I at first assumed it was 1/3. I was so confident that it was 1/3 that when it was revealed that the answer was 4/11, I spent the next half hour trying to process it. But then I came to a startling revelation. If we slightly rephrase the question, it would be 1/3. Instead of saying "atleast one dice rolls a three", what if we say "Dice 'A' rolls a three"? It completely changes the answer! The simple fact of knowing which specific dice was a guaranteed three, creates an entirely different possibility of outcomes!
@@nicholasogburn7746 If you're talking about the overlap, then you didn't catch his issue. The issue was using 6 instead of 6+6 in the sample space. The -1 less mind-boggling IMO.
i dont understand why its not 1/3. why does it matter which one rolls a 3 first. because either way arent u only going to have 6 possible numbers other than 3?
Question wording is wrong (it confuses) !!! This question must be asked in two ways : 1. Two dices are rolled, it is given that 3 appears on atleast one dice, find the probability of getting sum greater than 7. (Answer is 4/11). 2. Two dices are rolled, find the probability of getting 3 on atleast one of the dices and sum of dices is greater than 7. (Answer = 11/36 x 4/11 = 4/36 = 1/9)
Yes it does, and the result (4/11) includes this possibility. I think you might be confused because of the overlap. If you were to select the outcomes he marks (row 3 and column 3), and count them individually, you would also get 11.
@@beefy420 Exactly. Your options are (in terms of 1st die/2nd die) 1/3, 2/3, 3/3, 4/3, 5/3, 6/3, 3/1, 3/2, 3/4, 3/5, 3/6. You don't count 3/3 twice, because it's the same exact result, so it overlaps.
I like the videos and the teaching style here, but in this case, I disagree with the interpretation. If one of the dice is already known to be a 3, then all we’re looking for is the probability that the other die is greater than 4 (ie 5 or 6). That’s a pretty simple 2/6. In other words, one of the dot circles must be shrunken to encircle only a single dot because we know one of the dice is a 3.
It is assumed that the dice rolling 3 could be either the first one or the second one. If it were just the first one, the answer would be as you say [(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)] with last two combinations satisfying the condition. However, if you consider the possibility of the second dice rolling to 3 irrespective of the outcome of 1st dice, you will be looking at 6 combinations [(1,3),(2,3),(3,3),(4,3),(5,3),(6,3)] of which 2 satisfy this condition. If you combine these both, and exclude any element that is duplicated, I.E., (3,3), you will be left with 4 possibilities out of 11 combinations.
@@amulyam.339 the probability of rolling a dice at least once is 1. The probability of rolling a number which sum will be greater seven is 2/6. P(1) U P(2/6) = 2/6 The three is the first or the second throw so times 2. Its 4/6.
That’s what I thought for sure, too. It doesn’t matter which one is a 3, but we know that one of the dice is a 3, and the other one is 1 out of 6 different possibilities, and 2 of those 6 are greater than 7, so it seems like the answer should be 2 out of 6. In other words, if one of them is a 3, then the total will be 4, 5, 6, 7, 8, or 9, depending on what the other dice shows. If it shows a 5 or a 6, which makes 8 or 9, then it’s greater than 7. If it shows 4, 3, 2, or 1, then it’s not greater than 7, so 2 of the 6 possibilities is greater than 7. I realize that the 3 could have come up on either one of the 2 dice, or both, but I just don’t really get why that would make a difference. At least one of them is a 3, and there’s a 1/6 chance that the other one is, too. There’s a 5/6 chance that it’s not, but, regardless of that, there’s a 2/6 chance that the 2 dice add up to a number greater than 7. I’m sure I’m wrong, because I’m not an expert, like the teacher in the video, but I just really can’t understand why it wrong, and that makes me mad, so I’m going to just pretend like I’m right, and he’s wrong, because I’m a genius, and he apparently isn’t very bright. That may or may not be true, but it makes me feel better to think that, and feeling good is important to me.
I get it, if you roll two dice, there are 11 possible rolls (each of equal probibility) that meet the first requirement, so your sample size is 11 rolls. 4 of them will meet the second requirement.
I at first assumed it was 1/3. I was so confident that it was 1/3 that when it was revealed that the answer was 4/11, I spent the next half hour trying to process it. But then I came to a startling revelation. If we slightly rephrase the question, it would be 1/3. Instead of saying "atleast one dice rolls a three", what if we say "Dice 'A' rolls a three"? It completely changes the answer! The simple fact of knowing which specific dice was a guaranteed three, creates an entirely different possibility of outcomes!
