Lock Tie Scenario Math and Discussion
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Lock Tie Scenario Math and Discussion
Since this comes up several times a season and we always go through the same discussion every single time, it's probably best to have a thread where all data and discussion of such can be quickly accessed.
Variables
FJ right/wrong stats. I pulled data from the last three completed seasons (S35-37) ignoring 3rd place.
RR – 31%
RW – 26%
WR – 18%
WW – 25%
2nd place betting history. Found a post of mine from a 2016 game thread quoting stats provided by Keith Williams. viewtopic.php?p=209377#p209377
30 instances of lock tie scenarios at this time. 20 times(67%) 2nd place wagers everything. 3 times(10%) 2nd wagers all but $1. 7 times(23%) 2nd wagers some other number.
Tiebreaker(TB) question chances. One would think the leader has an advantage here because they accrued double the money of 2nd during the game. This stat doesn't make a ton of difference on the final stats, but it is a small factor. I will provide 4 different numbers based on the leader having a 50,55,60,65 percent chance of winning the TB clue.
The 7/12/2022 game box score shows the leader at 73% on the buzzer with 2nd at 55%. Very rough math shows the leader with a 57% chance on the TB clue assuming equal knowledge. Even though the TB clue is typically easy, 2nd has a slightly higher chance of not knowing the clue than does the leader. So the true odds based on this data is probably close to 60%.
Math
Leader wagers 0, 2nd wagers everything. 2nd answers correctly 49% of the time and we have a TB clue.
Leader 50% chance of winning TB clue = (.49)(.50) = .245 = leader wins 76%.
Leader 55% chance of winning TB clue = (.49)(.45) = .2205 = leader wins 78%
Leader 60% chance of winning TB clue = (.49)(.40) = .196 = leader wins 80%
Leader 65% chance of winning TB clue = (.49)(.35) = .1715 = leader wins 83%
Leader wagers 0, 2nd wagers something other than everything.
Leader wins 100% of the time.
Final numbers when the leader wagers 0:
Leader 50% TB = (.67)(.76) + .33 = 84%
Leader 55% TB = (.67)(.78) + .33 = 85%
Leader 60% TB = (.67)(.80) + .33 = 87%
Leader 65% TB = (.67)(.83) + .33 = 88%
Leader wagers 1, 2nd wagers everything.
Leader wins all scenarios except WR(18%), leader wins 82%.
Leader wagers 1, 2nd wagers everything but $1.
Leader wins all scenarios except WR(18%) where we have a TB.
Leader 50% TB = (.18)(.50) = .09 = leader wins 91%
Leader 55% TB = (.18)(.45) = .081 = leader wins 92%
Leader 60% TB = (.18)(.40) = .072 = leader wins 93%
Leader 65% TB = (.18)(.35) = .063 = leader wins 94%
Leader wagers 1, 2nd wagers some other number.
Leader wins 100% of the time.
Final numbers when leader wagers 1:
Leader 50% TB = (.67)(.82) + (.1)(.91) + .23 = .8704 = leader wins 87%
Leader 55% TB = (.67)(.82) + (.1)(.92) + .23 = .8714 = leader wins 87%
Leader 60% TB = (.67)(.82) + (.1)(.93) + .23 = .8724 = leader wins 87%
Leader 65% TB = (.67)(.82) + (.1)(.94) + .23 = .8734 = leader wins 87%
Standard wagers summary
So wagering either $0 or $1 from the lead shows the same 87% chance to win assuming the leader has a 60% chance to win the TB clue. If one thinks the leader has less than 60% chance, then wagering $1 shows very slightly better odds. If one thinks the leader has a greater than 60% chance, then wagering $0 shows very slightly better odds. But all things being equal, either wager is fine.
But now there's two other factors to consider, how the leader feels about the category and the player type of 2nd place. Category is self-explanatory, the better you feel about it the more likely you should wager $1 and vice-versa. 2nd place player type is a little more nuanced. Naturally it's next to impossible to tell that a player will wager all but $1 from 2nd place. But it is fairly easy to determine which players you wouldn't be worried about in a buzzer race with everything on the line. Some have the competitive nature to step into that situation with confidence, others do not. Tend to wager $0 against those that do not.
