Math Puzzles (from FiveThirtyEight - new puzzle every Friday) (1 Viewer)

[Aerial Assault]

Here's today's post, with a new puzzle and the solution to last week's.

The boardgame geek thread linked to in the solution is really interesting. Glad to see some genius there came up with the idea that the coach's departure was a metaphor for estimating probability.

 

[Short Corner]

Gotta look back at my topology and combinatoric stuff. Seems easiest to figure out the cases where you can't and then subtract the overlaps.

 

[fasteddie_21]

I love that BGG got referenced in the answer!

 

[Hoosier16]

1/4

 

[bostonfred]

Let's lay out the bridges like this

A.B.C

1D2E3

F.G.H

4I5J6

K.L.M

Where the letters are bridges and the numbers are islands. So you could cross using AFK, BGL or CHM. Makes it easier to discuss.

 

[bostonfred]

Step two, let's simplify the problem.

If we have one bridge, no islands, the odds it gets washed out are 50%.

If we have two bridges, no islands, the odds that both get washed out are 25%, so the odds that either or both survive are 75%.

If we have bridge - island - bridge, the odds of both bridges surviving are 25% which means the odds of either bridge being washed out are 75%.

Those are the easiest ones and the basis for solving the problem.

 

[bostonfred]

Step three let's add back some complexity. Two islands, 5 bridges in a H shape. We'll call them

A.B

1C2

D.E

where the islands are numbered 1 and 2 and the bridges are ABCD and E.

There are exactly four paths to cross the river. AD, BE, ACE or BCD. We don't need all of these paths to survive, just one or more.

The easy ones are AD and BE. The odds of AD surviving are 1 in 4 - there's a 3 in 4 chance that either A or D get washed out - and BE are also 1 in 4, so the odds that either survives are 7/16 (1 - (3/4 x 3/4)).

But there's also two more paths. And those are not independent events, because BCD and ACE both use some of the same bridges as AD and BE. That complicates things.

 

[Ignoratio Elenchi]

If we have two bridges, no islands, the odds that both get washed out are 25%, so the odds that neither gets washed out ate 75%.

If we have bridge - island - bridge, the odds of both bridges surviving are 25% which means the odds of either bridge being washed out are 75%.

Neither of those meet the requirements of the problem though. If there are N rows of islands, there would be N+1 islands in each row.

 

[culdeus]

Pretty sure I'm close on the fixed one. Not even sure where to start on the n+1 one.

 

[bostonfred]

We can lay out the problem using brute force.

A is washed out

B is washed out

A is washed out

B survives

A survives

B survives

A survives

B is washed out

That's going to get really complicated to type, though. So I'm going to refer to those instead as

00

01

11

10

Where the first 1 or 0 refers to A surviving or being washed out, and the second refers to B.

 

[culdeus]

I think the fixed one can be brute force fairly easy. There's only so many valid paths with so many terminals. So its just cumulative probability there. Problem is the n+1 way won't really work like that. You gotta figure out how each segment would break down how many valid paths.

 

[bostonfred]

There are 32 combinations for ABCDE (2^5)

00000

00001

00010

00011

00100

00101

00110

00111

01000

01001

01010

01011

01100

01101

01110

01111

10000

10001

10010

10011

10100

10101

10110

10111

11000

11001

11010

11011

11100

11101

11111

The only combinations that work are 1xx1x, x1xx1, x111x and 1x1x1. So that leaves 8 combinations of AD (1xx1x)

10010

10011

10110

10111

11010

11011

11110

11111

There's also 8 combinations of BE, but a lot of them overlap.

01001

01011

01101

01111

11001

11011 - overlap

11101

11111 - overlap

That leaves us with 14 out of 32 - or 7 out of 16, as I said above.

 

[bostonfred]

BCD x111x

01110

01111 - overlaps BE

11110 - overlaps AD

11111 - overlaps everything

ACE 1x1x1

10101

10111 - overlaps AD

11101 - overlaps BE

11111 - overlaps everything

That's just two additional paths. 16 valid paths and 16 invalid - exactly 50%.

