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Flyingcircus.
10th Sep 2001, 17:39
I have a mathematics question that has been bugging me for a while. It's not a complex question, some simple probability, but I get two different and (seemingly) correct answers. :eek:

Does anyone know of a website simillar to PPRuNe, but for the mathematically inclined?

Perhaps you guys can help me. The question is as follows.
Two ordinary dice are rolled. If it is known that one shows a 5, what is the probability that they total 8?

The Guvnor
10th Sep 2001, 17:49
Seems simple enough to me - if you already have a 5 showing, then there's only one possible 'show' to make 8 - and that's 3. There are six sides to a dice, and therefore your answer would be one in six.

Unless this is a trick question of some sort?

Flyingcircus.
10th Sep 2001, 18:01
That's what I thought. But consider all the possible combinations that involve one die showing 5.
(1,5);(2,5);(3,5);(4,5);(5,1);(5,2);(5,3);(5,4);(5,5);(5,6); (6,5)for a total of 11 combos. Now there are only two of those that add to 8, being (3,5) and (5,3). Therefore the probability being 2 in 11.

But at the same time I can't discount the other solution. It's driving me nuts!!!! :eek: :confused:

HugMonster
10th Sep 2001, 18:24
Not quite right, Flying Circus:- There are a total of 36 different throws of the dice, since each can be any of 1-6. Of those, one sixth of the throws will feature a 5 (Chances of either die being a 5 are 1/6). However, as you state the question, this is a given, therefore a certainty. The chance of the other die being a 3 is also 1/6.

Effectively you have been told that the probability of the first part is to be ignored, so you don't need to look at the likelihood of the first die being a 5 AND the likelihood of the second being a 3.

(The probability of throwing 8, with a 5 and a 3 is 2/36 = 1:18 - but that's not asked)

The mistake you're making is to consider the probability of the dice to be interdependent instead of separate.

Flyingcircus.
10th Sep 2001, 19:09
With all due respect, Hugmonster, there are eleven combinations of 2 dice that contain at least one 5, being those that I listed above, not 6 (one sixth of 36) as you suggested. A probability table follows, displaying the situation under consideration.

1st Dice....2nd Dice
...1...........5
...2...........5
...3...........5
...4...........5
...5...........1
...5...........2
...5...........3
...5...........4
...5...........5
...5...........6
...6...........5
As you can see, eleven combinations, two of which add to give 8, thus giving a probability of 2:11.

Before I go any further, I must say that I agree that the answer is 1:6, but the correct answer according to the exam marker was 2:11. When I took the matter up with him, he couldn't give me a satisfactory explanation as to why 1:6 is incorrect.

Another way of looking at it is that we must have a 5 one at least one dice. The prob. of it being the 1st dice is 1/2, and the prob. of it being the 2nd dice is also 1/2. Now, for a total of 8, the other dice must show a 3, which has a prob. of 1/6. So we have a tree which looks like this:

.............................2nd dice a 3
.......................1/6
.........1st dice a 5
....1/2 ...............5/6
.............................2nd dice not a 3

.............................1st dice a 3
....1/2.................1/6
.........2nd dice a 5
........................5/6
.............................1st dice not a 3

The probability of obtaining an 8, therefore, is 1/2*1/6 (the upper branch of the tree) + 1/2*1/6 (the third branch of the tree), which = 1/6. But I can't discount the other solution, and it's driving me nuts! :eek: :confused:

edited to space the table/tree out, as it was unreadable

[ 10 September 2001: Message edited by: Flying_Circus. ]

widgeon
10th Sep 2001, 19:19
question is misleading , if you say that die A is 5 then prob is 1 in 6 , if either die a or die b is 5 then the other answer is right . there are 2 ways to get 8 , 3 and 5 , 5 and 3. If you just leave one die showing 5 and only roll the other the answer is 1 in 6 .
Bur there again a clock that is stopped is statistically more accurate than one that gains a minute a day .

[ 10 September 2001: Message edited by: widgeon ]

FlyingForFun
10th Sep 2001, 19:33
Flying Circus,

At first I agreed with Guv and HM - the answer is obviously 1/6.

But, having thought about it, I think there are two different questions, with two different answers. (Edit - And widgeon apparently came to the same conclusion in the time it took me to type this explaination!)

The question which Guv and HM have answered is: "Given that the first dice shows a 5, what is the chance of them totalling 8?". In this case, the question could be re-phrased "What is the chance of the second dice showing a 3", and the answer, of course, is 1/6

The second question is: "Given that one or the other dice is showing a 5, what's the chances of them totalling 8?" This one's a bit harder to understand. Imagine you roll both dice, but don't look at them.

