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16 votes (50%)
12 votes (37.5%)
5 votes (15.62%)
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10 votes (31.25%)
3 votes (9.37%)
6 votes (18.75%)
5 votes (15.62%)
10 votes (31.25%)
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32 members have voted

Wizard
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Wizard
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June 19th, 2021 at 8:05:36 PM permalink
Let's assume the only goal, which is pretty much the case, is to beat the other player. It does not matter by how much.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
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June 19th, 2021 at 8:27:40 PM permalink
Quote: ChumpChange

Super Old School episode:
JEOPARDY! 1974-75 Nighttime Syndicated Season - YouTube
https://www.youtube.com/watch?v=GtyWD9QpFME
If they all won in Final Jeopardy, there would have been a tie. Would they have given out 2 cars and had 2 returning champions?


That version did not have returning champions, any more than the weekly syndicated versions of Family Feud or Match Game (which may air on Buzzr - look for "Match Game PM") did. I am under the impression that both players in a tie would have won the corresponding prize.
Wizard
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Wizard
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June 19th, 2021 at 8:56:53 PM permalink
I knew this would devolve into a discussion of Jeopardy rules.

Again, per the OP, each player may go only "high" or "low" per the amounts indicated. If we must discuss ties, assume both players LOSE in the event of a tie.
It's not whether you win or lose; it's whether or not you had a good bet.
ChumpChange
ChumpChange
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June 19th, 2021 at 9:09:21 PM permalink
The show has a track record of betting "high", no matter.
camapl
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June 19th, 2021 at 11:37:30 PM permalink
Quote: Wizard

I don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000
Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?



Iíve enjoyed reading through each problem on this thread, so thank you for that!


Bob should go high, as Amyís expected score is $10,000 regardless of how she bets. This maximizes his overall chance at a win.
Amy should go low, as Bobís expected score is $9,000 regardless of how he bets. This minimizes her overall chance at a loss.
* Actual results may vary.
unJon
unJon 
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June 20th, 2021 at 3:26:59 AM permalink
Quote: Wizard

I don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000
Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?



Ok, based on your latter clarification that Amy and Bob are only acting to maximize their chances of winning:


This is a simple game theory problem with no pure Nash equilibrium solution. Thatís because if Amy goes High, Bob is best off going Low (winning 50% for Bob). But if Bob goes low, then Amy does best going low (winning 100% for Amy). But if Amy goes low, Bob does best going high (winning 50%). But if Bob goes high, Amy does best going high (winning 75% for Amy).

In other words, Amy wants to do the same thing as Bob: go high or low together. And Bob wants to do the opposite of Amy: go low when she goes high and vice versa.

The mixed Nash equilibrium is found by choosing the probability of going High and Low such that the opponent is indifferent between the choice.

For Amy that means: 25x + 50(1 - x) = 50x + 0(1 - x)
x = 2/3

For Bob that means: 75y + 50(1 - y) = 50y + 100(1 - y)
y = 2/3

Both Amy and Bob should randomly choose high 2/3 of the time and low 1/3 of the time. That leads to Amy winning the show 2/3 of the time and Bob winning the show 1/3 of the time. Importantly, this is the mixed Nash equilibrium because neither Amy nor Bob can improve their chance of winning by varying their strategy so long as their opponent follows the 2/3 high and 1/3 low random strategy.

Note: that these %s would be different if Amy and Bob were trying to maximize the dollars they win based on Jeopardy rules.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
GM
GM
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June 20th, 2021 at 5:08:25 AM permalink
If you know elliptic curves, then I add that the point (-2,6) generates the group of rational points, all rational points are multiples of (-2,6) under group operation on the curve.