 

[bostonfred]

So how can we lay that out to solve the problem without using brute force? We can't just figure out the odds of each path, because there's a lot of overlap.

But we can calculate the odds of the independent paths and then add back the dependent paths. So in the case I laid out above, the odds of AD surviving were 1/4, and BE were 1 in 4, and the odds of either surviving were 7 in 16. Then we figured out the odds of ACE and BCD but not the overlap - in other words, we didn't care about all instances of ACE and BCD, only the ones where ACE survived and B and D both got washed out or BCD survived but A and E both got washed out, because any combination of ABCD or BCDE would satisfy AD or BE as well, and would therefore overlap. That's the key to solving this problem.

 

[bostonfred]

So now that we've done a simple one, let's do

A.B.C

1D2E3

F.G.H

8 bridges, 2^8 combinations or 256 total

Any pair of two bridges will survive 1/4 of the time. So there's 64 combinations where AF survived, 64 for BG and 64 for CG. But there's a lot of overlap. 1/4 of the time, AF survived. 3/4 it didn't. Of those 192 times AF wasn't available, 1/4 of the time BG was available, or 48. Of the 144 times AF and BG were unavailable, CH was available 1/4 of the time, or 36. So 148 out of 256 times you can just walk straight across.

To be continued

 

[Ignoratio Elenchi]

So now that we've done a simple one, let's do

A.B.C

1D2E3

F.G.H

8 bridges, 2^8 combinations or 256 total

Any pair of two bridges will survive 1/4 of the time. So there's 64 combinations where AF survived, 64 for BG and 64 for CG. But there's a lot of overlap. 1/4 of the time, AF survived. 3/4 it didn't. Of those 192 times AF wasn't available, 1/4 of the time BG was available, or 48. Of the 144 times AF and BG were unavailable, CH was available 1/4 of the time, or 36. So 148 out of 256 times you can just walk straight across.

To be continued

Don't want you to spend too much time going down the wrong road - this, too, doesn't stick to the conditions of the puzzle. If you have three islands across, then you'd need two rows of islands (as pictured in question 1 on fivethirtyeight).

 

[bostonfred]

Next let's take the left forks. We are interested in times the A D G are available and B and F aren't. Because we would never cross ADG if AF or BG were available. There are 8 combinations of ADG but not BF. They are

CEH

000

001

010

011

100

101

110

111

And we can't include 101 or 111, because we already counted those under CH. So that's 6 combinations.

Another way to think of it is that 1/4 of the time, CH is available. 3/4 of the time it isn't, or 192 instances. Of those, we need exactly ADG but not B or F. There's 2^5 or 32 combinations of ABDFG, and we only want 1 of them. 1/32 of 192 is 6 combinations.

We can do the same thing for BDF. That's 6 more combinations. So 12 combinations for the left forks.

 

[bostonfred]

The right forks CEG and BEH work almost the same way. The thing is some if those paths overlap with ADG and BDF.

So for CEG, we need combinations where AF isn't available and also where ADG isn't available and where CEG are available but B and H aren't. In others words, CEG but not BH with no AF and no AD. So of the 8 combinations of ADF, we see

ADF

000

001

010

011

100

101 overlap

110 overlap

111 overlap

So there's 5 combinations of CEG and 5 of BEH.

 

[bostonfred]

Last there are two combinations where we'd go all the way across. ADEH and CEDF. For those to work, all other paths would gave to be unavailable - otherwise you'd take a shorter path that we've already counted. There's only one combination where ADEH are available but BCFG aren't, and one where CEDF are available but ABGH aren't.

So for one row of three islands the odds that you will be able to cross are 64+48+36+12+10+2 / 256 or 172/256 or 43/64.

 

[bostonfred]

Now let's do the original problem.

A.B.C

1D2E3

F.G.H

4I5J6

K.L.M

Let's start with the same methodology we just used for the simpler examples. There are 2^13 total combinations, or 8192.

