T.R | Title | User | Personal Name | Date | Lines |
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909.1 | the *first* guesser has the advantage | CLT::GILBERT | | Thu Jul 28 1988 00:19 | 15 |
| Tell Susan that the first person has the advantage.
Suppose B is the bill. The first guesser should choose X very
close to the expectation's midpoint. (Below, I assume that the
pay-off is independent of the bill's amount). That is, choose X
so that:
P( B > X ) <= 1/2, and P( B < X ) <= 1/2.
Surely there is such an X. (why?)
Now suppose the second guesser chooses X+1; his probability of being
right is P(B>X), which is <= 1/2. Suppose the second guesser chooses
X-1; his probability of being right is P(B<X), which is also <= 1/2.
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909.2 | huh? how is she supposed to guess B exactly? | VIDEO::OSMAN | type video::user$7:[osman]eric.vt240 | Thu Jul 28 1988 19:59 | 15 |
| > P( B > X ) <= 1/2, and P( B < X ) <= 1/2.
>
> Surely there is such an X. (why?)
Right. X=B. Anything larger has 1/1 probability of being >X
and anything smaller has 1/1 probability of being <X.
So you're saying first guesser should choose B exactly. But how's
Susan supposed to do that ?
Anyway, how do you refute the .0 argument of saying second person
has larger range covered.
/Eric
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909.3 | Second choice wins, says gambler's instinct | POOL::HALLYB | The smart money was on Goliath | Fri Jul 29 1988 15:20 | 17 |
| > thinks bill is between $4 and $5 so she guesses $4.50. Her son thought
> bill was between $3.50 and $4.50 so when she says $4.50, he says $4.49.
I hope her son isn't named Bill, or this could get mighty confusing.
To me, it looks like Eric is right for the wrong reason. When Susan
says $4.50, Sonny should answer with the guess that maximizes the
amount in HIS range. It works out to be the same.
One could devise a computer game to simulate this. Have a random
"bill generator" select a number N, and two intervals (A, A+100)
and (B, B+100) where N is contained in both intervals.
Your opponent guesses N, then you get to guess N after seeing his guess.
I estimate the person who goes second will win 2/3 of the time.
Shouldn't be that tough to work thru the math if someone wanted to.
John
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909.4 | More abstract games. | PBSVAX::COOPER | Topher Cooper | Fri Jul 29 1988 15:42 | 49 |
| Let's look at a different game.
We have a source of random real numbers, following a "well-behaved"
distribution with a well defined but unknown mean. You can picture the
distribution as normal if you wish. The game is to guess the mean
as accurately as possible. Player 1, is given a single output,
while player 2 is given two outputs from the source. Who has the
advantage? Clearly player 2 does. By averaging the two numbers
player 2 can, on the average, do better in the game by a factor
of about sqrt(2) (exactly that if normal). Player 1's best strategy
is to guess the number given to him/her.
What if one of the numbers given to 2 is the same as the one given
to 1? Does that change anything? Answer: no, not if the goal is
to guess the mean as accurately as possible. But it *does* change
things if the goal is to simply guess the mean more accurately than
the opponent and if player 2 knows which number is the same as that
received by player 1.
Let's assume that the second sample, and thus the average of the
two samples, is larger than the first sample. If player 2 guesses
the average of the two samples, (s)he wins if the actual result
is greater than that average, or if the actual result is greater
than the half-way point between his/her guess and player 1's guess.
By moving his/her guess down toward player 1's guess, the size
of that latter region is increased without losing any other area.
So the best strategy for player 2 is to make a guess as close as
possible to player 1's guess on the same side as the average of
the two samples, which is the same (since one sample is the same
as player 1's best guess) as the side of player 1's guess that
player 2's second sample is on.
