Jeopardy betting algorithm
Ray Perkins
In the program "Jeopardy", is there an algorithm for determining
the optimum bet for "final Jeopardy". Assuming you expect a 50% chance
of answering correctly and you believe all the other players have an
equal chance to do so.
I'm no mathematician, but it seems most contestants bet according to
some formula. Obviously the equasion would contain your (and their?)
expected chances to get the answer right.
--
Ray Perkins
Steve Sosensky
As a former Jeopardy winner, I can tell you that the odds of getting the
question right are irrelevant. You only get money if you win, so you have
to assume you'll get the answer. If you're wrong, you end up paying taxes
on the prize you win, so cash comes out of your pocket, even if you don't
want the prize.
The amount you bet depends on where you stand going into Final Jeopardy.
The leader must bet to win. If s/he has more than twice the money of the
next competitor, the bet is to bring you to $1 more than twice the nearest
amount, so you still win if you're wrong. If the leader can be caught, the
bet is the amount to double the nearest amount plus $1. (If you're about to
be a 5-time champ, it is good sportsmanship to allow second place to tie
you, as Merv/Sony has to spend twice as much).
If you're in second place, you need to bet everything you have if you can
reach the leader. This is to keep the leader's bet honest. If the leader
doesn't bet enough, you'll win. If the leader is wrong, you'll win with
your highest possible total. If the leader beats you by a buck, you gave it
your all. If you're wrong, you wouldn't have won anyway.
If you're in third, things get a little more complicated. You can only win
if the others are both wrong. You must figure out the amounts the others
will have if they bet correctly and are wrong. If you can beat both those
amounts, bet to do so. If you're already there, bet $0. If you really want
that 2nd place prize, bet to win that.
Steve
Steve Sosensky, photographer http://www.loop.com/~steves
10834 Blix Street #213 818-508-4946
Toluca Lake, CA 91602 34*09'02" N, 118*22'47" W
Audubon in So. California http://www.audubon.org/chapter/ca/socal
San Fernando Valley AS http://www.audubon.org/chapter/ca/sfvas
Matthew T. Russotto
In article <3321E7...@nortel.ca>, Ray Perkins <ra...@nortel.ca> wrote:
}In the program "Jeopardy", is there an algorithm for determining
}the optimum bet for "final Jeopardy". Assuming you expect a 50% chance
}of answering correctly and you believe all the other players have an
}equal chance to do so.
}
}I'm no mathematician, but it seems most contestants bet according to
}some formula. Obviously the equasion would contain your (and their?)
}expected chances to get the answer right.
The assumption is, unfortaunely, terrible. There are a number of easy
cases, though:
1) Runaway. If you have more than double what your closest opponent
has, bet the difference between double your opponent's total and your
current total, minus $1.
2) Opponent runaway: Doesn't matter.
3) Third place contestant, other players are much higher but close to
each other: Figure out what the first place opponent will have if he
loses a bet that, if he won, would be $1 more than double the second
place opponent. If you have less than that, bet it all. If you have
more than that, bet so that if you lose, you still have more than that.
--
Matthew T. Russotto russ...@pond.com
"Extremism in defense of liberty is no vice, and moderation in pursuit
of justice is no virtue."
Jamie Dreier
ste...@loop.com wrote:
> On Sat, 08 Mar 1997 17:26:24 -0500, Ray Perkins <ra...@nortel.ca> wrote:
>
> >In the program "Jeopardy", is there an algorithm for determining
> >the optimum bet for "final Jeopardy". Assuming you expect a 50% chance
> >of answering correctly and you believe all the other players have an
> >equal chance to do so.
> >
> >I'm no mathematician, but it seems most contestants bet according to
> >some formula. Obviously the equasion would contain your (and their?)
> >expected chances to get the answer right.
> >--
> >Ray Perkins
> As a former Jeopardy winner, I can tell you that the odds of getting the
> question right are irrelevant. You only get money if you win, so you have
> to assume you'll get the answer. If you're wrong, you end up paying taxes
> on the prize you win, so cash comes out of your pocket, even if you don't
> want the prize.
Surely this is wrong.
You have a chance to win even if you get the question wrong. If you think
your chances of getting the question right extremely high, say, .99999,
then clearly you want to bet a lot, even take some chance of losing the
game.
> The amount you bet depends on where you stand going into Final Jeopardy.
>
> The leader must bet to win. If s/he has more than twice the money of the
> next competitor, the bet is to bring you to $1 more than twice the nearest
> amount, so you still win if you're wrong.
This does seem reasonable, though if you think your chances of getting the
correct answer (uh, question) are very small, you ought to be nothing.
> If the leader can be caught, the
> bet is the amount to double the nearest amount plus $1. (If you're about to
> be a 5-time champ, it is good sportsmanship to allow second place to tie
> you, as Merv/Sony has to spend twice as much).
