Probability Question - Can you solve this tricky Logic Puzzle?
video description
Let's assume that we place x red marbles and y blue marbles in the first bowl.
In the other bowl there should be 50-x reds and 50-y blues.
So when we pick randomly from a bowl the probability for the marble to be red would be
P(red)=0, 5-x/(x+y) + 0, 5-(50-x)/(100-x-y)
This is a 2 variable function. In order to get rid of variable y we can do the following procedure.
This quantity is the sum of two positive quantities.
a) We notice that the first quantity gets its higher value if y=0 and x>=1.
If y=0 then P(red)=0, 5 + 0, 5(50-x)/(100-x). This is a descending function, so it gets its highest value for x=1. so we get P(red)max=0, 5 + 0, 5-49/99 = 74. 7%
b) If we notice the second quantity, it gets its higher value if y=50 and x
Date: 2023-11-15
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Comments and reviews: 29
Whisky
The picture shown at 2: 00 is what I had in my head. I expected that we needed to split the marbles into even counts in each jar to make them fit (though this wasn-t actually stated. My answer was -it doesn-t matter - the chance is always the same if you have the same number of marbles in each jar; how many red or blue in each makes no difference-. But as soon as I saw Jar 2 full of blue marbles and Jar one being not full, I knew instantly that the image shown was a misrepresentation of the problem. If Jar 1 and Jar 2 could each hold 100 marbles and that was shown or stated, then the problem would be a fair test. As it is, we have to ignore our eyes to come up with the right answer.
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The picture shown at 2: 00 is what I had in my head. I expected that we needed to split the marbles into even counts in each jar to make them fit (though this wasn-t actually stated. My answer was -it doesn-t matter - the chance is always the same if you have the same number of marbles in each jar; how many red or blue in each makes no difference-. But as soon as I saw Jar 2 full of blue marbles and Jar one being not full, I knew instantly that the image shown was a misrepresentation of the problem. If Jar 1 and Jar 2 could each hold 100 marbles and that was shown or stated, then the problem would be a fair test. As it is, we have to ignore our eyes to come up with the right answer.
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stevekerp1
It's a probability question, but would be interesting to approach it as a likelihood question, though that would make it much more complex. If you put one blue marble in a jar and then put twenty red marbles on top of it, the probability of choosing the blue one is one out of 21 or about 0. 047. But the blue marble is much less likely to be selected because of its location. So putting one red in one jar, all the blues into the second jar and then all the remaining reds on top of all the blues would give you much better than 74. 74% odds.
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It's a probability question, but would be interesting to approach it as a likelihood question, though that would make it much more complex. If you put one blue marble in a jar and then put twenty red marbles on top of it, the probability of choosing the blue one is one out of 21 or about 0. 047. But the blue marble is much less likely to be selected because of its location. So putting one red in one jar, all the blues into the second jar and then all the remaining reds on top of all the blues would give you much better than 74. 74% odds.
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Arthur
There is nothing clarifying that if a selection is attempted from an empty jar that the marble would be selected from the other jar. If an empty jar is an option you aren't accounting for the no-marble probability. After selecting the first red marble in the jar with one marble there's a 50% chance of picking no-marble and a 25% chance of picking red or blue from the other jar if you don't mix it up.
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There is nothing clarifying that if a selection is attempted from an empty jar that the marble would be selected from the other jar. If an empty jar is an option you aren't accounting for the no-marble probability. After selecting the first red marble in the jar with one marble there's a 50% chance of picking no-marble and a 25% chance of picking red or blue from the other jar if you don't mix it up.
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Ben
Anyone giving a solution of putting the red marbles into the jars on top of the blue marbles is ignoring the fact that the marble must be picked RANDOMLY from the jar. So if you're going to ignore that requirement of the puzzle, why not just cut to the chase and put all red marbles in the same jar and just pick that jar? Oh wait, there's also a requirement that the jar is picked RANDOMLY.
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Anyone giving a solution of putting the red marbles into the jars on top of the blue marbles is ignoring the fact that the marble must be picked RANDOMLY from the jar. So if you're going to ignore that requirement of the puzzle, why not just cut to the chase and put all red marbles in the same jar and just pick that jar? Oh wait, there's also a requirement that the jar is picked RANDOMLY.
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Shivan
place the blue marbels first, than place the red marbels above it - pick the marbel your hand touched first.
as he did not specied the capacity, i would consider it as 50 according to first image the jar was full of marbels. but he did said it was up to me about the placement of marbels, the fact also depends on the density and size. there are too many variables in the question --
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place the blue marbels first, than place the red marbels above it - pick the marbel your hand touched first.
as he did not specied the capacity, i would consider it as 50 according to first image the jar was full of marbels. but he did said it was up to me about the placement of marbels, the fact also depends on the density and size. there are too many variables in the question --
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Miguel
Pausing to answer:
If we are allowed to -place the marbles however [we] like- in the jars, wouldn't placing half of the red marble in each jar, but being sure to place them on top of the blue marbles maximize our chance to pick a red marble -at random-?
