# More Dice

## Main Question or Discussion Point

The probability of rolling a 7 with a pair of ordinary dice is 1/6 because, of the 36 possible (and equally likely) combinations, six of them sum to 7. Now, suppose the dice are rolled out of one's sight and some honest person who can see the result tells us that at least one of the dice came up 6. This restricts the total number of possible combinations to the following 11 pairs:

(1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,5), (6,4), (6,3), (6,2), (6,1)

of which exactly two sum to 7. So we would now calculate the probability of a 7 as 2/11, slightly more than 1/6. Of course, the same analysis would give a probability of 2/11 if our honest friend had sais 'at least one 5' instead of 'at least one 6'. Or if he had said 'at least one 4'. In fact, we would arrive at the same probability if he had said 'at least one n' for ANY value of n. Obviously we have 'at least one n' for SOME value of n, so why not just assume this at fist without waiting for our friend to tell us? It would seem that this improves our a priori odds of rolling a 7 from 1/6 to 2/11.

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AKG
Homework Helper
I hate probability problems like this. Anyways, I think that the answer may lie in the fact that although you know that you have at least one n for some n, the fact that you don't know which specific n prevents you from using that reasoning.

Because your numbers are all assumptive with the number 6. If the honest person had of said 5 it would have been:

(1,5)(2,5)(3,5)(4,5)(5,5)(6,5)(5,6)(5,4)(5,3)(5,2)(5,1)

Although it is still 2/11 the number of 6s has deceased from when the honest person said 6. So assuming would be inaccurate to the number 6. However you can assume 2/11 probability to every set of numbers.

If the partner has to say "I see at least one of n" where n is the value of one of the two dice that were cast and then you have to say "the sum equals 7" or "the sum does not equal 7", what strategy could you and your partner decide upon, before starting the game, that would optimize your chance of being right most often, on any given roll of the pair of dice? And what is the overall probablity, using that method, that you would be right?

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AKG
Homework Helper
My initial guess (probably wrong) is that your strategy is to always say, "the sum is not 7" and you'll be right with a probability of 9/11. Actually, how about this. If it is 7, then there is an even number and an odd number. If he says an even number, then you say "it is 7." You'll be right in every case where there are not 2 even numbers. A pair of evens occurs 9 times out of 36, so this would have you be right 3/4 times, which is worse than the previous one. Oh, wait a minute. That first strategy is wrong, and it highlights the fact that something is wrong with saying "It would seem that this improves our a priori odds of rolling a 7 from 1/6 to 2/11." Right?

I don't agree with your 9/11
perhaps you are forgetting about doubles.

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AKG
Homework Helper
What doubles? If your friend says, "there is at least one 4" then there dice will be in the following possible permuations:

{1,4} {2,4} {3,4} {5,4} {6,4} {4,4} {4,1} {4,2} {4,3} {4,5} {4,6}.

If you are to say that there are 36 possible ways the dice could be cast, then order clearly matters, and so {3,4} is distinct from {4,3}. If anything, we can assume the dice are of different colours, in which case we can visibly distinguish a {3,4} from a {4,3}, yet still the person can say, "there is at least one 4" and we can conclude that 2 out of 11 permutations will sum to 7.

AKG said:
What doubles?
If you always call 'not summing to 7, then you will be right every time a double is thrown.

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AKG
Homework Helper
Ursole said:
If you always call 'not summing to 7, then you will be right every time a double is thrown.
But if the person says, "there is one n", then the permutation {n,n} is one of 9 (out of 11) cases when you'll be right, i.e. the double is included.

AKG said:
But if the person says, "there is one n", then the permutation {n,n} is one of 9 (out of 11) cases when you'll be right, i.e. the double is included.
But your strategy should also include you and your partner agreeing always to call the higher [or lower] number.

Then I make make the probability of being right equal to

(5/6)(4/5) + (1/6)(6/6) = 5/6

Do you agree?

.

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AKG
Homework Helper
Ursole said:
But your strategy should also include you and your partner agreeing always to call the higher [or lower] number.

Then I make make the probability of being right equal to

(5/6)(4/5) + (1/6)(6/6) = 5/6

Do you agree?
Yes, I agree.

However, it is possible to improve on 5/6!

AKG
Homework Helper
Ursole said:
However, it is possible to improve on 5/6!
Well, if the numbers don't sum to 7, and there is a die with value < 4, then he should call that number. So, this way, you will always be right if there is one die less than 4, and the sum is not seven, which is the case 21 out of 36 times. The remaining cases are
1,6*
2,5*
3,4*
4,5*
4,6*
5,6*
4,4
5,5
6,6

*and it's "reverse"

Now, he should say "4" to imply either that the sum is 7, or that there are two 4's (he's forced to say "4" in that case). If he says "4" guess that the sum is 7, and you'll be right 2 out of 3 times. There remain 12 cases, 4 of which sum to 7, and he can say 5 or 6 (whatever's appropriate) in these cases, and you should guess "the sum is not 7". You'll be right 8 times out of 12. So, we have 21 + 2 + 8 = 31/36, one better than 5/6 = 30/36 that we had before. Can it be further improved?