>
> If you're in second place, you need to bet everything you have if you can
> reach the leader. This is to keep the leader's bet honest. If the leader
> doesn't bet enough, you'll win. If the leader is wrong, you'll win with
> your highest possible total. If the leader beats you by a buck, you gave it
> your all. If you're wrong, you wouldn't have won anyway.
Consider this case, though. Leader has 9000. Second place now has 7000.
According to the above strategy, Second Place should bet everything, and
leader should bet 5001.
But supposing that Leader does bet 5001, surely Second Place ought to be
nothing, or maybe something less than 3001. That way, if they are both
wrong, Second Place still wins.
The best bet depends on your judgment of what the others will bet -- at
least often.
> If you're in third, things get a little more complicated. You can only win
> if the others are both wrong. You must figure out the amounts the others
> will have if they bet correctly and are wrong. If you can beat both those
> amounts, bet to do so. If you're already there, bet $0. If you really want
> that 2nd place prize, bet to win that.
Again: if you can be sure that the two leaders will bet as Steve suggests,
then you have a considerably better chance of winning than you do if you
just bet in the dark. But betting $0 is risky -- maybe Second Place will
bet 0, on the assumption that Leader will bet enough to ensure a win if
s/he's correct....
Tough game, Jeopardy! Certainly players sometimes make obvious errors in
their final bets.
-Jamie
Guy Bozlinski
ste...@loop.com (Steve Sosensky) wrote:
>If you're in second place, you need to bet everything you have if you
can
>reach the leader. This is to keep the leader's bet honest. If the
leader
>doesn't bet enough, you'll win. If the leader is wrong, you'll win with
>your highest possible total. If the leader beats you by a buck, you gave
it
OK, I never was on Jeopardy, but I disgree. I would assume that my
opponents would always to bet smart- that leader will protect his lead.
IE he has 8000 me- 7000 and 3rd 250. I have to figure the leader will
bet 6001. I also have to hope he gets the bloody question wrong. If he
does, he, in my example would have 1999 left. Therego, i would bet 5000
so if we both get it wrong, i would still win- if he didn't bet enough, i
still have a good chance. Note- I will assume that contestants never
play for ties unless the leader will be a 5-game champ. And you do have
to keep an eye on third place.
>If you're in third, things get a little more complicated. You can only
win
>if the others are both wrong. You must figure out the amounts the
others
>will have if they bet correctly and are wrong. If you can beat both
those
>amounts, bet to do so. If you're already there, bet $0. If you really
want
>that 2nd place prize, bet to win that.
Correct third place strategy comes into play near the end of DJ. You
rarely see it, but the 3rd place sometimes has to shut up and stand still
to win. If you have a measily amount your only goal should be hoping
that first and second place are as close to each other as possable=
keeping in mind you do have to have at least half their score. The
closer they are to each other at the end of DJ- the more both players
have to bet in FJ, increasing the chances of you sneaking in with a win.
Many times I have seen 3rd place ansering questions, so first place is
far enough of second place he can clinch a win over 3rd place with his
wager.
And yes, If it was a runaway, I'm in 2nd place but really wanted the 3rd
place prize, i would bet all and miss intentionally since I couldn't win
the game anyway.
*Guy*
Steve Sosensky
On 11 Mar 1997 01:14:27 GMT, VVT...@prodigy.com (Guy Bozlinski) wrote:
>
>Correct third place strategy comes into play near the end of DJ. You
>rarely see it, but the 3rd place sometimes has to shut up and stand still
>to win. If you have a measily amount your only goal should be hoping
>that first and second place are as close to each other as possable=
>keeping in mind you do have to have at least half their score. The
>closer they are to each other at the end of DJ- the more both players
>have to bet in FJ, increasing the chances of you sneaking in with a win.
>Many times I have seen 3rd place ansering questions, so first place is
>far enough of second place he can clinch a win over 3rd place with his
>wager.
This is a great point which most people miss.
Steve Sosensky
On 10 Mar 1997 22:50:55 GMT, pl43...@brownvm.brown.edu (Jamie Dreier)
wrote:
>ste...@loop.com wrote:
>
>> On Sat, 08 Mar 1997 17:26:24 -0500, Ray Perkins <ra...@nortel.ca> wrote:
>>
>> >In the program "Jeopardy", is there an algorithm for determining
>> >the optimum bet for "final Jeopardy". Assuming you expect a 50% chance
>> >of answering correctly and you believe all the other players have an
>> >equal chance to do so.
>> >
>> >I'm no mathematician, but it seems most contestants bet according to
>> >some formula. Obviously the equasion would contain your (and their?)
>> >expected chances to get the answer right.
