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BleedingChipsSlowly
BleedingChipsSlowly
Joined: Jul 9, 2010
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May 5th, 2014 at 8:05:17 PM permalink
Quote: BleedingChipsSlowly

[To Sonuvabish:] Are you in?

Well now, I'll have to wait a week for your response I see. Anyone else is welcome to sit down for the same terms.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
1call2many
1call2many
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May 6th, 2014 at 5:38:26 AM permalink
Quote: Sonuvabish

I read it, but I don't agree. If you are saying you pull both coins out and look only at one, which is always white, then randomly guess which one of three sides is white, then there is a 2/3 chance you are right. Regardless, your thinking is obviously superior to most of the forum!



You may not agree but the 100 trials I did do agree (at least with the multi trial format), I was an avid 1/2 believer until I did the trials and changed to the 2/3 camp. P.S don't take bleedingchips challenge he will bleed your chips at a rate of $6 for him to every $5 you win!
endermike
endermike
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May 6th, 2014 at 5:59:36 AM 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.

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:
Quote: Wizard

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

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.

Understanding it this way there are a variety of explanations in this thread which show 2/3rds to be the correct answer to the question asked. My personal favorite is quoted below:

2/3. There are three coin-sides that are white, and black appears on the other side on only one of them. Therefore white appears on the other two sides.

This is equivalently demonstrated, but perhaps more intuitively, with a four-sided top (e.g. a dreidel) with three white sides and one black side. What are the chances that a spin of such a top, if it lands on a white side, has a white side face down? The same logic as above applies: there are three ways for a white side to land face-up, and in two of those, another white side will be face down.

The equivalence between the two coins and the four-sided top is as follows: label the black side of the coin 1 and the opposite white side 3. Label the white/white coin 2/4. Then label the black side of the top 1 and the other three sides 2/3/4.


Imagine we are going to pull out one coin and then the other and note the colors of their sides. When we pull out the first coin, we note the color of the side facing us (W for white and B for Black). Next, before pulling the second coin, we flip over the first and note the color of the second side (W or B). Now we pull the second coin and repeat the noting of side colors with the second coin (WB, WW, or BW).

This process gves us four possible four letter sequences: WWWB, WWBW, WBWW, and BWWW (full enumeration of the sample space). In the absence of additional information (knowledge the color of a side for any pull) those sequences clearly have equal probability, 1/4th. The probability question "What is the probability the other side of that coin is white?" Is now equivalent to what is the chance W is in the second position (WWWB, WWBW, or BWWW).

When we are given that the first side observed is "W," that eliminates the possibility of BWWW from the sample space. That then leaves the sample space as WBWW, WWBW, and WWWB. Now the answer is clearly the chance of WWWB and WWBW out of WWWB, WWBW, and WWBW, given all 3 events are equally likely. Clearly the answer in this case is 2 out of 3.
endermike
endermike
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May 6th, 2014 at 6:07:09 AM permalink
Quote: kubikulann

To play the lawyer: it does not say how you choose the side you observe. Everybody assumed it was at random. But it could be a "Monty Hall"-like problem, where the operator deliberately chooses the white side to show to you, in the event you drew the bicolor coin.

Then the answer is not the same.

This was an interesting observation. While I think any plain text reading of the problem in the OP or the link makes it clear that the observed side should be at random, this was an interesting tangent. Thank you.
1call2many
1call2many
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May 6th, 2014 at 7:33:24 AM permalink
.Sorry I messed up see below for correction I hope
1call2many
1call2many
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May 6th, 2014 at 7:38:53 AM permalink
Imagine we are going to pull out one coin and then the other and note the colors of their sides. When we pull out the first coin, we note the color of the side facing us (W for white and B for Black). Next, before pulling the second coin, we flip over the first and note the color of the second side (W or B). Now we pull the second coin and repeat the noting of side colors with the second coin (WB, WW, or BW).

This process gves us four possible four letter sequences: WWWB, WWBW, WBWW, and BWWW (full enumeration of the sample space). In the absence of additional information (knowledge the color of a side for any pull) those sequences clearly have equal probability, 1/4th. The probability question "What is the probability the other side of that coin is white?" Is now equivalent to what is the chance W is in the second position (WWWB, WWBW, or BWWW).

When we are given that the first side observed is "W," that eliminates the possibility of BWWW from the sample space. That then leaves the sample space as WBWW, WWBW, and WWWB. Now the answer is clearly the chance of WWWB and WWBW out of WWWB, WWBW, and WWBW, given all 3 events are equally likely. Clearly the answer in this case is 2 out of 3.



Endermike, I thought the problem states you only draw one coin for possible letter sequences of WW WW( if coin 1) and WB BW (if coin 2) since each coin is equally likely to be drawn (1/2 chance), and white was observed, the combinations are WW if coin 1 or WB if coin 2. The options are not WW, WW (coin 1)and WB (coin 2) or else you are saying coin 1 is twice as likely to be drawn as coin 2.

Note that the four combinations you refer to come into play when you perform multiple trials because you have 3 ways to see white first with two of those ways having W on the flip side. In the OP you only have two ways to see the white first, either on coin 1 or coin 2 with equal chance. Not two ways on coin 1 and one way on coin 2.
dwheatley
dwheatley
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May 6th, 2014 at 7:38:54 AM permalink
For anyone that thinks it's 1/2, try it backwards:

You randomly draw a coin, and observe one side. It's black. What is the probability you drew the white/white coin? If you say 1/2, I will reach through the internet and slap you.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
1call2many
1call2many
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May 6th, 2014 at 7:54:52 AM permalink
Quote: dwheatley

For anyone that thinks it's 1/2, try it backwards:

You randomly draw a coin, and observe one side. It's black. What is the probability you drew the white/white coin? If you say 1/2, I will reach through the internet and slap you.



Thank you. The answer is obviously zero! Once information is gained (seeing white) the chances do change that it is a particular coin. once you see white you are twice a likely to have drawn the WW coin, even though the orinal selection was 50/50.

Please disregard my above post to endermike thinking a one trial answer was different to a multiple trial approach. Firmly back on the 2/3 bandwagon. I have flipped more times than the coins in this problem!
BleedingChipsSlowly
BleedingChipsSlowly
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May 6th, 2014 at 5:41:50 PM permalink
Quote: 1call2many

You may not agree but the 100 trials I did do agree (at least with the multi trial format), I was an avid 1/2 believer until I did the trials and changed to the 2/3 camp. P.S don't take bleedingchips challenge he will bleed your chips at a rate of $6 for him to every $5 you win!

Ah, you have done well grasshopper! The fact that one person has changed their thinking makes all my postings worthwhile.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
BleedingChipsSlowly
BleedingChipsSlowly
Joined: Jul 9, 2010
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May 6th, 2014 at 6:08:33 PM permalink
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.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia

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