Two coin puzzle

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October 18th, 2013 at 5:09:33 PM permalink

My guess...

I voted 1/2.
There are only 2 coins and by flipping the selected coin over once (1 time) you now have created a 1 in 2 chance that the other side is white.

I know you posted this one a while back but it was a bit different, with heads and tails and one coin having two heads or something like this. I didn't remember the answer which is why I am participating here and giving my honest guess.

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ThatDonGuy

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October 18th, 2013 at 5:12:12 PM permalink

Label the sides on the all-white coin A and B, and the white side on the other coin C.

You have an equal chance of seeing A, B, or C.

If it is A or B, the other side is white; if it is C, the other side is black. 2/3.

Buzzard

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October 18th, 2013 at 5:13:14 PM permalink

Gee, I am math illiterate and even I know this one. Of course I has to substitute a,b,c&d to solve it. But I did.

Shed not for her the bitter tear Nor give the heart to vain regret Tis but the casket that lies here, The gem that filled it Sparkles yet

MathExtremist

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October 18th, 2013 at 5:20:45 PM permalink

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
.

You should rephrase that second sentence.

But here's the

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.

"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

HotBlonde

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October 18th, 2013 at 5:21:22 PM permalink

You forgot the option

"I am Jerry Logan"

lol

OFFICIALLY and justifiably reclaimed my title as SuperHotBlonde!

HotBlonde

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October 18th, 2013 at 5:35:46 PM permalink

Quote: MathExtremist
But here's the

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.

I think you are wrong and this is why...

When you have selected 1 coin the other coin is now disregarded and has no part in this any more. The other coin has now been eliminated and we're talking about just the one coin. Since we know both coins have one side that is always going to be white (and this is what we ended up observing when we selected a coin showing the white side) we now know that it is only a 50/50 chance that the other side is either going to be black or it's going to be white. Which is why the answer is 1/2.

Demonstration:

Coin 1

White Side
X

Black Side
still remaining

Coin 2

White Side
X

White Side
still remaining

See, once we select 1 coin and we see that it is white, the only options we have left are that the other side is either black or white. 1 white out of 2 options remain.

OFFICIALLY and justifiably reclaimed my title as SuperHotBlonde!

s2dbaker

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October 18th, 2013 at 5:37:55 PM permalink

Quote: Wizard
What is the probability the other side of that coin is white?

This sounds familiar so if I get it right, it's not because of any genius. It's probably rotting in some memory bank in my brain somewhere. Let's see if I can pull it out intact.

There are four possibilities:

White Heads/White Tails
White Tails/White Heads
White/Black
Black/White

We discard all possibilities that have us draw the black side of the coin so:

White Heads/White Tails
White Tails/White Heads
White/Black
Black/White

That leaves three allowed possibilities:

White Heads/White Tails
White Tails/White Heads
White/Black

So the answer is 2/3

Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez

Buzzard

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October 18th, 2013 at 5:39:21 PM permalink

Hot Blonde once again proves she is a natural blonde.

Shed not for her the bitter tear Nor give the heart to vain regret Tis but the casket that lies here, The gem that filled it Sparkles yet

s2dbaker

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October 18th, 2013 at 5:45:40 PM permalink

Quote: HotBlonde
I think you are wrong and this is why...
When you have selected 1 coin the other coin is now disregarded and has no part in this any more. The other coin has now been eliminated and we're talking about just the one coin. Since we know both coins have one side that is always going to be white (and this is what we ended up observing when we selected a coin showing the white side) we now know that it is only a 50/50 chance that the other side is either going to be black or it's going to be white. Which is why the answer is 1/2.

Demonstration:

Coin 1

White Side
X

Black Side
still remaining

Coin 2

White Side
X

White Side
still remaining

See, once we select 1 coin and we see that it is white, the only options we have left are that the other side is either black or white. 1 white out of 2 options remain.

You left out one important item HB

When you draw the coin that is black on one side, then you are discarding its result half of the time. That then weights the coin with two white sides more heavily.

Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez

MathExtremist

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October 18th, 2013 at 5:52:47 PM permalink

Quote: HotBlonde

See, once we select 1 coin and we see that it is white, the only options we have left are that the other side is either black or white. 1 white out of 2 options remain.

If the dreidel example didn't convince you, let's color the sides. Instead of white, white, white, and black, make it red, yellow, blue, and black. Except you don't know which of the colors are on the opposite side of the black side, and there's a gremlin in the bag that repaints the three colors after each turn. The question is now:
You reach into a bag and pull out a coin, colored side up. What is the probability that the other side is colored?

This is exactly the same problem. Does your answer change?