>> >--
>> >Ray Perkins
>
>
>> As a former Jeopardy winner, I can tell you that the odds of getting the
>> question right are irrelevant. You only get money if you win, so you have
>> to assume you'll get the answer. If you're wrong, you end up paying taxes
>> on the prize you win, so cash comes out of your pocket, even if you don't
>> want the prize.
>
>Surely this is wrong.
>
>You have a chance to win even if you get the question wrong. If you think
>your chances of getting the question right extremely high, say, .99999,
>then clearly you want to bet a lot, even take some chance of losing the
>game.
>
But you still must assume the opponents will respond correctly unless they
have both been guessing wrong in similar categories. The week I was on
started out with a 3-time champ betting too little on a FJ that was easy
and he lost even though he responded correctly.
>> The amount you bet depends on where you stand going into Final Jeopardy.
>>
>> The leader must bet to win. If s/he has more than twice the money of the
>> next competitor, the bet is to bring you to $1 more than twice the nearest
>> amount, so you still win if you're wrong.
>
>This does seem reasonable, though if you think your chances of getting the
>correct answer (uh, question) are very small, you ought to be nothing.
>
As per the case above, only if you're *really* pessimistic about your
knowledge in the category.
>> If the leader can be caught, the
>> bet is the amount to double the nearest amount plus $1. (If you're about to
>> be a 5-time champ, it is good sportsmanship to allow second place to tie
>> you, as Merv/Sony has to spend twice as much).
>>
>> If you're in second place, you need to bet everything you have if you can
>> reach the leader. This is to keep the leader's bet honest. If the leader
>> doesn't bet enough, you'll win. If the leader is wrong, you'll win with
>> your highest possible total. If the leader beats you by a buck, you gave it
>> your all. If you're wrong, you wouldn't have won anyway.
>
>
>Consider this case, though. Leader has 9000. Second place now has 7000.
>
>According to the above strategy, Second Place should bet everything, and
>leader should bet 5001.
>
>But supposing that Leader does bet 5001, surely Second Place ought to be
>nothing, or maybe something less than 3001. That way, if they are both
>wrong, Second Place still wins.
Unless first place follows second's logic and bets nothing as well. In that
case, second place should bet, because that is the only chance of winning.
This arguement creates its own circular logic.
>
>The best bet depends on your judgment of what the others will bet -- at
>least often.
>
>> If you're in third, things get a little more complicated. You can only win
>> if the others are both wrong. You must figure out the amounts the others
>> will have if they bet correctly and are wrong. If you can beat both those
>> amounts, bet to do so. If you're already there, bet $0. If you really want
>> that 2nd place prize, bet to win that.
>
>Again: if you can be sure that the two leaders will bet as Steve suggests,
>then you have a considerably better chance of winning than you do if you
>just bet in the dark. But betting $0 is risky -- maybe Second Place will
>bet 0, on the assumption that Leader will bet enough to ensure a win if
>s/he's correct....
>
Again: beware of circular logic. You can always come up with a set of
circumstances where not betting by my method works, but I don't think they
are very likely.
>Tough game, Jeopardy! Certainly players sometimes make obvious errors in
>their final bets.
It certainly is. The test to qualify, BTW, is 50 questions in 50 different
categories each of $800 or $1000 difficulty. Alex's voice reads the clue
which appears on a TV screen and the testees have 10 seconds to respond
(not in the form of a question). You can't go back and repeat the tape.
They don't tell you, but the general consensus is that it takes 45 right to
pass. That qualifies you until the end of the season, and you may not get
called. I passed the first time and didn't get called. I did get called the
second time and then didn't get to play. A week's worth of shows get filmed
each day. 2 challengers times 5 days is 10 contestants who get to play
unless there is a 5-time champ who retires undefeated without a tie. Then
11 play, and there is a 12th potential contestant in case of emergency. If
you are one of the 2 left out, you do get to come back with a guarantee of
playing the next time. Both times I took the test, there were about 75
people taking it. The first time, 3 passed. The second time it was 5 or 6.
Remember that every contestant (except the celebrities) made it through
this process. Some of them get on and do well. Others don't even make it to
FJ. The signaling button is tricky too.
Try out if you get the chance. It's a fun experience.
Geoff Sheffield
Jamie Dreier wrote:
>
> ste...@loop.com wrote:
>
> > On Sat, 08 Mar 1997 17:26:24 -0500, Ray Perkins <ra...@nortel.ca> wrote:
> >
> > >In the program "Jeopardy", is there an algorithm for determining
> > >the optimum bet for "final Jeopardy". Assuming you expect a 50% chance
> > >of answering correctly and you believe all the other players have an
> > >equal chance to do so.
> > >
> > >I'm no mathematician, but it seems most contestants bet according to
> > >some formula. Obviously the equasion would contain your (and their?)
> > >expected chances to get the answer right.
