Jump to content

Portfolio Type II -- The Dice Game


Skyline

Recommended Posts

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?

Link to post
Share on other sites

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.

Link to post
Share on other sites

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?

Link to post
Share on other sites

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.

Link to post
Share on other sites

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 :)

Link to post
Share on other sites

  • 1 month later...
  • 2 weeks later...

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

Link to post
Share on other sites

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 by IBCONQUERER
Link to post
Share on other sites

  • 3 weeks later...

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.

Link to post
Share on other sites

How to find the probabilities of such large number of combinations..... :dontgetit:. 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. :eek:

Link to post
Share on other sites

  • 2 weeks later...

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 :/

Link to post
Share on other sites

  • 5 weeks later...

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.

Link to post
Share on other sites

  • 1 month later...

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 :(

Link to post
Share on other sites

  • 1 month later...

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...