The problem whether the sum of consecutive k-th powers can be the power of an integer (not necessarily a k-th power) has been studied, but I have not found anything about the problem you posed. I will ask someone who may know something.
Last edited by: GM on Jun 20, 2021
charliepatrick
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June 20th, 2021 at 5:10:03 AM permalink
Firstly consider how B plays depending on how A makes their decision with a random probability of P of going Hi.
A's possible winnings18,00110,0001,999
ProbabilityP/21-PP/2

Now look at the probability of B winning when B goes Hi or Lo.

B goes Hi: wins if (i) B correct & A goes Lo OR (ii) B correct & A goes Hi & A wrong: Pr = (1-P)/2 + P/4 = 1/2 - P/4.
B goes Lo: wins if (i) A goes Hi & A wrong: P/2.

So A has to choose P so B's chances to win are equal, so P = 2/3.
This makes B indifferent to Hi or Lo and has a 1/3 chance of being the game's winner.


Now check how A plays if they know B is picking with a random probability of S going Hi.
B's possible winnings16k8k0
ProbabilityS/21-SS/2

Now look at the probability of A winning when A goes Hi or Lo.

A goes Hi: wins if (i) correct OR (ii) wrong & B goes hi & B wrong: Pr = 1/2 + S/4.
A goes Lo: wins unless B goes Hi and correct: Pr = 1 - S/2.
B has to make Hi and Lo options for A the same, so S = 2/3.
This makes A indifferent to Hi or Lo and has a 2/3 chance of being the game's winner.

Also one can see than if either side alters their probabilities from 2/3, the other party can improve their chances by going Hi or Lo as appropriate.
teliot
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June 20th, 2021 at 6:51:02 AM permalink
Quote: GM

If you know elliptic curves, then I add that the point (-2,6) generates the group of rational points, all rational points are multiples of (-2,6) under group operation on the curve.

The problem whether the sum of consecutive k-th powers can be the power of an integer (not necessarily a k-th power) has been studied, but I have not found anything about the problem you posed. I will ask someone who may know something.

Yep I know about the group structure, so thanks. I appreciate any asking you can do. I'm going to ask one friend who is an expert in this area, just to see if this is known. Thanks again.
Last edited by: teliot on Jun 20, 2021
Poetry website: www.totallydisconnected.com
Wizard
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Wizard
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Thanks for this post from:
charliepatrick
June 20th, 2021 at 1:32:12 PM permalink
Quote: unJon



This is a simple game theory problem with no pure Nash equilibrium solution. Thatís because if Amy goes High, Bob is best off going Low (winning 50% for Bob). But if Bob goes low, then Amy does best going low (winning 100% for Amy). But if Amy goes low, Bob does best going high (winning 50%). But if Bob goes high, Amy does best going high (winning 75% for Amy).

In other words, Amy wants to do the same thing as Bob: go high or low together. And Bob wants to do the opposite of Amy: go low when she goes high and vice versa.

The mixed Nash equilibrium is found by choosing the probability of going High and Low such that the opponent is indifferent between the choice.

For Amy that means: 25x + 50(1 - x) = 50x + 0(1 - x)
x = 2/3

For Bob that means: 75y + 50(1 - y) = 50y + 100(1 - y)
y = 2/3

Both Amy and Bob should randomly choose high 2/3 of the time and low 1/3 of the time. That leads to Amy winning the show 2/3 of the time and Bob winning the show 1/3 of the time. Importantly, this is the mixed Nash equilibrium because neither Amy nor Bob can improve their chance of winning by varying their strategy so long as their opponent follows the 2/3 high and 1/3 low random strategy.

Note: that these %s would be different if Amy and Bob were trying to maximize the dollars they win based on Jeopardy rules.



I agree! With Charlie as well.

I think on the actual show in this situation, where player #2 has more than 2/3 of player #1, both players tend to go high too often.

Here are some actual statistics from this situation, from seasons 30 to 34:

Both correct: 26.89%
Both wrong: 29.42%
High player right, low player wrong: 24.72%
High player wrong, low player right: 18.97%
It's not whether you win or lose; it's whether or not you had a good bet.

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