> > >--
> > >Ray Perkins
>
>
> > As a former Jeopardy winner, I can tell you that the odds of getting the
> > question right are irrelevant. You only get money if you win, so you have
> > to assume you'll get the answer. If you're wrong, you end up paying taxes
> > on the prize you win, so cash comes out of your pocket, even if you don't
> > want the prize.
>
> Surely this is wrong.
>
> You have a chance to win even if you get the question wrong. If you think
> your chances of getting the question right extremely high, say, .99999,
> then clearly you want to bet a lot, even take some chance of losing the
> game.
>
> >
> > The leader must bet to win. If s/he has more than twice the money of the
> > next competitor, the bet is to bring you to $1 more than twice the nearest
> > amount, so you still win if you're wrong.
>
> This does seem reasonable, though if you think your chances of getting the
> correct answer (uh, question) are very small, you ought to be nothing.
>
> > bet is the amount to double the nearest amount plus $1. (If you're about to
> > be a 5-time champ, it is good sportsmanship to allow second place to tie
> > you, as Merv/Sony has to spend twice as much).
> >
> > If you're in second place, you need to bet everything you have if you can
> > reach the leader. This is to keep the leader's bet honest. If the leader
> > doesn't bet enough, you'll win. If the leader is wrong, you'll win with
> > your highest possible total. If the leader beats you by a buck, you gave it
> > your all. If you're wrong, you wouldn't have won anyway.
>
> Consider this case, though. Leader has 9000. Second place now has 7000.
>
> According to the above strategy, Second Place should bet everything, and
> leader should bet 5001.
>
> But supposing that Leader does bet 5001, surely Second Place ought to be
> nothing, or maybe something less than 3001. That way, if they are both
> wrong, Second Place still wins.
>
> least often.
This is an example of the most interesting situation, in which 2nd place
has more than 2/3 the amount of 1st place. If second place has less
than
1/2 of 1st place, it is a lock for 1st. Between 1/2 and 2/3, second
place must get the answer right to win. But at greater than 2/3, first
place is forced to bet enough that getting the answer wrong results in a
total less than second place's amount before the bet. Second place then
has to guess how easy the question is going to be, and probably should
bet
0 if the question looks like it's going to be difficult.
This assumes that the first place player will bet enough to protect
his/her
lead. I believe that this is a safe assumption, because it would be
hard
to live with yourself if you answered the question correctly, but let
second place pass you because you didn't bet enough.
>
> > If you're in third, things get a little more complicated. You can only win
> > if the others are both wrong. You must figure out the amounts the others
> > will have if they bet correctly and are wrong. If you can beat both those
> > amounts, bet to do so. If you're already there, bet $0. If you really want
> > that 2nd place prize, bet to win that.
>
> Again: if you can be sure that the two leaders will bet as Steve suggests,
> then you have a considerably better chance of winning than you do if you
> just bet in the dark. But betting $0 is risky -- maybe Second Place will
> bet 0, on the assumption that Leader will bet enough to ensure a win if
> s/he's correct....
>
> Tough game, Jeopardy! Certainly players sometimes make obvious errors in
> their final bets.
>
> -Jamie
phil_bentley
On Tue, 11 Mar 1997 07:38:55 GMT, ste...@loop.com (Steve Sosensky)
wrote:
>
>It certainly is. The test to qualify, BTW, is 50 questions in 50 different
>categories each of $800 or $1000 difficulty. Alex's voice reads the clue
>which appears on a TV screen and the testees have 10 seconds to respond
>(not in the form of a question). You can't go back and repeat the tape.
>They don't tell you, but the general consensus is that it takes 45 right to
>pass. That qualifies you until the end of the season, and you may not get
>
quite a bit, and it seems that there are a few contestants each week
who don't belong there at all... but if they passed a test like that,
they must be pretty sharp. I wonder if it's nerves, or just real bad
luck that they hit a series of questions they don't know?
Thanks,
Phil
Guy Bozlinski
>> Tough game, Jeopardy! Certainly players sometimes make obvious errors
in
>> their final bets.
>>
>> -Jamie
Another time the 2nd place contestant makes a betting error. He gets the
last clue which is a Daily Double. The bet is most always wrong. If 2nd
place can't catch 1st- s/he should his/her total will be exactly one-
half of the leader. The leader will be locked into betting $0- since the
win is guaranteed- second place will win just by getting FJ right.
If second place can pass the leader- you want to wager enough so if you
are right- you are in the lead in FJ and if you are wrong you're total
will be exactly one half.
IE 8000 7000 7000 should bet 3000 nothing else. If he's right
going into FJ he has 10,000 to 8000 if wrong he is trailing 8000-4000
first place would not throw away a guaranteed win (and $8000) all second
place has to worry is getting the clue right and not what the leader
wagers.