"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

Mission146

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October 18th, 2013 at 6:10:17 PM permalink

Simple Effect of Removal.

If there were three coins that were white on both sides and one coin that was black on both sides, then having drawn a white coin, the probability of drawing another white coin is 2/3.

The probability of drawing a coin white side up is 3/4, having done that, you've eliminated one White side and one possibility, leaving 2/3.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

Mosca

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October 18th, 2013 at 6:31:03 PM permalink

There are four possible sides you could be looking at. Since you know white, you could be looking at the top of the w/w coin, the bottom of the w/w coin, or the top of the w/b coin. So, 2/3..

A falling knife has no handle.

Wizard
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October 18th, 2013 at 6:43:40 PM permalink

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

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Dalex64

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October 18th, 2013 at 7:58:17 PM permalink

I think it is 1/2 because of how the question is phrased. The drawn coin always has a white side up/on top. You grabbed either the coin with a black bottom, or you grabbed the coin with the white bottom. 50/50

98Clubs

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October 18th, 2013 at 8:43:51 PM permalink

I don't understand the controversey, hammer the problem using random and discard all non-white initial selections.

2/3 because there are 4 sides on 2 coins, one side of 1 coin is not white. By selecting a white side of any coin there are three remaining possibilities according to the number of sides remaining. One coin will be not white on the other side, and one coin has two possible permutations, both solving as white. therefore the chance of the other side of either coin being white when viewing the white side is 2 in 3.

Some people need to reimagine their thinking.

HotBlonde

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October 18th, 2013 at 9:29:12 PM permalink

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

Aw, phewy! :-P

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seattledice

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October 18th, 2013 at 9:30:32 PM permalink

50% - it's a given you pulled a coin and looked at a white side so you no longer have to consider probabilities about the first coin. There are only two choices left - the other side is either white or black.

- edit - Oops

MrCasinoGames

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October 19th, 2013 at 2:01:18 AM permalink

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 think To make the puzzle more fun and more difficult, the question may be put down as:
There are 3 coins in a bag. One is white on one side and black on the other. The other 2 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?

Stephen Au-Yeung (Legend of New Table Games®) NewTableGames.com

98Clubs

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October 19th, 2013 at 2:09:59 AM permalink

4 in 5.

Some people need to reimagine their thinking.

SOOPOO

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October 19th, 2013 at 2:48:09 AM permalink

On August 8, 2010, I proposed essentially this very question, which is my favorite 'simple problem'. I titled the thread, SIMPLE PROBLEM. It goes something like this.... Assumption... boys and girls are born with equal frequncies.

Question.... A woman has two children, (not identical twins). One of them is a boy, what are the odds that the other one is a boy? Of course the answer is 1/3, but you can imagine that virtually everyone you ask will blurt out 1/2.

Dalex64

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October 19th, 2013 at 9:26:30 AM permalink

I finally figured it out in a way that I understand, to get the answer of 2 in 3.
You pull a coin with the white side facing up out of the bag.
you are either looking at:

the only white side of the white/black coin. the bottom is black.
one white side of the white/white coin. the bottom is white.
the other white side of the white/white coin. the bottom is white.

ThatDonGuy

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October 19th, 2013 at 2:18:41 PM permalink

Quote: SOOPOO
On August 8, 2010, I proposed essentially this very question, which is my favorite 'simple problem'. I titled the thread, SIMPLE PROBLEM. It goes something like this.... Assumption... boys and girls are born with equal frequncies.

Question.... A woman has two children, (not identical twins). One of them is a boy, what are the odds that the other one is a boy? Of course the answer is 1/3, but you can imagine that virtually everyone you ask will blurt out 1/2.

That's because the question can be interpreted in different ways. If, by "one of them is a boy," you are referring to a specific child (e.g. the older of the two), then the answer is 1/2. This is what causes the Monty Hall Problem to be so confusing - the intended answer assumes that, when you say, "Monty opens a door and shows (something besides the grand prize)," his choice of door was arbitrary (to be sure it would not be the prize) and not random.

Also, when you say, "One of them is a boy," it can be inferred that "only one of them is a boy," in which case the answer is zero.

dwheatley

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October 19th, 2013 at 2:38:19 PM permalink

I was asked this question during an interview at Goldman Sachs.

Wisdom is the quality that keeps you out of situations where you would otherwise need it

pacomartin

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October 20th, 2013 at 6:52:21 AM permalink

The question "What is the probability that you have chosen a two color coin or a single color coin?" is compelling, but irrelevant. But it is not the question.

