correct. However, your analysis also points out that the advantage to B of betting just $5000 is so strong that, if I were player A, I would be much less inclined to bet to cover than under ordinary circumstances. It makes so much sense for B (and possibly C) to aim for $20,000 that I'd advise player A to bet just $4,001.

Rex
Just curious, is this suggestion for the show? I understand the logic of it, but I wonder if it really bears out that enough people make good wagers. Justin sausville tried to outsmart his opponents (I think) on his Oscar fj, and only got away with it bc the big bettor missed. I thought it might have been a brilliant bet by Justin, but if I recall he didn't seem to get much credit for it here (don't quote me on that).

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Re: Final Jeopardy wagering theory

Post by GMB »

Just curious, is this suggestion for the show? I understand the logic of it, but I wonder if it really bears out that enough people make good wagers. Justin sausville tried to outsmart his opponents (I think) on his Oscar fj, and only got away with it bc the big bettor missed. I thought it might have been a brilliant bet by Justin, but if I recall he didn't seem to get much credit for it here (don't quote me on that).
Typically, the first place contestant wagering small is not suggested on the show, because, like you said, the second place (and sometimes third place) contestants more often than not overwager in Final Jeopardy! There is one scenario where the online wagering calculator suggests that the first place person bet nothing called Faith Love, where A = 2 (B - C), because the second place person (Player B) only has reason to bet to tie with the leader (Player A's) 0 bet, since that bet is the same maximum bet to keep Player C locked out. However, this still puts A in the position of getting FJ right and losing if B wagers irrationally.

On another subject, now that I've thought about, I think another problem with the wagering calculator is that it does not list when breakpoints overlap. I think that the breakpoint strategies should be different at breakpoint, when it reaches another breakpoint.

The "Evenly Spaced Scores with Two Thirds for First" is the same scenario on the calculator as long as Player B has at least 2/3rds of Player A. This scenario encourages A and B to go for the tie in covering their trailing opponents, so they can't lose to each other by a dollar if they both answer FJ incorrectly, and for C to bet everything to tie with B if they both get FJ right and A gets it wrong. This scenario I think works great when C < 2(A - B), and C has to get it right to win anyways and A and B can't possibly lose to C on the triple stumper.

However after the "First = Second + Half of Third" breakpoint for evenly spaced scores, where C is eligible on the triple stumper, the Evenly spaced scores scenario does not make much sense. First, even though by going for the tie Player B will not lose to Player A if they both get it wrong, C can overtake them both A and B if they get it wrong. Second, by betting all their money to tie with B, C eliminates their chances of winning on a triple stumper, which is probably their best shot at winning.

What I propose is a new scenario called "Evenly Spaced Scores with Three-Quarters for First"
Player A should bet the same in the other Evenly Spaced Scores scenarios and go for tie.
Player B should either bet all their money (and tie for the win if A bets correctly) or bet small enough to beat C on a triple stumper
Player C should bet as the normally would in a non-breakpoint scenario.

Does anyone agree with my theory on "evenly spaced scores."

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Re: Final Jeopardy wagering theory

Post by webster2000 »

This is my first post to this Board. I have been thinking about FJ Wagers for a couple of years and most intensely in the last two months. My point of departure has been my working knowledge of Game Theory. I have noticed that none of the FJ Wagering articles I have uncovered ever mention the need for mixed strategies, however the most usual combination of holdings at Final appears to require that the optimizing player avoid a rigid choice and resort to randomizing to insure their opponents do not anticipate his/her strategy. I finally found some basis for resolving my curiosity about a Game Theory solution to FJ and have obtained some basic results.My results are close to conventional wisdom except for the need for the top player to only bet high 2/3 of the time. If the players used this theoretically optimum strategy, the top player would win 1/2 the time while the second highest player would win 1/3, with the lowest winning 1/6 the time.

It is unlikely that Jeopardy players would ever or should ever or could ever use these heavily mathematical results in their heat of battle, but it seems instructive to know what theory implies. Given an expression of interest, I will post some of these results on Dropbox Public for your amusement and comments.

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Re: Final Jeopardy wagering theory

Post by creedofhubris »

webster2000 wrote:This is my first post to this Board. I have been thinking about FJ Wagers for a couple of years and most intensely in the last two months. My point of departure has been my working knowledge of Game Theory. I have noticed that none of the FJ Wagering articles I have uncovered ever mention the need for mixed strategies, however the most usual combination of holdings at Final appears to require that the optimizing player avoid a rigid choice and resort to randomizing to insure their opponents do not anticipate his/her strategy. I finally found some basis for resolving my curiosity about a Game Theory solution to FJ and have obtained some basic results.My results are close to conventional wisdom except for the need for the top player to only bet high 2/3 of the time. If the players used this theoretically optimum strategy, the top player would win 1/2 the time while the second highest player would win 1/3, with the lowest winning 1/6 the time.

