The Student Room Group

Help with Stats? Dice probability

Scroll to see replies

Reply 20
Original post by BabyMaths
There are 91=635391=6^3-5^3 outcomes that include at least one 4.
________________________________

For the case with exactly one six:

Those with two fours and one six contribute three successes.
Those with one four, one six and one other contribute twenty four (6x4) successes.

Result 27/91.
________________________________

For the case with exactly two sixes:

Those with one four contribute three successes.

Result 3/91.
________________________________

At least three posters so far with three different answers. :eek: What is f38 coming to?


This is worse than the Monty Hall problem!

I'm glad I'm rubbish at probability so I can stay out of the argument :biggrin:
Reply 21
Original post by BabyMaths
There are 91=635391=6^3-5^3 outcomes that include at least one 4.
________________________________

For the case with exactly one six:

Those with two fours and one six contribute three successes.
Those with one four, one six and one other contribute twenty four (6x4) successes.

Result 27/91.
________________________________

For the case with exactly two sixes:

Those with one four contribute three successes.

Result 3/91.
________________________________

At least three posters so far with three different answers. :eek: What is f38 coming to?



Yup, I agree with these answers.
Reply 22
Original post by davros


I'm glad I'm rubbish at probability so I can stay out of the argument :biggrin:


I'm rubbish at it too but that's not going to stop me. :tongue:
Original post by cole-slaw
A
ssuming the dice are independent, if I tell you I have 3 dice, 1 is showing a 4, then you DO only need to consider the other two. You can just forget about that dice because its no longer relevant. So its 5/36 and 1/36.


Bolded the error in the above. It is meaningless to talk about 'the other two' dice when the only information that you have is that at least one dice is a four.

You have answered a problem along the lines of "I roll three dice one at a time. I rolled a four on the first die. What is the probability I rolled exactly one six? What is the probability I rolled exactly two sixes?" (Except that, even then, you got the "Exactly one six" wrong, because it should be 10/36.)
(edited 10 years ago)
Reply 24
Original post by Swayum
The reason why this is incorrect is that you've assumed that the 4 was rolled in the first roll. Effectively you're saying "ok, I just rolled a 4, now what's the probability I roll 2 sixes or 1 six?". It's like you're ignoring the 4 all together, it's not affecting your answer (your answer is the same as the question "what's the probability of rolling exactly 1 six in 2 rolls?").

That's not the same thing - the best way to see it is to draw out a tree diagram.

OK, I didn't bother Bayesing it or anything - I just went with intuition, which is in hindsight a bad idea because humans are generally bad at probability. System 2, I choose you!
Reply 25
Original post by BabyMaths
There are 91=635391=6^3-5^3 outcomes that include at least one 4.
________________________________

For the case with exactly one six:

Those with two fours and one six contribute three successes.
Those with one four, one six and one other contribute twenty four (6x4) successes.

Result 27/91.
________________________________

For the case with exactly two sixes:

Those with one four contribute three successes.

Result 3/91.
________________________________

At least three posters so far with three different answers. :eek: What is f38 coming to?


hehehe you nailed it BabyMaths :borat:
Reply 26
Original post by Forum User
Bolded the error in the above. It is meaningless to talk about 'the other two' dice when the only information that you have is that at least one dice is a four.

You have answered a problem along the lines of "I roll three dice one at a time. I rolled a four on the first die. What is the probability I rolled exactly one six? What is the probability I rolled exactly two sixes?" (Except that, even then, you got the "Exactly one six" wrong, because it should be 10/36.)


If at least one dice is a four, then we know one dice is a four and the others we still have no information about.
You can give it a name if you like, say X, and the two dice that we don't know anything about are "the other two".

The order is irrelevent. All we are told is that one dice is a four.


oh yes, I meant 10/36 and 1/36. stupid mistake.
Reply 27
I'm going to try this.

I have 3 dice, 2 are normal dice and one only has 4s on it. I will roll them, I will be guaranteed to get at least 1 four, and the other two I have no information about.
Reply 28
Okay so my brother just came home and has helped me figure this out!

