# Math Help - Probability and Dice Loading

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, $d_1, d_2, d_3$ such that the numbers showing once rolled are $n_1, n_2, n_3$ then you are trying to find where

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

3. Originally Posted by pickslides
Let me confirm this.

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

$n_1> n_2> n_3> n_1$ ? 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: $n_i > n_j$ means "at least a 50% chance that $n_i$ is bigger than $n_j$

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

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

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

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: $n_i > n_j$ means "at least a 50% chance that $n_i$ is bigger than $n_j$

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

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

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:

$n_3 > n_2$
and
$n_2 > n_1$

is sufficient condition for
$n_3 > n_1$

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