It is unlikely that Jeopardy players would ever or should ever or could ever use these heavily mathematical results in their heat of battle, but it seems instructive to know what theory implies. Given an expression of interest, I will post some of these results on Dropbox Public for your amusement and comments.
If third place player is hopelessly out of it, I think the game theory is relatively simple. You can ignore cases where one of the top two players gets FJ correct and one player gets it wrong; it doesn't matter who bet what, the player who gets it right wins. The games that matter are where both players get it right and where both players get it wrong. As first place player, you bet high the percentage of time both you and your primary opponent expect to get FJ right (RRR+RRW) over all of the relevant games (RRR+RRW+WWW+WWR). Ignoring player skill it ends up being ~56% over the last 4 seasons though it fluctuates quite a bit. That's unexploitable, but since as I understand it 2nd place opponents generally bet high you're probably better off exploiting them and covering.

If third place player is competitive, it makes things messier.

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Judy5cents
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Re: Final Jeopardy wagering theory

Post by Judy5cents »

My strategy for Final Jeopardy was that if I were in third place, I'd bet everything I had. If I got it wrong, I'd still be in third place. If I were in first place, I'd bet enough to cover the second place opponent betting everything he or she had. If second place person gets it right and I get it wrong, well then he or she deserved to win and I didn't.

I know a lot of Jeopardy! fans love to come up with the perfect wager to best cover all outcomes. On the J! Archive, it was suggested that I bet $9600 in my first game. Glad I didn't have them on hand to ask for advice, as I would have come in second instead of winning.

There's always the possibility of a triple stumper and whoever wagers the least will win. But I made all my wagers assuming that I knew the answers and at least one of my opponents did too. Which is what happened two out of three times.

My advice is wager everything. It's not real money and you don't get to keep what's left over. You're there to have fun. It is a game show, after all. Bet the farm and win big, or go down in a blaze of glory. Coming in third with $20,000 or $2000 still amounts to $1000 check (minus the California state taxes). Just pay attention when you cross out the wrong answer.

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Woof
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Re: Final Jeopardy wagering theory

Post by Woof »

Judy5cents wrote:
My advice is wager everything. It's not real money and you don't get to keep what's left over. You're there to have fun. It is a game show, after all. Bet the farm and win big, or go down in a blaze of glory. Coming in third with $20,000 or $2000 still amounts to $1000 check (minus the California state taxes). Just pay attention when you cross out the wrong answer.
Sorry, but I can't agree with you here. In the case of the second-place player, the size of the wager can make the difference between winning and not. It's somewhat ironic, therefore, that it's the second-place player's wager that is most likely to be wrong with the major problem being ill-advised all-in bets. Since the leader almost always makes the cover bet, in second place you should always wager to win if the leader gets FJ wrong. Anything other than that just diminishes your chances of winning, which is after all what you're there to do, no?

Bob78164
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Re: Final Jeopardy wagering theory

Post by Bob78164 »

webster2000 wrote:This is my first post to this Board. I have been thinking about FJ Wagers for a couple of years and most intensely in the last two months. My point of departure has been my working knowledge of Game Theory. I have noticed that none of the FJ Wagering articles I have uncovered ever mention the need for mixed strategies, however the most usual combination of holdings at Final appears to require that the optimizing player avoid a rigid choice and resort to randomizing to insure their opponents do not anticipate his/her strategy. I finally found some basis for resolving my curiosity about a Game Theory solution to FJ and have obtained some basic results.My results are close to conventional wisdom except for the need for the top player to only bet high 2/3 of the time. If the players used this theoretically optimum strategy, the top player would win 1/2 the time while the second highest player would win 1/3, with the lowest winning 1/6 the time.

It is unlikely that Jeopardy players would ever or should ever or could ever use these heavily mathematical results in their heat of battle, but it seems instructive to know what theory implies. Given an expression of interest, I will post some of these results on Dropbox Public for your amusement and comments.
My initial post on the old Board explicitly discussed mixed strategies. I planned to use the second hand of my watch as a randomizer to determine which branch of a mixed strategy I should use. There is a paper on the subject that can be found on the Internet. --Bob

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