bonanova Posted October 11, 2008 Report Share Posted October 11, 2008 (edited) The marbles in John's collection were of 13 different colors; each marble was of a single color. His sisters, Jane and Joanne, especially liked the red ones, and asked if they might have one each. Feeling generous, but not wanting to give them everything they asked, John let them each take one marble from the black velvet bag where he kept his collection, but he did not let them look inside. It is an even money bet that both sisters got their wish and pulled out a red marble. How many red marbles might have been in John's collection before his sisters chose? Edit: Of the 13 possible colors, obviously some were red. There may have been zero of some of the other 12 colors. Edited October 12, 2008 by bonanova clarify the color distribution possibilitites Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 11, 2008 Report Share Posted October 11, 2008 Reveal hidden contents so there are 12 different colours (other than red). lets call this number x. So after both draws from the bag the number of red marbles must average at x for it to be an even bet. now we just have to find out the combinations as the maths is above. Got Geography C/w to do so if its still unsolved then ill figure it out later Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 11, 2008 Report Share Posted October 11, 2008 Reveal hidden contents In order for it be the same probability for both to pick a red marble, all the marbles, or none of them must be red, but seen as there are 13 different coloured marbles, there must be no red ones... right? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 11, 2008 Author Report Share Posted October 11, 2008 Asque said: Reveal hidden contents In order for it be the same probability for both to pick a red marble, all the marbles, or none of them must be red, but seen as there are 13 different coloured marbles, there must be no red ones... right? Nope. The OP specifies just one probability, not two. It is an even money bet that both sisters got their wish .... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 11, 2008 Report Share Posted October 11, 2008 (edited) a bit more of the maths Reveal hidden contents (x/y)*((x-1)*(y-1))=0.5 where x is the number of red marbles and y is the total number of marbles edit: typo Edited October 11, 2008 by reaymond Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 11, 2008 Author Report Share Posted October 11, 2008 reaymond said: a bit more of the maths Reveal hidden contents (x/y)*((x-1)*(y-1))=0.5 where x is the number of red marbles and y is the total number of marbles edit: typo Yup. That's the question. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 11, 2008 Report Share Posted October 11, 2008 is there only 1 answer, or are there an infinite amount of answers? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 11, 2008 Author Report Share Posted October 11, 2008 reaymond said: is there only 1 answer, or are there an infinite amount of answers? I don't know how many there are. But if you find more than two, you can stop. Quote Link to comment Share on other sites More sharing options...
0 unreality Posted October 11, 2008 Report Share Posted October 11, 2008 Reveal hidden contents (x/y)*((x-1)*(y-1))=0.5 where x is the number of red marbles and y is the total number of marbles so: (x/y)(x-1)(y-1) = 0.5 (x-1)(y-1) = xy-x-y+1 Multiply that by x/y and we have: x2y - x2 - xy + x all divided by y = 0.5 subtract 0.5 from both sides x2y - x2 - xy + x -------------------------------------------------- - 0.5 = 0 y split it up into different parts: x2y/y - x2/y - xy/y + x/y - 0.5 = 0 x2 - x2/y - x + x/y - 0.5 = 0 combine some like terms x2 - x + (x2 + x)/y - 0.5 = 0 x2 - x + (x+1)x/y - 0.5 = 0 divide all sides by x x - 1 + (x+1)/y - 0.5/x = 0 0.5/x = (1/2)/x = 1/(2x) so x - 1 + (x+1)/y - 1/(2x) = 0 Now x and y must both be positive integers, and y is equal to or larger than 13, and x is equal to or larger than 1 x + (x+1)/y - 1/(2x) = 1 (x+1)/y - 1/(2x) = 1 - x Because 1-x must be negative, 1/(2x) must be bigger than (x+1)/y 1/(2x) > (x+1)/y make both denominators equal to 2xy: y/(2xy) > (2x2+2x)/(2xy) which means y is bigger than 2x2 + 2x y > 2x(x+1) ... Right now we have: (x+1)/y - 1/(2x) = 1 - x 2x(x+1)/(2xy) - y/(2xy) = 1 - x (2x(x+1) - y)/(2xy) = 1 - x both sides of the equation are negative right now, but if we multiplied both by -1.. (y - 2x(x+1))/(2xy) = x - 1 Well I have to go, I'll think more about it later I know I didn't really do much lol, though I suspect the knowledge of 13 different colors must weed out certain solutions.... hmmm... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 11, 2008 Report Share Posted October 11, 2008 You are most likely looking for a somewhat simple solution. However, Reveal hidden contents John began with 10000000000 marbles, of which 7071067812 were red. That's a lot of marbles. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 11, 2008 Report Share Posted October 11, 2008 (edited) rossbeemer said: You are most likely looking for a somewhat simple solution. However, Reveal hidden contents John began with 10000000000 marbles, of which 7071067812 were red. That's a lot of marbles. Reveal hidden contents the OP doesn't atually say that there were exactly 13 marbles in the bag Edited October 11, 2008 by lemonymelon Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 11, 2008 Author Report Share Posted October 11, 2008 rossbeemer said: You are most likely looking for a somewhat simple solution. However, Reveal hidden contents John began with 10000000000 marbles, of which 7071067812 were red. That's a lot of marbles. That's a correct answer, and you are the first solver. Now let's introduce a sanity check by stipulating normal sized marbles and the ability for John to hand the bag to his sisters. Are there other solutions? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 11, 2008 Report Share Posted October 11, 2008 I think (but I'm not sure) that there are other solutions, but they might be quite challenging to find... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 11, 2008 Report Share Posted October 11, 2008 My quick answer..shot in the dark..haha Reveal hidden contents 39 total marbles, 26 of which are red...If I read the question correctly Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 11, 2008 Author Report Share Posted October 11, 2008 Hi tgreene83, Reveal hidden contents .44. We need .5 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 11, 2008 Report Share Posted October 11, 2008 Alright, I've never liked finding integer solutions to problems, so I just put this into excel and started searching. Reveal hidden contents 85 red marbles, 35 of other colors 85/120*84/119=7140/14280=1/2 I think you can hold a bag of 120 marbles. Not exactly elegant, but I don't think there are any smaller solutions. Can anyone find a non-brute force way to find that solution? Quote Link to comment Share on other sites More sharing options...
