Rolling two dice. You see one of them is a 4, what’s the probability that the other is a 6?
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- Опубликовано: 14 апр 2018
- Rolling two dice at the same time. You see one of them is a 4, what’s the probability that the other is a 6? This is a must-know probability problem for your statistic or probability class.
The answer is really NOT 1/6. Please see this video: • two essential conditio...
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The answer is really NOT 1/6. Please see this video: ruclips.net/video/pOUwaj2y6u4/видео.html
this is just another video where you didn't ask the question precisely. 1/6 is absolutely a valid answer to the question.
Thanks for this man. At 44yo finally someone explained me a thing that I learned many years ago but never _really_ understood.
My pleasure!!! Thank you for your comment!
You don’t understand how helpful your videos are. I love how you give more than one method to solve the question, it truly enables us, the viewers, to understand the topic in depth.
Isaiah Isijola awww thank you!!!!! I am very happy to see your nice comment, yay!!
hey black pen red pen, I was thinking about how 4 in this case is just an arbitrary number, so say that given that the first dice is not a six the probability of the second dice being a six is still 2/11?
The probability is the same for any specific number other than 6. That is, it's the same answer for 1, 2, 3, or 5. But if you are asking if at least one of the dice is _anything but a 6_ then the probability that the other dice is a 6 is 10/35.
Proof: There is only one case out of 36 permutations of 2 dice where it is NOT true that one of the dice is not a 6, namely 6-6. So, out of the 35 remaining permutations 10 satisfy the condition that the other dice is a 6.
Nice to see som varied content on this channel! Keep up the good work
Yay!
blackpenredpen, I am an advanced math undergraduate who does some freelance tutoring on the side. I started watching your videos back in the day when you mostly very basic algebra computations because of your clarity and style so that I could become a better tutor by watching you. You have gotten better over the years, and I've enjoyed watching your channel grow. Recently, you've started branching out into some really nifty computations, and I often found myself spending an hour or two on some of your trickier problems, only to find myself floored by the simplicity of the methods you use to effortlessly arrive at the same elegant conclusion. I no longer watch your videos just to observe your style; you have captivated my attention, and I eagerly await your new videos with fervent anticipation. Keep on doing a great job and being a real champion of the fundamentals--I'm humbled by your talent for presenting stunningly simple computations with supreme clarity.
Alexander Sanchez
Wow, thank you so much for your awesome comment!! I wish you the best luck in everything!!
I thought you only love integrals :D
Anonymous sometimes ;)
The dices don't get replaced or not get replaced and here it says "given b is 4" and here says what's the probability the other is 6 and simply they are independent events because when we roll a dice we won't get a section removed and to solve it we use P(A|B) And They Are Independent So The Answer Is Just 1/6
The 1st roll being 4 doesn’t impact the second roll. They’re independent. This is asking the probability that the second one is a 6, so isn’t the first bit of information useless ?
I was never good at probability either.
Edit: I’m about to watch the “why isn’t is 1/6” video
Edit 2: it makes sense now. I was assuming additional conditions where I shouldn’t have.
the wording is super troll. I also got 1/6 and after watching the video, I still believe that 1/6 is a correct answer to a poorly worded problem.
What my teacher failed to do in 150 lectures, he did in 15 mins.
Truly amazing...
I thought the two dice should be independent of each other which would make
P(A and B) = P(A)P(B) "if independent"
P(A I B) =P(A and B)/P(B) = P(A)
The reason why I would argue this question should be treated as the dice rolls are independent of each other would be that you have already observed that 1 dice has a set value making its probability fix, making your chart a 1 by 6 instead of a 6 by 6.
That's when you roll a die twice, what's the prob. that the 2nd time is a 6 given thae the 1st was a 4.
and the answer is just 1/6
It doesn't seem intuitive to me that the events are dependent because of when they are rolled, can you clarify this? Because the result of the second roll shouldn't be affected by observing the first
I agree I would like some clarification on this. I always enjoy watching BPRP's videos.
he means you're rolling them both at the same time, not one after the other. that's what makes it dependent
It is worth mentioning that the word or has a different interpretation in English than math. In English it is usually exclusive e.g. "let's eat pizza or Chinese food" means you will do one or the other but not both. In a probability context it is usually inclusive e.g. roll a 6 on the first or second die includes a 6-6. In programming we use or (inclusive) and xor (exclusive) to distinguish.
