The Wizard will bank this bet: 1/6 vs 1/11

[regnis]

So the other day on another thread Alan posted a picture of a sequential royal that he hit. He was dealt 4 pieces and drew the fifth. Now let me phrase this for the 1-11's:

I had a deck of cards and dropped four cards. They were the AKQJ of hearts. If I now flip over or select or draw one more card, what are the odds of drawing the royal?

[arcimede$]

Different situation, regnis ... there aren't two different A of hearts (or K or Q or J or T)..

[regnis]

Not so sure arc---there was one deuce remaining on the other die. Here there is one card remaining for the royal.

[Alan Mendelson]

Good point regnis because PROBABILITY says the card answer is 1/47 so why can't the answer for dice be 1/6 ??

[regnis]

Yes-unless these cards have the ability to change their faces like the dice that the 1-11's are using. Because I'm waiting for them to argue that we don't know which of the 52 cards was the Ace and therefore it could be any of the 52 x some multiplier or some convoluted answer.

P.S.--1 of 48 because I only dropped 4 cards--there are 48 remaining.

[Alan Mendelson]

Yes-unless these cards have the ability to change their faces like the dice that the 1-11's are using. Because I'm waiting for them to argue that we don't know which of the 52 cards was the Ace and therefore it could be any of the 52 x some multiplier or some convoluted answer.

P.S.--1 of 48 because I only dropped 4 cards--there are 48 remaining.

You're right -- you dropped four cards.

And I am still waiting to see a video proving the answer is 1/11 without changing the value of the die that shows a 2. The Wizard on his forum says he is waiting for the 1/6ers to show up with money. I'm waiting for his video.

[arcimede$]

Good point regnis because PROBABILITY says the card answer is 1/47 so why can't the answer for dice be 1/6 ??

Because it's a different situation as I explained above. Duh.

[arcimede$]

You're right -- you dropped four cards.

And I am still waiting to see a video proving the answer is 1/11 without changing the value of the die that shows a 2. The Wizard on his forum says he is waiting for the 1/6ers to show up with money. I'm waiting for his video.

That's still not how you compute probability. You will never get what you want because it doesn't apply to the question being asked.

[Alan Mendelson]

Arc, you are hopeless. Do you really think this is a question of probability -- or simply a question involving looking at two dice?

For you to use your probability answer you have to do funny things with two dice -- like rotate them to show different numbers. In a court of law that would be called tampering with the evidence.

[arcimede$]

Arc, you are hopeless. Do you really think this is a question of probability -- or simply a question involving looking at two dice?

Both. Here is the question ...

"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?"

How difficult is it to see that both are in the original question? I've bolded them to make it obvious

For you to use your probability answer you have to do funny things with two dice -- like rotate them to show different numbers. In a court of law that would be called tampering with the evidence.

No, you don't do anything to the two dice. The "probability" asks what are the odds across an infinite number of independent throws. You can estimate what this might be by throwing the dice yourself a large number of times. Recording 100-200 throws with a 2 showing will get you close.

Why do you keep denying what the question asks? Simple, you are emotionally invested in your incorrect answer.

[Alan Mendelson]

Arc I am going to defer to redietz who responded to your question in previous posts. He addressed the question and definition of "probability" and the wording of the question.

Frankly -- you can maintain that this involves multiple throws of two dice but I will never see it that way.

By the way, don't ask for multiple chances at a craps table either -- won't happen.

[arcimede$]

Arc I am going to defer to redietz who responded to your question in previous posts. He addressed the question and definition of "probability" and the wording of the question.

Yup, and I showed that his interpretation was silly.

Frankly -- you can maintain that this involves multiple throws of two dice but I will never see it that way.

By the way, don't ask for multiple chances at a craps table either -- won't happen.

No one is asking for "multiple chances" when they asked about "probability". For example, what is the probability you will be dealt 3 of a kind from a randomly shuffled deck of 52 cards? No one is asking about one hand being dealt. When "probability" is inserted into the question it directly implies that we are looking at a set number of events. Not a single event.

Hence to compute the "probability" you look at the number of distinct situations of 3oak and compare that to the total number of possible deals.

http://en.wikipedia.org/wiki/Classic...of_probability

Same applies to the dice question. You look at the total number of distinct situations of a pair of 2s (1) compared to the total number of possibilities that contain at least one 2 (11)

[Alan Mendelson]

I am glad you wrote this:

No one is asking for "multiple chances" when they asked about "probability". For example, what is the probability you will be dealt 3 of a kind from a randomly shuffled deck of 52 cards? No one is asking about one hand being dealt. When "probability" is inserted into the question it directly implies that we are looking at a set number of events. Not a single event.

Because the dice in a cup question is a single event.

[arcimede$]

No, Alan, the dice cup question ... "What is the probability that both dice are showing a 2?" .... is a probability question. It is no different than my question about 3oak. You are intentionally ignoring the real question because you KNOW that it leads to the correct answer of 1-11.

