Lobezno Posted April 29, 2012 Report Share Posted April 29, 2012 I figured out the formula for the general case where Ann rolls x times and Bob rolls y times. Here's some help:First, let's consider Bob's score.There are six possible cases for Ann's number, 1 to 6. Say her score is k. What is the probability that Ann's score is equal to k (let's just call it P(A=k) for now) AND that all of Bob's rolls are less than k? This tells us the probability that Ann will win given her score is k. Your final answer is the sum from k = 1 to 6.To actually find P(A=k), consider the case when k=1. There's only one possibility that all of her rolls are equal to one.Now find P(A=2). This means that we find that all of her rolls have the outcome P(1 or 2) (and we do this x times). However, we need to ensure that at least one of her rolls is equal to 2. So we need to subtract the case when all of her rolls are equal to 1.And we repeat this process until P(A=6). You should probably see the telescoping series. The formula isn't that bad, but it takes a little bit of thinking.It's easy to do this by just counting the different permutations, but how can I do this explicitly/algebraically? Reply Link to post Share on other sites More sharing options...
Ryan Giggs Posted April 30, 2012 Report Share Posted April 30, 2012 Hello guys, I'm having trouble understanding this portfolio, can anyone please point me in the right direction by helping me out with the first example?Thanks beforehand. Reply Link to post Share on other sites More sharing options...
Ryan Giggs Posted April 30, 2012 Report Share Posted April 30, 2012 Ok so I did the first one but I don't know how to do the others, can anyone help me in these? Reply Link to post Share on other sites More sharing options...
Ryan Giggs Posted April 30, 2012 Report Share Posted April 30, 2012 It seems I'm just replying to myself...Anyway I did the second part but I'm having a lot of trouble in the third one, I really have no clue of how to obtain it... I don't know where to begin....Please help! Reply Link to post Share on other sites More sharing options...
Absolutely Positively Posted April 30, 2012 Report Share Posted April 30, 2012 It seems I'm just replying to myself...Anyway I did the second part but I'm having a lot of trouble in the third one, I really have no clue of how to obtain it... I don't know where to begin....Please help!Yeah, our teacher said that she wrote down every single possible outcome (1296!).If you do so, you should see a pattern.I didn't want to do that so I did a graph which helped me. Reply Link to post Share on other sites More sharing options...
Ryan Giggs Posted April 30, 2012 Report Share Posted April 30, 2012 It seems I'm just replying to myself...Anyway I did the second part but I'm having a lot of trouble in the third one, I really have no clue of how to obtain it... I don't know where to begin....Please help!Yeah, our teacher said that she wrote down every single possible outcome (1296!).If you do so, you should see a pattern.I didn't want to do that so I did a graph which helped me.Well writting every possible outcome is a good way to make sure you get the right answer, but I really preffer getting it through a much more mathematical way instead of doing it the lonng way. If you want to, could you tell me which kind of graph you made, cause I really don't know how you'd be able to graph what the third part of the portfolio asks? Reply Link to post Share on other sites More sharing options...
Absolutely Positively Posted April 30, 2012 Report Share Posted April 30, 2012 My mistake, it's more like a table. Reply Link to post Share on other sites More sharing options...
Ryan Giggs Posted April 30, 2012 Report Share Posted April 30, 2012 My mistake, it's more like a table.Thanks I'll start devoloping it myself from here. Reply Link to post Share on other sites More sharing options...
Ryan Giggs Posted May 2, 2012 Report Share Posted May 2, 2012 Hey guys, I need some help on finding the general ecuation for the probabilities, and some help with number 4 regarding a game with multiple players. Can anyone help me? Reply Link to post Share on other sites More sharing options...
