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.
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.
Notation: means "at least a 50% chance that is bigger than
proposition: is impossible
remove the middle steps, and just think about 2 dice to start with.
You should be able to show that this is impossible.
Now add 1 more die:
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!
I think you're looking for nontransitive dice.