5 views

Skip to first unread message

Jul 22, 2002, 8:28:16 PM7/22/02

to

==> decision/high.or.low.p <==

I pick two numbers, randomly, and tell you one of them. You are

supposed to guess whether this is the lower or higher one of the two

numbers I picked. Can you come up with a method of guessing that does

better than picking the response "low" or "high" randomly (i.e.

probability to guess right > .5) ?

==> decision/high.or.low.s <==

Pick any cumulative probability function P(x) such that a > b ==> P(a)

> P(b). Now if the number shown is y, guess "low" with probability

P(y) and "high" with probability 1-P(y). This strategy yields a

probability of > 1/2 of winning since the probability of being correct

is 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)), which is > 1/2 by

assumption.

I pick two numbers, randomly, and tell you one of them. You are

supposed to guess whether this is the lower or higher one of the two

numbers I picked. Can you come up with a method of guessing that does

better than picking the response "low" or "high" randomly (i.e.

probability to guess right > .5) ?

==> decision/high.or.low.s <==

Pick any cumulative probability function P(x) such that a > b ==> P(a)

> P(b). Now if the number shown is y, guess "low" with probability

P(y) and "high" with probability 1-P(y). This strategy yields a

probability of > 1/2 of winning since the probability of being correct

is 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)), which is > 1/2 by

assumption.

#1. 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)) <-- that doesn't

follow.

#2. P(b) - P(a) < 0, by assumption.

#3. 1/2 * ((1 - P(a) + P(b)) is not the right probability equation.

Shouldn't it be 1/2 * ((1-P(a) + P(a))??

Jul 23, 2002, 4:59:50 AM7/23/02

to

Siqi Chen (aznb...@yahoo.com) wrote:

: ==> decision/high.or.low.p <==

: ==> decision/high.or.low.p <==

: I pick two numbers, randomly, and tell you one of them. You are

: supposed to guess whether this is the lower or higher one of the two

: numbers I picked. Can you come up with a method of guessing that does

: better than picking the response "low" or "high" randomly (i.e.

: probability to guess right > .5) ?

: supposed to guess whether this is the lower or higher one of the two

: numbers I picked. Can you come up with a method of guessing that does

: better than picking the response "low" or "high" randomly (i.e.

: probability to guess right > .5) ?

: ==> decision/high.or.low.s <==

: Pick any cumulative probability function P(x) such that a > b ==> P(a)

: > P(b). Now if the number shown is y, guess "low" with probability

: P(y) and "high" with probability 1-P(y). This strategy yields a

: probability of > 1/2 of winning since the probability of being correct

: is 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)), which is > 1/2 by

: assumption.

: #1. 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)) <-- that doesn't

: follow.

Sure it does.

You are right that the solution is confused, though. Not only does it

seem to suddenly switch a and b, but it also has the probability of

guessing "low" increase as the number increases, which is clearly wrong.

The following would be better:

==> decision/high.or.low.s <==

Pick any cumulative probability function P(x) such that a > b ==> P(a)

> P(b). Now if the number shown is y, guess "high" with probability

P(y) and "low" with probability 1-P(y). This strategy yields a

probability > 1/2 of winning since the probability of being correct

is 1/2 * (P(a) + (1-P(b))) = 1/2 + (P(a)-P(b)), which is > 1/2 by

assumption.

Jonathan Dushoff

dushoff at eno.princeton.edu

Jul 23, 2002, 6:23:53 AM7/23/02

to

On Tue, 23 Jul 2002 08:59:50 +0000 (UTC), dus...@gate.sinica.edu.tw

(Jonathan Dushoff) wrote:

(Jonathan Dushoff) wrote:

>

>: #1. 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)) <-- that doesn't

>: follow.

>

>Sure it does.

No it doesn't

1/2 * ((1-P(a)) + P(b)) = 1/2 + 1/2 * (P(b)-P(a))

which leads to the same conclusion.

It looks to me like the puzzle statement is wrong thoug. Instead of

"I pick two numbers, randomly, and tell you one of them."

it should start

"I pick two numbers and tell you one of them, chosen randomly."

--

Nis Jorgensen

Amsterdam

Please include only relevant quotes, and reply below the quoted text. Thanks

Jul 23, 2002, 3:45:35 PM7/23/02

to

It's still wrong. The correct strategy should be the opposite of what

the solution said. The correct strategy should be guess high with

probability F(y), where F(y) is any CPF function and y is the number

shown.

the solution said. The correct strategy should be guess high with

probability F(y), where F(y) is any CPF function and y is the number

shown.