The odds of any 3 bridges all being available are 1 in 8. So the odds you can go straight across AFK are 1/8, or 1024 combinations. BGL is 1/8 of the remaining 7168, and CHM is 1/8 of the remaining 7/8 of 7/8.

So we can make a formula for the number of straight across paths.

There are 2^n possible combinations, of which

2^(n-r) x (((2^r-1) / (2^r)) ^ c-i)

+

2^(n-r) x (((2^r-1) / (2^r)) ^ c-i)... repeating c times where i starts at c-1 and i=i-1 for each iteration.

Where n is the total number of bridges

R is the number of rows

C is the number of columns

I'm sure n can be expressed as a function of c and r and that the formula above can be expressed more cleanly than i starts at c-1 but I can't remember what that thing is called so screw it.

 

[bostonfred]

2^(n-r) x (((2^r-1) / (2^r)) ^ c-i)

+

2^(n-r) x (((2^r-1) / (2^r)) ^ c-i)... repeating c times where i starts at c-1 and i=i-1 for each iteration.

In the original problem where n=13, r=3 and c=3, that means

2^10 +

2^10 (7/8) +

2^10 (7/8) (7/8)

Or 1024 + 896 + 784 =

2804 of the 8192 combinations have a path that goes straight across.

 

[culdeus]

Now let's do the original problem.

A.B.C

1D2E3

F.G.H

4I5J6

K.L.M

Let's start with the same methodology we just used for the simpler examples. There are 2^13 total combinations, or 8192.

The odds of any 3 bridges all being available are 1 in 8. So the odds you can go straight across AFK are 1/8, or 1024 combinations. BGL is 1/8 of the remaining 7168, and CHM is 1/8 of the remaining 7/8 of 7/8.

So we can make a formula for the number of straight across paths.

There are 2^n possible combinations, of which

2^(n-r) x (((2^r-1) / (2^r)) ^ c-i)

+

2^(n-r) x (((2^r-1) / (2^r)) ^ c-i)... repeating c times where i starts at c-1 and i=i-1 for each iteration.

Where n is the total number of bridges

R is the number of rows

C is the number of columns

I'm sure n can be expressed as a function of c and r and that the formula above can be expressed more cleanly than i starts at c-1 but I can't remember what that thing is called so screw it.

Problem doesn't say you have to go straight across.

 

[bostonfred]

I'm not only working on the original problem. I'm working on the overall solution for any number of islands, while doing random chores. These posts are intended to be read together sequentially on the way to the answers to both.

 

[culdeus]

I'm not only working on the original problem. I'm working on the overall solution for any number of islands, while doing random chores. These posts are intended to be read together sequentially on the way to the answers to both.

Oops. Meant to delete the above as I skimmed what people were doing

 

[BroadwayG]

40%

 

[Ignoratio Elenchi]

1/4

40%

:showyourwork:

 

[Ignoratio Elenchi]

I mentioned after posting this week's puzzle that my solution requires basically no math. It seems like a solution would require a ton of brute force or some intensive formulas, but there's a convenient feature of the puzzle that makes it pretty easy to see (if you think about it the right way) what the probability must be without really doing any calculations.

I mentioned this feature to bfred once or twice the other day.

 

[hagmania]

Here's today's post, with a new puzzle and the solution to last week's.

nice work by the crew to get to that 2/3, especially all the long math guys. new puzzle required viewing the image so probably should just look at the link.

For the record in case all the math scared off some folks, I'm pretty sure I got the answer to the new puzzle today, and my solution requires basically no math. Just have to think about the problem the right way. We'll see.

Was your conclusion influenced on how the problem needs to be set up with n islands by n+1 islands?

 

[Ignoratio Elenchi]

Here's today's post, with a new puzzle and the solution to last week's.

nice work by the crew to get to that 2/3, especially all the long math guys. new puzzle required viewing the image so probably should just look at the link.

