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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.Quote:BleedingChipsSlowly[To Sonuvabish:] Are you in?

Quote:SonuvabishI 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!

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

The post that started it all:

I believe the sentence:Quote:WizardThis 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

Is best interpreted as:Quote:WizardYou randomly draw one coin, and observe one side only, which is white.

"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:

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.

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.

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.Quote:kubikulannTo 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 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.

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.

Quote:dwheatleyFor 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!

Ah, you have done well grasshopper! The fact that one person has changed their thinking makes all my postings worthwhile.Quote:1call2manyYou 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!

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.Quote:endermikeI 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:I believe the sentence:Quote:WizardThis 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

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.