My flawed logic? Yes. Cheating? Also yes. Great video, as always? Absolutely yes.
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Pausing to answer:
If we are allowed to -place the marbles however [we] like- in the jars, wouldn't placing half of the red marble in each jar, but being sure to place them on top of the blue marbles maximize our chance to pick a red marble -at random-?
My flawed logic? Yes. Cheating? Also yes. Great video, as always? Absolutely yes.
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Furry
My answer before seeing solution: put 1 red marble in one jar, and all the rest in the other. Thus, you have a 50% chance of automatically getting a red marble by choosing jar 1. If you pick jar 2, you have a 49/99 chance of getting a red marble. calculating the probabilities together, we get a 74/99 chance of getting a red marble
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My answer before seeing solution: put 1 red marble in one jar, and all the rest in the other. Thus, you have a 50% chance of automatically getting a red marble by choosing jar 1. If you pick jar 2, you have a 49/99 chance of getting a red marble. calculating the probabilities together, we get a 74/99 chance of getting a red marble
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Bikash
I can enhance the probability from 0. 74 to 1. Since I have no constraints on how I can arrange the marbles, I will evenly distribute red and blue marbles to each jar and place all the red marbles on the upper part. So even if I blindly pick one jar and one marble, I will pick from the top and it will be a red one.
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I can enhance the probability from 0. 74 to 1. Since I have no constraints on how I can arrange the marbles, I will evenly distribute red and blue marbles to each jar and place all the red marbles on the upper part. So even if I blindly pick one jar and one marble, I will pick from the top and it will be a red one.
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Arnab
First we divide 50 blue balls into 25+25 and place them in two jars. Next we divide 50 red balls equally and place them in both jars on top of blue balls. So irrespective of the jar we chose we get a red ball because they r on top. Will that work because it results in 100% chances of a red ball.
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First we divide 50 blue balls into 25+25 and place them in two jars. Next we divide 50 red balls equally and place them in both jars on top of blue balls. So irrespective of the jar we chose we get a red ball because they r on top. Will that work because it results in 100% chances of a red ball.
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Jake
1 red one in a jar and the other jar 99 mixed.
OR top of both jars red.
when you say PICK it's natural to assume you are reaching in the jar. How do you select a marble buried in the middle of a jar. right?
you need to think physically. you need to define select and how it's done.
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1 red one in a jar and the other jar 99 mixed.
OR top of both jars red.
when you say PICK it's natural to assume you are reaching in the jar. How do you select a marble buried in the middle of a jar. right?
you need to think physically. you need to define select and how it's done.
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Deist
The pictures leading up to the problem didn't help. Each jar was clearly shown incapable of holding anywhere near both sets of balls. It may not be logically impossible to do so, just physically so. We could also turn the blue balls into dust making it 100% probable that a red ball is drawn.
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The pictures leading up to the problem didn't help. Each jar was clearly shown incapable of holding anywhere near both sets of balls. It may not be logically impossible to do so, just physically so. We could also turn the blue balls into dust making it 100% probable that a red ball is drawn.
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Tehom
See, you have to nail it down by saying things like -The jars are shaken until they are fully randomized-, otherwise us Math Meanies say things like -Put the red marbles on top in the 49/50 jar, and pack the marbles densely to maximize the probability that they will be unchanged by shaking-
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See, you have to nail it down by saying things like -The jars are shaken until they are fully randomized-, otherwise us Math Meanies say things like -Put the red marbles on top in the 49/50 jar, and pack the marbles densely to maximize the probability that they will be unchanged by shaking-
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Nitesh
The puzzle doesn't say we need to shake the jar before picking. Now then we can place blue marbles at bottom of both jars and then fill the top with red marbles. Then just randomly pick marble but don't dip hand towards the bottom half. Just pick randomly from the top half.
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The puzzle doesn't say we need to shake the jar before picking. Now then we can place blue marbles at bottom of both jars and then fill the top with red marbles. Then just randomly pick marble but don't dip hand towards the bottom half. Just pick randomly from the top half.
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Aniran
The if I put 25 Marbles in a first which is blue and 25 Marbles in the second Ja which is also blue and also if after that I place 25 red Marbles on in the the first Ja and keep 25 Marbles in the second then the maximum probability of drawing a red marble is the most
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The if I put 25 Marbles in a first which is blue and 25 Marbles in the second Ja which is also blue and also if after that I place 25 red Marbles on in the the first Ja and keep 25 Marbles in the second then the maximum probability of drawing a red marble is the most
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shabroz
I mean what -, can you make some less complications and try to solve in an easier way. Just simply put 50% Blue marble in jar 1 and 50% of another Blue marble in jar 2 and rest the remaining red marbel one the top of both the jars hence 0. 5+0. 5 gives prob = 1
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I mean what -, can you make some less complications and try to solve in an easier way. Just simply put 50% Blue marble in jar 1 and 50% of another Blue marble in jar 2 and rest the remaining red marbel one the top of both the jars hence 0. 5+0. 5 gives prob = 1
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darktemplar1984
I'm agree with Shivan, I put in both jars 25 blue then the 25 red then the first you touch you grab, the the possibility is of 100 on take a red, in the video he said take your time (or something like that) to think LOGICALLY, no think mathematically
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I'm agree with Shivan, I put in both jars 25 blue then the 25 red then the first you touch you grab, the the possibility is of 100 on take a red, in the video he said take your time (or something like that) to think LOGICALLY, no think mathematically
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Jae
The solution is good, but still incomplete. It explains -when Jar 1 has only red marbles and all blue marbles are in Jar 2, - putting only one marble in Jar 1 is the best. But it does not explain if -putting no blue marbles to Jar 1- is the best policy.