It said in the given answer that the optimal strategy is to guess low

with probability F(y) and high with probability 1 - F(y).

Now, to find the value for the probability of a correct guess using

our

method, we use the following, correct?:

P(correct guess) = P(we were shown the higher number H) * P(we guessed

"high" given H) + P(we were shown the lower number L) * P(we guessed

"low"

given L)

=> (plugging in from the original selection method)

P(correct guess) = (1/2) * (1 - F(H)) + (1/2) * F(L)

=>

P(correct guess) = (1/2) (1 - (F(H) - F(L)))

=> Since F(H) - F(L) > 0 (by assumption)

P(correct guess) = (1/2) (1 - (a positive value))

=>

Therefore,

P(correct guess) < 1/2.

The method given, as you can see, yields a worth than .5 chance of

being

correct. It's easy to see that only if we reverse our guessing

strategy, we

get a > .5 chance of being correct.

Jul 24, 2002, 12:18:29 AM7/24/02

to

In article <ahj5u6$1g9m$1...@news1.sinica.edu.tw>,

Jonathan Dushoff <dus...@gate.sinica.edu.tw> wrote:

>Siqi Chen (aznb...@yahoo.com) wrote:

>: #1. 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)) <-- that doesn't

>: follow.

>

> Sure it does.

Jonathan Dushoff <dus...@gate.sinica.edu.tw> wrote:

>Siqi Chen (aznb...@yahoo.com) wrote:

>: #1. 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)) <-- that doesn't

>: follow.

>

> Sure it does.

[ ... ]

>==> decision/high.or.low.s <==

> Pick any cumulative probability function P(x) such that a > b ==> P(a)

> > P(b). Now if the number shown is y, guess "high" with probability

> P(y) and "low" with probability 1-P(y). This strategy yields a

> probability > 1/2 of winning since the probability of being correct

> is 1/2 * (P(a) + (1-P(b))) = 1/2 + (P(a)-P(b)), which is > 1/2 by

> assumption.

This still doesn't fix what I assume Siqi was objecting to, namely:

1/2*(P(a) + (1 - P(b))) = 1/2 + (P(a) - P(b)) / 2

not what is written above. The conclusion is unaffected, of course.

Cheers,

Geoff.

-----------------------------------------------------------------------------

Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --

ft...@maths.usyd.edu.au | Gameplayer by vocation.

-----------------------------------------------------------------------------

Jul 25, 2002, 5:01:38 AM7/25/02

to

>

> This still doesn't fix what I assume Siqi was objecting to, namely:

>

> 1/2*(P(a) + (1 - P(b))) = 1/2 + (P(a) - P(b)) / 2

>

> not what is written above. The conclusion is unaffected, of course.

>

> This still doesn't fix what I assume Siqi was objecting to, namely:

>

> 1/2*(P(a) + (1 - P(b))) = 1/2 + (P(a) - P(b)) / 2

>

> not what is written above. The conclusion is unaffected, of course.

>

No, I disagree. See my "decision/high.or.low" thread. The conclusion

is wrong in that the correct strategy should be the exact opposite of

what was posted.

Jul 25, 2002, 8:28:06 AM7/25/02

to

Siqi Chen (aznb...@yahoo.com) wrote:

: No, I disagree. See my "decision/high.or.low" thread. The conclusion

: is wrong in that the correct strategy should be the exact opposite of

: what was posted.

No, we all agree. Geoff/Fred meant that the conclusion of the

(partially) corrected version he was responding to was unaffected by the

fact that I dropped a factor of 1/2 from one of the terms.

We all agree that the archive version is backwards. The question is,

does anybody know how it can be fixed?

Jul 25, 2002, 3:46:07 PM7/25/02

to

> No, we all agree. Geoff/Fred meant that the conclusion of the

> (partially) corrected version he was responding to was unaffected by the

> fact that I dropped a factor of 1/2 from one of the terms.

> (partially) corrected version he was responding to was unaffected by the

> fact that I dropped a factor of 1/2 from one of the terms.

Ahh, you're right. My fault.

Reply all

Reply to author

Forward

0 new messages

Search

Clear search

Close search

Google apps

Main menu