So 91 is the total sample space of getting at least one 6. To get 91 he said:
Imagine the first dice is a 6, the total number of options of getting a 6 in the other two dices is 6x6 = 36

Now, imagine the first dice isn't a 6, we then have the options: 61, 62, 63, 64, 65, 66, 16, 26, 36, 46, 56 - these are 11 different combinations of ways of getting at least one 6 with other other two dice.

We times this 11 by 5 as if the first dice isn't a 6, there are 5 options it can take. 55+36 makes a sample space of 91.

He started off with getting exactly two 6s
To find the sample space of getting two or more sixes -

If the first dice is a 6 and we need one or more 6s
If the second dice is a 6 we have 6 available options for the third dice
OR if the second dice still isn't a 6, in this case we have 1 available option for the third dice, we times this by 5 again as there are 5 different ways the second dice isn't a 6

Now, if the first dice isn't a six. We only have one option, and that is for the two of dices to be 6s. Again there are 5 different ways of the first dice not being a six, so 5x1 = 5

11+5 = 16 and this is our sample space of the total number of outcomes being two or more 6s.
We minus one from this as we are looking for EXACTLY two 6s, and the only option that isn't exactly two 6s are the dice 666, which appears once, this give us 15/91 for the second part.

Back to the first part, to get exactly one 6, this is the number of combinations of getting 1 or more 6s minus that of getting 2 or more 6s, this gives 91-16, which is 75 and the answer 75/91.

To be honest, I understand his workings out but if this was to pop up in an exam I'd have no clue what was going on. :/ Have completely given up on it now - really hard question!
Reply 29
Original post by atsruser
This isn't much help, I'm afraid, but for the first question, I get P(614142)=3/10P(6_1 | 4_1 \cup 4_2) = 3/10 , where 616_1 means "a 6 occurs precisely once", and so on.

I'll put up my working if you want.


Here it is, for part 1.

We want [text]P(6_1| 4_1 \cup 4_2).

Note that 41,424_1, 4_2 are mutually exclusive so P(4142)=P(41)+P(42)P(4_1 \cup 4_2) = P(4_1) + P(4_2).

For P(41)P(4_1) we are considering events of the form "4XX", which has probability 165656=25216\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} = \frac{25}{216} but since the 4 can occur in 3 positions, we have P(41)=3×25216=2572P(4_1) = 3 \times \frac{25}{216} = \frac{25}{72}

A similar argument gives P(42)=572P(4_2) = \frac{5}{72} and so P(4142)=512P(4_1 \cup 4_2) = \frac{5}{12}

Now P(614142)=P(61(4142))P(4142)=P((6141)(6142))5/12=P(6141)+P(6142)5/12P(6_1| 4_1 \cup 4_2) = \frac{P(6_1 \cap(4_1 \cup 4_2))}{P(4_1 \cup 4_2)} = \frac{P((6_1 \cap 4_1) \cup (6_1 \cap 4_2))}{5/12} = \frac{P(6_1 \cap 4_1) + P(6_1 \cap 4_2)}{5/12}

For P(6141)P(6_1 \cap 4_1) we are considering events of the form "64X" with probability 161646=154 \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{4}{6} = \frac{1}{54} but we have 6 permutations of each such event giving P(6141)=654=19P(6_1 \cap 4_1) = \frac{6}{54} =\frac{1}{9}.

Similarly we have P(6142)=3×161616=172P(6_1 \cap 4_2) = 3 \times \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{72} since we're considering events of the form "644" with 3 perms.

Hence we have [text]P(6_1| 4_1 \cup 4_2) = \frac{1/9 + 1/72}{5/12} = \frac{3}{10}.
Original post by cole-slaw
If at least one dice is a four, then we know one dice is a four and the others we still have no information about.
You can give it a name if you like, say X, and the two dice that we don't know anything about are "the other two".