Yes. you are right!
I first made a comment agreeing with you, but I think you missed the point.
Subtracting the probability that both of them are 6 from the total is not an issue of inclusion but of double counting, as he explained.
There is a 6/36 probability that the first one is 6 and also a 6/36 probability that the second one is six, but in one of those 6 cases, the other die is 6 as well. (We already counted that case when we counted the cases for the first die).
So, in conclusion, the probability of either being a 6 is 11/36. The probability of either being 6, but not both, would, in fact, be *10*.
@@cynicviperyes, but why did he subtract the probability of both being 6 and still got 11/36?? I understand it made us not count that special case (both being 6) twice, but in “11/36” both being 6 is still a case we're taking into account! So why did we still take into account that case even though we subtracted its probability of happening?
@salva1519 Maybe I wasn't clear enough, I hope I can be clearer now.
The case where both dice are 6 is in common to both dicerolls.
Imagine we both have a number of books ourselves, let's say each 2 for felicity. In addition, we have 1 book in common. I can say I have 3 books and you can say you have 3 books, but if we count the total number of books (including the one that we have in common), it's 5, i.e. 3 + 3 - 1 (adding each of our books, then subtracting one to remove the double counting of the book in common)
Or we could find the sum by adding the books we have in private and then adding the one we have in common
2 + 2 + 1.
(With the dice case, it'd be 5/36 + 5/36 + 1/36)
I used to think of the sample spaces in the form of Venn Diagrams. This tabular method changed my perspective, from now on. It's very helpful!
Thanks!!!!
"Friendship ended with Venn Diagrams, now sample space is my new best friend."
@@m0rtez713 Indeed lol
Thank you, I have never heard of the sample space method. It helped me so much! It's a great way to visualise the problem, and thanks to it i now have a much better and clearer understanding of the problem.
Wow, this guy is a master on everything in math.👍👍👍
I'm late to the party, but considering how the question is worded, the answer should be 1/6. If you didn't yet observe that one dye was a 4, but instead asked what the probability the other dye would be a six if one of them was a 4, then this explanation is correct. However, that wasn't the question asked. KNOWING the outcome of a dye vs. PREDICTING the outcome (and the subsequent dependency of that prediction) are not the same.
I had this in school and never truly got it. I could do the calculations correctly but never knew why it worked. The last explanation with the "space" made it click for me. Thank you! :)
Habbopingvinen I am glad :)
That was awesome!
As a freshman, in high school, taking Algebra 2, I can definitely say that probability is the most terrifying thing there.
OMG!!! that's awesome, we love you, make more videos, please
Excellent explanation! Cheers.
Probability is my weak point of maths. This video really helped me!
Love you man never stop uploading
The last part of the video make me finally understand this formula ! Thanks !
That's the secret weapon, sample space!!
But the problem of that is, say once we roll 10 dices, we can't draw that anymore...
blackpenredpen if you have a good 10D visualisation, no problem
Abathur but I can't draw lll
Lol*
The wording of the problem is ambiguous. If we assume that the person rolling the dice saw BOTH dice, then he we wouldn't say "one is a 4" if both were 4. So there are then only 10 cases, not 11 where one die is a 4 and the chance that the other is a 6 is 2/10 = 1/5. On the other hand, if the person saw only one of the two dice, the chance that the unseen one is a 6 is 1/6. The answer 2/11 applies in the case where the person sees BOTH dice and AT LEAST one of them is a 4. That is clearly how BPRP interpreted the problem, but when I saw the problem statement I asked myself what other reasonable interpretations could apply, and came up with 3 possibilities in total with answers 1/5, 1/6, and 2/11.
Amazing part of the box on the end!!
Practically the answer is 1/6
If someone tells you "one of the dice is 4" it means that he looked at a die and saw a 4
The probability of the other being 6 is just 1/6
great video!
Great explanation. Really enjoyed the video.
Thank you!
The original question could have been stated more clearly. There is an essential difference between 3 interpretations of it
Q1. 'We can see only one of them, it's a 4. What is the chance for the other one to be a 6?'
Q2. 'We can see them both and exactly one of them is a 4. What is the chance that the other is a 6?'
Q3. 'We can see them both and at least one of them is a 4. What is the chance that there is a 6?'