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The solution is good, but still incomplete. It explains -when Jar 1 has only red marbles and all blue marbles are in Jar 2, - putting only one marble in Jar 1 is the best. But it does not explain if -putting no blue marbles to Jar 1- is the best policy.
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Krushn
I have better solution for this.
First place 25-25 blue marbles in each jar.
Not place 25-25 red marbles in each jar.
Now which ever jar is picked there are - % chances that only red marble will be picked up
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I have better solution for this.
First place 25-25 blue marbles in each jar.
Not place 25-25 red marbles in each jar.
Now which ever jar is picked there are - % chances that only red marble will be picked up
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Terry
Place 1 red marble in 1 jar (jar A) and place all blue marbles and the rest of the red marbles in the other jar (jar B. If jar A is chosen, the odds are 100%. If jar B is chosen the odds are nearly 50% (49/99 = 49. 5%.
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Place 1 red marble in 1 jar (jar A) and place all blue marbles and the rest of the red marbles in the other jar (jar B. If jar A is chosen, the odds are 100%. If jar B is chosen the odds are nearly 50% (49/99 = 49. 5%.
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BlueMoon
oh this was interesting. I had a bias in my thinking i somehow assumed it should always be 50 marbles in each bowl. Which you clearly didn't say. Interesting how the mind always assumes stuff that isn't there.
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oh this was interesting. I had a bias in my thinking i somehow assumed it should always be 50 marbles in each bowl. Which you clearly didn't say. Interesting how the mind always assumes stuff that isn't there.
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Ahmet
Very nice! I am a bit frustrated though because the jars that you show are full with 50 marbles. This made it mandatory to put maximum 50 marbles in one jar. Quite misleading. But still, your videos are awesome!
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Very nice! I am a bit frustrated though because the jars that you show are full with 50 marbles. This made it mandatory to put maximum 50 marbles in one jar. Quite misleading. But still, your videos are awesome!
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Red
Your illustrations ruined this one. You make it look like each jar can only fit 50 marbles, so it looks like a trick question where not matter how you arrange them the probability stays at 50%.
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Your illustrations ruined this one. You make it look like each jar can only fit 50 marbles, so it looks like a trick question where not matter how you arrange them the probability stays at 50%.
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Stephen
If you have more jars (say 20, you can get the probability of selecting a red marble close to 1. Same technique, i. e, put one red marble in each of 19 jars and the rest in the 20th jar.
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If you have more jars (say 20, you can get the probability of selecting a red marble close to 1. Same technique, i. e, put one red marble in each of 19 jars and the rest in the 20th jar.
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education
I manage to make 100% probability. Split evenly. Put the blue marbles at the bottom of each jars, then add the red marbles ON TOP of the blue marbles. Et voila
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I manage to make 100% probability. Split evenly. Put the blue marbles at the bottom of each jars, then add the red marbles ON TOP of the blue marbles. Et voila
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octavmandru
To all saying you should place the blue marbles at the bottom and red on top: what happens if you agitate the jar before? Or just pick a marble from bottom?
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To all saying you should place the blue marbles at the bottom and red on top: what happens if you agitate the jar before? Or just pick a marble from bottom?
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Vaisnava
You can take it further. Place all the blue marbles on the bottom first. Then red marbles on top. Your odds will be 100% if you take off the top; )
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You can take it further. Place all the blue marbles on the bottom first. Then red marbles on top. Your odds will be 100% if you take off the top; )
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David
One red marble in Jar 1. 49 red marbles and 50 blue marbles in Jar 2, if jars aren't shaken up, with red marbles after the blues have all been placed.
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One red marble in Jar 1. 49 red marbles and 50 blue marbles in Jar 2, if jars aren't shaken up, with red marbles after the blues have all been placed.
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albert
If the 25-25 blue ones were kept at the bottom of both the jars, and half-half of red ones were upwards, then the probability would be 100%
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If the 25-25 blue ones were kept at the bottom of both the jars, and half-half of red ones were upwards, then the probability would be 100%
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Md
I will fill bottom half of the jars with blue marbles and upper half with red ones. This way I will have - % chance of getting red marbles -
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I will fill bottom half of the jars with blue marbles and upper half with red ones. This way I will have - % chance of getting red marbles -
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