It was two years ago today...

[slingshot]

To everyone who thinks I'm wrong: please get yourselves two dice, and a cup, shake rattle and roll, and then figure what the odds are for the second die to also be a two.

That's what the question is.

Don't bother me with side bets, don't bother me with anything else. It's stupid. Read and comprehend English. Act out the question.

Alan- you're using common sense. You oughta know by now that doesn't fly on this forum.

[Alan Mendelson]

Of course it's common sense. It's also listening and interpreting the question that was asked.

The question was not about the number of dice combinations containing a two (11) and how many of those combinations can contain 2-2 (1). It was a simple question that asked you to consider the six faces on a single die.

[RS__]

Alan's never going to change his opinion. He's just too stubborn.

[Alan Mendelson]

Alan's never going to change his opinion. He's just too stubborn.

That's right because I read the question and properly interpreted it.

You read the same question but failed to understand the problem that was presented and instead launched into the answer to a different question that is taught in statistics classes.

[Alan Mendelson]

For those of you who want the answer to be 1/11 here is your question:

Given two dice, with a 2 showing on at least one of the two dice, what are the odds that both dice will show 2-2?

The answer to this question is 1/11.

But when you tell me that at least one die is already showing a two, and then you are asking me what are the odds that the second die will also show a 2, I only consider ONE die and the answer is 1/6.

And if you don't understand the difference between these two questions, I can't help you.

[RS__]

Alan, we don't want your "help", if you can even call it that.

[mickeycrimm]

To everyone who thinks I'm wrong: please get yourselves two dice, and a cup, shake rattle and roll, and then figure what the odds are for the second die to also be a two.

Alan, your suggestion here is quite revealing. You are telling us to do something that you have not done yourself. Because if you did you would be quite surprised by the results.

All that's needed is a spreadsheet with two columns. After every roll where "at least one of the dice is a two" make a mark in the 1st column. Then take note of whether the other die is a two. If it is then make a mark in the 2nd column. Keep collecting those stats. You will start to see a pattern developing. Through about 500 decisions you will find that the frequency of the other die being a 2 is a hell of a lot closer to 1 in 11 than 1 in 6.

My prediction, you heard it here first, is that through 500 decisions the other die will be a 2 somewhere between 40 and 50 times. If the frequency is 1 in 11 then the average would be 45 times.

By your logic the other die would be a two on average 83 times though 500 decisions.

I'm used to collecting emperical evidence like this. I do it on video line games all the time where I don't have information on the frequency of line pays. So this is what I'm going to do. I'm going to take your suggestion. I just need to find a store that sells dice cups and dice. I will make this emperical study as free time permits. And I will publish my results here.

Alan, why don't you do the same and publish your results.

[a2a3dseddie]

For those of you who want the answer to be 1/11 here is your question:

Given two dice, with a 2 showing on at least one of the two dice, what are the odds that both dice will show 2-2?

The answer to this question is 1/11.

But when you tell me that at least one die is already showing a two, and then you are asking me what are the odds that the second die will also show a 2, I only consider ONE die and the answer is 1/6.

And if you don't understand the difference between these two questions, I can't help you.

Both your questions mention 2 dice. Both mention a 2 on at least one of the 2 dice. Your first question is how the original problem was worded. Your second statement is not what was originally asked.

Once again, here is the original problem from the Wizard's Site:

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?

Where in this original problem do you see:

then you are asking me what are the odds that the second die will also show a 2, I only consider ONE die and the answer is 1/6.

?

[regnis]

And that's the typical response from someone that doesn't understand conditional probability. Same reason you work for a living and can't gamble for one. Another typical casino loser.

To put the record straight.

1. I gambled for a living for many years.
2. I can and presently do make a ridiculous amount of money working with no chance of loss, and working only about 20 hours a week max. That still leaves me the ability to play the horses full time.
3. I don't ever have to sit in a smoke filled casino unless I want to--and then I can leave any time I want.
4. If one die is a 2, the chance of the other die being a 2 is 1 of 6.

I should add that while Kentucky derby day has historically not been a good day for me, this year I hit 12 of 14 races, a load of pick 3's, two pick 4's and a pick 5. I made more in that one day than you can possibly make as a full time gambler. You can look up what those exotic bets paid if you want to be sick. That being said, I still prefer the safety of my part time job with no risk of loss. Do I hate dealing with clients--often yes. But I also hated gambling at times and dealing with the people involved in gambling.

Try this riddle: If you and me are the only 2 people in a room, and I am one of those 2 people, what are the odds that you are the other person.

[Count Room]

If I tell you at least one of the dice is a "2", but both dice stay behind the screen, the odds are 1/11 that both dice are a "2".

If I pull a "2" die from behind the screen and show it to you, only then will the odds become 1/6 that both are a "2".

EDIT: This is very similar to poker in many respects. The odds change when you have more or less information available to you.

[regnis]

Count-you have sufficient info. They told you one die is a 2. That only leaves 6 other choices (i.e. 1 die). The one that shows a 2 can't change--at least not in an honest game.

But I think we should bury this again for 2 years unless it has value to Dan in the traffic it brings. Let's put it in Rob's storage shed.

[Count Room]

To everyone who thinks I'm wrong: please get yourselves two dice, and a cup, shake rattle and roll, and then figure what the odds are for the second die to also be a two.

Alan, your suggestion here is quite revealing. You are telling us to do something that you have not done yourself. Because if you did you would be quite surprised by the results.