Now, take a quick peek at the first. Is it a 5? Yes? Then what's the chances of the other one being 3? 1/6, obviously.

And if the first one isn't a 5? Now you're allowed to look at the second one, but you have to forget what the first one said. Is this one a 5? No? Well, in that case, the bet's off. But if it is a 5? Well, what's the chances of the first one being a 3? Think about this one for a second - we already know the first one isn't a 5, so the chances of it being a 3 are now increased to 1/5. Following?

So the answer to the second question is actually going to be somewhere in between 1/5 and 1/6 - and that's your answer of 2/11.

(To go into a bit more detail:

We've got the probabilities of 1/6 and 1/5 above, but how do we get from there to 2/11? At first I thought take the average - but this is not correct.

The reason you can't take the average is that we're more likely to be looking for a 3 on the second dice than we are to be looking for a 3 on the first dice. This is because we don't even bother to look for a 3 on the first dice if the first dice shows a 5.

We need to take a weighted average, to account for the fact that we're only 5/6 as likely to be looking for a 3 on the first dice as on the second. Thus, the answer is:

1/6 (the chances of the 3 on the second dice) * 1 (the weighting factor)

Plus:

1/5 (the chances of the 3 on the first dice) * 5/6 (the weighting factor)

All divided by:

11/6 (the sum of the two weighting factors)

Which gives the answer of 2/11)

Very good question!

FFF
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[ 10 September 2001: Message edited by: FlyingForFun ]

[ 10 September 2001: Message edited by: FlyingForFun ]

Flyingcircus.
10th Sep 2001, 20:21
FFF. Thankyou for your solution. I should have realised that there would have to be two questions for there to be two solutions. I must admit, though, I haven't come across the "Weighted Average" method before, but it seems to work. I shall have to reserch that in future, when I've got a spare month or so. (That's probably how long it'll take me to get back into it!!!)

Thanks again.

The Guvnor
10th Sep 2001, 20:23
After reading Flying Circus's post I needed to go and lie down as my head hurt - and I had horrible flashbacks involving one of my former maths teachers that I am convinced was a Gestapo officer in a former life. :D :D :D

Said flashback mainly revolved around being whacked over the head with a ruler (one of the long blackboard ones) and being told "Read the questions carefully you bloody boy" (He wasn't one of the more PC teachers) - and indeed I see that I have fallen into a trap with the correct answer being 2/11 as it doesn't specify which die has 5 on it.

Note, however, that I have given myself a bit of a get-out with my closing sentence in my first post!

Well done Flying_Circus and Flying For Fun!

FlyingForFun
10th Sep 2001, 20:41
Weighted averages have lots of uses.

Good example is working out an average share price. Imagine a particular stock which has been bought by two dealers in a given day. (Not very popular, this stock!)

The first dealer bought 1000 shares at \$1.00 each. The second dealer bought 2000 shares at \$1.10 each.

What's the average price the stock has been bought at?

((\$1.00 * 1000) + (\$1.10 * 2000)) / 3000

Which gives you \$1.067. (Not \$1.05, which is the average price before you consider the weighting.)

Note that the second dealer's price has more "weight" than the first, because he bought more stock.

If you're trying to understand weighted averages, I think this example is much easier than the dice, because it's easier to visualise.

FFF
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[ 10 September 2001: Message edited by: FlyingForFun ]

Icarus
10th Sep 2001, 21:58
This is an example of Conditional Probability; which tells us the probability that a particular event will occur if we already know that another specific event has occurred.

Pr(A | B) = Pr(A n B)/Pr(B)

The question can be phrased as;

A = Event of getting a total of 8 on a pair of dice.
B = Event of getting a 5 on first roll; Pr(B) = 1/6
Calculate Pr(A | B).

The solution is,

A n B = event of getting a 5 on first roll and 8 total.
A n B can occur only if we roll (5,3), so Pr(A n B) = 1/36.

Thus,

Pr(A | B) = 1/36/1/6 = 6/36 = 1/6

:p

[ 10 September 2001: Message edited by: Icarus ]

golden_hands
11th Sep 2001, 00:04
Icarus, I'm not sure that you are correct! It seems you roll the dice after each other, while in the question the dice were rolled at the same time. In your case you leave out the possibility that the second may become also a 5. While rolling them at the same time gives an extra opportunity for the second to become 5 and the first one to be a 3.

The chance if you have 8 where a 5 is involved after rolling the dice is 1/6*1/6 + 1/6*1/6 = 1/18.

