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After Monty opens a door with goat, if I am still able to switch my initial choice into this opened door with goat.
No, that is not allowed.
If Monty claimed that opened door with goat is not allowed to be selected, for the second choice, there is only 2 doors available for selection (Switch or Stay). "1/2" chance to get the CAR is closer to probability in practice instead of "2/3"
Therefore, I assume that the opened door with goat is still considered as a choice for selection in order to calculate "2/3" probability in Monty Hall Problem, if this is the case, then everything is much clear on probability calculation.
On the other words, in reality, you would never select the opened door with goat, but in Monty Hall Problem, Monty still considers this opened door with goat as a valid choice for selection under Monty Hall Problem.
Do you 100% confirm that selecting the opened door with goat is not allowed under Monty Hall Problem?
Thanks, to everyone very much for any suggestions :>
1) After Monty opens a door with goat, am I still able to switch my
initial choice into this opened door with goat? It may sound stupid to
pick an opened door with goat instead of 2 unopened doors, but it
could be the condition under Monty Hall Problem.
2) After Monty opens a door with goat, I cannot switch my initial
choice into this opened door with goat.
I would like to confirm which one is correct condition under Monty
Hall Problem.
Does anyone have any suggestions?
Thanks, to everyone very much for any suggestions :>
This is the question related to this topic...
Second choice : how does the probability of second choice depends
on the initial choice? Making second choice is after initial
choice in this order, but it does not make any depended condition
on calculating probability for second choice. Both choices are
independent to each other, there is missing connection on how to
both choices related to each other mathematically.
Do you have any suggestions?
Thanks in advance for any suggestions
Monty Hall problem
http://mathforum.org/dr.math/faq/faq.monty.hall.html
1) You make an initial guess.
2) You're told: In a moment, Monty is going to show you
a goat behind one of the other doors.
When he does that, will you want to switch to the
remaining door, or not?
You have to answer now.
3) You decide whether to switch or not.
4) Monty shows you a goat.
5) You switch or not, as you decided in step 3.
Questions:
1) Initial choice : "2/3" probability of initially selecting a goat.
This is clear with no problem.
2) Before opening any door with 3 unopened doors available for
selection, the probability of initial choice is still valid at
this moment.
After opening a door by Monty and only 2 unopened doors available
for selection. The problem is starting here on calculating
probability for second choice.
3) Second choice : how does the probability of second choice depends
on the initial choice? Making second choice is after initial
choice in this order, but it does not make any depended condition
on calculating probability for second choice. Both choices are
independent to each other, there is missing connection on how to
both choices related to each other mathematically.
Does anyone have any suggestions?
Thanks in advance for any suggestions
I can not download that file. Can you cut and paste the data here.
Here you go ...
Thanks you very much for any suggestions
2013 2014
4% 46%
83% 26%
30% 92%
89% 74%
38% 21%
54% 13%
51% 64%
47% 87%
7% 13%
77% 95%
57% 19%
24% 20%
3% 1%
73% 93%
85% 28%
4% 14%
39% 19%
74% 71%
59% 16%
91% 51%
51% 64%
45% 54%
67% 7%
87% 81%
9% 39%
30% 41%
15% 31%
50% 89%
61% 25%
46% 86%
70% 69%
77% 16%
25% 20%
81% 10%
7% 95%
67% 95%
23% 51%
56% 52%
88% 4%
43% 23%
22% 69%
36% 60%
43% 78%
26% 81%
65% 29%
55% 70%
53% 12%
80% 96%
68% 96%
7% 72%
33% 14%
20% 4%
54% 52%
68% 87%
24% 95%
31% 72%
53% 22%
45% 91%
34% 37%
3% 16%
23% 5%
87% 38%
33% 29%
71% 20%
62% 98%
22% 60%
61% 48%
30% 59%
56% 83%
32% 40%
80% 59%
36% 29%
2% 63%
55% 1%
72% 90%
69% 49%
20% 64%
9% 67%
50% 24%
6% 70%
67% 44%
36% 45%
76% 62%
34% 64%
27% 95%
47% 24%
86% 43%
74% 40%
18% 18%
61% 30%
67% 2%
29% 57%
46% 77%
65% 81%
63% 29%
28% 5%
89% 96%
68% 8%
53% 26%
27%
97%
34%
44%
59%
50%
21%
14%
41%
32%
71%
40%
2%
48%
36%
19%
27%
22%
84%
8%
47%
41%
92%
69%
67%
25%
54%
82%
77%
27%
81%
50%
10%
79%
40%
92%
34%
45%
90%
39%
21%
40%
98%
96%
14%
5%
33%
56%
25%
79%
There are 2 T tests. The variances are calculated form the data. Can you provide the data in text form?
