# More Dice

1. Jul 4, 2004

### Ursole

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.

2. Jul 4, 2004

### AKG

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.

3. Jul 5, 2004

### The Bob

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.

4. Jul 5, 2004

### Ursole

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?

.

5. Jul 5, 2004

### AKG

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?

6. Jul 5, 2004

### Ursole

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

Last edited: Jul 5, 2004
7. Jul 5, 2004

### AKG

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.

8. Jul 5, 2004

### Ursole

If you always call 'not summing to 7, then you will be right every time a double is thrown.

.

9. Jul 5, 2004

### AKG

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.

10. Jul 5, 2004

### Ursole

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?

.

Last edited: Jul 5, 2004
11. Jul 5, 2004

### AKG

Yes, I agree.

12. Jul 5, 2004

### Ursole

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

13. Jul 5, 2004

### AKG

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?

14. Jul 5, 2004

### Ursole

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

.

Last edited: Jul 5, 2004
15. Jul 6, 2004

### AKG

Can you prove this?

I suggest using the [ code ] [ / code ] tags.
Code (Text):
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

16. Jul 6, 2004

### vikasj007

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.

17. Jul 6, 2004

### Ursole

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

Thanks for the tip.

.

18. Jul 6, 2004

### NateTG

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).

19. Jul 7, 2004

### vikasj007

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)

20. Jul 8, 2004

### NateTG

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.