Can it be further improved? No, seems like we can't escape from having 4 false negatives and 1 false positive.

The formatting below didn't come out well here.
The totals are the sum of Prob.

Let probability partner says n = PPSn
Let probability of guessing right = PGR
Let PGR * PPSn = Prob

ALWAYS GUESS NOT 7:

n PPSn PGR Prob

6 11/36 9/11 9/36
5 9/36 7/9 7/36
4 7/36 5/7 5/36
1,2,3 9/36 1 9/36
------
30/36

IMPROVED: (Always call Not 7 except when partner sees (1,6), (6,1), or (6,6) and calls 6)

n PPSn PGR Prob

6 3/36 2/3 2/36
5 9/36 7/9 7/36
4 15/36 13/15 13/36
1,2,3 9/36 1 9/36
------
31/36

n PPSn PGR Prob

6 6/36 2/3 4/36
5 6/36 2/3 4/36
4 3/36 2/3 2/36
1,2,3 21/36 1 21/36
-------
31/36

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AKG
Homework Helper
Ursole said:
Can it be further improved? No, seems like we can't escape from having 4 false negatives and 1 false positive.
Can you prove this?

The formatting below didn't come out well here.
I suggest using the [ code ] [ / code ] tags.
Code:
ALWAYS GUESS NOT 7:

n                             PPSn          PGR              Prob

6                            11/36          9/11              9/36
5                              9/36           7/9              7/36
4                             7/36            5/7              5/36
1,2,3                        9/36             1                9/36
------
30/36

IMPROVED: (Always call Not 7 except when partner sees (1,6), (6,1), or (6,6) and calls 6)

n                             PPSn          PGR               Prob

6                             3/36          2/3                2/36
5                             9/36          7/9                7/36
4                             15/36       13/15              13/36
1,2,3                         9/36           1                  9/36
------
31/36

n                            PPSn           PGR               Prob

6                             6/36           2/3               4/36
5                             6/36           2/3               4/36
4                             3/36           2/3               2/36
1,2,3                       21/36           1               21/36
-------
31/36

Ursole said:
.....we have 'at least one n' for SOME value of n, so why not just assume this at fist without waiting for our friend to tell us? It would seem that this improves our a priori odds of rolling a 7 from 1/6 to 2/11.
well iwas thinking on this, and perhaps instead of trying to improve the chances, we all should work on the question, that why do we not take the probability of a total of 7 to be 2/11?

i seemed to develop my own theory on this, but as you will see, it is not without errors, so plz go through it and comment.

Here we observe that the probability for the sum to be 7 comes out to be 2/11, when we know that one of the die has a particular number(n)

Let us take the possible cases for different values for n

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

If here, we take the overall probability, (favorable cases)/(total cases),
It comes out as 12/66.
Here you would also observe that all the possible outcomes of the throw are in duplicate, except where both the dice show the same number. There are six such cases, so if we add those six cases to the total number o cases, then the probability comes out to be 12/72, which is equal to 1/6.

Let us take a single case (say for n=1)

1- (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (3,1) (4,1) (5,1) (6,1) [1,1]

Although the total possible outcomes here are 11, I believe that ideally they should be counted as 12. What I am trying to say here is that if we take the cases for the two die separately, firstly where 1 comes on the first die, they come out to be 6, where 1 such case satisfies the condition of the sum being 7, and when 1 comes on the second die, there are 6 cases again. On adding them, we find that there are 2 favorable cases out of 12 possible outcomes, giving the probability of 2/12, or 1/6.

But as the case where 1 comes on both die comes twice, we take it to be a single case instead of two different cases (which I believe is a major error in probability), and believe that for any value of n, we have 11 possible combinations out of which 2 fulfill the condition, whereas in reality there are 12 cases (not 11) out of which 2 combinations fulfill the condition giving us the probability of 1/6.

Although this leaves us with the question that in original 36 cases we were counting the doubles as a single case, but i am trying to do something about it, and any help is welcome.

No doubt it is provable, but I don't know of one.

Thanks for the tip.

.

NateTG
Homework Helper
Ursole said:
The probability of rolling a 7 with a pair of ordinary dice is 1/6 because, of the 36 possible (and equally likely) combinations, six of them sum to 7. Now, suppose the dice are rolled out of one's sight and some honest person who can see the result tells us that at least one of the dice came up 6. This restricts the total number of possible combinations to the following 11 pairs:

(1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,5), (6,4), (6,3), (6,2), (6,1)

of which exactly two sum to 7. So we would now calculate the probability of a 7 as 2/11, slightly more than 1/6. Of course, the same analysis would give a probability of 2/11 if our honest friend had sais 'at least one 5' instead of 'at least one 6'. Or if he had said 'at least one 4'. In fact, we would arrive at the same probability if he had said 'at least one n' for ANY value of n. Obviously we have 'at least one n' for SOME value of n, so why not just assume this at fist without waiting for our friend to tell us? It would seem that this improves our a priori odds of rolling a 7 from 1/6 to 2/11.
How about the following: I bet, at 14:2 that the sum of two dice thrown is going to be 11. Then someone comes along, and gives the value of one of the dice. Then I'll bet you 2:11 that the two dice are not 11.

