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Shadow's Adventure

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.


Puzzle ID:#28292
Fun:*** (2.38)
Difficulty:*** (2.69)
Submitted By:Question_MarkAsx*****




Shadow the Adventurer can't get enough of adventures. So he decides to embark on a great adventure. This is his adventure map.

...................4a - 5a - 6a - HOME
.....2a - 3a - 4b - 5b - 6b - HOME
1 - 2b - 3b - 4c - 5c - 6c - HOME
.....2c - 3c - 4d - 5d - 6d - HOME
..................4e - 5e - 6e - HOME

His main intention is to start from point 1, go to point 6c and go home. He may only move one space on the map at a time. He may not go backwards, or up and down. So from point 1, he might go to 2a, then to 3b, then to 4d and so on. When he reaches points 6a, 6b, 6c, 6d or 6e, he can take a long route, or a short route home.

What is the probability of Shadow reaching point 6c and taking the short route home?

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Jan 27, 2006

I am completely lost. But I'm sure it's a good one anyway.
Feb 11, 2006

I got 121/994
Feb 21, 2006

The instructions need to explain that you can go from 4c to either 5b, 5c, or 5d but not to 5a or 5e.
This wasn't clear.

Here is the table counting the possible paths to each point.
For example, the 17 in the middle is for point 5c. You can get there from 4b (5), 4c (7) or 4d(5). 5+7+5=17.
The 45 is point 6c which gives the answer 45/326.


Feb 27, 2006

i mistook wat u wanted the answer to be! oops well anyways the name of the riddle caught my attention
Mar 01, 2006

God puzzle but I screwed up by allowing for diagonal moves.
Mar 03, 2006

Yes. MarcM's got the idea.
(user deleted)
Mar 31, 2006

The answer to this puzzle is incorrect. the actual answer Sold be 1/8. The subtle error made has to do with the fact that not every path occurs with the same probability. Let's simplify this to just the third level to fully understand. There are 7 distinct paths through the third level (1,2a,3a : 1,2a,3b : 1,2b,3a : 1,2b,3b : 1,2b, 3c : 1,2c,3b : 1,2c,3c). Since two of those paths include 3a, if you were to use the logic of the proposed solution the probability of passing through 3a would be 2/7. In actuality the probabilty of passing through 3a is 5/18. Here is the logic. To pass through 3a Shadow must pass through 2a or 2b. Sitting at position 1 it makes sense that 2a, 2b, and 2c will occur with 1/3 probabilty each. From 2a there are only 2 choices however (3a and 3b) therefore each will occur with probabilty 1/2. Following this correct logic the probabilty of passing through 3a equals 1/2*probability of passing through 2a + 1/3*probability of pasing through 2b or (1/2)*(1/3) + (1/3)*(1/3) = 5/18
May 26, 2006

I think, now that I understand the question, that I agree with the "multiply the probabilities" method given by Cassius.
Jun 21, 2006

Multiplying the probabilities using Cassius's method gave me 277/3888 = ~14.25%.
Sep 26, 2007

Aug 11, 2008

Very confusing question. I think he should go simple "1 - 2b - 3b - 4c - 5c - 6c - short route HOME" and just ignore other possible routes. This will be possible 100% of the times, at least if he's just moving his finger on the map. If he follows he map in the real world we have to consider the possibility of any danger that could occour. Point 2b could for example be at the lion cage at the zoo, which will lead to very complicated math.
(user deleted)
Nov 28, 2008

I did it using the method same as Causius's. and the result was 121/486
* * 5/54 41/324 297/1944
1/3 5/18 13/54 77/324 473/1944
1/3 8/18 18/54 88/324 484/1944
1/3 5/18 13/54 77/324 473/1944
* * 5/54 41/324 297/1944
(user deleted)
Nov 28, 2008

sorry, 121/972, as he has half chance to go the short route home.
Dec 09, 2008

Way to go live! I'm glad to see that after 2.5+ years that someone finally got the correct answer: 121/972 ~ 12.45%
Jan 29, 2010

I was prepared to write exactly the same comment but somebody already did!

The sollution is WRONG! The paths don't have all the same probability to be taken.
Oct 04, 2010

This problem is still ambiguous in that I interpreted a "long route" home as "not going directly east" so there would be 2 ways to do that at 6c. From this interpretation, the answer is 121 / 486 / 3.

But if you meant it as once he reaches points 6a, 6b, 6c, 6d, or 6e, he has the option of taking routes unrelated to the diagram: a "long route" or "short route," then the answer is 121 / 486 / 2.

Though I will still admit this problem was pretty cool.
Aug 09, 2013

Very confusing, nebulous, poorly-explained teaser. The mathematical challenge was a total write-off because I had no idea what the actual rules were. But it seems like it would have been a fun and interesting one if explained better.
Aug 11, 2013

What opgpop and eighsse said. "Long route home" and "Short route home" need to be defined. According to the map there's a 2/3 chance of taking the long route home, so I agree with opgpop and got 121/1458.

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