[Justin7]

Warm-up problem:
Roll 2 fair 6-sided dice. How often will at least 1 of them be a "1"?

Intermediate problem:
Roll 3 fair 6-sided dice. How often will at least 2 of them be a "1"?
Since there are 216 possible ways to roll 3 dice, give your answer in terms of "x outcomes out of a possible 216".

Advanced problem to follow later.

Ganchrow and RickySteve, please wait 24 hours to post.

[Nicky Santauro]

rolling 2 dices and having at least one of the two to be a 1..

Let's do one at a time. your first answer is



30.61% that one of them is a 1..

_____________

[picoman]

Originally Posted by Justin7

Warm-up problem:
Roll 2 fair 6-sided dice. How often will at least 1 of them be a "1"?

1-1,1-2,1-3,1-4,1-5,1-6,2-1,3-1,4-1,5-1,6-1

11/36 = 31%

Intermediate problem:
Roll 3 fair 6-sided dice. How often will at least 2 of them be a "1"?
Since there are 216 possible ways to roll 3 dice, give your answer in terms of "x outcomes out of a possible 216".

41/216 = 19%

Advanced problem to follow later.

Ganchrow and RickySteve, please wait 24 hours to post.

what is the advance problem?

[Justin7]

And with the warm-up, give it in terms of X out of 36...

[Justin7]

Picoman,

Your answer on the intermediate problem is incorrect. Perhaps if you gave your analysis on it, other people could help.

Instead of counting individual incomes for the warm-up, you might consider using a different approach that works better as the complexity goes up.

[Nicky Santauro]

so my 30.62% was right..


you guys do it the complicating way. all you have to do is


(0.833)x(0.833) = 0.6938 that there won't be any 1's..

so that means that 30.62% there will be at least one 1..


it takes one second to figure this out. no need to be doing all that work for nothing.

[picoman]

Originally Posted by Justin7

Picoman,

Your answer on the intermediate problem is incorrect. Perhaps if you gave your analysis on it, other people could help.

Instead of counting individual incomes for the warm-up, you might consider using a different approach that works better as the complexity goes up.

16/216=5%

[picoman]

i think you can still count it. the 11 earlier with 3rd as 1 then add 1-1-2, 1-1-3, 1-1-4, 1-1-5, 1-1-6

[DrunkenLullaby]

Agree with Pico, 16/216.

[Ganchrow]

Originally Posted by picoman

16/216=5%

Correct. (I'll assume that now that the question's been correctly answered it's OK for me to chime in.)

Obviously there are many ways to figure the answer. Following would be the generic symbolic solution.

Define the following events:

Let D1 = Die# 1 is a 1
Let D2 = Die# 2 is a 1
Let D3 = Die# 3 is a 1
Let A = D1 ∩ D2
Let B = D1 ∩ D3
Let C = D2 ∩ D3

Our goal is to find P(A ∪ B ∪ C).

We know from the axioms of probability that for any events A, B, and C:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

We have:

P(D1) = P(D2) = P(D3) = ¹/₆

P(A) = P(D1 ∩ D2) = P(D1) * P(D2) = ¹/₆ * ¹/₆ = ¹/₃₆ (independent events)
P(B) = P(D1 ∩ D3) = P(D1) * P(D3) = ¹/₆ * ¹/₆ = ¹/₃₆
P(C) = P(D2 ∩ D3) = P(D2) * P(D3) = ¹/₆ * ¹/₆ = ¹/₃₆

P(A ∩ B) = P(A ∩ C) = P(A ∩ C) = P(A ∩ B ∩ C) = P(D1 ∩ D2 ∩ D3) = ¹/₆ * ¹/₆ * ¹/₆ = ¹/₂₁₆

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C)
= ¹/₃₆ + ¹/₃₆ + ¹/₃₆ - ¹/₂₁₆ - ¹/₂₁₆ - ¹/₂₁₆ + ¹/₂₁₆
= ¹⁶/₂₁₆

[Ganchrow]

Here's a follow-up question:

A fair coin is flipped 100 times and for every head, a red marble is paced in a bag, while for every tail a black marble is placed in the same bag.

You randomly select 10 marbles (without replacement) from the bag and find 9 to be red and 1 and to be black.

Q: What would be the fair odds on an over/under of 60 red marbles (out of 100) in the bag?

[square1]

I'm getting: Over 60, +1030
Under 60, -1030

[Ganchrow]

Originally Posted by square1

I'm getting: Over 60, +1030
Under 60, -1030

[Ganchrow]

square's spreadsheet is right on point.

This question is provides a straightforward application of Bayesian inference where the availability of additional evidence is used to refine the prior probability distribution. I briefly explained the concept in this post and provided a couple of examples and a spreadsheet.

All those interested should check the above post and peruse the spreadsheets. As always feel free to ask any questions.

I'll try to post another problem of Bayesian inference more obviously related to sports betting at a later date.