Non-standard wagers from the lead
Many tend to think that since 2nd will always wager everything, there's nothing wrong with wagering more than $1 in order to maximize winnings. Obviously we now know 2nd does not always wager everything, but let's look at two non-standard wagers and how they affect winning chances.
In the 7/12/2022 game the leader had exactly $20k, 2nd had exactly $10k, and 3rd had $3200. For the time being, let's keep things simple and pretend 3rd place doesn't exist. Let's examine wagers from the lead of $9999 and $19999, two wagers that would seem to be plausible. In the future I will also examine a wager of $13599 from the lead which locks out 3rd. The hard part with such will be estimating what do to with the "other" wagers from 2nd place in the WW scenario.
Leader wagers 9999, 2nd wager irrelevant.
Leader wins all scenarios except WR(18%), leader wins 82%.
Leader wagers 19999, 2nd wagers everything.
Leader wins all scenarios except WR(18%), leader wins 82%.
Leader wagers 19999, 2nd wagers everything but $1.
Leader wins RR(31%) and RW(26%) scenarios.
Leader loses WR(18%) scenario.
WW(25%) scenario results in a TB clue.
Leader 50% TB = .57 + (.25)(.50) = .695 = leader wins 70%
Leader 55% TB = .57 + (.25)(.55) = .708 = leader wins 71%
Leader 60% TB = .57 + (.25)(.60) = .720 = leader wins 72%
Leader 65% TB = .57 + (.25)(.65) = .732 = leader wins 73%
Leader wagers 19999, 2nd wagers some other number.
Leader wins RR(31%) and RW(26%) scenarios.
Leader loses WR(18%) and WW(25%) scenarios.
Final numbers when leader wagers $19999.
Leader 50% TB = (.67)(.82) + (.1)(.71) + (.23)(.57) = .7515 = leader wins 75%
Leader 55% TB = (.67)(.82) + (.1)(.72) + (.23)(.57) = .7525 = leader wins 75%
Leader 60% TB = (.67)(.82) + (.1)(.73) + (.23)(.57) = .7535 = leader wins 75%
Leader 65% TB = (.67)(.82) + (.1)(.74) + (.23)(.57) = .7545 = leader wins 75%
Summary
Standard wagers of $0 and $1 from the lead show winning chances of 87%
Passive non-standard wagers from the lead show winning chances of 82%
Aggressive non-standard wagers from the lead show winning chances of 75%
Obviously non-standard wagers greatly reduce winning chances. Some will argue that the extra money is worth the risk. But since the more the leader wagers the less chance they have of winning, it's almost certainly not worth losing the chance to return to continue earning money.
The FJ stats are slightly different from when these numbers were last calculated resulting in slightly higher winning chances for the leader.
Hopefully the math and logic above are accurate. Corrections if necessary are welcomed.
This concludes Golf's Lecture(TM), now feel free to take a nap.
Variables
FJ right/wrong stats. I pulled data from the last three completed seasons (S35-37) ignoring 3rd place.
RR – 31%
RW – 26%
WR – 18%
WW – 25%
2nd place betting history. Found a post of mine from a 2016 game thread quoting stats provided by Keith Williams. viewtopic.php?p=209377#p209377
30 instances of lock tie scenarios at this time. 20 times(67%) 2nd place wagers everything. 3 times(10%) 2nd wagers all but $1. 7 times(23%) 2nd wagers some other number.
Tiebreaker(TB) question chances. One would think the leader has an advantage here because they accrued double the money of 2nd during the game. This stat doesn't make a ton of difference on the final stats, but it is a small factor. I will provide 4 different numbers based on the leader having a 50,55,60,65 percent chance of winning the TB clue.
