Poll

1 vote (1.72%)
No votes (0%)
14 votes (24.13%)
39 votes (67.24%)
No votes (0%)
1 vote (1.72%)
No votes (0%)
1 vote (1.72%)
No votes (0%)
2 votes (3.44%)

58 members have voted

FinsRule
FinsRule
Joined: Dec 23, 2009
  • Threads: 123
  • Posts: 3782
May 6th, 2014 at 6:34:12 PM permalink
This is ridiculous:

There are 4 sides on two coins:

Side 1 on Coin 1
Side 2 on Coin 1

Side 3 on Coin 2
Side 4 on Coin 2

------------------------
Side 1 is B for Black
Side 2 is W for White

Side 3 is W for White
Side 4 is W for White

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

There are 4 possibilities:

B / W
W / B
W / W
W / W

When you draw, take away the B/W possibility. That means there are 3 left, and one has the B on the other side.

2/3. Done.
BleedingChipsSlowly
BleedingChipsSlowly
Joined: Jul 9, 2010
  • Threads: 21
  • Posts: 962
May 6th, 2014 at 6:37:35 PM permalink
I have no argument about the logic you posted. The argument I have is your logic doesn't address the OP question.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
endermike
endermike
Joined: Dec 10, 2013
  • Threads: 7
  • Posts: 584
May 6th, 2014 at 7:08:03 PM permalink
Quote: BleedingChipsSlowly

I have no argument about the logic you posted. The argument I have is your logic doesn't address the OP question.

Quite simply, you are wrong. The intent of the question is as a conditional probability question.

1) A plain text reading of the OP shows that:
"You randomly draw one coin, and observe one side only, which is white." This clearly is meant to say that all sides are equally probable, but you know that the black one was not seen. If you choose to ignore the color information, the sentence construction is less logical than standard English. "You randomly draw one coin." Would be the reasonable way of writing what your interpretation requires.

2) As does the wizard's thoughts on the most common solution in this thread:
Quote: Wizard

I forgot that I posted this before. Evidently this is a bright group, as the majority are right.

3) As does his answer on the mathproblems site.

Maybe you do not agree. In that case, best of luck with your interpretation, I will agree to disagree.
kenarman
kenarman
Joined: Nov 22, 2009
  • Threads: 28
  • Posts: 966
May 6th, 2014 at 7:33:28 PM permalink
Quote: BleedingChipsSlowly

Quote: endermike

I do agree that issue has caused significant disagreement. However, I take issue with your suposition that the questions asked was the one you answered in the first spoiler. I belive the question asked was the one you addressed in your second spoiler.

The post that started it all:

Quote: Wizard

This is by far the most controversial problem on my mathproblems. I don't post it here because I don't know the answer, but for the edification and education of the rest of the forum.

There are two coins in a bag. One is white on one side and black on the other. The other is white on both sides.

You randomly draw one coin, and observe one side only, which is white.

What is the probability the other side of that coin is white?

Please put solutions in

Have a nice day
.

I believe the sentence:
Is best interpreted as:
"You randomly draw one coin from the bag. You observe nothing about the coin you did not draw. The coin you do draw, you pull out in a closed fist (so you can't observe anything about the coin visually). When you open your hand the only side visible is white." I believe it is meant to tell you that you only observe a side of coin and nothing more.

What the bloody HELL!! "Best interpreted as?" How about sticking to what was stated. Clearly the OP logical proposition is based on a coin having already been selected. If The Wizard intended otherwise his statement of the problem is on error. I sense I sharp intake of breath among the forum readers: yes, I am calling out The Wizard. To illustrate the difference, consider my bar bet proposition modified so that the bets are placed AFTER a coin has been drawn and a white side is observed. In that case I'm still willing to go head-to-head with any takers - provided I get to wager the $5 the flip side being black against a $3 wager it is white.