For the record in case all the math scared off some folks, I'm pretty sure I got the answer to the new puzzle today, and my solution requires basically no math. Just have to think about the problem the right way. We'll see.

Was your conclusion influenced on how the problem needs to be set up with n islands by n+1 islands?

Yes - specifically the fact that there are always n rows of n+1 islands each puts the bridges in a particular arrangement that makes the problem very easy to solve.

 

[BroadwayG]

 

[Hoosier16]

My new answer is 847/2048 = 41.36%

I looked at it as a 3 level problem. You must advance to each level to succeed. The probability to advance to the first row of islands is 7/8. The probability of advancing to the 2nd set is 11/16. The probability for getting to the South shore is also 11/16 (same arrangement of islands and bridges).

Work:

For 1st, the probability of NOT advancing is (1/2)^3 or 1/8 so prob of advancing is 7/8.

For the 2nd and 3rd, the probability of NOT advancing from each end island is 1/2 * 3/4 * 7/8, or 21/64. For the middle island, it's 3/4 * 1/2 * 3/4, or 9/32. The total prob of NOT advancing is 1/3[(21/64)+(21/64)+(9/32)] = 60/192=5/16. So, the prob of advancing is 5/16.

For advancing through all 3 levels, it's (7/8)*(11/16)*(11/16)

My sticking point is whether I need to account for the probability of arriving at each island in the above calc. Going from the 1st set of islands has an equal chance of being on any of the 3 islands. Going from the 2nd set of islands assumes equal probability, and that may be where I go wrong.

It feels like the solution should be simpler. I'd love to hear the simple solution because I don't see it at all.

 

[Ignoratio Elenchi]

It feels like the solution should be simpler. I'd love to hear the simple solution because I don't see it at all.

There's definitely a simpler way to visualize the solution. I gave a hint above - the fact that there are always n+1 islands in n rows means the bridges are in a particular configuration that they otherwise wouldn't be.

The solution I have in mind relies on basically no math at all. Think about boats.

I'll write up my solution and post it in a spoiler shortly.

 

[Ignoratio Elenchi]

To add more to the hint I alluded to in the previous post:

Spoiler

Assume the bridges are low enough to the water that a boat couldn't pass beneath them. After the storm rolls through, what's the probability that a boat will be able to pass through the islands from west to east?

 

[JayMan]

It feels like the solution should be simpler. I'd love to hear the simple solution because I don't see it at all.

There's definitely a simpler way to visualize the solution. I gave a hint above - the fact that there are always n+1 islands in n rows means the bridges are in a particular configuration that they otherwise wouldn't be.

The solution I have in mind relies on basically no math at all. Think about boats.

I'll write up my solution and post it in a spoiler shortly.

Looking at it from a boat going through west-east only if maze of bridges are gone is the exact same configuration as a walker going through north-south only if a maze of bridges remains (thus this 2 by 3 problem yields the same results for a n by (n+1) configuration).

A walker can go through if and only if a boat can’t and vice versa. The probability of an event and its complement always sums to 1, thus a walker has a 50% chance of getting to the south shore.

 

[Hoosier16]

It feels like the solution should be simpler. I'd love to hear the simple solution because I don't see it at all.

There's definitely a simpler way to visualize the solution. I gave a hint above - the fact that there are always n+1 islands in n rows means the bridges are in a particular configuration that they otherwise wouldn't be.

The solution I have in mind relies on basically no math at all. Think about boats.

I'll write up my solution and post it in a spoiler shortly.

Looking at it from a boat going through west-east only if maze of bridges are gone is the exact same configuration as a walker going through north-south only if a maze of bridges remains (thus this 2 by 3 problem yields the same results for a n by (n+1) configuration).

Agree with this so the probability of a walker making it is the same as a boat not making it. I understand that completely.

A walker can go through if and only if a boat can’t and vice versa. The probability of an event and its complement always sums to 1, thus a walker has a 50% chance of getting to the south shore.

