Lock Tie Scenario Math and Discussion
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- Second Banana
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Re: Lock Tie Scenario Math and Discussion
I'm not seeing where the repeated failure of contestants to go all in from second makes an appreciable difference for $0 over $1. Wouldn't this only matter if second held back a single dollar?
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Re: Lock Tie Scenario Math and Discussion
You're not missing anything.
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Re: Lock Tie Scenario Math and Discussion
As of March 22, 2016, Keith Williams's site gave the following numbers regarding the lock tie scenario.
30 total occurrences
20 wager everything
3 wager all but $1
7 wager some other amount
The most common wager from 2nd is obviously everything. The 2nd most common wager would be all but $1. Therefore it would make sense to model it.
30 total occurrences
20 wager everything
3 wager all but $1
7 wager some other amount
The most common wager from 2nd is obviously everything. The 2nd most common wager would be all but $1. Therefore it would make sense to model it.
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Re: Lock Tie Scenario Math and Discussion
I don't know what numbers Keith is using, but my numbers—derived from J! Archive, do not match.Golf wrote: ↑Mon Feb 20, 2023 5:32 pm As of March 22, 2016, Keith Williams's site gave the following numbers regarding the lock tie scenario.
30 total occurrences
20 wager everything
3 wager all but $1
7 wager some other amount
The most common wager from 2nd is obviously everything. The 2nd most common wager would be all but $1. Therefore it would make sense to model it.
I currently have:
50 total occurrences
38 wager everything
2 wager all but $1 (and none in the last 25 years)
10 wager some other amount
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Re: Lock Tie Scenario Math and Discussion
I’m ok with your numbers, it’s more complete vs what we had in 2016. Just happy we have the data to use for calculations.
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Re: Lock Tie Scenario Math and Discussion
So in 24% of the scenarios the trailer doesn’t bet it all. Fascinating.
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Re: Lock Tie Scenario Math and Discussion
There is a major problem with this entry. It states:GoodStrategy wrote: ↑Sun Feb 19, 2023 8:21 pmThere'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.
"For much of the show's history, the existence of co-champions could be reconciled with that law by precluding players who knew each other previously from playing against one another."
This sentence misses the fact that even before the internet existed in any form, Jeopardy! itself brought many contestants into contact before taping. There were always contestants who met at audition and then played on the same show. And, there were many contestants who were held over from tape days due to ties or other reasons.
Or, the very worst, returning co-champions themselves at the end of a taping week. There is no indication they were imprisoned to prevent contact.
So, the fact is if the risk players might agree to offer ties makes it a crime to allow ties, then it did so from the beginning of Jeopardy! history.
On the other hand, the other paragraph lays out the logic behind the Chu-Williams strategy quite well. And it does bring up serious questions as to whether too many would use that strategy if ties were brought back.
In the end, I'm not necessarily saying that Jeopardy! should bring back ties. Only that it is not a criminal issue. And if it were, it means Jeopardy! was on the verge of a criminal scandal for thirty years. I choose not to believe that.
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Re: Lock Tie Scenario Math and Discussion
It's worth mentioning here that losing on a tiebreaker probably increases your probability of being invited to participate in the Second Chance Competition, like Erica Weiner-Amachi. (Based on the Inside Jeopardy podcast, my understanding is that Second Chance will be an annual event.)
This factor militates in favour of wagering 0 when you have the lead with exactly double your opponent's score, if it is your first game.
This factor militates in favour of wagering 0 when you have the lead with exactly double your opponent's score, if it is your first game.
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Re: Lock Tie Scenario Math and Discussion
Here is a formula that may be useful for further discussion in this thread.
Suppose you lead with exactly double the score of your closest opponent, entering FJ. Then to maximize your probability of winning the game:
wager 0 if x − y ≤ z, and
wager 1 if x − y ≥ z, where:
x is the quotient of (the probability that you and the opponent both get FJ) divided by (the probability that you miss FJ and the opponent gets it);
y is the quotient of (the probability that the opponent wagers all-but-one) divided by (the probability that the opponent wagers all-in);
z is the quotient of (the probability that you would win a tiebreaker) divided by (the probability that the opponent would win a tiebreaker).
Suppose you lead with exactly double the score of your closest opponent, entering FJ. Then to maximize your probability of winning the game:
wager 0 if x − y ≤ z, and
wager 1 if x − y ≥ z, where:
x is the quotient of (the probability that you and the opponent both get FJ) divided by (the probability that you miss FJ and the opponent gets it);
y is the quotient of (the probability that the opponent wagers all-but-one) divided by (the probability that the opponent wagers all-in);
z is the quotient of (the probability that you would win a tiebreaker) divided by (the probability that the opponent would win a tiebreaker).
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Re: Lock Tie Scenario Math and Discussion
Nicely done.Mathew5000 wrote: ↑Thu Mar 23, 2023 11:48 pm Here is a formula that may be useful for further discussion in this thread.