The 7/12/2022 game box score shows the leader at 73% on the buzzer with 2nd at 55%. Very rough math shows the leader with a 57% chance on the TB clue assuming equal knowledge. Even though the TB clue is typically easy, 2nd has a slightly higher chance of not knowing the clue than does the leader. So the true odds based on this data is probably close to 60%.
Math
Leader wagers 0, 2nd wagers everything. 2nd answers correctly 49% of the time and we have a TB clue.
Leader 50% chance of winning TB clue = (.49)(.50) = .245 = leader wins 76%.
Leader 55% chance of winning TB clue = (.49)(.45) = .2205 = leader wins 78%
Leader 60% chance of winning TB clue = (.49)(.40) = .196 = leader wins 80%
Leader 65% chance of winning TB clue = (.49)(.35) = .1715 = leader wins 83%
Leader wagers 0, 2nd wagers something other than everything.
Leader wins 100% of the time.
Final numbers when the leader wagers 0:
Leader 50% TB = (.67)(.76) + .33 = 84%
Leader 55% TB = (.67)(.78) + .33 = 85%
Leader 60% TB = (.67)(.80) + .33 = 87%
Leader 65% TB = (.67)(.83) + .33 = 88%
Leader wagers 1, 2nd wagers everything.
Leader wins all scenarios except WR(18%), leader wins 82%.
Leader wagers 1, 2nd wagers everything but $1.
Leader wins all scenarios except WR(18%) where we have a TB.
Leader 50% TB = (.18)(.50) = .09 = leader wins 91%
Leader 55% TB = (.18)(.45) = .081 = leader wins 92%
Leader 60% TB = (.18)(.40) = .072 = leader wins 93%
Leader 65% TB = (.18)(.35) = .063 = leader wins 94%
Leader wagers 1, 2nd wagers some other number.
Leader wins 100% of the time.
Final numbers when leader wagers 1:
Leader 50% TB = (.67)(.82) + (.1)(.91) + .23 = .8704 = leader wins 87%
Leader 55% TB = (.67)(.82) + (.1)(.92) + .23 = .8714 = leader wins 87%
Leader 60% TB = (.67)(.82) + (.1)(.93) + .23 = .8724 = leader wins 87%
Leader 65% TB = (.67)(.82) + (.1)(.94) + .23 = .8734 = leader wins 87%
Standard wagers summary
So wagering either $0 or $1 from the lead shows the same 87% chance to win assuming the leader has a 60% chance to win the TB clue. If one thinks the leader has less than 60% chance, then wagering $1 shows very slightly better odds. If one thinks the leader has a greater than 60% chance, then wagering $0 shows very slightly better odds. But all things being equal, either wager is fine.
But now there's two other factors to consider, how the leader feels about the category and the player type of 2nd place. Category is self-explanatory, the better you feel about it the more likely you should wager $1 and vice-versa. 2nd place player type is a little more nuanced. Naturally it's next to impossible to tell that a player will wager all but $1 from 2nd place. But it is fairly easy to determine which players you wouldn't be worried about in a buzzer race with everything on the line. Some have the competitive nature to step into that situation with confidence, others do not. Tend to wager $0 against those that do not.
Non-standard wagers from the lead
Many tend to think that since 2nd will always wager everything, there's nothing wrong with wagering more than $1 in order to maximize winnings. Obviously we now know 2nd does not always wager everything, but let's look at two non-standard wagers and how they affect winning chances.
In the 7/12/2022 game the leader had exactly $20k, 2nd had exactly $10k, and 3rd had $3200. For the time being, let's keep things simple and pretend 3rd place doesn't exist. Let's examine wagers from the lead of $9999 and $19999, two wagers that would seem to be plausible. In the future I will also examine a wager of $13599 from the lead which locks out 3rd. The hard part with such will be estimating what do to with the "other" wagers from 2nd place in the WW scenario.
Leader wagers 9999, 2nd wager irrelevant.
Leader wins all scenarios except WR(18%), leader wins 82%.
Leader wagers 19999, 2nd wagers everything.