Without pen and paper, I would say that A1=1/6, A2=1/5, A3=2/11.
Here's something: What is the amount of unique combinations after rolling 3 dice at once? Ex. (1,1,2) (1,2,1) (2,1,1) are the same.
Thank you SO much at explaining it. You, Khan academy, 3B1B and Examsolutions.net have saved me from giving up and looking another time.
really good mate thank you
If you are watching and learning from this video, you are making good use of youtube. Thanks Steve 👍
The end part was quite insightful! :) Love your videos!
252 subs to go
I realized the answer partway through, but thought a really nice reason for the equation being that way while you were showing the sample space. Since the probability is a fraction where the numerator is the amount of successes, and the denominator is the number of outcomes, there is a really easy way to think abut it.
For the numerator, the P(A and B) counts every time A happened while B happened. This is the number of successes in the conditional probability since for all the times A happened, did it occur in the sample space of the conditional probability function.
For the denominator, the P(B) gives the number of total outcomes, or the size of the new probability space. This is because a conditional probability is just limiting the sample space to meet a condition. This means he size of the sample space is P(B).
Both the numerator and denominator parts of the conditional probability function end up being a fraction, because they are probabilities, but the only parts that matter is the numerator. In the example given the denominators are the same and got canceled. I noticed that this will always happen because this is the size of the total sample space. Since both probabilities happen of the same sample space, they have the same denominator. This may require changing the bases of the probabilities so that they both have a denominator equal to the size of the sample space, since it may have been simplified.
Great video, i learned something. But i have a little question. How do we know when to add and when to multiply when we are playing with probabilities. I think it is very easy to make a mistake here, so i just wanted it to be clarified if possible
Thank you so very much sir.
Nice explanation.
I solved it using the second method you've shown. There are 11 cases in which you see a 4, out of those, 2 have the other one as a 6, so 2/11 :D
EDIT : it actually depends on what you call "one of them is a four". Is it "only one of them is a four" or "one of them is a four, and the other might be as well" ? Is the first case, my answer is the correct one, in the second case, you answer is the correct one. It is not clear throughout the video though.
I disagree with P(1st=4 or 2nd=4) :
P(no 4 at all) = 5/6 * 5/6 = 25/36
P(at least one 4) = 1-25/36 = 11/36
P(1st=4 or 2nd=4) = P(at least one 4) - P(both 4) = 11/36 - 1/36 = 10/36
Final answer should then be 1/5
On your drawing at the end you shouldn't count the (4,4) box, because you already excluded it ! So there are 10 boxes corresponding to "one of them is a 4", and two of which are "there is a 6", so 2/10 = 1/5.
If u see all the results of rolling 2 die in 11 outcomes u can see 4 and out of them in 2 u can see 6
I find it really important to have clear wordings in such "riddles". Here "one of them is a four" has the meaning "at least one of them is a four" or "you see one of them and that one is a four". If you see that "exactly one of them is a four" (which can be implied by the text in some sense) the answer would be trivial 1/5.
Another fact that comes into place: here the order is not mentioned. If it were mentioned "the first is a 4" the second would be independent of that throw.
Now: A family has two kids. One (at least) of them is a boy. What is the probability for the other being a girl? ( P(X=boy)=50% )
"You see one of them and that is a four" is too much information that would result in the probability 1/6 for the other die. The important part is that you don't know which one is a 4 so you can't distinguish between them. Knowing "this one is a 4" is strictly more information than "One of them is 4". The wordings could be interpreted as the first one which a lot of people did as one can see in the comments but the second one is the correct one if you are a critical thinker does only uses information that is actually given. I agree that riddles should be clearly worded.
Same problem, reworded: The Mudra family has 2 kids. You have met neither of them; you know (at least) one of them is male. You are told one of the Mudra children will visit. The probability of that child being female is 1/3. Agree or disagree.
would you agree that critical thinking is more about making good inferences about things not expressly stated, as opposed to following the letter of a question? Like if I asked a simple kinematics physics problem, I assume you would ignore quadratic air resistance, length contraction, gravitational time dilation, differences in gravitational strength due to non-uniform earth density and shape, etc. because that clearly isn't practical and therefore by inference isn't probably the answer I'm looking for. Even though that would be a more corret answer given the parameters of the question.