Suppose you lead with exactly double the score of your closest opponent, entering FJ. Then to maximize your probability of winning the game:
wager 0 if x − y ≤ z, and
wager 1 if x − y ≥ z, where:
x is the quotient of (the probability that you and the opponent both get FJ) divided by (the probability that you miss FJ and the opponent gets it);
y is the quotient of (the probability that the opponent wagers all-but-one) divided by (the probability that the opponent wagers all-in);
z is the quotient of (the probability that you would win a tiebreaker) divided by (the probability that the opponent would win a tiebreaker).
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Re: Lock Tie Scenario Math and Discussion
You’ve got a problem when X-y=zMathew5000 wrote: ↑Thu Mar 23, 2023 11:48 pm Here is a formula that may be useful for further discussion in this thread.
Suppose you lead with exactly double the score of your closest opponent, entering FJ. Then to maximize your probability of winning the game:
wager 0 if x − y ≤ z, and
wager 1 if x − y ≥ z, where:
x is the quotient of (the probability that you and the opponent both get FJ) divided by (the probability that you miss FJ and the opponent gets it);
y is the quotient of (the probability that the opponent wagers all-but-one) divided by (the probability that the opponent wagers all-in);
z is the quotient of (the probability that you would win a tiebreaker) divided by (the probability that the opponent would win a tiebreaker).
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Re: Lock Tie Scenario Math and Discussion
wager Σ (-1)^n from 0 to ∞ then.Woof wrote: ↑Fri Mar 24, 2023 9:59 amYou’ve got a problem when X-y=zMathew5000 wrote: ↑Thu Mar 23, 2023 11:48 pm Here is a formula that may be useful for further discussion in this thread.
Suppose you lead with exactly double the score of your closest opponent, entering FJ. Then to maximize your probability of winning the game:
wager 0 if x − y ≤ z, and
wager 1 if x − y ≥ z, where:
x is the quotient of (the probability that you and the opponent both get FJ) divided by (the probability that you miss FJ and the opponent gets it);
y is the quotient of (the probability that the opponent wagers all-but-one) divided by (the probability that the opponent wagers all-in);
z is the quotient of (the probability that you would win a tiebreaker) divided by (the probability that the opponent would win a tiebreaker).
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Re: Lock Tie Scenario Math and Discussion
Or, for that matter, betting to tie in some of the other scenarios that would've called for such back in the day.Mathew5000 wrote: ↑Thu Mar 23, 2023 11:24 pm It's worth mentioning here that losing on a tiebreaker probably increases your probability of being invited to participate in the Second Chance Competition, like Erica Weiner-Amachi. (Based on the Inside Jeopardy podcast, my understanding is that Second Chance will be an annual event.)
This factor militates in favour of wagering 0 when you have the lead with exactly double your opponent's score, if it is your first game.
Last edited by GoodStrategy on Fri Mar 24, 2023 9:44 pm, edited 1 time in total.
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Re: Lock Tie Scenario Math and Discussion
You can win 4(?)+ games to get to the ToC the old, hard way or you can win 2-3 games max to get to the ToC the new way. I know which way I'm picking
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Re: Lock Tie Scenario Math and Discussion
There's no problem. If x − y = z then your probability of winning the game is maximized by wagering 0 and is also maximized by wagering 1.Woof wrote: ↑Fri Mar 24, 2023 9:59 amYou’ve got a problem when X-y=zMathew5000 wrote: ↑Thu Mar 23, 2023 11:48 pm Suppose you lead with exactly double the score of your closest opponent, entering FJ. Then to maximize your probability of winning the game:
wager 0 if x − y ≤ z, and
wager 1 if x − y ≥ z, where:
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Re: Lock Tie Scenario Math and Discussion
So, a Schroedinger's cat scenario, then?Mathew5000 wrote: ↑Fri Mar 24, 2023 2:11 pmThere's no problem. If x − y = z then your probability of winning the game is maximized by wagering 0 and is also maximized by wagering 1.Woof wrote: ↑Fri Mar 24, 2023 9:59 amYou’ve got a problem when X-y=zMathew5000 wrote: ↑Thu Mar 23, 2023 11:48 pm Suppose you lead with exactly double the score of your closest opponent, entering FJ. Then to maximize your probability of winning the game:
wager 0 if x − y ≤ z, and
wager 1 if x − y ≥ z, where:
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Re: Lock Tie Scenario Math and Discussion
I'm not sure whether that's a joke. I'll answer seriously. This has nothing to do with quantum mechanics; it's a math formula. If x − y = z, it means that for the particular values used as inputs to the formula, the probability of winning the game would be the same whether you wager 0 or 1 (and also, there is no other wager that would give you a greater probability of winning the game).Woof wrote: ↑Fri Mar 24, 2023 4:07 pmSo, a Schroedinger's cat scenario, then?Mathew5000 wrote: ↑Fri Mar 24, 2023 2:11 pm
There's no problem. If x − y = z then your probability of winning the game is maximized by wagering 0 and is also maximized by wagering 1.