SOOPOO

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October 20th, 2013 at 7:37:06 AM permalink

Quote: ThatDonGuy
That's because the question can be interpreted in different ways. If, by "one of them is a boy," you are referring to a specific child (e.g. the older of the two), then the answer is 1/2. This is what causes the Monty Hall Problem to be so confusing - the intended answer assumes that, when you say, "Monty opens a door and shows (something besides the grand prize)," his choice of door was arbitrary (to be sure it would not be the prize) and not random.

Also, when you say, "One of them is a boy," it can be inferred that "only one of them is a boy," in which case the answer is zero.

I don't know why you would think when I say 'one of them is a boy' that would in any way infer I was talking about a specific child. Quite the opposite, that is why I specifically used the words I did. And if I meant the facts to be that "only' one of them was a boy, I would have used the word 'only'.
I stand by the question.... and the answer....

Wizard
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October 20th, 2013 at 8:16:18 AM permalink

Quote: MrCasinoGames
I think To make the puzzle more fun and more difficult, the question may be put down as:
There are 3 coins in a bag. One is white on one side and black on the other. The other 2 is white on both sides.

That is the way many people tell it. However, it is easy to eliminate the all-black coin as a possibility. Adding it seems to be unnecessarily adding useless information.

Quote: SOOPOO
Question.... A woman has two children, (not identical twins). One of them is a boy, what are the odds that the other one is a boy? Of course the answer is 1/3, but you can imagine that virtually everyone you ask will blurt out 1/2.

It is asked that way a lot too. Actually, it is a slightly different problem. Both can still be solved simply using Bayes.

If you go this way I think you should make it more clear it is a random woman drawn out of the population of all women with two children, except those with two girls.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

pacomartin

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October 20th, 2013 at 9:01:08 PM permalink

Quote: Wizard
I think you should make it more clear it is a random woman drawn out of the population of all women with two children, except those with two girls.

I would say that the answer is 1/2 if you don't specify that statement.

MathExtremist

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October 20th, 2013 at 9:27:44 PM permalink

Quote: pacomartin
I would say that the answer is 1/2 if you don't specify that statement.

Isn't it technically neither 1/2 nor 1/3 but based on the actual gender split at the time the question is being asked? It's a qualitatively different question than the coin question. In the coin question, you know the whole population. But in the child question, you're assuming that the probabilities for a woman having two children (in order) are p(boy,girl) = p(girl,boy) = p(girl,girl), which is almost certainly not true.

"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

kubikulann

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October 21st, 2013 at 6:50:07 AM permalink

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

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.

Reperiet qui quaesiverit

Wizard
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October 21st, 2013 at 7:02:06 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.

You sound like the kind of annoying frequentist statisticians who are never satisfied with anything I write when asking a probability question. They will never concisely explain how to correctly ask a question, but send pages and pages explaining why a simple statement was in error.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

kubikulann

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October 21st, 2013 at 7:14:08 AM permalink

Quote: Wizard
You sound like the kind of annoying frequentist statisticians who are never satisfied .

???
Sorry if I upset you.

I don't mean any harm. Just thinking out loud, developing an interesting question in new directions.

Your statement is in no way "in error". I'm just pointing to the fact that there are two possibilities. This is, by the way, the reason why many people have difficulties in grasping the Monty Hall problem: they assume Monty draws at random.

I am VERY SATISFIED with everything you write. And hurt that you have felt hurt. You make me think, and I just share it with you.

Reperiet qui quaesiverit

kubikulann

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October 21st, 2013 at 7:18:40 AM permalink

Oh, and by the way, I am more sympathetic to the Bayesian approach than the frequen tist one. :-)

Reperiet qui quaesiverit

Wizard
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October 21st, 2013 at 7:33:01 AM permalink

Quote: kubikulann
I am VERY SATISFIED with everything you write. And hurt that you have felt hurt. You make me think, and I just share it with you.

Sorry if my tone wasn't clear. I wasn't annoyed at all. I thought you were deliberately being argumentative in a lawyerly kind of way, so I responded in jest. No hard feelings at all.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ewjones080

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October 23rd, 2013 at 3:46:35 AM permalink

I too got

1/2

initially, because the wording of the puzzle seems almost ambiguous. However I understand the logic behind

2/3

Sometimes I think it's all about how the person understands the question.

1call2many

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May 2nd, 2014 at 9:17:51 AM permalink

I know this is an old thread that probably no one cares about anymore, but I liked it and want to throw in my 2 cents.

I like the 1/2 answer.

Once the coin is drawn and white is seen ( a 3/4 chance), then that probability is done. If it is carried on to determine the color of the other side it leads to the 2/3 answer. I feel that these are separate events.

