Is it possible to load three dice so that the first will probably show a higher number than the second, the second will probably show a higher number than the third, and the third will probably show a higher number than the first?

Loading dice means you change the probability that each number will be rolled, but they all still add up to one.

Please only give me hints. Don't tell me how to solve the problem. Thanks.

2. Originally Posted by seuzy13
Is it possible to load three dice so that the first will probably show a higher number than the second, the second will probably show a higher number than the third, and the third will probably show a higher number than the first?
Let me confirm this.

If you have 3 dice, $\displaystyle d_1, d_2, d_3$ such that the numbers showing once rolled are $\displaystyle n_1, n_2, n_3$ then you are trying to find where

$\displaystyle n_1> n_2> n_3> n_1$ ? I don't think this is possible.

3. Originally Posted by pickslides
Let me confirm this.

If you have 3 dice, $\displaystyle d_1, d_2, d_3$ such that the numbers showing once rolled are $\displaystyle n_1, n_2, n_3$ then you are trying to find where

$\displaystyle n_1> n_2> n_3> n_1$ ? I don't think this is possible.
Sort of, but not quite. Dice one has a probability of showing a higher number than dice two, etc. So if dice one has, for instance, a 1/2 probability of getting a six and dice two has a 1/4 probability of getting a 5 and a 1/4 probability of getting a 4 (the rest of the probabilities being evenly distributed over the remaining numbers 1-6), then dice one will probably show a higher number than dice two. So is there a way to continue this such that the last dice will probably be higher than the first dice?

It seems like I wouldn't be asked this question if it weren't possible.

4. Notation: $\displaystyle n_i > n_j$ means "at least a 50% chance that $\displaystyle n_i$ is bigger than $\displaystyle n_j$

proposition: $\displaystyle n_1 > n_2 > n_3 > n_1$ is impossible

remove the middle steps, and just think about 2 dice to start with.
$\displaystyle n_1 > n_2 > n_1$
You should be able to show that this is impossible.

Now add 1 more die:
$\displaystyle n_1 > n_2 > n_3 > n_1$

You can probably generalise your method from the 2 dice problem. i also suspect that its possible to show that the 2 dice version is a necessary condition for the 3 dice version. Either of these approaches would complete the proof

I hope that isn't too much info!

5. Originally Posted by SpringFan25
Notation: $\displaystyle n_i > n_j$ means "at least a 50% chance that $\displaystyle n_i$ is bigger than $\displaystyle n_j$

remove the middle steps, and just think about 2 dice to start with.
$\displaystyle n_1 > n_2 > n_1$
You should be able to show that this is impossible.

Now add 1 more die:
$\displaystyle n_1 > n_2 > n_3 > n_1$

Try and see a way that your 2 dice version of the problem is a necessary condition for the 3 dice version to work. Since you have already established that 2 dice version is impossible, this will complete your proof.
I know it really does seem impossible, and if I were working with ordinary numbers instead of probability distributions, I know I could easily prove that, but is it possible that this is different?

Ah, well, I'll think about it a little more. Thanks.

6. An alternative, and probably faster approach is to tackle your proposition directly. Try and show that:

$\displaystyle n_3 > n_2$
and
$\displaystyle n_2 > n_1$

is sufficient condition for
$\displaystyle n_3 > n_1$

This would make the proposition impossible.

7. I think you're looking for nontransitive dice.

8. Originally Posted by undefined
I think you're looking for nontransitive dice.
Yes, that's definitely it. Thanks.

9. that explains why i couldn't finish those proofs then