Leader wins all scenarios except WR(18%), leader wins 82%.
Leader wagers 19999, 2nd wagers everything but $1.
Leader wins RR(31%) and RW(26%) scenarios.
Leader loses WR(18%) scenario.
WW(25%) scenario results in a TB clue.
Leader 50% TB = .57 + (.25)(.50) = .695 = leader wins 70%
Leader 55% TB = .57 + (.25)(.55) = .708 = leader wins 71%
Leader 60% TB = .57 + (.25)(.60) = .720 = leader wins 72%
Leader 65% TB = .57 + (.25)(.65) = .732 = leader wins 73%
Leader wagers 19999, 2nd wagers some other number.
Leader wins RR(31%) and RW(26%) scenarios.
Leader loses WR(18%) and WW(25%) scenarios.
Final numbers when leader wagers $19999.
Leader 50% TB = (.67)(.82) + (.1)(.71) + (.23)(.57) = .7515 = leader wins 75%
Leader 55% TB = (.67)(.82) + (.1)(.72) + (.23)(.57) = .7525 = leader wins 75%
Leader 60% TB = (.67)(.82) + (.1)(.73) + (.23)(.57) = .7535 = leader wins 75%
Leader 65% TB = (.67)(.82) + (.1)(.74) + (.23)(.57) = .7545 = leader wins 75%
Summary
Standard wagers of $0 and $1 from the lead show winning chances of 87%
Passive non-standard wagers from the lead show winning chances of 82%
Aggressive non-standard wagers from the lead show winning chances of 75%
Obviously non-standard wagers greatly reduce winning chances. Some will argue that the extra money is worth the risk. But since the more the leader wagers the less chance they have of winning, it's almost certainly not worth losing the chance to return to continue earning money.
The FJ stats are slightly different from when these numbers were last calculated resulting in slightly higher winning chances for the leader.
Hopefully the math and logic above are accurate. Corrections if necessary are welcomed.
This concludes Golf's Lecture(TM), now feel free to take a nap.
Last edited by Golf on Thu Jul 14, 2022 2:47 pm, edited 1 time in total.
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Re: Lock Tie Scenario Math and Discussion
viewtopic.php?p=353945#p353945
viewtopic.php?p=313084#p313084
The tie breaker was implemented in 2016 so choosing to use only data from before 2016 is a pretty questionable choice.
Nobody has wagered all but a dollar in a lock-tie scenario since the tie breaker was implemented.
You're cherry picking data and assumptions to arrive at the conclusion that wagering $0 or $1 gives the same chance of success.
viewtopic.php?p=313084#p313084
The tie breaker was implemented in 2016 so choosing to use only data from before 2016 is a pretty questionable choice.
Nobody has wagered all but a dollar in a lock-tie scenario since the tie breaker was implemented.
You're cherry picking data and assumptions to arrive at the conclusion that wagering $0 or $1 gives the same chance of success.
I had a dream that I was asleep and then I woke up and Jeopardy! was on.
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Re: Lock Tie Scenario Math and Discussion
(Original post quoted. Golf, I think you should feel free to make changes to the top post as you adapt and respond to objections and tweak your reasoning. That way we won't have to read through the whole thread to find what your final thinking is. But I also think we should have the original version to consult.)Golf wrote: ↑Thu Jul 14, 2022 11:02 amSpoiler
Since this comes up several times a season and we always go through the same discussion every single time, it's probably best to have a thread where all data and discussion of such can be quickly accessed.
Variables
FJ right/wrong stats. I pulled data from the last three completed seasons (S35-37) ignoring 3rd place.
RR – 31%
RW – 26%
WR – 18%
WW – 25%
2nd place betting history. Found a post of mine from a 2016 game thread quoting stats provided by Keith Williams. viewtopic.php?p=209377#p209377
30 instances of lock tie scenarios at this time. 20 times(67%) 2nd place wagers everything. 3 times(10%) 2nd wagers all but $1. 7 times(23%) 2nd wagers some other number.
