1
 |
Stroggoz New Zealand. Sep 21 2016 08:46. Posts 5296 | | |
i got this problem and thought of LP
The Chevalier de M´er´e (1650s) won alot of money gambling on throwing at
least one 6 after four throws of a single fair six-sided die. Is the probability of this event occuring
greater than or less than 1/2
?
spoiler:
my working:
Ax =x sixes per 4 dice rolls
our sample space
n(s)=1296, n(A1)= (5x5x5x1)x(4c1)=500
n(A2)= (5x5x1x1)x(4c2)=150
n(A3)= (5x1x1x1)x(4c3)=20
n(A4) = (1x1x1x1)x(4c4) =1
P(A1UA2UA3UA4) =n(A1UA2UA3UA4)/n(A) =.5177
|
|
| One of 3 non decent human beings on a site of 5 people with between 2-3 decent human beings | |
|
|
1
 |
bigredhoss Cook Islands. Sep 21 2016 10:41. Posts 8648 | | |
yeah, you can also just do 1 - (5/6)^4 which gets the same answer.
edit to remove extra unnecessary parens |
|
| Truck-Crash Life | Last edit: 21/09/2016 10:51 |
|
|
1
|
Stroggoz New Zealand. Sep 21 2016 11:29. Posts 5296 | | |
lol dam i didnt even think of that. These guys who invented probability theory must have been making sick bank |
|
| One of 3 non decent human beings on a site of 5 people with between 2-3 decent human beings | Last edit: 21/09/2016 11:31 |
|
|
1
 |
failsafe United States. Sep 22 2016 01:11. Posts 1037 | | |
yea but imagine you're at SAO level 163 or something and someone approaches you with this question. the reasoning is razor thin and you take the flip and win what does it mean about your chances lateron |
|
|
1
| |
p(A1uA2uA3uA4) = p1 + p2 + p3 + p4 = 1 - (1 - p1 * p2 * p3 * p4) = 1 - p(A1nA2nA3nA4)
you can also intuitively answer that any given number on a die takes 3.5 rolls to get, so rolling four times is + EV
| | On September 22 2016 00:11 failsafe wrote:
yea but imagine you're at SAO level 163 or something and someone approaches you with this question. the reasoning is razor thin and you take the flip and win what does it mean about your chances lateron |
the events are independent so the chances are still the same |
|
| | Last edit: 22/09/2016 12:35 |
|
|
1
|
failsafe United States. Sep 24 2016 09:28. Posts 1037 | | |
so 1 is all (or every of each) and the individual probabilities are each of possibility x. nothing prevents misreading so for instance your notation could never be 1+sum(p) even if you subtracted 1 afterward. also you could never begin assigning a 0 probability (read none of any) to the situation itself coming about.
so for instance you avoid erroneous information like the off-putting challenge. |
|
|
1
| |
| | On September 24 2016 08:28 failsafe wrote:
for instance your notation could never be 1+sum(p) even if you subtracted 1 afterward. |
sure it can
all probabilities = 1 = p + q
probability of something = p
probability of something not happening = 1 - p = q
probability of something not happening, not happening = 1 - (1 - p) = 1 - q = p |
|
| | Last edit: 27/09/2016 07:29 |
|
|
|