So, if the two dice are 11, then you pay me net 3, and if two dice total to 11, then you pay me net 0. Seems like a pretty good deal to me, and according to you, both bets are fair... or are they?

This brain teaser relies on the fact that the mechanics are a bit vauge, so it's not at all clear whether the odds are actually 2 in 11 or 1 in 7.

If you roll the dice until there is a 6 showing, then the odds that the total is 7 are 2 in 11.
If you roll the dice and have a stranger pick one of the dice at random and anounce the face value, then the odds are 1 in 7.

The trick is that the former option collapses two occurances of 6: (6,6) into one case, while the other does not - when the dice are showing doubles, this 'honest person' is twice as likely to indicate the value on the face of the die.

The problem doesn't indicate whether the honest person chose one of the values on the dice at random (in this case the odds are still 1/7), or we or that the roll was one that contained a 6 (so the odds are 2/11).

vikasj007 said:
Although this leaves us with the question that in original 36 cases we were counting the doubles as a single case, but i am trying to do something about it, and any help is welcome.

i think i made a mistake here, if we are taking 36 cases, then there are no 2 cases identical, so just forget it..

but if you seriously think on what i have said in post 16, i think you all will also agree with me that this is major error in probability.

with this i think i have made my first achievment, in my conquest to prove that maths is not perfect, and it has errors and loopholes using which we can prove the impossible as correct and verified.
(something like 1=2)

NateTG
Homework Helper
vikasj007 said:
but if you seriously think on what i have said in post 16, i think you all will also agree with me that this is major error in probability.
Looking at post 16, it's odd that it doesn't bother you that instead of coming up with 36 possible cases for rolling two dice, you're coming up with 66. Perhaps you're counting some cases twice? For example, the (1,2) case shows up in both the first and second rows of your table.

Probability depends on what you know, that is just the idea of probability: you assign equal chances to the things you do not know. Therefore the chance of 7 is higher when you know the number of one of the dice (you will have to share the total probability of 1 between less possible events).

The following shows that if you do not know what number one of the die throws gives, you will get chances of 1/6 for 7 even when you follow the logic where this brainteaser started with:

Probability of 7 WHEN one die throw gave 1 = 2/11
Probability that one die throw gave 1 = 11/36
Thus probability of sum seven AND one die throw was 1 = (2/11) * (11/36) = 2/36
Because we do not care whether 1 was thrown first or second, (1,6) and (6,1) are the same events and the chance on that event is 2/36.
Proof:
- Event A is "sum seven WHEN one die throw gave 1";
- Probability that one die throw gave 6 in case of event A = 1
- Probability of event A = 2/36
-- Thus, probability of "event A AND one die throw gave 6" = 1 * 2/36

The same reasoning applies to the other two events:
"(2, 5) or (5, 2)" and "(3, 4) or (4, 3)".

Thus one of these events happens ones in six times: 2/36 + 2/36 + 2/36 = 6/36 = 1/6

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NateTG said:
Looking at post 16, it's odd that it doesn't bother you that instead of coming up with 36 possible cases for rolling two dice, you're coming up with 66. Perhaps you're counting some cases twice? For example, the (1,2) case shows up in both the first and second rows of your table.
i thin you have not understoo the point. i have pointed the same fact there, that we are considering all the possible cases for every value of 'n'. and i have also mentioned that the cases are coming in duplicate (although they do not effect probability in any way), but the point to be noticed here is that all the cases except the cases of doublets are duplicate, and this is what i was trying to tell. if all the cases are in duplicate then why are the doublets not so.

if we take the doublets also in duplicate, then the probability will become 1/6, instead of 2/11.

so plz go through the post again and this time give it a bit more time.

Just would like to put in my two cents...

The problem is in fact a misunderstanding of "conditional probability". In probability theory, "a priori" infomation received will in general change the probability of a given event.

There is a similar paradox I heard of. Suppose that two out of three prisoners (call them A, B, C) are to be executed, but we don't know which two of them wil be. Now consider a particular prisoner A. The Pr(A dies) = 2/3. However, now suppose it is certain that B will be killed. Then Pr(A dies| B will be killed) = 1/2. (For two events C and D, Pr(C|D) is pronounced as "probability of C given D").

Why is that a paradox? Because some people confuse the probability of Pr(A dies) with Pr(A dies| B will be killed) and think that they should be the same.

Coming to your paradox, I think your assertion is based on the fact that $$Pr(A) = \Sigma_{i} Pr(A|B_{i})Pr(B_{i})$$ where {B_i} is a set of *mutually disjoint and exhaustive* events. But when you think about it are the events {at least 1 four} and {at least 1 six} disjoint (i.e. can they happen at the same time)? No, as the outcome (4,6) shows. So your reasoning does not stand.

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