I agree that this is also an important trait. Still a question cannot rightly be answered if it is not stated accurate enough. And I also think that in probability questions things can change a lot with one word more, or less in the task.
Thanks bro it’s really help me
wow probability by bprp ! I guess it's the first time
LEGEND_CLASH probably? Lol
But it's certainly not the last time
More on probability and polynomial/root function and graphs plz!
More and more probability videos please.
Will do!
Very insightful, but I guess that's to be expected from your videos
Thank you!!!!
Maybe you could do a video explaining where does the P(a|b) formula, keep up the good job!
Wu Hoo
Is there a way to hold 3 different markers?
Chrnan6710 with three hands, all is possible
He tried it in a recent video but it didn’t go too well.
The thing that makes statistics such a dubious practice is that it has no power to predict future events. It can only model events that have already occurred. Imagine the question was, "If you roll two dice and one falls on the floor where you can't see it. What's the probability that the other one is a six?" Then the answer would be 11/36, the number of ways either die could be a six. See how, just by observing one die, the probability changes? This is because we're not talking about rolling one die then asking what WILL happen to the next one. Conditional probability only determines the likelihood of finding a particular record of past events based on knowledge of part of that record.
awesome thanks
gr8 video
doing a 100,000 subscribers special?
You'll see!
countdown to 100,000 subscribers!
Is the or for the probability of both of the dice to be a six an exclusive or?
I hate the wording of this question. the "look" part makes this super troll. If there are two dice and I look at one of them and it's a four the chances the other is a six is 1/6.
In your question someone actually looked at both of them and reported back that at least one of them is a four. Given THAT what are the chances that one is a six.
And of course if I looked at both dice then there aren't any chances anymore. I know if one of them is a six or not. It's collapsed.
Beautiful presentation. I live Kolkata, India. I eagerly wait tor your videos for outstanding presentation. I have a request. Please make a video on the intgral from negative infinity to positive infinity e tohe power negative x squre dx.
Thank you.
P(one of them is 4) should contemplate both as 4 as a previous condition, giving a 12 possibilities sample space, in which 2 of them is 4,6 or 6,4.... prove me wrong please!
Besides, if this is true, the answer is 2/12 or 1/6..... making perfect sense since the possibility of a dice giving any number is 1/6 no matter what other dice gave as a result...no?
Hi bprp! I really liked video on probability and combinatorial. I have thought a problem: if I choose two numbers from 0 to 100 randomly, what is the probability that the sum of the numbers is 75?
(2*75)/(101*101)
nCr(76,1)/nCr(101,2)
The two numbers are independently chosen so we can look at them seperately. The first number needs to be
Luca Zara I prob can make a video on this by next week.
Wrong answer, see my comment for the correct one. But I am curious how you got to that answer cause Torgeir Kruke wrote the same, can you explain?
Does that mean that knowing the position of the 4 drastically changes the chances of the other die being a 4?
So if you know that the second die is a 4 then the possibility space is the fourth column and the chances that the first die is a 4 is one in six. Same if you are given that the first die was a 4. But if you're only given that one of the dice was 4, not knowing if it's the first or second, changes the probability of the other one being a 4 to one in eleven. Is this right or am i missing something here?
You have it right there :) If we would for example know that the first die rolled a 4 then we would know nothing about the second one -> independence. But here we know that one of them is a 4 and not which die rolled it, so if we look at the first die and see a 3 for example than we would know that the second one must be the 4 -> dependent
Logically speaking, that or is exclusive ;)
Please more stats :D
10:54 it doesn't even matter
hope you get the reference. otherwise you've missed some great music
Yup, linkin park
coool
I use the sample space wherever possible to check my work. 😁
do a video on sequence and series
Eerav Sameer I have tons of them already. Check out my calc playlists
Shouldn’t the question specify that the dices are different? Because if they are equal, which is a standard assumption, the answer is just 1/6, no?
Two non-different dice are one die. All two dice are two different dice, so specifying that wouldn‘t make a difference. The probability is not 1/6 *because* you can‘t distinguish the one that rolled 4 from the other one as it not given which die is it. If it were mentioned which die is the guaranteed 4 then it would be 1/6.
