Question

Question with probability!

Answer 1

part a, i got 1/32

Answer 2

@dan815 @ganeshie8

Answer 3

I'm kinda lost for part b, do we just take the individual probabilities of each money amount and then average?

Answer 4

your part a looks wrong

Answer 5

really? how do i revise it? or where am I going wrong ?

Answer 6

i thought it'd be just 1/2*x, where x equals each level.

Answer 7

probabilities must add up to 1

in your case they don't

Answer 8

you have 6 bins and you're saying that each probability is 1/32

since all the probabilities must add up to 1 and
1/32 + 1/32 + 1/32 + 1/32 + 1/32 + 1/32 is not 1

your answer cannot be correct

Answer 9

Consider bin2
can you tell me how you got 1/32 ?

Answer 10

Oh, i see what you're saying. So does that mean, i'll have 3 cases? For i = 1,6; i = 2,5; i = 3,4? sorry, for the delayed response.

Answer 11

but in that case, we still have to model it for i, so how wuld we do that? cause if it was to be modelled by the dolar amount, that'd make sense.

Answer 12

something like that

Answer 13

Look at bin2
how many routes are there form top arrow to bin2 ?

Answer 14

each route consists of "four left" and "one right", yes ?

Answer 15

um, sorry, i'm not sure what you mean

Answer 16

oh wait, yes, i get it now.

Answer 17

similarily, there are 4 right and 1 left for bin 5.

Answer 18

so how many total routes are there from top arrow to bin2 ?

Answer 19

um, 2^5?

Answer 20

remember the grid problem that we did few days ago ?

Answer 21

mhmm

Answer 22

4L and 1R

so a route may look like this :
LLLRL

Answer 23

how many different strings are possible that contain 4L and 1R ?

Answer 24

So 32C2?

Answer 25

really ?
try again

Answer 26

5C1

Answer 27

or 5C4 right?

Answer 28

similar for bin 5.

Answer 29

and then for bin 3, we need 3 left and 2 right. So, 5C3 or 5C2 right?

Answer 30

and for bin 1 and 6, it's just 1 for each right?

Answer 31

Yes!

Answer 32

so 32 total paths. and then for 1,6, we have 1/32 each. and for 2,5, we have 5/32 each. and then for 3,4, we have 10/32 each

Answer 33

that makes wonderufl sense! thank you so much! Ganeshie, i've struggled for probability for so long, adn you're really helpful, thank you!

Answer 34

So, based on this new probability, i'd have to analyze the amount of money they would win righ?

Answer 35

Looks good !

Answer 36

so do i just say 15.625%*2 chance of winning 500$?

Answer 37

and so on? for the other amounts?
15.625 = 5/32

Answer 38

Yes,

for part b :

expected money = 1000*1/32 + 500*5/32 + 100*10/32 + 100*10/32 + 500*5/32 + 1000*1/32

Answer 39

ANd for part c, do we just recalculate part a, with the maximum number of steps decreased by one for each combinatorics operation and the total steps being 16, instead of 32? and then having calculated this, we recalculate part b for these new values right?

Answer 40

Exactly !

Answer 41

thank you!

Answer 42

part d is that it takes left twice.

Answer 43

Lastly, for modeling the dice question, i posted my solution so far and I believe it works without any endless loops (regardless of how small the probability of that happeneing is.) just a heads up for that one also.

Answer 44

part d should be :

1/8 for i=1,4
(3C1)/8 for i=2,3
1000*(1/8) + 500(3/8) + 100(3/8) + \(\color{red}{100}\)(1/8) = \(\color{red}{362.5}\)

Answer 45

right ?

Answer 46

oh right! yes, part d should be that! thank yoU!

Answer 47

So do you think my part c and d are coherent?

Answer 48

yes everything else looks good

Answer 49

part b and part c have same answers : 281.25

this is a bit surprising yeah

Answer 50

but i'm not actually miscalculating anything am I? perhaps the change in value amount that we're not adding in is just the right amount to not add in to get the same answer if we had one extra option to turn.

Answer 51

yeah i thought the same...