100 replies to this topic

[Skyline]

Is anyone doing this portfolio?
Does someone have an idea how to solve the last three tasks...?

[harshil]

nope!!!! but i was wondering if you could help me out with the previous tasks...... thanks

Edited by Austin Glau, Jun 16, 2011 - 03:28.
no text speak

[Kavishk Gandhi]

Can anyone please help me out with the part in the portfolio 'The Dice Game' where Ann can roll the dice twice while Bob can only once. The portfolio has to be submitted this week and I can't get past this Question. Please help !!!!

[Absolutely Positively]

Can anyone please help me out with the part in the portfolio 'The Dice Game' where Ann can roll the dice twice while Bob can only once. The portfolio has to be submitted this week and I can't get past this Question. Please help !!!!

Ahm, yeah.
I calculated the probability that Bob doesn't lose.
So, If he throws a 1 then Ann has to throw 11 - 1 combination.
If he throws a 2, then Ann has to throw 11, 21, 12 or 22 - 4 combinations...
And so on till 6. Therefore Ann has 36 possible combinations.
In the end it's 1 - (some kind of formula you should get) | And the 1 in the beginning is important as you want it for Ann.

[Kavishk Gandhi]

Can anyone please help me out with the part in the portfolio 'The Dice Game' where Ann can roll the dice twice while Bob can only once. The portfolio has to be submitted this week and I can't get past this Question. Please help !!!!

Ahm, yeah.
I calculated the probability that Bob doesn't lose.
So, If he throws a 1 then Ann has to throw 11 - 1 combination.
If he throws a 2, then Ann has to throw 11, 21, 12 or 22 - 4 combinations...
And so on till 6. Therefore Ann has 36 possible combinations.
In the end it's 1 - (some kind of formula you should get) | And the 1 in the beginning is important as you want it for Ann.

So the thing is that in the question it is said that she 'can' throw once more, so there is a probability that she doesnt throw it the second time, that will only be when she wins ( the probability of her winning in d first try was easily calculated in the previous question being 15/36) so if she looses that ( probability of 21/36) will she win when she throws it d second time ?

[Mohit Tandon]

Can anyone please help me out with the part in the portfolio 'The Dice Game' where Ann can roll the dice twice while Bob can only once. The portfolio has to be submitted this week and I can't get past this Question. Please help !!!!

Ahm, yeah.
I calculated the probability that Bob doesn't lose.
So, If he throws a 1 then Ann has to throw 11 - 1 combination.
If he throws a 2, then Ann has to throw 11, 21, 12 or 22 - 4 combinations...
And so on till 6. Therefore Ann has 36 possible combinations.
In the end it's 1 - (some kind of formula you should get) | And the 1 in the beginning is important as you want it for Ann.

So the thing is that in the question it is said that she 'can' throw once more, so there is a probability that she doesnt throw it the second time, that will only be when she wins ( the probability of her winning in d first try was easily calculated in the previous question being 15/36) so if she looses that ( probability of 21/36) will she win when she throws it d second time ?

hey kg!! wassup??

she "can" throw twice! so she will throw twice! what has she got to lose if she throws? you are anyways gonna take the highest number whether thats on the first die or the second or the 200th !!
Absolutely Positively is absolutely right!!

Edited by Austin Glau, Jun 18, 2011 - 17:13.
no text speak

[Kavishk Gandhi]

Can anyone please help me out with the part in the portfolio 'The Dice Game' where Ann can roll the dice twice while Bob can only once. The portfolio has to be submitted this week and I can't get past this Question. Please help !!!!

Ahm, yeah.
I calculated the probability that Bob doesn't lose.
So, If he throws a 1 then Ann has to throw 11 - 1 combination.
If he throws a 2, then Ann has to throw 11, 21, 12 or 22 - 4 combinations...
And so on till 6. Therefore Ann has 36 possible combinations.
In the end it's 1 - (some kind of formula you should get) | And the 1 in the beginning is important as you want it for Ann.

So the thing is that in the question it is said that she 'can' throw once more, so there is a probability that she doesnt throw it the second time, that will only be when she wins ( the probability of her winning in d first try was easily calculated in the previous question being 15/36) so if she looses that ( probability of 21/36) will she win when she throws it d second time ?

hey kg!! wassup??

she "can" throw twice! so she will throw twice! wat has she got to lose if she throws? u r anyways gonna take the highest number whether thats on the first die or the second or the 200th !!
Absolutely Positively is absolutely right!!

So what if she wins the first time and doesnt want to throw twice ?? ??

Edited by Austin Glau, Jun 18, 2011 - 17:13.
no text speak

[Absolutely Positively]

So wat if she wins the first time and doesnt wanna throw twice ?? ??

You are - of course - allowed to make it more complicated than it already is. . .
But I don't think that this 'can' is really that relevant, or did you get a different result then?

[jaisaha94]

Can anyone please help me out with the part in the portfolio 'The Dice Game' where Ann can roll the dice twice while Bob can only once. The portfolio has to be submitted this week and I can't get past this Question. Please help !!!!