Betting in FJ helps determine the winner- but very often the game is
decided on ply in the final stages of DJ
*Guy*
Scott D. Rhodes
>>It certainly is. The test to qualify, BTW, is 50 questions in 50 different
>>categories each of $800 or $1000 difficulty. Alex's voice reads the clue
>>which appears on a TV screen and the testees have 10 seconds to respond
>>(not in the form of a question). You can't go back and repeat the tape.
>>They don't tell you, but the general consensus is that it takes 45 right to
>>pass. That qualifies you until the end of the season, and you may not get
>>
>quite a bit, and it seems that there are a few contestants each week
>who don't belong there at all... but if they passed a test like that,
>they must be pretty sharp. I wonder if it's nerves, or just real bad
>luck that they hit a series of questions they don't know?
One other possibility: Note that in the test, they have 10 seconds to answer.
How often do you get that much time on the show? If Contestand A can reliably
get every answer in 7 seconds, but Contestant B can get about 75% of the
questions right in 3 seconds, than Contestant A is still going to get creamed
on the show despite doing better on the test.
Russell Blackadar
Geoff Sheffield (geoff@m_l.com) wrote:
[snipped all quoted material, see original]
> This is an example of the most interesting situation, in which 2nd place
> has more than 2/3 the amount of 1st place. If second place has less
> than
> 1/2 of 1st place, it is a lock for 1st. Between 1/2 and 2/3, second
> place must get the answer right to win. But at greater than 2/3, first
> place is forced to bet enough that getting the answer wrong results in a
^^^^^^ ??
> total less than second place's amount before the bet.
But first place is not forced to bet stupidly. Clearly,
if you're in first, and your assessment is that *nobody*
will answer correctly, then you should bet zero. Otherwise,
you stand to lose to the following strategy:
> Second place then
> has to guess how easy the question is going to be, and probably should
> bet
> 0 if the question looks like it's going to be difficult.
> This assumes that the first place player will bet enough to protect
> his/her
> lead. I believe that this is a safe assumption, because it would be
> hard
> to live with yourself if you answered the question correctly, but let
> second place pass you because you didn't bet enough.
True, but wouldn't it also be hard to live with yourself if
you answered wrong and lost, because of a stupid bet you'd
made contrary to your better instincts?
The flaw in your logic is that you assume 1st will choose
to bet; then you go on to develop an opposing strategy
such that it is *not* in 1st's self-interest to bet. You
can't rely on 1st to be that stupid. In fact, at least if
you think 1st is a good game theoretician, you (2nd) probably
*should* bet, despite considerable doubts about your ability
to answer, because that's the only way to beat a 1st that
bets zero. (And as I said, 1st *should* bet zero in the
above case.)
At the opposite extreme, when the subject looks to be easy for
both, then clearly both should bet. Between these two extremes,
the situation is murky and it becomes a guessing game. All that
you can easily say, in general, is that 2nd should (will) be less
cautious than 1st, and 1st should take this into account, i.e.
1st should be less cautious than otherwise.
--
Russell Blackadar, rus...@mdli.com
phil_bentley
On Fri, 14 Mar 1997 07:21:10 GMT, rho...@ipass.net (Scott D. Rhodes)
wrote:
>>>It certainly is. The test to qualify, BTW, is 50 questions in 50 different
>>>categories each of $800 or $1000 difficulty. Alex's voice reads the clue
>>>which appears on a TV screen and the testees have 10 seconds to respond
>>>(not in the form of a question). You can't go back and repeat the tape.
>>>They don't tell you, but the general consensus is that it takes 45 right to
>>>pass. That qualifies you until the end of the season, and you may not get
>>>
>>quite a bit, and it seems that there are a few contestants each week
>>who don't belong there at all... but if they passed a test like that,
>>they must be pretty sharp. I wonder if it's nerves, or just real bad
>>luck that they hit a series of questions they don't know?
>
>One other possibility: Note that in the test, they have 10 seconds to answer.
>How often do you get that much time on the show? If Contestand A can reliably
>get every answer in 7 seconds, but Contestant B can get about 75% of the
>questions right in 3 seconds, than Contestant A is still going to get creamed
>on the show despite doing better on the test.
>
have 10 seconds to think, and then can give your answer orally, you
have a very good point. However, if you have to complete your answer
within 10 seconds, then I think you would be fine on the show... by
the time alex reads the question and calls your name, it's already
been about 5 seconds, and I've seen him let people hem and haw for
another several seconds fairly often.
Dean Huffman
Guy Bozlinski wrote:
>
> >> Tough game, Jeopardy! Certainly players sometimes make obvious errors
> in
> >> their final bets.