Do you have any suggestions?
Thank you very much for any suggestions :>
Text file for download:
http://1drv.ms/1o5pHJU
Hi;
Take a good look at this. I am going to use it for your problem.
https://onlinecourses.science.psu.edu/stat500/node/50
The method does not seem to care about the different sizes. What are the variances of both populations?
I get no idea on variances of both populations, can I still apply T Test in my case?
Thanks you very much for any suggestions
A 2 sample T test might be what you need. Are the two samples independent?
Two samples are independent, could you please show me on how to apply T test in this case? in order to determine which years students are doing better.
Thanks you very much for any suggestions
In 2013, there are 100 students taking EXAM
In 2014, there are 150 students taking EXAM
In term of percentage of students within 2014 year, their scores are getting improvement comparing with 2013 year. The main issue is not knowing on how to define the benchmark on 2013.
For example, John's score is 64 on 2013, Mary's score is 73 on 2014, then Mary in 2014 is getting better than John in 2013. Should I use average score on 2013 be benchmark in this case?
Does anyone have any suggestions on how to do statistical comparisons on unequal size groups? so I can compare between 2013 and 2014, in order to determine which year students are doing better.
Does anyone have any suggestions?
Thanks in advance for any suggestions
Download file:
http://1drv.ms/1nvIxOA
I can get 3% annually return on fixed deposit.
If a house is worth $10000, and I can get 70% mortgage from Bank at 4% interest rate,
I need $3000 as initial capital and borrow $7000 from bank to purchase this house,
Luckly, if I can get $500 rental annually, then I get 5% ROI on rental property, but I need to pay 4% mortage rate, so I get 5% - 4% = 1% net income from rental property.
If the price of house keeps unchanged for a period of time, should I place $3000 on fixed deposit or should I purchase $10000 house for 1% net income annually?
Does anyone have any suggestions on how to do this math?
Thanks in advance for any suggestions :>
hi oem7110
...
But what is the price of a cash coupon?
...
Bob
Cash coupon can be used like cash within specific store with 5 years period before it expires.
It can be any gift to your friends during christmas season, so they can purchase what they want.
Does it seems to the right decision to purchase the $20000 cash coupon?
Thanks you very much for any suggestions
Installment Plan A
If I purchase items over $100,000, there is 0% interest for 48 months installment plan, but there is $65 monthly service fee.
Installment Plan B
If I purchase items over $50000, there is 0% interest for 48 months installment plan, but there is $72 monthly service fee.
If I need to purchase an item about $80000 then I have to choose Installment Plan B,
If I need to purchase an item about $80000, and purchase cash coupon $20000 then I can choose Installment Plan A,
Should I purchase cash coupon for $20000 right now for installment plan A? or
Should I only purchase the item at $80000 for installment plan B?
Does anyone have any suggestions on which one is better in term of money management?
Thanks in advance for any suggestions
Thank everyone very much for suggestions with detailed explaination
Do I need to every possible X in order to determine n? such as X = 1050, but when I try different X within the formula, could you please tell me when I should stop and find the right n?
There is one issue about selecting the right sample size I concern.
For example,
If I take 30 sample size for evaluation based on Normal distribution, the number occurrence of 6 is 20.
If I take 120 sample size for evaluation based on Normal distribution, the number occurrence of 6 is 20.
If I take 300 sample size for evaluation based on Normal distribution, the number occurrence of 6 is 20.
So from different point of views based on sample sizing,
for the first 30 sample, the number occurrence of 6 is 20, after that, no 6 occur in futher sampling.
the number occurrence of 6 seems too much based on 30 sample size,
the number occurrence of 6 seems fair based on 120 sample size,
but the number of occurence of 6 seems too little based on 300 sample size.
Therefore, to determine whether the die is fair or biased depends on how we select the sample size to fit what we want, and that is the problem I concern.
Does anyone have any suggestions on any approach on Math to solve the issue of sample sizing?
Thanks everyone very much for any suggestions
If I should use normal distribution, do I need at least 300 sample size in order to apply 2/3 standard deviations?
Will this sample size be the minimum requirement for normal distribution?
What about binomial distribution? Can this sample size apply to binomial distribution too?
Thanks everyone very much for any suggestions
I understand your approach on determining the expected value, but I don't understand what I need to take care because I cannot just add the variances.