Tiebreaker(TB) question chances. One would think the leader has an advantage here because they accrued double the money of 2nd during the game. This stat doesn't make a ton of difference on the final stats, but it is a small factor. I will provide 4 different numbers based on the leader having a 50,55,60,65 percent chance of winning the TB clue.
The 7/12/2022 game box score shows the leader at 73% on the buzzer with 2nd at 55%. Very rough math shows the leader with a 57% chance on the TB clue assuming equal knowledge. Even though the TB clue is typically easy, 2nd has a slightly higher chance of not knowing the clue than does the leader. So the true odds based on this data is probably close to 60%.
Math
Leader wagers 0, 2nd wagers everything. 2nd answers correctly 49% of the time and we have a TB clue.
Leader 50% chance of winning TB clue = (.49)(.50) = .245 = leader wins 76%.
Leader 55% chance of winning TB clue = (.49)(.45) = .2205 = leader wins 78%
Leader 60% chance of winning TB clue = (.49)(.40) = .196 = leader wins 80%
Leader 65% chance of winning TB clue = (.49)(.35) = .1715 = leader wins 83%
Leader wagers 0, 2nd wagers something other than everything.
Leader wins 100% of the time.
Final numbers when the leader wagers 0:
Leader 50% TB = (.67)(.76) + .33 = 84%
Leader 55% TB = (.67)(.78) + .33 = 85%
Leader 60% TB = (.67)(.80) + .33 = 87%
Leader 65% TB = (.67)(.83) + .33 = 88%
Leader wagers 1, 2nd wagers everything.
Leader wins all scenarios except WR(18%), leader wins 82%.
Leader wagers 1, 2nd wagers everything but $1.
Leader wins all scenarios except WR(18%) where we have a TB.
Leader 50% TB = (.18)(.50) = .09 = leader wins 91%
Leader 55% TB = (.18)(.45) = .081 = leader wins 92%
Leader 60% TB = (.18)(.40) = .072 = leader wins 93%
Leader 65% TB = (.18)(.35) = .063 = leader wins 94%
Leader wagers 1, 2nd wagers some other number.
Leader wins 100% of the time.
Final numbers when leader wagers 1:
Leader 50% TB = (.67)(.82) + (.1)(.91) + .23 = .8704 = leader wins 87%
Leader 55% TB = (.67)(.82) + (.1)(.92) + .23 = .8714 = leader wins 87%
Leader 60% TB = (.67)(.82) + (.1)(.93) + .23 = .8724 = leader wins 87%
Leader 65% TB = (.67)(.82) + (.1)(.94) + .23 = .8734 = leader wins 87%
Standard wagers summary
So wagering either $0 or $1 from the lead shows the same 87% chance to win assuming the leader has a 60% chance to win the TB clue. If one thinks the leader has less than 60% chance, then wagering $1 shows very slightly better odds. If one thinks the leader has a greater than 60% chance, then wagering $0 shows very slightly better odds. But all things being equal, either wager is fine.
But now there's two other factors to consider, how the leader feels about the category and the player type of 2nd place. Category is self-explanatory, the better you feel about it the more likely you should wager $1 and vice-versa. 2nd place player type is a little more nuanced. Naturally it's next to impossible to tell that a player will wager all but $1 from 2nd place. But it is fairly easy to determine which players you wouldn't be worried about in a buzzer race with everything on the line. Some have the competitive nature to step into that situation with confidence, others do not. Tend to wager $0 against those that do not.
Non-standard wagers from the lead
Many tend to think that since 2nd will always wager everything, there's nothing wrong with wagering more than $1 in order to maximize winnings. Obviously we now know 2nd does not always wager everything, but let's look at two non-standard wagers and how they affect winning chances.
In the 7/12/2022 game the leader had exactly $20k, 2nd had exactly $10k, and 3rd was negative and out of the picture. Let's examine wagers from the lead of $9999 and $19999, two wagers that would seem to be plausible.
