[Guest]

Two thousand dice are rolled simultaneously.

What is the probability of the sum of all the numbers rolled:

Equaling one thousand?

Equaling four thousand?

Being less than thirteen thousand?

Being less than five thousand?

What is the probability that one fifth of the numbers rolled will be odd

Edited October 27, 2010 by matthew laming

[Guest]

Probability of 0 for the sum to equal one thousand since two thousand dice are rolled and the lowest number is on each is one

Probability of 1 for the sum being less than 13000 since 2000 sixes would only be 12000

Probability of 3.87*10^-121 that one fifth will be odd if we don't care what the other 1600 numbers are, 8.71*10^-603 if the other numbers have to be even

I am still working on the other two questions since they involve quite a few different combinations of numbers to equal 4000 or be less than 5000

[Guest]

Probability of 0 for the sum to equal one thousand since two thousand dice are rolled and the lowest number is on each is one

Probability of 1 for the sum being less than 13000 since 2000 sixes would only be 12000

Probability of 3.87*10^-121 that one fifth will be odd if we don't care what the other 1600 numbers are, 8.71*10^-603 if the other numbers have to be even

I am still working on the other two questions since they involve quite a few different combinations of numbers to equal 4000 or be less than 5000

Do you just come up with random probability questions?

[Guest]

P(Sum = 1K) = 0; as minimum sum is 2K

P(Sum >13K) = ; as maximum sum is 12K

For the other two parts, I have a theory to propose. Not sure if it is correct though.

Lets say the sum of rolls of dice will be a normal distribution. Further, the expected number on each roll of dice is 3.5 (calculated as 1/6 + 2/6 + 3/4 + 4/6 + 5/6 + 6/6 = 3.5) Then for 2K rolls, the mean of all sums would be 3.5 x 2K = 7K

Now, because the max sum is 12K and min sum is 2K and a 3 sigma variation covers almost all the range, then sigma for the normal distribution of all sums of 2K rolls should be 5K/3

Now for the questions:

P(sum = 4K) = 0

As the point probability in a normal distribution is zero

P(sum <5K) = 11.5%

(Mean = 7K, Std deviation = 5K/3 and random normal variable = 5K)

[bonanova]

P(Sum = 1K) = 0; as minimum sum is 2K

P(Sum >13K) = ; as maximum sum is 12K

For the other two parts, I have a theory to propose. Not sure if it is correct though.

Lets say the sum of rolls of dice will be a normal distribution. Further, the expected number on each roll of dice is 3.5 (calculated as 1/6 + 2/6 + 3/4 + 4/6 + 5/6 + 6/6 = 3.5) Then for 2K rolls, the mean of all sums would be 3.5 x 2K = 7K

Now, because the max sum is 12K and min sum is 2K and a 3 sigma variation covers almost all the range, then sigma for the normal distribution of all sums of 2K rolls should be 5K/3

Now for the questions:

P(sum = 4K) = 0

As the point probability in a normal distribution is zero

P(sum <5K) = 11.5%

(Mean = 7K, Std deviation = 5K/3 and random normal variable = 5K)

Nice approach

P(sum = 4K) = 0.

What if all the dice showed 2?

[Guest]

Don't you just love coming up with random things to give people headaches? mmmmmmm

[bonanova]

Two thousand dice are rolled simultaneously.

What is the probability of the sum of all the numbers rolled:

Equaling four thousand?

Being less than five thousand?

What is the probability that one fifth of the numbers rolled will be odd

Simulation - the lazy person's way of calculating probability - shows that

all three of these probabilities are vanishingly small.']

The average value of a single die is 3.5, and 2000 trials will achieve very close to that average.

Thus the total of 2000 fair dice will be very close to 7000.

In 1 million rolls of 2000 dice, the sum ranged from 6642 [avg - 358] to 7363 [avg + 363]

So the chances of even 6500 are less than one in a million.

Achieving 6000 is astronomically improbable; 5000 and 4000 are science fiction.

Similarly, in 2000 dice we can expect very nearly 1000 odd numbers.

In 1 million rolls of 2000 dice, the odd dice ranged from 882 [44.10%] to 1102 [55.1%]

So the probability of having as few as 900 [45%] tails is only slightly more than 1 in a million.

Achieving 400 tails [20%] has a calculable probability, but as fast as you could roll the dice,

it won't happen before the big crunch.

Change 2000 dice to 20 dice and the numbers are realistic, but still unlikely [less than 1%].

From 10 million rolls of 20 dice,

The total is less than 50 about 0.33% of the time.

The total is equal to 40 about 0.0015% of the time.

One-fifth of dice are odd about 0.46% of the time.

[dark_magician_92]

Two thousand dice are rolled simultaneously.

What is the probability of the sum of all the numbers rolled:

Equaling one thousand?

Equaling four thousand?

Being less than thirteen thousand?

Being less than five thousand?

What is the probability that one fifth of the numbers rolled will be odd

1). 0

2). 1/(12*1999)

3). 1

4) and 5). thinking