Dinstruction Posted May 2, 2012 Report Share Posted May 2, 2012 I figured out the formula for the general case where Ann rolls x times and Bob rolls y times. Here's some help:First, let's consider Bob's score.There are six possible cases for Ann's number, 1 to 6. Say her score is k. What is the probability that Ann's score is equal to k (let's just call it P(A=k) for now) AND that all of Bob's rolls are less than k? This tells us the probability that Ann will win given her score is k. Your final answer is the sum from k = 1 to 6.To actually find P(A=k), consider the case when k=1. There's only one possibility that all of her rolls are equal to one.Now find P(A=2). This means that we find that all of her rolls have the outcome P(1 or 2) (and we do this x times). However, we need to ensure that at least one of her rolls is equal to 2. So we need to subtract the case when all of her rolls are equal to 1.And we repeat this process until P(A=6). You should probably see the telescoping series. The formula isn't that bad, but it takes a little bit of thinking.It's easy to do this by just counting the different permutations, but how can I do this explicitly/algebraically?I'm not sure what else you want me to explain. I believe my method is deductive and doesn't require listing every permutation. Reply Link to post Share on other sites More sharing options...
arra Posted May 2, 2012 Report Share Posted May 2, 2012 I figured out the formula for the general case where Ann rolls x times and Bob rolls y times. Here's some help: First, let's consider Bob's score. There are six possible cases for Ann's number, 1 to 6. Say her score is k. What is the probability that Ann's score is equal to k (let's just call it P(A=k) for now) AND that all of Bob's rolls are less than k? This tells us the probability that Ann will win given her score is k. Your final answer is the sum from k = 1 to 6. To actually find P(A=k), consider the case when k=1. There's only one possibility that all of her rolls are equal to one. Now find P(A=2). This means that we find that all of her rolls have the outcome P(1 or 2) (and we do this x times). However, we need to ensure that at least one of her rolls is equal to 2. So we need to subtract the case when all of her rolls are equal to 1. And we repeat this process until P(A=6). You should probably see the telescoping series. The formula isn't that bad, but it takes a little bit of thinking. It's easy to do this by just counting the different permutations, but how can I do this explicitly/algebraically? I'm not sure what else you want me to explain. I believe my method is deductive and doesn't require listing every permutation. I also deductively found the general formula. I think what he's saying -- if he had the same problem that I had -- is that his "general" formula is still in series notation. However, with a little bit of algebraic manipulation you can - I am 100% sure -- bring this into a nice and closed equation Reply Link to post Share on other sites More sharing options...
Adele L Posted June 14, 2012 Report Share Posted June 14, 2012 I have the formula. You DO NOT need a denominator. Everyone who was obnoxious about this, you are terrible people. I'm not gonna explain how I got this, but pretty much... Pa(X) is probability that Anne rolls "x", and same for Pb(Y). Yeah. Reply Link to post Share on other sites More sharing options...
rsingh3 Posted June 22, 2012 Report Share Posted June 22, 2012 I have the formula. You DO NOT need a denominator.Everyone who was obnoxious about this, you are terrible people.I'm not gonna explain how I got this, but pretty much...Pa(X) is probability that Anne rolls "x", and same for Pb(Y).Yeah.I have the formula. You DO NOT need a denominator.Everyone who was obnoxious about this, you are terrible people.I'm not gonna explain how I got this, but pretty much...Pa(X) is probability that Anne rolls "x", and same for Pb(Y).Yeah.But you know I did the same what u mentioned here and got 70/216 but the correct answer is 125/216, can elaborate little bit more may one or two elements of your matrix Reply Link to post Share on other sites More sharing options...
IBCONQUERER Posted June 25, 2012 Report Share Posted June 25, 2012 (edited) This portfolio was a struggle! I ended up with a very very weird and I mean WEIRD approach which involved a pascals triangle and high order polynomials and I needed to manipulate that information to fit my model..I am so glad I was able to do it! Edited June 25, 2012 by IBCONQUERER Reply Link to post Share on other sites More sharing options...
iHubble Posted July 17, 2012 Report Share Posted July 17, 2012 I have the formula. You DO NOT need a denominator.Everyone who was obnoxious about this, you are terrible people.I'm not gonna explain how I got this, but pretty much...Pa(X) is probability that Anne rolls "x", and same for Pb(Y).Yeah.Yes, you need a denominator for the general statement. I got 18/20. Reply Link to post Share on other sites More sharing options...
taknev Posted July 24, 2012 Report Share Posted July 24, 2012 How to find the probabilities of such large number of combinations..... . Could someone please help me out, about how to find it, if there is any other simple method to try out. If there is no simple method, please post the longer method.Please reply as quickly as possible.Thank you. Reply Link to post Share on other sites More sharing options...