> >>
> >> -Jamie
>
> Another time the 2nd place contestant makes a betting error. He gets the
> last clue which is a Daily Double. The bet is most always wrong. If 2nd
> place can't catch 1st- s/he should his/her total will be exactly one-
>
>
> *Guy*
What I could never understand is why the contestant who is ahead (A)
doesn't bet an amount of money such that, if he gets his answer correct
and the second guy (B) bets everything and also gets the answer correct,
then they both have an EQUAL and WINNING amount of money. Some say they
should not do that because the other guy (B) might beat A the next day.
The way I see it is that A knows "he could have won", so he should have
a better advantage playing B the next day than playing an unknown
quantity.
==================================================================
Dean G. Huffman Phone: 217 787-3745 FAX 217 787-0680 (Home)
peri...@slip.net 217 782-9666 217 524-2432 (Office)
This is a court of law, young man, not a court of justice.
-- Oliver Wendell Holmes, Jr. H
==================================================================
QuikSand
In the several responses posted here, there seems to be yet another very
reasonable assumption that is being missed. All Jeopardy contestants are
fairly smart, and though they may have "specialties" it would be pretty
safe to draw a conclusion or two from that. I think it's reasonable to
assume:
-There will be some FJ questions that EVERYONE (or nearly so) would get
-There will be some FJ questions that NO ONE (or nearly so) would get
-There will be some in between - perhaps that only someone with particular
knowledge of the subject would be likely to get, or ones with multiple
"guess" answers suggested by the clue's context
The suggestion here is that the likelihood of any individual contestant
knowing the answer is not an INDEPENDENT event form another contestant's
chance of knowing. Specifically - if the ? is tough enough to stump the
leader, then it's fairly likely to be tough enough to stump the 2nd and
3rd place finishers, too.
Which leads to a slight wrinkle in strategy - especially if you're not in
the lead. Consider the scenario posted earlier on this thread:
LEADER has $9000, YOU have $7000, 3RD is ignored.
Previous discussions have suggested that the LEADER will bet $5001 - an
assumption with which I agree. But someone else said that YOU should "bet
everything - to keep the leader honest." I don't buy that-- that means
that the only way you win is if the LEADER misses it, and you get it
right. (Recall, this leader didnt get to be the leader by dumb luck - she
beat you in the 10 categories so far...) And further, you can't put any
"pressure" on other people's bets, which are made simultaneously.
I suggest that the optimal bet here is $4000. This affords you the win in
either of 2 scenarios-- the "leader wrong, you right" and the "stumper."
Bottom line, I don't see any wisdom in hanging your hat on the chance that
YOU will be the only one of 3 contestants to get the answer right.
Personally, I would guess that that scenario (a 2nd or 3rd place person
being the only right answer) is fairly uncommon-- maybe 10% of the time.
(?)
Obviously, this theory (that the question is more likely to be a
"creampuff" or "stumper" than somewhere in between) leads to a revised
thinking of how to bet if you are in 3rd place, as well. Also, the
position of hte 3rd place person affects the 2nd place person's thinking,
or it should.
I think that generally, when you're not in the lead, you almost have to
play expecting to lose. Make your bets so that if the question is a
"stumper" you'll come out on top. I think your odds of getting a
"stumper" are better than the odds of getting a question that you know,
but the leader does not. (Of course, if your specialty category comes up,
maybe you bet the farm)
QuikSand
Jamie Dreier
VVT...@prodigy.com (Guy Bozlinski) wrote:
> Another time the 2nd place contestant makes a betting error. He gets the
> last clue which is a Daily Double. The bet is most always wrong. If 2nd
> place can't catch 1st- s/he should his/her total will be exactly one-
> half of the leader. The leader will be locked into betting $0- since the
> win is guaranteed- second place will win just by getting FJ right.
Hm, that's a good point, I agree with you.
-Jamie
Dale Johannesen
|>It certainly is. The test to qualify, BTW, is 50 questions in 50 different
|>categories each of $800 or $1000 difficulty. Alex's voice reads the clue
|>which appears on a TV screen and the testees have 10 seconds to respond
|>(not in the form of a question). You can't go back and repeat the tape.
|>They don't tell you, but the general consensus is that it takes 45 right to
|>pass. That qualifies you until the end of the season, and you may not get
I imagine that other factors count beyond how many questions you get
right. Contestants who look like they're having fun make for a much
better TV show, which, after all, is their main goal.
Douglas J. Zare
QuikSand <quik...@aol.com> wrote:
>In the several responses posted here, there seems to be yet another very
>reasonable assumption that is being missed. All Jeopardy contestants are
>fairly smart,
No, they are not. Our definitions might differ, but I challenge you to
explain the three-way tie at $0 which happened in a seniors' tournament a
few years ago. (AT turned to the camera. "I'm sorry, there are no
winners...") The contestants displayed neither knowledge nor intelligence.