I don't see this at all. Just because the compliments sum to one doesn't mean they are equal. The walker making it and the boat making it are compliments, not what is bolded above. The bolded are equal, but are not compliments.

 

[bostonfred]

A walker can't always make it across when a boat can't pass, if the side bridges are down.

A bridge BnobridgeC

. ------ N ------ N

B ------ O ------ O

R ------ B ------ B

I ------ R ------ R

D ------ I ------ I

G ------ D ------ D

E ------ G ------ G

. ------ E ------ E

AnobridgeB bridge C

N ------ . ------ N

O ------ B ------ O

B ------ R ------ B

R ------ I ------ R

I ------ D ------ I

D ------ G ------ D

G ------ E ------ G

E ------ . ------ E

 

[JayMan]

It feels like the solution should be simpler. I'd love to hear the simple solution because I don't see it at all.

There's definitely a simpler way to visualize the solution. I gave a hint above - the fact that there are always n+1 islands in n rows means the bridges are in a particular configuration that they otherwise wouldn't be.

The solution I have in mind relies on basically no math at all. Think about boats.

I'll write up my solution and post it in a spoiler shortly.

Looking at it from a boat going through west-east only if maze of bridges are gone is the exact same configuration as a walker going through north-south only if a maze of bridges remains (thus this 2 by 3 problem yields the same results for a n by (n+1) configuration).

Agree with this so the probability of a walker making it is the same as a boat not making it. I understand that completely.

A walker can go through if and only if a boat can’t and vice versa. The probability of an event and its complement always sums to 1, thus a walker has a 50% chance of getting to the south shore.

I don't see this at all. Just because the compliments sum to one doesn't mean they are equal. The walker making it and the boat making it are compliments, not what is bolded above. The bolded are equal, but are not compliments.

I see what you mean - but the fact that the walker going through is the same configuration as the boat going through... and that one making it means the other doesn't... can't lead to the walker having an 80% chance and the boat a 20% chance, no? (if they're the same configuration from their point of view).

 

[Ignoratio Elenchi]

A walker can't always make it across when a boat can't pass, if the side bridges are down.

A bridge BnobridgeC

. ------ N ------ N

B ------ O ------ O

R ------ B ------ B

I ------ R ------ R

D ------ I ------ I

G ------ D ------ D

E ------ G ------ G

. ------ E ------ E

AnobridgeB bridge C

N ------ . ------ N

O ------ B ------ O

B ------ R ------ B

R ------ I ------ R

I ------ D ------ I

D ------ G ------ D

G ------ E ------ G

E ------ . ------ E

Hard to see the formatting there, maybe it's just my mobile, but it looks like a boat can pass there. Enter from the west on the bottom, then go up, then go right.

The other poster is right, and that's part of the key to the solution. If a particular configuration of bridges would allow a boat to pass from west to east, then the man can't cross from north to south, and vice versa. Every possible permutation of bridges will allow exactly one of those outcomes - either the man can cross, or a boat can.

 

[Ignoratio Elenchi]

It might take a little convincing, I can write it up in more detailed fashion later this morning, but in general the complete solution involves a few observations:

- if a man can cross, a boat can't, and vice versa.

- because there are n x n+1 islands, the bridges are arranged in concentric squares.

- because of this square arrangement (and the fact that each bridge has an independent probability of 50% of being destroyed) the question "what is the probability the man can cross?" is the same exact question as "what is the probability that a boat can cross?"

- these two questions have the same answer, and their answers must sum to 100%, so the answer is 50%.

 

[Hoosier16]

It feels like the solution should be simpler. I'd love to hear the simple solution because I don't see it at all.

There's definitely a simpler way to visualize the solution. I gave a hint above - the fact that there are always n+1 islands in n rows means the bridges are in a particular configuration that they otherwise wouldn't be.

The solution I have in mind relies on basically no math at all. Think about boats.

I'll write up my solution and post it in a spoiler shortly.

Looking at it from a boat going through west-east only if maze of bridges are gone is the exact same configuration as a walker going through north-south only if a maze of bridges remains (thus this 2 by 3 problem yields the same results for a n by (n+1) configuration).