+Gonçalo Ferreira:
Regardless of whether you can tell the difference, there ARE two dice. This makes 36 possible outcomes. Looking at which outcomes have at least one 4 narrows this down to 11 possibilites. Two of those possibilities also have a 6. Hence the answer is 2/11. That's the answer, regardless of whether you can tell the difference between the dice or not.
Ironically, if you can tell the difference--say one is red and one is white, and I tell you that the red one is a 4, the probability that the white one is 6, is 1/6. BlackPenRedPen shows this in his followup video. This is the opposite of what you suggest.
Good gracious if the math teacher didnt know.
The first part where you calculate the prob of one of the dice being 6 is wrong. 11/36 is the probability that one or both are six: 1 - (5/6)(5/6). The probability that only one of the dice is a six will be (11/36)-(1/36) = 10/36. This is easy to see also if you consider that there are 36 ways to combine two dice, and 10 ways where only one dice is a six: Either The first is a six and the other has five ways to not be a six, or the second is a six and the first has five ways of not being a six.
The logical or is always either or both, always. What you mean is the "exclusive or" (xor) which wasn't asked here.
Supremebubble In part I do agree with you, but the way he explains his reasoning he surly uses xor as an explanation. He gives this as the explicit reason why he subtracts 1/36.
Torgeir Kruke
Nope he subtracts 1/36 because of double counting. the case (6,6) would be accounted for twice if he does not subtract it. I explained this thoroughly in other comments already so check them out if you want to know more details
Is it common in statistics for 'or' to mean xor?
A or B means "just A, just B, both A&B"
that is andor. which is common is mathematics in general
2/11 uses the probability that at *least* one of them js four. What about *exactly* one of them is four? Because there are 11 scenarios where at least one is four, but only 10 where it has exactly one four, with 1/5 as the resulting probability. Either way, by some magic, the chance of the other being six is more than 1/6
sir can u plz add more videos on conditional probability and bayes theorem
plzzzz
i'm finding it difficult to understand
ふぁーーー
その書き方始めて知りました 便利だけど苦手なジャンルやわ～
Take A and B to be two separate events. Why would you subtract P(A and B) when trying to find P(A or B) ??? Written out as P(A or B) = P(A) + P(B) - P(A and B) ???? Does "A or B" exclude "A and B" ? As in, if A is true and B is true, then A or B would be true I'm specifically referring to 3:47, why did you subtract "P(both=6)" ?
Double counting. The event that both dice rolled 6 are within P(A) and P(B) so without subtracting we ould have counted it twice. Or let me put it differently:
The event that the first die is 6 has 6 possibilities namely (6,1), (6,2), (6,3), (6,4), (6,5) and (6,6)
That is why the probability is 6/36 = 1/6
The event that the second die is 6 als has 6 possibilities namely (1,6), (2,6), (3,6), (4,6), (5,6) and (6,6)
That is why the probability is 6/36 = 1/6
But the event that (at least) one of them is a 6 has only 11 possibilities namely (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (1,6), (2,6), (3,6), (4,6) and (5,6). It does not have 12 because the outcome (6,6) (when both dice are 6) would be counted twice. So we have 6+6-1=11 and so the probability is 11/36. Hope that helps
Supremebubble Ahhhh I see what you mean. In this case, (6,6) is counted twice. So the full list of (A=6 or B=6) is; (1,6)(2,6)(3,6)(4,6)(5,6)(6,6) and (6,1)(6,2)(6,3)(6,4)(6,5) , which in total is 11 possible cases. Thank you
No problem :)
I’m hoping someone can clarify something for me please. And to preface this - I’m not a mathematician, so there may be something very fundamental I don’t get. From watching this, I feel like the pretest and post test probabilities don’t change. By this I mean, it seems that the chances of a six being rolled doesn’t seem to be affected by whether or not you already have a four. I had though this would be akin to the Monty Haul problem, in that once one variable was shown or “fixed”, I would have thought (logically, not mathematically) that the chances changed. I guess another way of asking this would be how is this not the same as only rolling one die, and simply placing the other one as a four, since we aren’t calculating the probabilities until AFTER we know one is a four? Thanks for any insight!
I made a follow up video today ruclips.net/video/pOUwaj2y6u4/видео.html
blackpenredpen Saw the follow up - it was very helpful. It seems that the key here really is understanding the grammar. The math doesn’t seem to complicated, but it’s more a matter of knowing which parts of the question are truly fixed vs subject to probability. Thanks again for the explanation!