Now if we are only allowed to count the times when a 5 is involved things change a little. The nr of possibilities a 5 is involved is 11 out of 36. The required nrs are 3,5 or 5,3 which is a chance of 2/36 the chance of 8 when only a 5 and 3 are involved is 2/36 / 11/36 = 2/36 * 36/11 = 2/11.

Hope I got it right :confused:

Mac the Knife
11th Sep 2001, 01:59
Do you know that there are right and left handed dice?
That casino dice are normally RH:000 dice ?
That Sicherman dice are spotted wrong but have the same probabilities as normal dice.
http://users.erols.com/ee/dice.htm http://www.learn-gambling-betting.com/dice-cheats/ http://www.gamecabinet.com/rules/Dice.html http://users.erols.com/ee/d-im-col.htm http://www.quixotic-solutions.com/Cheating_Internet.htm

are some of the sites I wasted time reading following this thread

I learned quite a bit

Icarus
11th Sep 2001, 09:03
The question was:

"Two ordinary dice are rolled. If it is known that one shows a 5, what is the probability that they total 8?"

It is irrelevant if they are rolled together or one at a time. One die has been observed to show a 5 (either because they were rolled separately, or because one die (the 5) stopped rolling before the other and then we work out the probability of the other die being a 3 before it stops rolling); therefore it is Conditional Probability as explained above.

Mr Creosote
11th Sep 2001, 09:29
The wording of the original question was potentially misleading.
What if we re-word it to read:

Two ordinary dice are thrown. What is the probability that they total 8, with one of them being a 5 ?

There are clearly only 2 ways out of 11 combinations that this can happen.

HugMonster
11th Sep 2001, 12:57
Sorry, Mr. Creosote, it's not misleading at all.

As the question is put, there is only one answer - 1/6.

If he wanted the answer 2/11, he should have asked the question "What is the probability of rolling a total of eight, with one die being a 5"?

FlyingForFun
11th Sep 2001, 14:37
Sorry Hug, but I disagree.

The answer to the question "What is the probability of rolling a total of eight, with one die being a 5?" is 2/36 (or 1/18). If you roll the dice 360 times, approximately 20 of those times you will get a total of 8, with one die being 5.

The question is "What is the probability of rolling a total of eight, given that one die is a 5 (but we don't know which).

Now, out of the 360 times you roll the dice, only 110 of them (approximately) will show one die being a 5, so you can discount the other 250. And of those 110, only 20 will add up to 8. Hence 2/11.

(I think most of us now agree this is the answer, it's only the wording of the question to make it as clear as possible which is still in dispute.)

FFF
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Icarus
11th Sep 2001, 15:57
The phrase"If it is known that one shows a 5" clearly indicates that one specific event has occured (Conditional Probability).

Based on the mathematics for Conditional Probability the answer or 1/6.

Any other answer is arguable depending upon your (mis-)interpretation of the question!

Flyingcircus.
11th Sep 2001, 16:59
Conditional probability. Weighted averages. AGHHHH....What have I done?? :eek: :eek: :D

I may have to look up my maths teacher again and show him this thread (in hardcopy form of course, as I'm quite sure he doesn't trust those 'new-fangled' computer thingies!). The answer, as I think I have posted, was 2/11. I lost a few marks in a really important exam due to this particular question, and it's great to have it finally answered (as much as I can understand, anyway).

I think that this website is a great resource, giving access to the opinions and experience from people all over the world. I'm sure this wasn't quite what Danny envisaged when he set this up, but it's often the unexpected side effects that provide the greatest benefit.

Keep it up guys, this is great! :)

[ 11 September 2001: Message edited by: Flying_Circus. ]

FlyingForFun
11th Sep 2001, 17:09
Icarus,

We do not know that one specific event has occurred. We know that one of two events has happened.

Let's assume that the dice are coloured: red and green.

Now, the question is "What's the probability of a total of 8, given that either the red die is showing 5, or the green one is showing 5."

With that, you can either draw up a list of all 11 possibile ways the "either the red die is showing 5, or the green one is showing 5" as flying_circus did, note that there are 11 possibilities (R5G1, R5G2, R5G3, R5G4, R5G5, R5G6, R1G5, R2G5, R3G5, R4G5, R6G5) and that only 2 of those add up to 8. Or you can take the more theoretical approach that I took, and get the same answer.

But, to re-iterate, the way in which most of us have interpreted the question, we do not know that one particular event has happened - merely that one of two events has happened.

FFF
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captainowie
11th Sep 2001, 17:49
Sorry to interrupt guys, but I just had to share these proofs.
1)
All Women are Evil.