Yup. It’s the potential for double-counting that can get ya!
@@nicholasogburn7746 If you're talking about the overlap, then you didn't catch his issue. The issue was using 6 instead of 6+6 in the sample space. The -1 less mind-boggling IMO.
i dont understand why its not 1/3. why does it matter which one rolls a 3 first. because either way arent u only going to have 6 possible numbers other than 3?
0:39
Seems to have nothing to do with the 10 Socks Probability Problem.
As a craps dealer for 10 years it drove me nuts how he drew the three on the die drawing.
Awesome explanation
Mind blown.
Wow!! Is this related to the first two parts though? Great videos though! Keep it up!
The guy rolling the dice is the one wearing the socks
@@lipranditoys great comment
@@lipranditoys 😂
Question wording is wrong (it confuses) !!!
This question must be asked in two ways :
1. Two dices are rolled, it is given that 3 appears on atleast one dice, find the probability of getting sum greater than 7. (Answer is 4/11).
2. Two dices are rolled, find the probability of getting 3 on atleast one of the dices and sum of dices is greater than 7. (Answer = 11/36 x 4/11 = 4/36 = 1/9)
Superb.
great work!!
A three appears on at least one of the dice. Doesn' this mean that the second one can have a three too, since the wording is "At least one"?
Yes it does, and the result (4/11) includes this possibility. I think you might be confused because of the overlap. If you were to select the outcomes he marks (row 3 and column 3), and count them individually, you would also get 11.
@@beefy420 Exactly. Your options are (in terms of 1st die/2nd die) 1/3, 2/3, 3/3, 4/3, 5/3, 6/3, 3/1, 3/2, 3/4, 3/5, 3/6. You don't count 3/3 twice, because it's the same exact result, so it overlaps.
@@AzureKyle Thank you,. This explains.
@@beefy420 bnk
@@rrangana11 I'm
I like the videos and the teaching style here, but in this case, I disagree with the interpretation. If one of the dice is already known to be a 3, then all we’re looking for is the probability that the other die is greater than 4 (ie 5 or 6). That’s a pretty simple 2/6. In other words, one of the dot circles must be shrunken to encircle only a single dot because we know one of the dice is a 3.
It is assumed that the dice rolling 3 could be either the first one or the second one. If it were just the first one, the answer would be as you say [(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)] with last two combinations satisfying the condition. However, if you consider the possibility of the second dice rolling to 3 irrespective of the outcome of 1st dice, you will be looking at 6 combinations [(1,3),(2,3),(3,3),(4,3),(5,3),(6,3)] of which 2 satisfy this condition. If you combine these both, and exclude any element that is duplicated, I.E., (3,3), you will be left with 4 possibilities out of 11 combinations.
@@amulyam.339 the probability of rolling a dice at least once is 1. The probability of rolling a number which sum will be greater seven is 2/6. P(1) U P(2/6) = 2/6 The three is the first or the second throw so times 2. Its 4/6.
That’s what I thought for sure, too. It doesn’t matter which one is a 3, but we know that one of the dice is a 3, and the other one is 1 out of 6 different possibilities, and 2 of those 6 are greater than 7, so it seems like the answer should be 2 out of 6.
In other words, if one of them is a 3, then the total will be 4, 5, 6, 7, 8, or 9, depending on what the other dice shows. If it shows a 5 or a 6, which makes 8 or 9, then it’s greater than 7. If it shows 4, 3, 2, or 1, then it’s not greater than 7, so 2 of the 6 possibilities is greater than 7.
I realize that the 3 could have come up on either one of the 2 dice, or both, but I just don’t really get why that would make a difference.
At least one of them is a 3, and there’s a 1/6 chance that the other one is, too. There’s a 5/6 chance that it’s not, but, regardless of that, there’s a 2/6 chance that the 2 dice add up to a number greater than 7.
I’m sure I’m wrong, because I’m not an expert, like the teacher in the video, but I just really can’t understand why it wrong, and that makes me mad, so I’m going to just pretend like I’m right, and he’s wrong, because I’m a genius, and he apparently isn’t very bright. That may or may not be true, but it makes me feel better to think that, and feeling good is important to me.
@@chriswebster24 😂
That is what I reasoned too and 2/6=0.333
4/11=0.363
So pretty close
I get it, if you roll two dice, there are 11 possible rolls (each of equal probibility) that meet the first requirement, so your sample size is 11 rolls. 4 of them will meet the second requirement.
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