DrThot It's not really because of the grammar, it's more about the setups and situations.
P (1 die = 6)
1/6 x 5/6 + 5/6 × 1/6 = 5/18
Gucci gang
Rolling two dice should be an independent event, issinit?
I made a follow up video today ruclips.net/video/pOUwaj2y6u4/видео.html
But if you see a 4 already in one die, you could only get 1,2,3,4,5,6 for the other one... Shouldn't the probability of getting a six be 1/6? The fact that you got a 4 in one die shouldn't affect the probability of the other one... Can someone enlighten me? c:
I was thinking the same thing.
We know that one die is a 4. So getting a 4 on one die doesn't affect the other die. However *not* getting a 4 does affect the other result because the second one now must roll a 4 -> not independent
Because seeing a 4 eliminates the possibility of both dice being 1s,2s,3s,5s and 6s which increases the probability that at least one is a 6.
There's also the possibility that the second one's the four.
Perhaps the actual problem should be phrased as "...you KNOW that of the dice is a 4..."
I like probability
I am stuck with the following differential equation, could you please provide some hint or technique to solve it?
f'(x)=e^{-x(f(x))}
EDIT: With f(1)=1
:-)
Find the largest power of 2 that divides 200!/100!
Someone help pls: when he calculated the prob of one of them is a certain number out of 6. Doesnt he just has to say 1/6 (thats the prob of the first obe beeing the certain number) + 1/5 times 1/6 for the second dice because the first dice cant be a six and the second one has to be. Im actually pretty good ib math, but i just dont get that dude.
where does 1/5 come from? maybe you mean 5/6 times 1/6 because if you do you have the right idea. You could also add the probability of the first one being a 6 ( 1/6 ) with the probability of the second one being a 6 *given* that the first one is not a 6 (so 5/6 times 1/6) which gives us 1/6 + 5/6 x 1/6 = 6/36 + 5/36 = 11/36 the same and correct result. He probably avoided this way because this is also conditional probability
When card counting
I was about to say ">inb4 that "bear" is your channel icon", but lo' and behold, I was too late to the punch, APPARENTLY...
How come the or/and operations with probability is the same as the one with sets?
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
{A ∪ B} = {A} + {B} - {A ∩ B}
1/5 you missed keys words which meant sample space did not include the option for two fours
The probability that one die is 4 is 5/18. The probability that at least one die is 4 is 11/36.
Pierre Abbat 😊
i saw the thumbnail and got scared because my name is ethan lol
God i can't stand probabilities, id much rather do a nasty integral that find out the probability of drawing cards in a specific order. Thank god i don't need a class on this for computer science.
This is mind blowing and I still.dot get it. it just seems like it should be 1/6.
Let me make this the most clear I can for you, so you have the full picture:
Let's say we know that the first die is a 4. What is the probability that the second die rolls a 6?
Given the information that the first die rolled a 4 we know that all the possible outcomes are (4,1), (4,2), (4,3), (4,4), (4,5) and (4,6). Each of them is equally likely and the only outcome that has a 6 as the second die is (4,6). So one outcome is the one we want and we have 6 equally likely possibilities that makes 1/6 as our chance. This was the intuitive part.
Let's now come to the actual question. We now know that one of the dice rolled a 4. What is the probability that the other one rolled a 6?
Given this information we know that from the original 36 possible outcomes for the two dice (6 times 6) we only have the following 11 outcomes as possible outcomes:
(1,4), (2,4), (3,4), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,4) and (6,4)
Why? Because these are all the outcomes where one of the dice is a 4. This is what we know about our rolls, this is the given information and all of these outcomes are obviously equally likely because dice are always fair in our exercises. So how many of these outcomes have a 6 as the other die? There are (4,6) and (6,4) so we have 2 out of 11 equally likely possibilities which gives us the probability of 2/11.
This is the elementary explanation of what is going on. Hope it helps :)
Supremebubble Thanks so much.
Can you say why I'm wrong here: Probability (getting 1 six) = 1/6 x 5/6 + 5/6 x 1/6 = 10/36
I thought of it as: probability of getting six in first roll but not a six in second roll + probability of getting 6 in second roll but not in first roll.