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Re: Lock Tie Scenario Math and Discussion
His equation cuts out a lot of fluff, like what happens when both players miss, or what happens when second place wagers to guarantee a loss. Those aren't factored in. It only uses the probabilities that are relevant to deciding whether to wager $0 or $1. There's been some discussion about what those probabilities actually are throughout this thread, and they can be estimated in different ways depending on what assumptions you use. This can make it a lot simpler to see understand what happens when you change your assumptions and use different probabilities.
So for example....
Using the RR/WR statistics from Golf's first post (viewtopic.php?p=379465#p379465) we get 31%/18% --> X = 1.72
Using the second place wager info from my post (viewtopic.php?p=353945#p353945) we get 1/12 ---> Y = .083
Assume the chance of winning a tiebreaker is an even split. 50/50 --> Z = 1
X-Y > Z, therefore bet $1.
Golf's original post would have had Y at 10/67 = .149, and Z at 60/40 = 1.5
So for example....
Using the RR/WR statistics from Golf's first post (viewtopic.php?p=379465#p379465) we get 31%/18% --> X = 1.72
Using the second place wager info from my post (viewtopic.php?p=353945#p353945) we get 1/12 ---> Y = .083
Assume the chance of winning a tiebreaker is an even split. 50/50 --> Z = 1
X-Y > Z, therefore bet $1.
Golf's original post would have had Y at 10/67 = .149, and Z at 60/40 = 1.5
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Re: Lock Tie Scenario Math and Discussion
It was indeed a joke. Sorry for not including a smiley or something. The Schrödinger’s cat reference was the superposition of two states (1 and 0) at the same time (X-y=z). Presumably, it resolves into one or the other when the wager is revealed, but thankfully no cats were harmed in that gedanken experiment.Mathew5000 wrote: ↑Fri Mar 24, 2023 10:54 pmI'm not sure whether that's a joke. I'll answer seriously. This has nothing to do with quantum mechanics; it's a math formula. If x − y = z, it means that for the particular values used as inputs to the formula, the probability of winning the game would be the same whether you wager 0 or 1 (and also, there is no other wager that would give you a greater probability of winning the game).Woof wrote: ↑Fri Mar 24, 2023 4:07 pmSo, a Schroedinger's cat scenario, then?Mathew5000 wrote: ↑Fri Mar 24, 2023 2:11 pm
There's no problem. If x − y = z then your probability of winning the game is maximized by wagering 0 and is also maximized by wagering 1.
I do get the point you were driving at and will ponder more deeply on the subject
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Re: Lock Tie Scenario Math and Discussion
I am curious about what the third lock-tie game was, that Keith counted, in which second-place wagered all but a dollar. The URL of the page from Keith's site that Golf quoted was presumably "thefinalwager.co/2016/03/22/final-jeopardy-lock-tie-tiebreaker-wager/" but that is not operative now and not archived by the Wayback Machine. The corresponding video is still on YouTube but in the video Keith doesn't discuss the statistics in detail (timecode 2:09).OntarioQuizzer wrote: ↑Mon Feb 20, 2023 8:22 pmI don't know what numbers Keith is using, but my numbers—derived from J! Archive, do not match.Golf wrote: ↑Mon Feb 20, 2023 5:32 pm As of March 22, 2016, Keith Williams's site gave the following numbers regarding the lock tie scenario.
30 total occurrences
20 wager everything
3 wager all but $1
7 wager some other amount
The most common wager from 2nd is obviously everything. The 2nd most common wager would be all but $1. Therefore it would make sense to model it.
I currently have:
50 total occurrences
38 wager everything
2 wager all but $1 (and none in the last 25 years)
10 wager some other amount
Those all-but-a-dollar wagers must be related to the contestant briefing, where they are told that you cannot win with a zero final score.
What's relevant is the quotient of the chance of your opponent betting all-but-a-dollar over the chance of your opponent betting everything (y in my formula). Sometimes you might be against a player you're pretty sure is rational, in which case just take y=0. Also if the host has made a comment that your score is exactly double, or if it's obvious even to someone not mathematically inclined (for example, if the scores are 20,000 to 10,000), this would increase the probability that your opponent bets everything, reducing y to almost 0. But what if you have reason to believe that your opponent does not have a good grip on wagering theory or basic arithmetic, and the scores are perhaps 15,600 to 7,800, and the host has not hinted that "the wagering will be very important today". Even there, I believe it would be unwarranted to take y as more than 0.1. (Unless this is some contrived situation, like it's a returning champion whom you've seen bet all-but-a-dollar in previous games on the taping day.)
In summary, y will be small compared to x and z, so small in fact that you can safely ignore it. The true value of y is certainly less than the margin of error on your estimates of x and z (except perhaps in that rare case I mentioned, where you have already seen this opponent wager all-but-a-dollar in FJ in previous games).