Could you please give me any further suggestions?
Thank you very much for your suggestions
Could you please tell me how much sample size should be collected in order to determine whether I should use binomial or normal distribution?
When I look up the value 3.46 in normal tables (0.0003369), how can I know whether the value indicates a biased or fair die?
If I draw a line on accepting 5% error, then when the number of 6 was 1050, and the determined value 1.73 equals to 0.9582 in normal tables,
the probability of 4.18% is considered to be fair die within 5% error, am I on the right track?
Thank you very much for your suggestions
I find an interesting topic about dice on the web, but don't know the answer, does anyone have any suggestions on how to work out this math?
You are given 3 dice. One dice is normal unbiased, one dice is biased for even numbers and the third is biased for odd number.
Biased for even(odd) numbers means that when the dice is rolled, the probability that the number facing up will be even(odd) is 4/5. Also, each of the even(odd) numbers have equal probability of being on top and so do the odd(even).
That is, if you roll the dice biased for even numbers, the probability of having 2 on top is 4/5 x 1/3 = 4/15, while the probability of having 1 on top is 1/5*1/3 = 1/15
In the unbiased dice, each number (1 to 6) has equal probability of being on the top face.
The three dice are rolled together over and over, and their sum (of numbers on the top faces of the three dice) "S" noted each time. What value of "S" is expected to be most frequent?
Thanks in advance for any suggestions
If the die is fair, and I roll a die 6000 times, then the occurrence for each number (1,2,3,4,5,6) will be 1000 times.
In reality, the number of occurrence will not be the same. Is there any approach to determine whether the die is based or not? If the total outcome of 6 is 3500 times, and the outcome for the rest of numbers are 500 times, then I can see there is a clear biased. I get no idea on how to draw the line on defining how much occurrence will indicate a biased die.
Does anyone have any suggestions?
Thanks in advance for any suggestions
Oh I see, (Sum of all [Probability x payoffs]) / Total numbers of combination, right?
I try it.
Thank you very much for your suggestions
I just finished on adjust the payoffs for all the numbers.
Do you have any suggestions?
Thank you very much for any suggestions
Since I bet $5 for each game on 5 different numbers (betting $1 on each number), if the sum of the dices equal to any numbers between 4 and 8, then I will collect my original $1 back on winning number, but will lose $4 on the other number.
If sum of the dices is 3,9,10,11,12,13,14,15,16,17,18 among 3 dices, then I lose $5 totally.
If sum of the dices is 4 among 3 dices, then my profit is $62 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $62 - $4 = $58.
If sum of the dices is 5 among 3 dices, then my profit is $31 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $31 - $4 = $27.
If sum of the dices is 6 among 3 dices, then my profit is $18 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $18 - $4 = $14.
If sum of the dices is 7 among 3 dices, then my profit is $12 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $12 - $4 = $8.
If sum of the dices is 8 among 3 dices, then my profit is $8 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $8 - $4 = $4.
Do you have any suggestions?
Thank you very much for any suggestions
Hi Bobbym:
Thank you very much for suggestions
I get another similar issue with little different conditions:
Three dices are rolled, and I bet $1 on 4,5,6,7,8 numbers with different profits, so total bets is $5.
If sum of the dices is 4 among 3 dices, then my profit is $62 and collect my original $1 back, but I lose $4 on the other numbers.
If sum of the dices is 5 among 3 dices, then my profit is $31 and collect my original $1 back, but I lose $4 on the other numbers.
If sum of the dices is 6 among 3 dices, then my profit is $18 and collect my original $1 back, but I lose $4 on the other numbers.
If sum of the dices is 7 among 3 dices, then my profit is $12 and collect my original $1 back, but I lose $4 on the other numbers.
If sum of the dices is 8 among 3 dices, then my profit is $8 and collect my original $1 back, but I lose $4 on the other numbers.
I get no idea on how to determine the expected value for this combination, since only one condition can be met, and I must lose bets on another 4 numbers.
Do you have any suggestions?
Thanks in advance for any suggestions
Eric
Three dices are rolled, and I bet $1 on 6,
If there is no 6 among 3 dices, then I lose $1.
If there is three 6 among 3 dices, then my profit is $12 and collect my original $1 back.
If there is two 6 among 3 dices, then my profit is $2 and collect my original $1 back.
If there is one 6 among 3 dices, then my profit is $1 and collect my original $1 back.
I get no idea on how to determine the expected value for this combination, since only one condition can be met.
Does anyone have any suggestions?
Thanks in advance for any suggestions
Eric
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