I am willing to take that bet for anything over 20 trials. Trials to be independently verified.
Be careful when you follow the masses, the M is sometimes silent.
BleedingChipsSlowly
BleedingChipsSlowly
Joined: Jul 9, 2010
  • Threads: 21
  • Posts: 962
May 6th, 2014 at 7:53:42 PM permalink
Quote: endermike

Quite simply, you are wrong. The intent of the question is as a conditional probability question.

1) A plain text reading of the OP shows that:
"You randomly draw one coin, and observe one side only, which is white." This clearly is meant to say that all sides are equally probable, but you know that the black one was not seen. If you choose to ignore the color information, the sentence construction is less logical than standard English. "You randomly draw one coin." Would be the reasonable way of writing what your interpretation requires.

2) As does the wizard's thoughts on the most common solution in this thread:
3) As does his answer on the mathproblems site.

Maybe you do not agree. In that case, best of luck with your interpretation, I will agree to disagree.

We agree: The intention is clear but the problem statement is flawed. Perhaps this problem would not have been so controversial if stated properly. This is The Wizard's world so I can understand why you would pitch your tent in his camp. I have no compunction stating "The Wizard has no clothes" for this one.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
MathExtremist
MathExtremist
Joined: Aug 31, 2010
  • Threads: 88
  • Posts: 6526
May 6th, 2014 at 8:22:51 PM permalink
Quote: BleedingChipsSlowly

Posters are answering two questions on this thread: the one asked and one that is pretty close but not the one asked.


The question asked:

You randomly draw one coin and observe one side only which is white. What is the probabilty that the other side of that coin is white? [My emphasis.]

The question many posters thnk was asked but was not:

If you randomly draw a coin and observe one side only as white, what is the probability the other side will be white?


I'm not following. What do you think is the substantive difference in information between these two questions, such that the answer would be different? What's the difference between "one coin" and "a coin"?
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1404
  • Posts: 23738
May 6th, 2014 at 8:34:17 PM permalink
Quote: BleedingChipsSlowly

To illustrate the difference, consider my bar bet proposition modified so that the bets are placed AFTER a coin has been drawn and a white side is observed. In that case I'm still willing to go head-to-head with any takers - provided I get to wager the $5 the flip side being black against a $3 wager it is white.



I'll take that bet.
It's not whether you win or lose; it's whether or not you had a good bet.
sodawater
sodawater
Joined: May 14, 2012
  • Threads: 64
  • Posts: 3321
May 6th, 2014 at 8:36:32 PM permalink
I haven't read this thread at all, but are there really seven pages of debate over the most well known conditional probability problem of all time?
BleedingChipsSlowly
BleedingChipsSlowly
Joined: Jul 9, 2010
  • Threads: 21
  • Posts: 962
May 6th, 2014 at 8:52:29 PM permalink
Quote: Wizard

Quote: BleedingChipsSlowly

To illustrate the difference, consider my bar bet proposition modified so that the bets are placed AFTER a coin has been drawn and a white side is observed. In that case I'm still willing to go head-to-head with any takers - provided I get to wager the $5 the flip side being black against a $3 wager it is white.



I'll take that bet.

And you would win, oh Great and Mighty Wizard! We are not worthy! Mea culpa, I love the taste of crow. There is a 1/2 chance that the flip side of a coin will be black, but the double-white coin is in play twice as often.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
mustangsally
mustangsally
Joined: Mar 29, 2011
  • Threads: 25
  • Posts: 2463
May 6th, 2014 at 8:55:18 PM permalink
Quote: sodawater

I haven't read this thread at all, but are there really seven pages of debate over the most well known conditional probability problem of all time?

maybe
seems like the basic problem for many is defining the sample space

I see to start
3 sides = white
1 side = black

the condition is
now I see 1 white side

that leaves 2 white sides I have not yet seen
and
1 black side I have not yet seen
this = 3 total sides not seen

1 side = black
so the probability of the other side being black = PB =
1 side / 3 total sides not seen

probability of the other side being white = 1 - PB

Sally
I Heart Vi Hart

  • Jump to: