If you roll two dice for one die to show a six won’t that be odds of 1 in 3?
You’re counting 6, 6 twice, which I initially thought had to be done. See the following:
http://www.math.hawaii.edu/~ramsey/Probability/TwoDice.html
As borschevsky is alluding to in the question he asked you, you will get a 1 and a 6 twice as often as a 6 and a 6.
I think you’re right that the answer you get depends very much on how the situation is framed.
Lay out 36 pairs of dice in a grid showing all 36 possible results.
Of the 6 pairs in which the die on the right side of the pair is six, the left die will also be 6 once (one chance in 6).
Of the 6 pairs in which the die on the left side of the pair is six, the right die will also be 6 once (one chance in 6).
Of the 11 pairs in which at least one of the dice in the pair is 6, the other die will be 6 in one case (1 in 11).
Take two real dice and see if you come up with 1/11.
Ok, I just rolled two dice for a good chunk of time. I discarded all rolls that did not include a 6 and I stopped after 20 double sixes. I had 132 rolls that included at least one 6.
Here are the ratios for the rolls.
4 22 5.5
5 26 5.2
6 29 4.833333333
7 30 4.285714286
8 35 4.375
9 41 4.555555556
10 85 8.5
11 86 7.818181818
12 96 8
13 98 7.538461538
14 99 7.071428571
15 111 7.4
16 112 7
17 117 6.882352941
18 118 6.555555556
19 129 6.789473684
20 132 6.6
I’m afraid you’ll have to explain to me what those numbers represent.
octopus, then you think with a roll of two dice, you have an equal chance of rolling two sixes as you do rolling a 1 and a 6?
Do you agree that the probability of rolling a 1 and a 6 is 1/18?
Of course it is. You have the chance of rolling the pair 1,6 or 6,1 assuming you can tell the rolls apart with some mechanism. Now the chance of the ordered pair 1,6 or 6,1 is actually 1/36. But notice that a 1,6 is the same chance as a 6,6 if you can discern the dice.
The sum of 1 and 6 is 7 and that is a 1/6 chance of occurring. So what are we looking for sums or particular probabilities of particular events. In that example he is summing events. If 1,6 is considered exactly equivalent to 6,1 than yes those 2 have twice the probability of 6,6.
Let’s make the math easier and use the coin example. Throw out all double tails. Use a penny and a nickel.
What are the outcomes
Penny Nickel
H T
T H
H H
How many possible ways can I show one head? I can’t show one head with double tails so that isn’t even in the denominator. It’s not counted. Correct?
So how many ways can I show you a head?
A) I can show you Penny’s head if Penny comes up head and Nickel comes up tails. Correct? That is one count.
B) I can show you Penny’s head if Penny comes up head Nickel comes up head. That is two.
C) I can show you Nickel’s head if Nickel comes up head and Penny comes up tails. That is three.
D) I can show you Nickel’s head if Nickel comes up head and Penny comes up head. That is four.
Ah - ha ha ha.
Four ways to show you a head in which at least one head was flipped. These are four permutations of a state in which one coin is revealed and one is concealed with the constraint that there will only be a revelation if at least one is a head. Correct?
Now how many of those permutations satisfy the second case where the concealed coin is also a head? Two events B and D. 2/4 is 1/2 which is what you’d expect with a fair coin.
Cool. Now please Google probability of rolling snake eyes.
Yes it is, it’s just more detailed. If the red die is a 6, the odds of a green 6 are 1/6. If the green die is a 6, the odds of a red 6 is 1/6. That covers all possibilities, and in every case, it is 1/6. It doesn’t mater if the dice are colored – it’s 1/6 in every possible case.
Then why are you claiming this?:
What I’m enjoying now are the folks writing simulations and doing manual dice throwing trials.
The ambiguity is in the problem statement. Once you decide on your preferred interpretation of the problem statement then your result (1/11 or 1/6) is foreordained. And can be proven with a few short lines of small words.
So any effort spent to simulate is foolish. It’s merely proving the trivial conclusion you already (perhaps unwittingly) assumed.
Snake eyes is 1/36.
1,1 is one ordered pair out of 36.
Simple. I’m not the one confusing what the actual constraints of the question are.
Shoot let’s say we only count the probability of the die roll after we rolled a 6. Instead of revealing a 6 and asking we are only counting 6s after we rolled a 6. What’s the probability of that? Why would the probability of the die change? It only has 6 possible states regardless of what was rolled before.
I’m confused. You agree the probabilty of rolling a 1 and a 6 is 1/18, and that rolling a 6 and a 6 is 1/36, yet you count 6,6 twice here:
Do you agree with the following?:
http://www.math.hawaii.edu/~ramsey/Probability/TwoDice.html
There are 36 equal rolls of two dice.
That’s not the OP’s problem. It didn’t say anything about BOTH dice, but what would happen IF one condition was already met. That condition is no longer a factor, as we are not considering multiple conditions, nor conditional conditions.
It doesn’t matter which is the “other die,” since this is not a conditional situation. All conditions (no 6’s found, etc.) are eliminated. All that matters is the odds of ONE toss of ONE die, with the possibilities of 1…6.
And just in case, running my program for 10,000,000 iterations confirms a 1:6 ratio, within 3 decimal places. I doubt if running it longer will change much, so it’s the premise that counts here.
I stand by my post #89, more than ever. Anyone want 10 million data points to look at?
Different problems.
You’d like to think so. But some are still arguing about questions unrelated to the ambiguity of the problem statement, like red vs green dice and ordered pairs. For those, a simulation can be handy to demonstrate why 1/11 is correct according to one interpretation.
Once again you’ve assumed one interpretation of the problem. And then spent effort tautologically proving your own assumption. Yes, Virginia, the odds of a 1:6 outcome really are 1:6.
FWIW, I prefer your interpretation of the problem too. But it’s just that: one interpretation of the subtly (and deliberately) ambiguous sentences in the original problem statement.
The fact the multiple choice answers have choices which are correct for each interpretation and for the common confusion of [is R6-G6 different from G6-R6?] pretty well proves the problem was carefully deliberately *designed *to produce these incompatible lines of thinking.
You’re falling into the same trap I did. Given your assumptions, your solution is exactly right and 1/11 is the correct answer. But the assumption is one I didn’t even realize I was making, that six was picked as the “magic” number before any roles, and therefore any roles that didn’t contain a six would be discarded. The other possibility was best explained to me by Chronos in post 68. You aren’t actually told that six was selected before any roles; it could just be a statement of what one of the dice showed. In that case, the fact that one of the dice is a particular number doesn’t add any new information to the problem, and the other die has a 1/6 chance of being that number.
Because what is being revealed and what is being concealed are different states. Look at it closely. There are 11 different pairs.
1,6
2,6
3,6
4,6
5,6
6,6
6,1
6,2
6,3
6,4
6,5
That’s 11 things right? 6,6 is 1/11 correct? That is out of this set of ordered pairs 6/6 occurs 1 time out of 11. Makes sense right?
But we have 11 ordered pairs that could exist. The other 25 are discarded. Each of these 11 have at least one 6. If you reveal the 6 you see what the other number could be. But look! There are two different 6s in the 6,6 example that could be revealed. The left or the right, the first or second, the red or green, etc. So there are actually 12 different sixes that could be revealed. And out of that 12 only 2 of the revealed have another 6 concealed and that is reveal red 6 conceal green 6 and reveal green 6 conceal red 6. That’s why the denominator is not 11. Because we aren’t dealing with a set of 11 ordered pairs. We are dealing with revealing a *particular *6 from a set of 11 ordered pairs.
But the question isn’t what is the probability of one of these to occur. The question is what is the probability if I am shown a 6 that the other number is a 6. Well with colored dice there are 12 combinations that can occur because what matters is what is the state of the dice. Are they revealed, concealed? What number is on them. So if I can reveal a green 6 with 6 equally likely permutations on a red die that’s 6 events. If I can reveal a red 6 with 6 equally likely permutations on a green die that’s 6 different events. Yes or no? Is a revealing a green 6 the same as revealing a red 6? No. isn’t.
If I show you a green 6 what are states the red die can be in? 1,2,3,4,5,6. Correct? Why would one number be favored over another. If I show you a red 6 what are the states the green die can be in? 1,2,3,4,5,6. Correct? Why should 6 be a lessor probability than the 1,2,3,4 or 5? There’s no logical reason to treat 6 differently.