Leader wagers 9999, 2nd wager irrelevant.
Leader wins all scenarios except WR(18%), leader wins 82%.
Leader wagers 19999, 2nd wagers everything.
Leader wins all scenarios except WR(18%), leader wins 82%.
Leader wagers 19999, 2nd wagers everything but $1.
Leader wins RR(31%) and RW(26%) scenarios.
Leader loses WR(18%) scenario.
WW(25%) scenario results in a TB clue.
Leader 50% TB = .57 + (.25)(.50) = .695 = leader wins 70%
Leader 55% TB = .57 + (.25)(.55) = .708 = leader wins 71%
Leader 60% TB = .57 + (.25)(.60) = .720 = leader wins 72%
Leader 65% TB = .57 + (.25)(.65) = .732 = leader wins 73%
Leader wagers 19999, 2nd wagers some other number.
Leader wins RR(31%) and RW(26%) scenarios.
Leader loses WR(18%) and WW(25%) scenarios.
Final numbers when leader wagers $19999.
Leader 50% TB = (.67)(.82) + (.1)(.71) + (.23)(.57) = .7515 = leader wins 75%
Leader 55% TB = (.67)(.82) + (.1)(.72) + (.23)(.57) = .7525 = leader wins 75%
Leader 60% TB = (.67)(.82) + (.1)(.73) + (.23)(.57) = .7535 = leader wins 75%
Leader 65% TB = (.67)(.82) + (.1)(.74) + (.23)(.57) = .7545 = leader wins 75%
Summary
Standard wagers of $0 and $1 from the lead show winning chances of 87%
Passive non-standard wagers from the lead show winning chances of 82%
Aggressive non-standard wagers from the lead show winning chances of 75%
Obviously non-standard wagers greatly reduce winning chances. Some will argue that the extra money is worth the risk. But since the more the leader wagers the less chance they have of winning, it's almost certainly not worth losing the chance to return to continue earning money.
The FJ stats are slightly different from when these numbers were last calculated resulting in slightly higher winning chances for the leader.
Hopefully the math and logic above are accurate. Corrections if necessary are welcomed.
This concludes Golf's Lecture(TM), now feel free to take a nap.
Nice work, Golf! One addition I'd recommend is a brief statement of your conclusion right at the top, maybe with a tiny synopsis of the reasoning that led there.
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Re: Lock Tie Scenario Math and Discussion
Thanks. I have edited that section appropriately.seaborgium wrote: ↑Thu Jul 14, 2022 12:59 pm 3rd had $3,200 (which, I assume, is why 2nd only bet $3,000—to keep her locked out). (3rd went negative in the following game.)
Please explain to me how the addition of the TB would alter 2nd's wager because I can't see how it would.MattKnowles wrote: ↑Thu Jul 14, 2022 2:03 pm viewtopic.php?p=353945#p353945
viewtopic.php?p=313084#p313084
The tie breaker was implemented in 2016 so choosing to use only data from before 2016 is a pretty questionable choice.
Nobody has wagered all but a dollar in a lock-tie scenario since the tie breaker was implemented.
You're cherry picking data and assumptions to arrive at the conclusion that wagering $0 or $1 gives the same chance of success.
But we can certainly add the data you linked post TB to the 2016 data I used. It would be helpful if a 3rd party with easy access could provide up to date data post TB.
Thanks. I will adjust the OP as necessary with corrections and/or updated data.opusthepenguin wrote: ↑Thu Jul 14, 2022 2:32 pm (Original post quoted. Golf, I think you should feel free to make changes to the top post as you adapt and respond to objections and tweak your reasoning. That way we won't have to read through the whole thread to find what your final thinking is. But I also think we should have the original version to consult.)
Nice work, Golf! One addition I'd recommend is a brief statement of your conclusion right at the top, maybe with a tiny synopsis of the reasoning that led there.
There will always be those that question the data or conclusions or whatever. But I think the general point should be how close wagers of $0 and $1 are and how much winning chances fall off with other wagers.