I think they averaged about $1,600 before Final Jeopardy.
The contestants make decisions which are definitely errors under the
mildest assumptions, such as desiring money. Even aside from Daily Doubles
and Final Jeopardy, frequently the nonleaders ring in on the last few
questions when their only hope is for the leader to get the question wrong
first.
>[...]
>LEADER has $9000, YOU have $7000, 3RD is ignored.
>[...]
Two person zero-sum games are essentially solved (I won't list the other
hypotheses). This is not a zero-sum game, but you treat it as such, and
your analysis is not correct for that assumption.
There are interesting situations which occur with 3-player finishes, even
assuming that the payoff is zero-sum. Do the contestants get to conspire
beforehand?
Douglas Zare
http://www.cco.caltech.edu/~zare/contest.html -- 2nd IMC
Brian and/or Margo Gordon
I know you have to bid an amount up to the score you already have, but is
it explicit in the rules that the bet must be nonnegative? I mean, if you
have 5000, you could bet -12000, which is less than your total, blow the
final jeopardy question on purpose, and lose -12000 to end up at 17000....
hmmmm....
who are three people who have never been in my kitchen?
BillRyan22
>Subject: Re: Jeopardy betting algorithm
>
>On Sat, 08 Mar 1997 , Ray Perkins <ra...@nortel.ca> wrote:
>
>>In the program "Jeopardy", is there an algorithm for determining
>>the optimum bet for "final Jeopardy". Assuming you expect a 50% chance
>>of answering correctly and you believe all the other players have an
>>equal chance to do so.
This is an interesting problem. For those of you who are not familiar
with Jeopardy, 3 players each accumulate "money" in two initial rounds.
Then there is one "final Jeopardy" question (actually they give you the
answer and you have to guess the question that results in this answer).
Before the final question, each player bets any amount up to the amount of
"money" he/she has accumulated. If you are correct in final jeopardy, the
amount of your bet is added to your previous total; otherwise it is
subtracted. Players write down secretly their bets for final Jeopardy
before seeing the final question, so no one knows how much the other two
have bet or how difficult the question may be. They know only the broad
category of the question.
Only the winner gets in cash the amount of his/her "money" after final
Jeopardy and gets to play again. There are also special bonuses for
five-time winners.
If two contestants are tied for the lead after final Jeopardy, both win
the amount of their money and get to come back for the next game.
Thus let us assume that the overriding goal is to win the game and that
one will not bet so much money that he/she will reduce the chances of
winning. This assumption holds throughout the following discussion.
I should note that we are not dealing with perfect logicians here. Often
their bets defy logic. A lot of a contestants' strategies will depend
upon what each thinks the others will bet. Much of the discussion below
is based upon informal empirical evidence from watching Jeopardy over the
years.
STRATEGIES - A VS. B ONLY
Let A be the person in the lead going into final jeopardy. He has A$. B
is in 2nd place with B$ and C is in 3rd with C$.
For now, let's assume that C has little or no "money" and has no chance of
winning, so that the contest boils down to A vs. B. If B$ is less than
A$/2, B has no chance of winning. Say A$=9000 and B$=4000. The most that
B can get is $8000, and A will bet something less than $1000 and be
guaranteed a win.
Now suppose that A$=9000 and B$=5000. The most B can get is $10,000. A
can bet $1001. He will win if (a) he is right or (b) both A and B are
wrong. Only if A is wrong and B is right will B win. Since B must be
right, she might as well bet her whole $5000 and hope that A is wrong.
She cannot win if A is right.
The situation changes, however if B$ > 2/3 of A$. Say A$=9000 and
B$=7000. For some reason, in practice B will most often bet all $7000
(even though she shouldn't). A most often guards against the likelihood
that B will bet all and be right. Accordingly, A will usually bet
something like $5001 in order to top the $14000 that B might get. A's
minimum bet will generally be 2B$-A$+1. If A is wrong, A will drop to
$3999, which is less than B had before betting. This is why B should not
bet. If A is wrong, A will end up with 2(A$-B$)-1 dollars. Call this
quantity Q$.
Let us call scenario (X) the situation where:
- B$>A$/2
- B bets everything even if B$> 2/3 of A$ (common)
- A bets 2B$-A$+1 in order to beat out B if B bets all and is right.
I believe that scenario (X) is the most common in Jeopardy.
Define scenario (Y) to mean betting per scenario (X) and both A and B are
wrong.
Obviously other scenarios can and do occur when B$ > 2/3 of A$. There are
a myriad of possibilities, depending upon what B thinks that A thinks that
B thinks, etc. And the reasoning gets more complicated if C is still in
the running
STRATEGIES FOR B AND C
PROBABILITY IF B AND/OR C BEING WRONG IF A IS WRONG
If A bets as described above, it is apparent that neither B nor C can win
unless A is wrong in final Jeopardy. So B and C should assume that A is
wrong and falls to Q$.