You cannot compute probability from a single event. Go read the reference I provided you.

[Alan Mendelson]

Arc here is the original question again 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?

There is no way that I can consider this to be more than a one roll, one time event.

There is no way that I can't "freeze" one die on its face showing a 2 and then consider the six faces of the second die.

Now if in your mind it is something different and you get to change the value showing on a die to accommodate your 1/11 answer, then be my guest.

[arcimede$]

Sorry Alan, your denial of the definition of "probability", which I provided you above, is silly. There is no such thing as probability (other than 1 or 0) for "one time events"

Here's the same type of question ... You are dealt 5 cards on a VP machine. A friend covers the screen so you can't see the cards. He tells you that the hand contains at least a one deuce. What is the probability you have at least two deuces?

Note the question does not specify which deuce is dealt. It could be hearts, diamonds, clubs or spades. Hence all possible combinations of a pair of deuces is possible. Your approach would limit it to just cases when the deuce is in hearts (or any one of the other suits) and all pairs would have to contain the heart deuce. Of course, that is obviously not what the question asks.

You are making this exact mistake with the dice.

[redietz]

Arci, why do you think the asking for "probability" should override the initial description of a single event? This is not an equation. This is a piece of writing, and one word at the end does not override a bunch of words in the beginning just because you say so. You are not the arbiter of logic, how this piece of writing works, how it should be interpreted, and so on. You have neither the authority nor the credentials to do that just because you understand basic probability theory.

Rarely have I gone head to head with arci over something. In this case, I feel better about my position partly because it's the humble position. It requires no self-designated expertise regarding what an author meant, nor how something should be read or interpreted. Arci assumes a lot here, and even acknowledges that most people may not see it his way. Well, if a piece of writing leads to 1/6 as an answer for most people, and 1/6 is wrong, that is not the readers' fault. Arci, go look up the author of the question and give him/her hell, because that is where the fault lies. It's the writer's responsibility if he leads readers astray.

And actually, yes, there is a definition of "probability" that can refer to single events. IT'S THE PRIMARY DEFINITION, for God's sake. Now everybody please go to their actual dictionary and look it up.

P.S. Arci, you've been semi-obnoxious and I do not know why, but let me get in an academic dig here. If you're using Wikipedia as your reference in a serious discussion, you're in trouble. And you know this as well as I do.

[Rob.Singer]

There's several reasons for his obnoxious attitude today, but let's just pick the easy, less obtrusive one: another gambling binge that just didn't work out so well.

[Zedd]

So the other day on another thread Alan posted a picture of a sequential royal that he hit. He was dealt 4 pieces and drew the fifth. Now let me phrase this for the 1-11's:

I had a deck of cards and dropped four cards. They were the AKQJ of hearts. If I now flip over or select or draw one more card, what are the odds of drawing the royal?

This question is analogous to setting one die to a deuce and then asking what is the probability of rolling a deuce with the other die.

It is NOT analogous to the question where we roll two dice until at least one of them is a two—then asking what is the probability both are deuces.

For that analogy to work, you would have to ask:

"I dropped five cards until at least four of the cards are part of a hearts Royal Flush. What is the probability that the fifth card completes the Royal?"

The answer to that question is not 1/47.

It's 1/235.

[arcimede$]

Arci, why do you think the asking for "probability" should override the initial description of a single event? This is not an equation. This is a piece of writing, and one word at the end does not override a bunch of words in the beginning just because you say so. You are not the arbiter of logic, how this piece of writing works, how it should be interpreted, and so on. You have neither the authority nor the credentials to do that just because you understand basic probability theory.

The single event is just describing the situation (just like I did in my VP example). It is not asking a question. How can one ask a question without some descriptive information? Look at where the question mark appears. The question is obvious. It has nothing to do with me. Is it a little tricky? Sure, but that is only if you don't take the time to parse it carefully.

Rarely have I gone head to head with arci over something. In this case, I feel better about my position partly because it's the humble position. It requires no self-designated expertise regarding what an author meant, nor how something should be read or interpreted. Arci assumes a lot here, and even acknowledges that most people may not see it his way. Well, if a piece of writing leads to 1/6 as an answer for most people, and 1/6 is wrong, that is not the readers' fault. Arci, go look up the author of the question and give him/her hell, because that is where the fault lies. It's the writer's responsibility if he leads readers astray.

And actually, yes, there is a definition of "probability" that can refer to single events. IT'S THE PRIMARY DEFINITION, for God's sake. Now everybody please go to their actual dictionary and look it up.

P.S. Arci, you've been semi-obnoxious and I do not know why, but let me get in an academic dig here. If you're using Wikipedia as your reference in a serious discussion, you're in trouble. And you know this as well as I do.

Use any other source you choose. They will all say the same thing for an obvious math question.