As we all know, women require time and money.
W=T*M
And as we also know, Time is money.
T=M
Therefore Women = money times money = money squared
W=M^2
Now, Money is the root of all evil.
M=E^1/2
Therefore M^2=E.
Equating M, we arrive at the solution that:
W=E. All Women are Evil

2)
Salary Theorem
The less you know, the more you make.
Proof:

Postulate 1: Knowledge is Power.
Postulate 2: Time is Money.
As every engineer knows: Power = Work / Time
And since Knowledge = Power and Time = Money
It is therefore true that Knowledge = Work / Money .
Solving for Money, we get:
Money = Work / Knowledge
Thus, as Knowledge approaches zero, Money approaches infinity, regardless of the amount of Work done.

Just a thought! :D

HugMonster
12th Sep 2001, 06:46
To prove that 2=1:-

Assume:- a=b

Therefore a^2=a*b (multiplying throughout by a)

a^2-b^2=ab-b^2 (subtracting b^2 from each side)

(a+b)*(a-b)=b(a-b) (factorising each side)

(a+b)=b (Removing (a-b) from each side by division)

From the first assumption we can sustitute a for b so:-
2b=b

Therefore
2=1 (Dividing throughout by b)

[ 12 September 2001: Message edited by: HugMonster ]

captainowie
12th Sep 2001, 16:09
Well done HugMonster. The flaw is when you divide both sides by (a-b). Since a=b, a-b=0, and division by zero is just not done, or else you get strange things like 2=1.

Another couple of Proofs.

1) All horses are the same colour.
Proof by mathematical induction.
One horse is one colour, so it has to be the same colour as itself.

Next, assume a group of K horses are all the same colour.

Then take a group of K+1 horses. Remove one horse from the group. What is left is then a group of K horses, which, from our assumption, are all the same colour.

Place the horse back in the group and remove a different horse. What is left is also a group of K horses, also all the same colour. Repeat this process for all K+1 horses.

Thus if one horse is the same colour, and K+1 horses are the same colour if K horses are the same colour, then all horses are the same colour.

2) All horses have an infinite no. of legs.
A horse must have an even number of legs. If not, it would overbalance in one direction or another.

Now, we all know that a horse has forelegs in the front, and two legs in the rear. 4+2=6. Six legs is certainly an odd number of legs for a horse to have.

The only number that is both even and odd is infinity. Therefore horses have an infinite number of legs.

If there were a horse that had a finite number of legs, well, that would be a horse of a different colour, and, from Proof 1), that dosen't exist!

Icarus
13th Sep 2001, 09:14
Just to get back on topic and the original question.

The answer can only be 1 in 6 (1/6).

FFF
The question is phrased, IF IT IS KNOWN, therefore one die (either of the two, BUT the first to be looked at) has been observed and it is a 5, hence one specific event has occurred and there now exists a 1 in 6 chance of a total of 8 across the two dice.

You cannot ‘conveniently’ forget the value on the other die if your first observation did not show a 5 as your earlier post said.
If you look at Die-A and it is a 5, the probability of a total of 8 from the two dice is 1/6.

If Die-A is not a five and you check the other die and it is (5) then the probability of a total of 8 is either 0 (zero) or 1 (one).

Mr Creosote
13th Sep 2001, 10:13
Icarus

Suppose I handed you a closed box containing 2 dice. I tell you that one of the dice shows a 5. What is the probability that the two dice add up to 8?

It is known that one dice is a 5 ('cos I told you) but not which ('cos you haven't seen it). :)

Icarus
13th Sep 2001, 11:30
Mr Creosote.

Ahh. Is the cat alive or is it dead?

From your point of view, if you have observed 1 value only (5) then the probability is 1/6.
If you have observed both it is either 1 or 0, but then why ask (yourself)the question if you already know the answer! Actually, if you have observed both, then no probabilities exist only an acuality.

From my point of view it is irrelevant because the actual probability (either 0, 1/6 or 1) has already been determined in the real word by a third party (you).

Any attempt by me to provide the correct answer is a futile exercise! ;)

Furthermore, the question says "It is known that one of them is a 5".

It does NOT say "It is known that ONLY ONE OF THEM is a 5."

Consequently that means R5:G5 and G5:R5 are valid options (remember I dont know which one was observed, red or green) to consider, which was not shown in the preceding example.

That gives 2/12 (1/6) and not 2/11 as the answer!

[ 13 September 2001: Message edited by: Icarus ]

Wizdum
13th Sep 2001, 11:45
If a tree falls in a forest and nobody hears it....

.... does anyone care?

Taking the word "pedantic" to uncharted territories here on PPrune... LMFAO