It is a bit like looking at two coins tossed, seeing one is heads and asking what is the chance that the other is heads? Is it 1/2 because each throw is independent, or is it 1/3 chance, because HH has 1/4 for two coins tossed and TT can be ruled out. I feel it is 1/2 because once given information about one result they become separate events.

Basically the drawing and seeing white has one probability, and then the color of the opposite side (once white seen) becomes a separate and independent probability.

To calculate the second probability I can ask which coin was chosen? Is it White with white or White with Black (only two coins in bag) so 1/2 is my answer.

The 2/3 answer implies that there are 2/3 ways to draw the white/white coin and only 1/3 ways to draw the white/black coin from a bag?

chrisr

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May 2nd, 2014 at 9:42:21 AM permalink

Quote: 1call2many
I know this is an old thread that probably no one cares about anymore, but I liked it and want to throw in my 2 cents.

I like the 1/2 answer.

Once the coin is drawn and white is seen ( a 3/4 chance), then that probability is done. If it is carried on to determine the color of the other side it leads to the 2/3 answer. I feel that these are separate events.

It is a bit like looking at two coins tossed, seeing one is heads and asking what is the chance that the other is heads? Is it 1/2 because each throw is independent, or is it 1/3 chance, because HH has 1/4 for two coins tossed and TT can be ruled out. I feel it is 1/2 because once given information about one result they become separate events.

Basically the drawing and seeing white has one probability, and then the color of the opposite side (once white seen) becomes a separate and independent probability.

To calculate the second probability I can ask which coin was chosen? Is it White with white or White with Black (only two coins in bag) so 1/2 is my answer.

The 2/3 answer implies that there are 2/3 ways to draw the white/white coin and only 1/3 ways to draw the white/black coin from a bag?

This is a conditional probability problem.. i.e. given one piece of infomation.. what can i infer?

it would be easy to confirm the 2/3 answer experimentally. Since this is a gambiling site this is my experiment.

bet $1

draw a coin and flip it if the black side shows the bet pushes. If the coin flip is white.. check the otherside.. you win 3:2 if it is black. and you lose if it is white.

This would be a bad bet.. fair odds would be 2:1. This is the kindof situation that we are trying to describe with the p(w|w)=2/3 answer.

1call2many

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May 2nd, 2014 at 10:58:24 AM permalink

I may try it experimentally.

I'm not sure i understand your experiment, you said "draw a coin and flip it". Did you mean look at it and only flip if it was white?

chrisr

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May 2nd, 2014 at 2:21:50 PM permalink

Quote: 1call2many
I may try it experimentally.

I'm not sure i understand your experiment, you said "draw a coin and flip it". Did you mean look at it and only flip if it was white?

draw the coin and flip it regardless.. if it is face up black that is a push. put the coin back in the bag and draw one of the two again.

if it lands face up white.. look at the other side of the coin and resolve the bet. white=lose black=pay. then put the coin back in the bag and draw again.

1call2many

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May 5th, 2014 at 7:06:00 AM permalink

I have performed 100 trials using the following method. I wrote W for White with white out on three sides of two quarters and left one side blank for black. I placed the coins in a pencil case and drew a coin randomly, holding it in my palm. I opened my hand to see which side was up, if black was seen I noted it on an excel spreadsheet with a 1 and returned the coin to the bag. If a W was seen I noted that then flipped the coin and noted the flip color on the spread sheet. I did not cut and paste the data as it got messed up when i posted

The results were B first 29, W first 71, W Flipside 51, B Flipside 20

The results are quite clear that I (and other 1/2 proponents) are wrong. Black was viewed first 29/100, close to the expected 1/4. Of the 71 times that white was seen first white was seen on the flip side 51/71 to blacks 20/71, pretty close to the 2/3 answer for flip side white. It became obvious as the trials went on that flip side black was difficult to get as there was only really a 1/4 chance of getting it (20/100 experimentally) because I had a 1/2 chance of choosing the W/B coin and then another 1/2 chance the W is up.

JoePloppy

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May 5th, 2014 at 8:12:16 AM permalink

Quote: 1call2many

To calculate the second probability I can ask which coin was chosen? Is it White with white or White with Black (only two coins in bag) so 1/2 is my answer.

The 2/3 answer implies that there are 2/3 ways to draw the white/white coin and only 1/3 ways to draw the white/black coin from a bag?

This is an awesome thread!

The problem is you don't know which coin was chosen---
You could be looking at side 1 of the white/white coin. = other side is white
You could be looking at side 2 of the white/white coin. = other side is white
You could be looking at side 1 of the white/black coin. = other side is black

2 out of 3 will be white.

It took me a little to fully grasp this concept. Thanks to all for explaining their thoughts. Kudos on doing the experiment Icall2many.