I made a follow up video today ruclips.net/video/pOUwaj2y6u4/видео.html
Anteater23 you are missing the case where both dice roll a six. this has the probability 1/6 x 1/6 = 1/36 which should be added to your 10/36 as it is another possible case giving you the result 11/36. Don‘t forget that the statement „one of them is 6“ always means „at least one of them“. You could have done the following also:
probability that the first die rolls a 6 times the probability that the second die rolls a 6 *if* the first one didn‘t roll a 6. This is plausible cause if the first one rolled a 6 you don‘t have to look at the second one cause your desired outcome already happened. If however the first one didn‘t give you a 6 (which has the probability 5/6) *then* you have the look at the second one as this one also has a 1/6 chance of you giving you a 6. Overall this leads to 1/6 + 5/6 x 1/6 = 6/36 + 5/36 = 11/36 which is again the correct result derived differently.
Supremebubble Well if it means probability at least one is a six then why did he takeaway 1/36 in the video?
Doraemon theme XD
Can somebody explain why the answer isn’t 1/6th, since you know the first die is a 4 you only have the second die’s probability to worry about, and the probability of the second die being a 6 is 1/6th please help
The thing is: You don't know that the first die is a 4. You know that *one* die is a 4. It could also be the second one and that is the key thing here. If you would know that the first one is a 4 then you are absolutely right about the second die having a 1/6 chance of being a 6. But we have less information than that. Let me make this the most clear I can for you, so you have the full picture:
Let's say we know that the first die is a 4. What is the probability that the second die rolls a 6?
Given the information that the first die rolled a 4 we know that all the possible outcomes are (4,1), (4,2), (4,3), (4,4), (4,5) and (4,6). Each of them is equally likely and the only outcome that has a 6 as the second die is (4,6). So one outcome is the one we want and we have 6 equally likely possibilities that makes 1/6 as our chance. This was the intuitive part.
Let's now come to the actual question. We now know that *one of the dice* rolled a 4. What is the probability that the other one rolled a 6?
Given this information we know that from the original 36 possible outcomes for the two dice (6 times 6) we only have the following 11 outcomes as possible outcomes:
(1,4), (2,4), (3,4), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,4) and (6,4)
Why? Because these are all the outcomes where one of the dice is a 4. This is what we know about our rolls, this is the given information and all of these outcomes are obviously equally likely because dice are always fair in our exercises. So how many of these outcomes have a 6 as the other die? There are (4,6) and (6,4) so we have 2 out of 11 equally likely possibilities which gives us the probability of 2/11.
This is the elementary explanation of what is going on. Hope it helps :)
Supremebubble yeah helped a lot thank you so much
I made a follow up video today ruclips.net/video/pOUwaj2y6u4/видео.html
Isn't the given "B" here ?
Casino card counting
why isnt the 4 useless...?
It makes me so mad when I get a probability greater than 1.
My brain is fucked
I don't get it:
The possible Solutions that ONE of them is four:
4/1, 4/2, 4/3, 4/5, 4/6, 1/4, 2/4, 3/4, 5/4, 6/4 => 10 possibilities out of 36.
Calculating:
P(1st one is 4, 2nd different) = 1/6*5/6
P(2nd one is 4, 1st different) = 5/6*1/6.
P(one of the upper 2 is happening): 5/36*2 = 10/36.
Where is my mistake?
If the question is to be interpreted in a different way, e.g.: You throw 2 dice, one of them is 4, the 2nd one can be 1,2,3,4,5,6 - what's the chance that it is 6? -> then the answer should be 1/6, as the 1st dice is irrelevant.
Where is my mistake?
Maybe it's just my poor english skills... But if not, I'd propose to make questions more "well-defined" without any chance for mis-interpretation. That would make it easier to follow.
no, it's not!
?
blackpenredpen DON’T question Dirac.
Deep inside, you know it’s not!
Unser liebe Herr würfelt doch nicht
0:36 it's not so much fun :D
And most parts aren't hard, they are only extremely counter intuitive. that's what makes them tricky
Heisenberg: "There is no god...."
This is so counter-intuitive. I know this answer must be correct, since the formula has been used correctly, but shouldn't common sense say the answer is just 1/6? Because the rolls of each dice are independent. I don't get why it is higher than that
I made a follow up video today ruclips.net/video/pOUwaj2y6u4/видео.html
Oh btw, I should also recommend you to check out "Monte Hall's problem"