>>11659795>>11659785Okay you are right. I tried 25
>>11659774 hoping it would be difficult. It's not that difficult but it took me quite some time to solve. Does this test have a time limit? If it's a time limit like only 1 hour, then the difficulty of the test is created by the time pressure, not the difficulty of the problems.
Give me like 5 hours and I will get all of them right for sure. If I only have 1 hour I probably will get only the first 15 or so of them.
Here's how I solved 25, took me more than 10 minutes.
Possible ways to get 7 are permutations of the sets (1,1,5),(1,2,4),(1,3,3),(2,2,3). Now the question is, when do we want to reroll exactly 2? The answer is when the probability of getting them right is higher than rerolling all 3 but lower than rerolling 1.
>reroll 2 instead of 1When you have 2 right, the probability is always 1/6 for the last. The highest chance of rerolling two right is when you keep a 1, then the chance becomes 5/36 based on the sets above (1x3,3+2X2,4+2x1,5 /36) which is below 1/6. That means it is always better to reroll just 1 if you have already 2 correct.
>reroll 2 instead of 3The chance of getting them right by rolling all 3 is the number of permutations of above sets, so 15/216=2.5/36. So to reroll 2, the number that we keep must have chance above that. Continuing from above, the chances are 4/36 for keeping 2, 3/36 for 3, 2/36 for 4 and 1/36 for 5. So rerolling 2 is only optimal if no more than 1 is right and we keep 1, 2 or 3.
>possibilitiesFrom the above we get the following possibilities.
1 with 6s -> 3
2 with 5s and 6s -> 12
3 with 4s, 5s and 6s -> 27
27+3+12=42. 42/6=7 -> 7/36 is the correct answer.
My solution feels very brute forced. Is there a more elegant way to solve it that takes under 10 minutes?