0 unreality Posted October 11, 2008 Report Share Posted October 11, 2008 Chuck Rampart said: Alright, I've never liked finding integer solutions to problems, so I just put this into excel and started searching. Reveal hidden contents 85 red marbles, 35 of other colors 85/120*84/119=7140/14280=1/2 I think you can hold a bag of 120 marbles. Not exactly elegant, but I don't think there are any smaller solutions. Can anyone find a non-brute force way to find that solution? Let me retry, scratching my old post on this topic. I'll make a new equation. Reveal hidden contents r = red marbles t = total marbles (r/t) * (r-1)/(t-1) = 1/2 looks better as: ®/(t) * (r-1)/(t-1) = 1/2 (r*(r-1)) / (t*(t-1)) = 1/2 multiple both sides by 2t(t-1) 2r(r-1) = t(t-1) or 2r(r-1) - t(t-1) = 0 Now we want to solve for one of the variables, with none of that variable on the other side of the equation This would involve a mixture of factoring and taking the square root if possible, though I don't see how yet Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 12, 2008 Author Report Share Posted October 12, 2008 Chuck Rampart said: Alright, I've never liked finding integer solutions to problems, so I just put this into excel and started searching. Reveal hidden contents 85 red marbles, 35 of other colors 85/120*84/119=7140/14280=1/2 I think you can hold a bag of 120 marbles.Not exactly elegant, but I don't think there are any smaller solutions. Can anyone find a non-brute force way to find that solution? John can certainly pass that bag. Nice. Can anyone find an even smaller solution? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 12, 2008 Report Share Posted October 12, 2008 bonanova said: John can certainly pass that bag. Nice. Can anyone find an even smaller solution? Uh-oh. The next smallest solution is not a solution at all, because it only has 6 non-red marbles. Is there a trick to the wording I'm missing? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 12, 2008 Report Share Posted October 12, 2008 (edited) how precise do you want the answer? Reveal hidden contents 10 red marbles and 14 total in bag = 0.495 chance 13 red marbles and 18 total in bag = 0.509 chance Reveal hidden contents 15 red marbles and 21 total in bag = even chance edit: add smaller solutions Edited October 12, 2008 by reaymond Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 12, 2008 Author Report Share Posted October 12, 2008 Reveal hidden contents Let's say that for the non-red marbles, there is no restrictions on their color or their number. I think it's clear now that 13 colors was a red herring [npi], and it may be affecting the search for a smaller solution. The OP did not intend to say there were at least 13 non-red marbles, but under some interpretations it does say that.So, now let the OP say ... The bag contains John's marble collection. Some of his marbles are red. It's an even bet that if his sisters each draw one marble, they will both be red. How many red marbles might there have been in the bag before the sisters chose theirs? Reveal hidden contents 15, 85 and 7071067812 Is there a solution smaller than 15 red marbles? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 12, 2008 Report Share Posted October 12, 2008 Reveal hidden contents total in bag 4 red before sisters pick 3 Quote Link to comment Share on other sites More sharing options...
0 Prime Posted October 12, 2008 Report Share Posted October 12, 2008 bonanova said: The bag contains John's marble collection. Some of his marbles are red. It's an even bet that if his sisters each draw one marble, they will both be red. How many red marbles might there have been in the bag before the sisters chose theirs? Reveal hidden contents 15, 85 and 7071067812 Is there a solution smaller than 15 red marbles? Reveal hidden contents Let's say that for the non-red marbles, there is no restrictions on their color or their number. I think it's clear now that 13 colors was a red herring [npi], and it may be affecting the search for a smaller solution. The OP did not intend to say there were at least 13 non-red marbles, but under some interpretations it does say that.So, now let the OP say ... To me the OP clearly implied 13 different colors present in the bag. For this new problem statement the minimal solution is: Reveal hidden contents 3 red marbles in a bag of total 4. (3/4)*(2/3) = 1/2 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 12, 2008 Report Share Posted October 12, 2008 just beat you to it Prime agreed the OP did imply the idea that there was at least 1 marble of the other 12 colours. Quote Link to comment Share on other sites More sharing options...
0 Prime Posted October 12, 2008 Report Share Posted October 12, 2008 reaymond said: just beat you to it Prime agreed the OP did imply the idea that there was at least 1 marble of the other 12 colours. So you did. While I went on rambling about why, what, and how. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 13, 2008 Author Report Share Posted October 13, 2008 reaymond said: Reveal hidden contents total in bag 4 red before sisters pick 3 That's it. Here I am wondering why this was never found. Nice going. Quote Link to comment Share on other sites More sharing options...
Question
bonanova
The marbles in John's collection were of 13 different colors; each marble was of a single color.
His sisters, Jane and Joanne, especially liked the red ones, and asked if they might have one each.
Feeling generous, but not wanting to give them everything they asked, John let them each take
one marble from the black velvet bag where he kept his collection, but he did not let them look inside.
It is an even money bet that both sisters got their wish and pulled out a red marble.
How many red marbles might have been in John's collection before his sisters chose?
Edit: Of the 13 possible colors, obviously some were red. There may have been zero of some of the other 12 colors.
Edited by bonanovaclarify the color distribution possibilitites
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