Ahm, yeah.
I calculated the probability that Bob doesn't lose.
So, If he throws a 1 then Ann has to throw 11 - 1 combination.
If he throws a 2, then Ann has to throw 11, 21, 12 or 22 - 4 combinations...
And so on till 6. Therefore Ann has 36 possible combinations.
In the end it's 1 - (some kind of formula you should get) | And the 1 in the beginning is important as you want it for Ann.

So the thing is that in the question it is said that she 'can' throw once more, so there is a probability that she doesnt throw it the second time, that will only be when she wins ( the probability of her winning in d first try was easily calculated in the previous question being 15/36) so if she looses that ( probability of 21/36) will she win when she throws it d second time ?

i think we have to assume she throws it twice and not that if she wins on the first try, she needn't throw it a second
it also makes the calculation alot easier

[rahuld3eora]

Just read the entire portfolio tasks quickly and go with your gut feeling on which route to take.....cause I heard that the person who writes the portfolio has some kind of word limit so he has to leave things ambiguous but if your gut is wrong you get a 0, so be careful about that .

[jaisaha94]

Just read the entire portfolio tasks quickly and go with your gut feeling on which route to take.....cause I heard that the person who writes the portfolio has some kind of word limit so he has to leave things ambiguous but if your gut is wrong you get a 0, so be careful about that .

haha beggar

[Kavishk Gandhi]

Just read the entire portfolio tasks quickly and go with your gut feeling on which route to take.....cause I heard that the person who writes the portfolio has some kind of word limit so he has to leave things ambiguous but if your gut is wrong you get a 0, so be careful about that .

haha beggar

joey u went with ur gut fr dat ??

[jaisaha94]

Just read the entire portfolio tasks quickly and go with your gut feeling on which route to take.....cause I heard that the person who writes the portfolio has some kind of word limit so he has to leave things ambiguous but if your gut is wrong you get a 0, so be careful about that .

haha beggar

joey u went with ur gut fr dat ??

hahahaha

[Desy Glau]

so who has finished this task and is able to give a little more guidance?

so in the fourth question, the player will win if their score is higher than the bank's and the player will lose (i.e. the bank will win) if their score is the same with the bank's? what will happen if the player's score is lower than the bank's?

wouldn't the best game be a fair one? like when the probability of winning for the player and for the bank are the same?

can anyone hint me on the last question with multiple players?

[Desy Glau]

I need some help with this IA...

what did you do for the last part? how did you decide who the winner is when you have multiple players? does everyone have to pay the player with the highest score or only the banker will pay him? what will happen if two players get the same number?

Thank you so much!!

[Keel]

I need some help with this IA...

what did you do for the last part? how did you decide who the winner is when you have multiple players? does everyone have to pay the player with the highest score or only the banker will pay him? what will happen if two players get the same number?

Thank you so much!!

If you have multiple players you can still keep it a player vs dealer game but I'm not sure whether that will meet the criteria of a game with "multiple players". If two players get the same score and both beat the bank, then the bank must pay both players individually and vise versa; just like in blackjack.

However, if you ask your teacher and this doesn’t really fit you need to think of some rules where the bank has a higher advantage. E.g. players only can have one roll but the bank gets two. It should be obvious that the more players there are, the less likely the bank is going to win. Alternatively, you could turn the bank into a player and have the rule 'winner takes all'. If there’s a draw, the pot is split; just like in poker.

[Desy Glau]

thank you!

what if it's like this:
banker: 2
player A: 1
player B: 2
player C: 3
player 5

should player A pay the banker since his score is lower?
should player A pay player D or will the banker be the one who pays? A's score is the lowest
player B gets the same with the banker...will he have to pay?
what happen to the rest? so they get their money back? or is it like everyone's bet will be given to player D? if so then does the banker bet too?

my teacher said we can come up with whatever rule as long as it's consistent

Edited by Desy Glau, Aug 29, 2011 - 09:21.

[Keel]

thank you!

what if it's like this:
banker: 2
player A: 1
player B: 2
player C: 3
player 5

should player A pay the banker since his score is lower?
should player A pay player D or will the banker be the one who pays? A's score is the lowest
player B gets the same with the banker...will he have to pay?
what happen to the rest? so they get their money back? or is it like everyone's bet will be given to player D? if so then does the banker bet too?

my teacher said we can come up with whatever rule as long as it's consistent

I think the IA allows a lot of flexiblity and I would go with your teachers advise.

If you use the blackjack method then player A would pay bank, and bank would pay player C and D. For player B you define the rules as to who wins (I mean you could even say that in the case of a tie no one wins!).

If it's winner takes all then banker, player A, B and C pay player D.

I mean in my opinion, as long as the game isn't biased towards the bank or players anything is acceptable.

[Desy Glau]

is there any way to determine the cost of playing and money won using the probability? or can we come up with random numbers?

[Desy Glau]

can anyone PLEASE guide me with the last two questions??? how to determine the costs and payout?? and how do you count the probability of winning when there are multiple players??? HELP HELP HELP

Edited by Desy Glau, Aug 31, 2011 - 06:56.