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Re: Lock Tie Scenario Math and Discussion
It’s a different scenario and people will react differently to it. The recommended wager for 2nd place shouldn’t change, it should be all-in with both scenarios, but because it is a different scenario they approach it differently and may come up with a different wager.
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Re: Lock Tie Scenario Math and Discussion
Another not everything wager from 2nd 2/14/23.
If I read the OP correctly, the non all-in wager from second is running about 1/3?
Yikes.
If we use that and accept that the Leader is a slight favorite in both a buzzer race and the FJ answer, then $0 becomes optimal? Probably not GTO but when the repeatable behavior of opponents is factored in….
If I read the OP correctly, the non all-in wager from second is running about 1/3?
Yikes.
If we use that and accept that the Leader is a slight favorite in both a buzzer race and the FJ answer, then $0 becomes optimal? Probably not GTO but when the repeatable behavior of opponents is factored in….
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Re: Lock Tie Scenario Math and Discussion
Prior to the no-ties rule a $0 wager was optimal because it guaranteed a return for the next day.danspartan wrote: ↑Wed Feb 15, 2023 8:26 am If we use that and accept that the Leader is a slight favorite in both a buzzer race and the FJ answer, then $0 becomes optimal? Probably not GTO but when the repeatable behavior of opponents is factored in….
What does everyone think about the proposal that after the no-ties rule a $0 wager is still optimal for the leader, irrespective of the stats on betting behavior from 2nd, because it guarantees a return for the next clue?
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Re: Lock Tie Scenario Math and Discussion
Incredibly hard disagree. This was solved seven years ago.Robert K S wrote: ↑Wed Feb 15, 2023 9:20 amPrior to the no-ties rule a $0 wager was optimal because it guaranteed a return for the next day.danspartan wrote: ↑Wed Feb 15, 2023 8:26 am If we use that and accept that the Leader is a slight favorite in both a buzzer race and the FJ answer, then $0 becomes optimal? Probably not GTO but when the repeatable behavior of opponents is factored in….
What does everyone think about the proposal that after the no-ties rule a $0 wager is still optimal for the leader, irrespective of the stats on betting behavior from 2nd, because it guarantees a return for the next clue?
https://thejeopardyfan.com/2016/01/stra ... e-win.html
$0 is only optimal when Final appears to be in a category where you think you'd be worse than the average Jeopardy! contestant.
Also, I was under the impression that Golf's conclusion was "either $1 or $0 is defensible here"—I'm not sure where you get that $0 is solely optimal under the circumstances.
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Re: Lock Tie Scenario Math and Discussion
I could see second holding back a dollar. They might assume the leader is going to bet a dollar so they get a tie on a W/R scenario and will at least have something left if they get it wrong. Not what I’d recommend, but it’s possible.
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Re: Lock Tie Scenario Math and Discussion
To be fair, that hasn't happened in over 25 years: https://www.j-archive.com/showgame.php?game_id=2617
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Re: Lock Tie Scenario Math and Discussion
Correct. Any other wager is not defensible as shown in the OP.OntarioQuizzer wrote: ↑Wed Feb 15, 2023 9:30 am Also, I was under the impression that Golf's conclusion was "either $1 or $0 is defensible here"—I'm not sure where you get that $0 is solely optimal under the circumstances.
The above math is based on the tie breaker format. Most people tend to think $1(or more) is GTO, and $1 would be slightly optimal if 2nd place always wagered everything. But of course this does not happen in actuality even though it should in theory.
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Re: Lock Tie Scenario Math and Discussion
Did anybody ever find the games where second place wagered all but a dollar?
It looks like we’ve only found one game, the one Andy linked to above. That was also a special scenario lock-tie where two players were tied in second place.
It’s a bit unnecessary to talk about the optimal wager when second place doesn’t wager everything. Either they never bet all but a dollar (in which case 0 and 1 are both automatic wins) or we have no empirical way to estimate that probability and the calculation is just magic.
