[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 604
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One Megamogg threw the dice 2011 times and the other Megamogg threw it 2012. What is the probability that the odd numbers of the second one rolled more times than those of the first one?
The problem is scored 3 points. No cool terver formulas are needed here. Just logic and simple richematics.
Dropped it in my personal message.
Reshetov: If it is the opponent's first move in Megamind's game, then the dice are marked non-transitively. Consequently, Megazmog has only to choose one of the two remaining dice, which gives a non-transitive advantage over the opponent's already chosen die. A bearded scam. Not interesting at all.
Yuri, I understand that I can google anything, the transitivity is written about in the discussion of the problem too. What remains is to do it practically.
The fact that this is a bearded scam, does not cancel the attempts to "honestly" solve the problem, i.e. without using the search. I don't know how to approach it yet. But this does not mean that the solution can immediately post here, copying from google.
The problem is only for those who will try to solve it on their own or don't know the solution yet.
Where's the answer?
The megabrain must make three cubes (A, B, C) that have the properties: A->B->C->A.
Where the sign -> means that the probability of winning with the die on the left is higher than with the die on the right.
Where the sign -> means that the probability of winning when playing the die on the left is higher than the die on the right.
drknn, this is for you, you love these:
Silent Guardian
You are standing at a fork in the road, one of the roads leads to the house (but you don't know which one). Fortunately, there is a sentinel at the fork, who is either a truth teller or a liar. Unfortunately, the guard is mute, but fortunately not deaf :-)) and understands you. The guard, when saying "yes" and "no", pronounces them as "woo" and "yoo", but which one means "yes" and which one means "no" is unknown. He cannot pronounce any other sounds and, besides, he is unable to make a gesture in the right direction (perhaps he is armless as well :-)). On top of that, he's also stupid: he can't understand longer questions over 15 words. What question to ask the guardian to find out which road will lead you to your home? You can only ask one question, and only one that the guard can answer.
The megabrain must make three cubes (A, B, C) that have the properties: A->B->C->A.
Where the sign -> means that the probability of winning when playing the die on the left is higher than the die on the right.
Sort of, yeah. Not really, though.
Right, A>A. That's the Reshetian transitivity.
It's not really a difficult problem.
It is enough to get such dice, two of which are the same (in terms of probability of winning) and the third is a loser.
For example:
1: 111222
2: 333666
3: 555444
But here it will be difficult to pick players who will pick the 1st die, so you have to confuse things a bit. For example, like this:
1: 111333
2: 222666
3: 555444
or like this:
1: 111444
2: 222666
3: 555333
Now the only thing required of the megabrain is not to take the 1st die.
11133333 or 111444 is too bad, expectation is 2 or 2.5, which is much worse than 3.5 on a standard cube.
Well, of course, there are all kinds of suckers, but I think Megamozig should be ashamed to beat them...
Great respect for the task, a good argument not in favour of those who think you can't win on random outcomes.
MegaBrain's opponent is not a dummy, so he will choose the die with the highest MO, or at least not the worst one. Since the dice are "fair" (the probability of falling out is the same), a rational opponent will choose the die with the maximum sum of points on its edges. The Megamind must either take advantage of his desire or level out the result of the choice by offering him dice with the same MO.
P.S. By the way, Megamozg has an interesting abbreviation ))
It's not really a difficult task.
It is enough to get such dice, two of which are the same (in terms of probability of winning), and the third one is a loser.
Now the only thing required of the megabrain is not to take the 1st die.