Some final Jeopardy questions are more difficult than others. If A is
wrong, it might be indicative of a difficult question, and I think the
probability of B also being wrong is 2/3. Ditto for C. See APPENDIX for
rationale.
GENERAL PRINCIPLES FOR B & C
Take into account the odds against you being right (if A wrong)
If you are < Q$, you must bet all.
If you are above Q$, do not risk a bet that would take you below Q$ ("bet
safely").
Do not assume that your opponent is a perfect logician.
MY STRATEGIES FOR B
If I were B and less than 2/3 of A$, I would bet all.
If C is between Q$/2 AND Q$, I would anticipate C betting all. If 2C$>B$,
I would bet 2C$-B$ if I could bet safely. Otherwise, I would bet $0 to
make C bear the risk of having to be right.
If B were greater than 2/3 if A$, B should assume that A would bet as
shown above and bet as follows:
1. If C$<B$/2, I would bet $0 because C is no threat.
2. Else if B$>3C$/2, I would bet 2C$-B$, if I could bet this safely.
Even if B is wrong, she still tops C$. For example, B=$8000; C=$5000.
Bet $2000 (or $2001)
3. If C$ is > 2/3 of B$, I would bet B$-C$ if I could bet this safely.
Limiting my bet to this amount negates C's $0 bet option.
Otherwise, I would bet $0 to put the burden of being right onto C.
MY STRATEGIES FOR C
If C has between Q$/2 and Q$, C must bet and be right and hope that
something like scenario (Y) takes place. If C has to be right, he might
as well bet all his "money".
If B$>3C$/2 (see B3 above), I would note that if B is wrong, she drops to
2(B$-C$). I would bet just enough to beat this amount if I could do so
safely.
In all other cases, I would bet $0 and hope for something like scenario
(Y), where both A and B make large bets and are wrong.
If some of the arithmetic seems burdensome, there are over four minutes of
advertising, etc. just before final Jeopardy in which to make
calculations.
I got into this problem a little late, I hope this posting is not
redundant.
Bill Ryan
APPENDIX RE: PROBABILITIES
In practice A must be wrong for B and/or C to have a chance. If A is
wrong, I think it increases the chances of B and C being wrong also. It
may be a tough final Jeopardy question if A is wrong.
Let us imagine 3 urns containing R (for Red or Right) and W (for White or
Wrong) marbles. Each urn contains 3 marbles. Since we must assume that A
is wrong, he must draw a W marble, meaning that no urn contains RRR
marbles.
The 3 urns then contain (1) WRR, (2) WWR, or (3) WWW marbles. We do not
know from which urn that A has drawn the assumed W marble. But note that
there are 6 W marbles out of the total of 9 marbles.
Thus, I think that the probabilities that A drew from a given urn are:
urn (1) 1/6, urn (2) 2/6 and urn (3) 3/6 based upon the respective number
of W's in each urn. Subtracting one W from each urn (A's drawing), leaves
a 1/6 probability that the remaining two are RR, a 1/6 probability of RW
(half of 2/6), a 1/6 probability of WR and a 3/6 (=1/2) probability of WW.
The probability that the next marble drawn is W is 2/3.
If the 2nd marble (answer) is W (Prob = 2/3), the probability that the 3rd
marble is also W becomes 3/4. Multiplying 2/3 x 3/4 gives 1/2 as the
probability of WW, which is consistent with the above probability of WW if
the first marble is W.
If the 2nd marble is R (Prob = 1/3), the probability that the 3rd will be
W (or R) becomes 1/2. Multiplying 1/3 x 1/2 = 1/6 = Probability of RW per
above.
I
.
America's Commissioner
BillRyan22 wrote:
> Now suppose that A$=9000 and B$=5000. The most B can get is $10,000. A
> can bet $1001. He will win if (a) he is right or (b) both A and B are
> wrong. Only if A is wrong and B is right will B win. Since B must be
> right, she might as well bet her whole $5000 and hope that A is wrong.
> She cannot win if A is right.
OK, in the above example, I am player A. I assume that player B is going to wager the entire
$5000, hoping that he is correct and I am incorrect. Now, a $1001 wager will ascertain my
winning if I am right, or if we are both wrong. On the other hand, a wager of $8999 will do the
same. If both of us are wrong, I win $1 to $0. If we are both correct, I win $17,999 to
$10,000. In the event that B is right and I am wrong, then I lose regardless of which wager I've
chosen.
ObMonkeyWrench: In the event that player B is psychic (never mind that the psychic only has
$5000 going into final Jeopardy) and knows/guesses I'm going to play this way, then he may bet $0
and pray that I miss the question. I'll take my chances that player B is neither psychic nor
that dumb.