All that's needed is a spreadsheet with two columns. After every roll where "at least one of the dice is a two" make a mark in the 1st column. Then take note of whether the other die is a two. If it is then make a mark in the 2nd column. Keep collecting those stats. You will start to see a pattern developing. Through about 500 decisions you will find that the frequency of the other die being a 2 is a hell of a lot closer to 1 in 11 than 1 in 6.

My prediction, you heard it here first, is that through 500 decisions the other die will be a 2 somewhere between 40 and 50 times. If the frequency is 1 in 11 then the average would be 45 times.

By your logic the other die would be a two on average 83 times though 500 decisions.

I'm used to collecting emperical evidence like this. I do it on video line games all the time where I don't have information on the frequency of line pays. So this is what I'm going to do. I'm going to take your suggestion. I just need to find a store that sells dice cups and dice. I will make this emperical study as free time permits. And I will publish my results here.

Alan, why don't you do the same and publish your results.

I agree with mickey's suggestion here. It would bury this matter 100%.

[mickeycrimm]

I should add that while Kentucky derby day has historically not been a good day for me, this year I hit 12 of 14 races, a load of pick 3's, two pick 4's and a pick 5. I made more in that one day than you can possibly make as a full time gambler. You can look up what those exotic bets paid if you want to be sick. That being said, I still prefer the safety of my part time job with no risk of loss. Do I hate dealing with clients--often yes. But I also hated gambling at times and dealing with the people involved in gambling.

Regnis, congratulations on your big Kentucky Derby score. But dare I ask, would your pile of losing tickets from over the years be big enough to fill a barn?

[redietz]

And that's the typical response from someone that doesn't understand conditional probability. Same reason you work for a living and can't gamble for one. Another typical casino loser.

To put the record straight.

1. I gambled for a living for many years.
2. I can and presently do make a ridiculous amount of money working with no chance of loss, and working only about 20 hours a week max. That still leaves me the ability to play the horses full time.
3. I don't ever have to sit in a smoke filled casino unless I want to--and then I can leave any time I want.
4. If one die is a 2, the chance of the other die being a 2 is 1 of 6.

I should add that while Kentucky derby day has historically not been a good day for me, this year I hit 12 of 14 races, a load of pick 3's, two pick 4's and a pick 5. I made more in that one day than you can possibly make as a full time gambler. You can look up what those exotic bets paid if you want to be sick. That being said, I still prefer the safety of my part time job with no risk of loss. Do I hate dealing with clients--often yes. But I also hated gambling at times and dealing with the people involved in gambling.

Try this riddle: If you and me are the only 2 people in a room, and I am one of those 2 people, what are the odds that you are the other person.

Excellent riddle that highlights my perspective on this.

[quahaug]

And that's the typical response from someone that doesn't understand conditional probability. Same reason you work for a living and can't gamble for one. Another typical casino loser.

To put the record straight.

1. I gambled for a living for many years.
2. I can and presently do make a ridiculous amount of money working with no chance of loss, and working only about 20 hours a week max. That still leaves me the ability to play the horses full time.
3. I don't ever have to sit in a smoke filled casino unless I want to--and then I can leave any time I want.
4. If one die is a 2, the chance of the other die being a 2 is 1 of 6.

I should add that while Kentucky derby day has historically not been a good day for me, this year I hit 12 of 14 races, a load of pick 3's, two pick 4's and a pick 5. I made more in that one day than you can possibly make as a full time gambler. You can look up what those exotic bets paid if you want to be sick. That being said, I still prefer the safety of my part time job with no risk of loss. Do I hate dealing with clients--often yes. But I also hated gambling at times and dealing with the people involved in gambling.

Try this riddle: If you and me are the only 2 people in a room, and I am one of those 2 people, what are the odds that you are the other person.

Excellent riddle that highlights my perspective on this.

I was going to leave this thread alone because everything that can be said has been and closed minds are closed. I would however like to know your analysis of the original question which I don't think you ever gave. The original wording of the question seems clear and concise to me.

[jbjb]

12 people are in a room. 6 are wearing green shirts. 6 are wearing red shirts. Alan is in the green group, I'm in the red. Two people are picked. One from green and one from red. If AT LEAST Alan or I am picked, what is the probability that Alan or I are the other one picked?

Again, it's 1 in 11 and is EXACTLY THE SAME as the two dice problem!

Edit: And if anyone says it's not the same, we're each holding a 2 also. Every other person is holding a different number.

If any of you still cannot fathom this, you're hopeless.

[RS__]

I think part of it is a lack of understanding conditional probability.

For example (in blackjack) when your first card is an Ace, you have a 4/13 chance to get a blackjack. When your first card is a ten, you have a 1/13 chance of getting a blackjack. Now, here's the question:

You are dealt a blackjack. Which card is more likely to have been your first card? A ten/face or an ace? Let's let Alan answer before anyone else does.

[RS__]

Here's another example of conditional probability.

Let's say you're looking at the NFL lines, Patriots vs Dolphins. The spread is Patriots -2.5 @ -110. The money line is Patriots -140, Dolphins +130. Let's assume these are the true odds with vig added equally on each side.

What is the probability the Patriots beat the spread? It's 50%.


If the Patriots win the game, what is the probability they beat the spread (won by 2.5 points or more)? It's NOT 50%. It's significantly higher than 50%.

[Alan Mendelson]

4. If one die is a 2, the chance of the other die being a 2 is 1 of 6.

It really was a simple question. But they couldn't see it.

[mickeycrimm]

While shopping in Walmart today I bought a yahtzee set just to pirate the cup and dice. So I'm ready to run the first sample space of 100 decisions. I'm conducting this experiment just as the question below dictates. This was the original question and it is the only one that matters for this empirical study. I'll get back to you with the results sometime tonight.

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?