No you can't, because with the way the original question was worded you don't have any information about any particular dice, and so there is no dice that you can label 'X'. The information you are given is about all three dice, not about any specific dice. You can see that it goes wrong by considering a simple example:

I flip two coins, a red coin and a blue coin. The possible results, with the result of the red coin shown first are {HH, HT, TH, TT}
I look at the coins that I have flipped and tell you that at least one of them is Tails. What is the probability they are both tails?
Your way gives the incorrect answer of 1/2. The correct answer is 1/3 because the information I have given you reduces the sample space to the equally likely {HT, TH, TT}.
Original post by destinyx
Okay so my brother just came home and has helped me figure this out!

etc


This isn't the question that you posed in the OP. You said that there was at least one 4, not at least one 6, which is why everyone got different answers to those in the book.
Original post by cole-slaw
I'm going to try this.

I have 3 dice, 2 are normal dice and one only has 4s on it. I will roll them, I will be guaranteed to get at least 1 four, and the other two I have no information about.


For those dice your revised numbers (1/36 and 10/36) are the correct probabilities for getting exactly two sixes and exactly one six.
But those are not the dice in the OP (at least we weren't told they were anything other than perfectly ordinary dice).
Reply 33
Original post by Forum User
No you can't, because with the way the original question was worded you don't have any information about any particular dice, and so there is no dice that you can label 'X'. The information you are given is about all three dice, not about any specific dice. You can see that it goes wrong by considering a simple example:

I flip two coins, a red coin and a blue coin. The possible results, with the result of the red coin shown first are {HH, HT, TH, TT}
I look at the coins that I have flipped and tell you that at least one of them is Tails. What is the probability they are both tails?
Your way gives the incorrect answer of 1/2. The correct answer is 1/3 because the information I have given you reduces the sample space to the equally likely {HT, TH, TT}.


Can I ask you a similar question, just to clarify my understanding.

I have three cards, one is black on both sides, one is white on both sides and one is black on one side and white on the other side.

So the sample space is {BB WW BW}

I pull out a card face up. the face is black.

The sample space is now {BB BW}

What is the probability it is the BB card?
Original post by cole-slaw
Can I ask you a similar question, just to clarify my understanding.

I have three cards, one is black on both sides, one is white on both sides and one is black on one side and white on the other side.

So the sample space is {BB WW BW}

I pull out a card face up. the face is black.

The sample space is now {BB BW}

What is the probability it is the BB card?


2/3.

What do I win?

Let me ask a question. I pull out a card from your selection of cards. I look at both sides of the card and then announce truthfully "at least one side is black". What is the probability it is the BB card?

Spoiler

(edited 10 years ago)
Reply 35
Original post by Forum User
2/3.

What do I win?


Opprobium.

The reason it is 2/3 is because BB is twice as likely as BW given that B>=1


For the same reason, when considering the information set

{HT, TH, TT} given T >= 1

the probabilities are (1/4, 1/4, 1/2)
Reply 36
Oh god guys sorry!! Totally didn't realise that mistake! DD:
Reply 37
I think this thread can be closed now. Sorry for the wasted time, can't believe a made a mistake when posting the question!
Original post by cole-slaw
Opprobium.

The reason it is 2/3 is because BB is twice as likely as BW given that B>=1



No, the reason it is 2/3 is because you are told information about one specific face, the one face that was looked at. That is not the same at all as being given the information "B>=1".

In fact your answer is wrong, because given B>=1 the chance the card is BB is 1/2. For example:

I draw one card from your selection of three cards. I look at both faces of the card. I tell you that at least one of the two faces is black. What is the probability it is the BB card? Now the answer is 1/2, and the information given is "B>=1".
(edited 10 years ago)
Reply 39
Original post by Forum User
2/3.

What do I win?

Let me ask a question. I pull out a card from your selection of cards. I look at both sides of the card and then announce truthfully "at least one side is black". What is the probability it is the BB card?

Spoiler



Incorrect. Its a question of sampling vs selection.

If you are sampling a card at random, and then telling me "at least one side is black/white" (whichever happens to be true) then the answer is 2/3.

If you are deliberately selecting a card with at least one black side, and then retrospectively announcing your selection policy, the answer is 1/2.

This is known as the Monty Hall problem. It confuses a lot of probability students.

The most valid assumption in OPs question is that the dice was rolled at random and the number 4 was then seen and then that imparted to the OP, rather than a set of results selected for the presence of a 4.

Quick Reply