Brother Code Posted August 4, 2012 Report Share Posted August 4, 2012 I figured out the formula for the general case where Ann rolls x times and Bob rolls y times. Here's some help:First, let's consider Bob's score.There are six possible cases for Ann's number, 1 to 6. Say her score is k. What is the probability that Ann's score is equal to k (let's just call it P(A=k) for now) AND that all of Bob's rolls are less than k? This tells us the probability that Ann will win given her score is k. Your final answer is the sum from k = 1 to 6.To actually find P(A=k), consider the case when k=1. There's only one possibility that all of her rolls are equal to one.Now find P(A=2). This means that we find that all of her rolls have the outcome P(1 or 2) (and we do this x times). However, we need to ensure that at least one of her rolls is equal to 2. So we need to subtract the case when all of her rolls are equal to 1.And we repeat this process until P(A=6). You should probably see the telescoping series. The formula isn't that bad, but it takes a little bit of thinking.It's easy to do this by just counting the different permutations, but how can I do this explicitly/algebraically?I'm not sure what else you want me to explain. I believe my method is deductive and doesn't require listing every permutation.I also deductively found the general formula. I think what he's saying -- if he had the same problem that I had -- is that his "general" formula is still in series notation. However, with a little bit of algebraic manipulation you can - I am 100% sure -- bring this into a nice and closed equation I figured out the formula for the general case where Ann rolls x times and Bob rolls y times. Here's some help:First, let's consider Bob's score.There are six possible cases for Ann's number, 1 to 6. Say her score is k. What is the probability that Ann's score is equal to k (let's just call it P(A=k) for now) AND that all of Bob's rolls are less than k? This tells us the probability that Ann will win given her score is k. Your final answer is the sum from k = 1 to 6.To actually find P(A=k), consider the case when k=1. There's only one possibility that all of her rolls are equal to one.Now find P(A=2). This means that we find that all of her rolls have the outcome P(1 or 2) (and we do this x times). However, we need to ensure that at least one of her rolls is equal to 2. So we need to subtract the case when all of her rolls are equal to 1.And we repeat this process until P(A=6). You should probably see the telescoping series. The formula isn't that bad, but it takes a little bit of thinking.It's easy to do this by just counting the different permutations, but how can I do this explicitly/algebraically?I'm not sure what else you want me to explain. I believe my method is deductive and doesn't require listing every permutation.I also deductively found the general formula. I think what he's saying -- if he had the same problem that I had -- is that his "general" formula is still in series notation. However, with a little bit of algebraic manipulation you can - I am 100% sure -- bring this into a nice and closed equation Hey guys, I'm having trouble in getting the general formula. Is it necessary to use permutation? my teacher said that we can see the pattern through the tables. somehow, I'm failed to figure it out. someone help me :/ Reply Link to post Share on other sites More sharing options...
saj Posted September 2, 2012 Report Share Posted September 2, 2012 Hello All:Please help with this part : I want to come up with a general equation.Investigate the game when both players can roll their dice twice, and also when both players can roll their dice more than twice, but not necessarily the same number of times.Thanks. Reply Link to post Share on other sites More sharing options...
xjgege Posted October 3, 2012 Report Share Posted October 3, 2012 What type of information are you including in the introduction?I mean, obviously we introduce the task and purpose, but in the sample (SL) papers I did we always had a chart or graph of some sort to include.My introduction page looks lonely Reply Link to post Share on other sites More sharing options...
darkusdragon Posted December 1, 2012 Report Share Posted December 1, 2012 No your conjecture is missing a main thing. It may work for smaller values but doesn't work for more higher values.And dont post your conjectures openly here. Reply Link to post Share on other sites More sharing options...
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