It looks like we’ve only found one game, the one Andy linked to above. That was also a special scenario lock-tie where two players were tied in second place.
It’s a bit unnecessary to talk about the optimal wager when second place doesn’t wager everything. Either they never bet all but a dollar (in which case 0 and 1 are both automatic wins) or we have no empirical way to estimate that probability and the calculation is just magic.
I had a dream that I was asleep and then I woke up and Jeopardy! was on.
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Re: Lock Tie Scenario Math and Discussion
I'd say $0 if you think you are weak in the FJ category and know that your opponent is strong. But if you feel your buzzer skills are weaker than your opponent, you might think about the $1 wager to avoid the coin flip.OntarioQuizzer wrote: ↑Wed Feb 15, 2023 9:30 amIncredibly hard disagree. This was solved seven years ago.Robert K S wrote: ↑Wed Feb 15, 2023 9:20 amPrior to the no-ties rule a $0 wager was optimal because it guaranteed a return for the next day.danspartan wrote: ↑Wed Feb 15, 2023 8:26 am If we use that and accept that the Leader is a slight favorite in both a buzzer race and the FJ answer, then $0 becomes optimal? Probably not GTO but when the repeatable behavior of opponents is factored in….
What does everyone think about the proposal that after the no-ties rule a $0 wager is still optimal for the leader, irrespective of the stats on betting behavior from 2nd, because it guarantees a return for the next clue?
https://thejeopardyfan.com/2016/01/stra ... e-win.html
$0 is only optimal when Final appears to be in a category where you think you'd be worse than the average Jeopardy! contestant.
Also, I was under the impression that Golf's conclusion was "either $1 or $0 is defensible here"—I'm not sure where you get that $0 is solely optimal under the circumstances.
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Re: Lock Tie Scenario Math and Discussion
There's also this one: https://www.j-archive.com/showgame.php?game_id=297MattKnowles wrote: ↑Wed Feb 15, 2023 10:12 am Did anybody ever find the games where second place wagered all but a dollar?
It looks like we’ve only found one game, the one Andy linked to above. That was also a special scenario lock-tie where two players were tied in second place.
It’s a bit unnecessary to talk about the optimal wager when second place doesn’t wager everything. Either they never bet all but a dollar (in which case 0 and 1 are both automatic wins) or we have no empirical way to estimate that probability and the calculation is just magic.
But basically, two occurrences in nearly 40 years.
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Re: Lock Tie Scenario Math and Discussion
I was referring to holding back a dollar in the TB era. Obviously, holding back when the leader had zero incentive to bet anything was just stupid.OntarioQuizzer wrote: ↑Wed Feb 15, 2023 9:55 amTo be fair, that hasn't happened in over 25 years: https://www.j-archive.com/showgame.php?game_id=2617
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Re: Lock Tie Scenario Math and Discussion
See the math and logic regarding this in the first post for detailed examples.
It's not exactly the thing to do nowadays either!
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Re: Lock Tie Scenario Math and Discussion
I wouldn’t mind going back to the old rule-except make them split the money. Tie at $20,000? You each get $10,000.
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Re: Lock Tie Scenario Math and Discussion
Either that or, if one issue is scheduling of contestants, let both keep their money but still do the tiebreaker to determine the returning champion.danspartan wrote: ↑Thu Feb 16, 2023 9:44 am I wouldn’t mind going back to the old rule-except make them split the money. Tie at $20,000? You each get $10,000.
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Re: Lock Tie Scenario Math and Discussion
There's likely another reason for tiebreakers as explained here - in the social media era, when it's more likely contestants would've had contact with each other, allowing only one winner reduces the incentive for collusion.yclept wrote: ↑Sun Feb 19, 2023 5:24 pmEither that or, if one issue is scheduling of contestants, let both keep their money but still do the tiebreaker to determine the returning champion.danspartan wrote: ↑Thu Feb 16, 2023 9:44 am I wouldn’t mind going back to the old rule-except make them split the money. Tie at $20,000? You each get $10,000.