Except for the fact that the game involves a discrete source of
random variables, rather than continuous, this latter game is
equivalent to the game in .0 assuming that the two players are
approximately evenly matched, and they are both "unbiased" guessers
at the price. (The former assumption doesn't really make any
difference if violated, but it complicates the analysis; the
latter adds an extra factor to the ideal strategy if the bias
is known). We can view the two players as sources of random numbers
with some distribution around the mean of the actual price. The
first player has their own a priori guess as a sample from that
source, while the second player has the first player's guess plus
their own guess made before they heard the first player's.
The second player has a clear advantage, because they have more
information.
Topher
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909.5 | When player 1 does have an advantage. | PBSVAX::COOPER | Topher Cooper | Fri Jul 29 1988 17:44 | 17 |
| I should add that there are circumstances, unlikely to apply in
the actual game, where the first player does have an advantage.
If the accuracy of the guesses is sufficiently high (variance of
the sampling sufficiently small) then there is a good chance that
the first player can guess the right amount exactly (this is possible
because the distribution is discrete; for a continuous distribution
the probability of getting *exactly* the right answer is zero).
The first player therefore has the advantage of being able to block
the second player from guessing the exact answer. To take advantage
of this the first player needs to get "right on the money" fairly
often. The exact accuracy needed depends (I think sensitively)
to assumptions about the distribution of guesses around the actual
value.
Topher
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909.6 | Mom gets the bad deal. | HPSTEK::XIA | | Fri Jul 29 1988 18:59 | 17 |
| I agree with the original noter that the second guesser has the
advantage. The reason is when it is time for the son to make a
guess, he has the extra informtion of what the mother's estimate
of the bill is. The reason there is a controvercy is because
the problem is not really well formulated in the original note;
hence, is open to different intepretations.
My assumptions are:
1. The mother believes that the distribution function for the bill is
a bell curve with maximum at $4.50.
2. The son believes that the distribution function for the bill
is a bell curve (same shape as his mother's) with maximum at $4.00.
3. The distribution of the actual bill is a bell curve with maximum
at $4.25.
Eugene
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909.7 | | CLT::GILBERT | | Fri Jul 29 1988 20:30 | 12 |
| re .2:
Hmmm. I gues I wasn't clear enough. In the following,
> P( B > X ) <= 1/2, and P( B < X ) <= 1/2.
>
> Surely there is such an X. (why?)
X is the value that's *chosen*, and B is the random variable.
I've started on a formal proof. Doesn't anybody else see why this works?
Moral support requested.
|
909.8 | | HPSTEK::XIA | | Fri Jul 29 1988 22:19 | 3 |
| I thought about the problem again, and I retract .6 . I need to
think about it more (perhaps not :-).
Eugene
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909.9 | Can you quantify intuition ? | CVG::PETTENGILL | mulp | Sun Jul 31 1988 01:03 | 7 |
| I would restate the competition to be as follows:
Player 1 guesses the bill
Player 2 guesses whether the bill is high or low
An algorithmic solution for player 2 is almost identical to that for player 1,
but in my opinion an intuitive solution is much easier for player 2.
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909.10 | Maybe we could arrange a bet... | CHALK::HALLYB | The smart money was on Goliath | Mon Aug 01 1988 13:09 | 10 |
| .7> Hmmm. I gues I wasn't clear enough. In the following,
.7>
.7>> P( B > X ) <= 1/2, and P( B < X ) <= 1/2.
.7>>
.7>> Surely there is such an X. (why?)
There is such an X since CDFs are continuous monotonics defined everywhere.
But the first guesser doesn't know P().
John
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909.11 | | BEING::POSTPISCHIL | Always mount a scratch monkey. | Tue Aug 02 1988 19:00 | 13 |
| Re .7:
Okay, you have moral support.
However, I think we need some more discussion about the knowledge each
of the players has. If the first player's believed distribution for
the price does not correspond to the actual distribution, they cannot
find the optimal guess that your proof requires. Can the second player
take advantage of that even if they do not have better information
about the distribution?
-- edp
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909.12 | Player 2 has the advantage. | PBSVAX::COOPER | Topher Cooper | Wed Aug 03 1988 20:24 | 88
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