I'm a bigot

2/3

BleedingChipsSlowly

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May 5th, 2014 at 8:35:26 AM permalink

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.]

One particular coin has already been selected. No mater whch one is selected there is a 1/2 chance the flip side is white if the observed side is white.

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?

This question asks about a probability based on observing one side of either coin. There are four coin faces to cosider, but the probability asked for does not cosider the case of observing a black face. If it did, the probabily of the other side being white would be 3/4. Since that posibility is not considered, the probabilty (based on observing one of three white sides) that the other side is white is 2/3.

�You don�t bring a bone saw to a negotiation.� - Robert Jordan, former U.S. ambassador to Saudi Arabia

Sonuvabish

Sonuvabish

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May 5th, 2014 at 9:05:28 AM permalink

I don't know how to use spoiler tags. There are two coins. Most people are saying there is a 2/3 chance that they drew one of two random coins? That makes no sense. You are all wrong. If you choose a random coin, and it is always the white side face up, there is a 50% the other side is white or black, based on the problem. There's a 100% chance the white side will be facing you, not 3/4.

1/1 (given) x 1/2 (two coins) = 50%

3/4 (white sides) x 2/3 (white sides) = 6/12 = 50% chances you would see two white sides chosen randomly

There is never a 2/3 chance of anything.

BleedingChipsSlowly

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May 5th, 2014 at 9:19:52 AM permalink

Quote: Sonuvabish
There is never a 2/3 chance of anything.

I present logic to the contrary in my post above, under the question not asked that many thought was.

�You don�t bring a bone saw to a negotiation.� - Robert Jordan, former U.S. ambassador to Saudi Arabia

Sonuvabish

Sonuvabish

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May 5th, 2014 at 5:00:15 PM permalink

Quote: BleedingChipsSlowly
I present logic to the contrary in my post above, under the question not asked that many thought was.

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!

Sonuvabish

Sonuvabish

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May 5th, 2014 at 5:07:55 PM permalink

Quote: JoePloppy
This is an awesome thread!

The problem is you don't know which coin was chosen---
You could be looking at side 1 of the white/white coin. = other side is white
You could be looking at side 2 of the white/white coin. = other side is white
You could be looking at side 1 of the white/black coin. = other side is black

I'm a bigot

Dude, you think there is a greater than 67% chance you chose a random coin out of 2 possible coins? If I have an all white quarter, and a quarter painted white only on heads, and i randomly choose 1, what are the chances I will flip two black tails in a row?

JoePloppy

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May 5th, 2014 at 5:40:11 PM permalink

Quote: Sonuvabish
Quote: JoePloppy
This is an awesome thread!

The problem is you don't know which coin was chosen---
You could be looking at side 1 of the white/white coin. = other side is white
You could be looking at side 2 of the white/white coin. = other side is white
You could be looking at side 1 of the white/black coin. = other side is black

I'm a bigot

Dude, you think there is a greater than 67% chance you chose a random coin out of 2 possible coins? If I have an all white quarter, and a quarter painted white only on heads, and i randomly choose 1, what are the chances I will flip two black tails in a row?

I think you need to reread the question, dude. And for the record, 2/3 is LESS than 67%

2/3

BleedingChipsSlowly

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May 5th, 2014 at 6:51:42 PM permalink

Quote: Sonuvabish
I read it, but I don't agree.

Perfencly fine, I'm not always right. Thank you for taking the time to read what I wrote.

Quote: Sonuvabish
Regardless, your thinking is obviously superior to most of the forum!

I am smart enough to know that that is not true. Several forum members could kick my ass, logically speaking, standing on one foot with one hand tied behind their back. If it's any solice I had to think long and hard about this problem. Actually, the basic logic was presented in another thread I posted to involving the probabilty that a second child would be a boy. Anyway, having spent some time teaching I do think I have a knack for presenting problems in a way that fosters insight.

Ok, big boy, let's you and me have it out. At the bar. No pythons, no pit bulls. Smoke if you would like. $100 buyin to make it interesting. We use a sack with two coins, one white on both sides, the other black on one side and white on the other. For each round you wager $3 and I wager $5. An impartial chimp draws one of the coins and slaps it down on the table. If the upside is black it's a push. If the upside is white the flip side decides the winner. If the flip side is white I win your $3. I f the flip side is black you get my $5. Are you in?

�You don�t bring a bone saw to a negotiation.� - Robert Jordan, former U.S. ambassador to Saudi Arabia

RS

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May 5th, 2014 at 7:13:46 PM permalink

It's 1/2. How is this controversial?

BleedingChipsSlowly