Agree with this so the probability of a walker making it is the same as a boat not making it. I understand that completely.

A walker can go through if and only if a boat can’t and vice versa. The probability of an event and its complement always sums to 1, thus a walker has a 50% chance of getting to the south shore.

I don't see this at all. Just because the compliments sum to one doesn't mean they are equal. The walker making it and the boat making it are compliments, not what is bolded above. The bolded are equal, but are not compliments.

I see what you mean - but the fact that the walker going through is the same configuration as the boat going through... and that one making it means the other doesn't... can't lead to the walker having an 80% chance and the boat a 20% chance, no? (if they're the same configuration from their point of view).

The walker succeeding and the boat failing are equal. The walker succeeding and the boat succeeding are compliments.

A simple check is to look at a 2 X 1 configuration. The answer isn't 50%, it's 15/32 (3/4 to get to the 1st set of islands by 5/8 to get to the other side).

 

[Hoosier16]

It might take a little convincing, I can write it up in more detailed fashion later this morning, but in general the complete solution involves a few observations:

- if a man can cross, a boat can't, and vice versa.

- because there are n x n+1 islands, the bridges are arranged in concentric squares.

- because of this square arrangement (and the fact that each bridge has an independent probability of 50% of being destroyed) the question "what is the probability the man can cross?" is the same exact question as "what is the probability that a boat can cross?"

- these two questions have the same answer, and their answers must sum to 100%, so the answer is 50%.

I think it's man can cross equals boat can't cross. They both cannot succeed.

 

[JayMan]

It might take a little convincing, I can write it up in more detailed fashion later this morning, but in general the complete solution involves a few observations:

- if a man can cross, a boat can't, and vice versa.

- because there are n x n+1 islands, the bridges are arranged in concentric squares.

- because of this square arrangement (and the fact that each bridge has an independent probability of 50% of being destroyed) the question "what is the probability the man can cross?" is the same exact question as "what is the probability that a boat can cross?"

- these two questions have the same answer, and their answers must sum to 100%, so the answer is 50%.

I think it's man can cross equals boat can't cross. They both cannot succeed.

No - IE meant that 'man can cross' is the same as 'boat can cross' - because of the concentric square configuration... And since one makes it when the other doesn't, and vice versa... the only possible answer is 50%.

 

[Ignoratio Elenchi]

It might take a little convincing, I can write it up in more detailed fashion later this morning, but in general the complete solution involves a few observations:

- if a man can cross, a boat can't, and vice versa.

- because there are n x n+1 islands, the bridges are arranged in concentric squares.

- because of this square arrangement (and the fact that each bridge has an independent probability of 50% of being destroyed) the question "what is the probability the man can cross?" is the same exact question as "what is the probability that a boat can cross?"

- these two questions have the same answer, and their answers must sum to 100%, so the answer is 50%.

I think it's man can cross equals boat can't cross. They both cannot succeed.

Well yes, in any configuration of islands, "man can cross" is synonymous with "boat can't cross" and vice versa. And "man can cross" + "boat can cross" = 100%, because they're complements, but we wouldn't know any more than that. But in this particular case, because the bridges are arranged in a square, the problem as viewed from north to south is the same exact problem as if it was viewed from west to east. The possible paths a man could take are identical to those a boat could take. Because of the square arrangement, not only does "man can cross" + "boat can cross" = 100%, we know that they equal each other.

Perhaps this will be more convincing: In the 2x3 islands problem, there are 2^13 possible permutations of bridge "states" (destroyed or whole). Take any random permutation that allows a boat to pass through. Rotate the entire picture 90 degrees clockwise, and flip the "state" of each bridge. Now you have a path the man can use to cross.

So we have a one-to-one mapping of outcomes where the boat can cross to outcomes where the man can cross. That is, for every possible outcome that allows a boat to cross, there is exactly one unique corresponding outcome that allows the man to cross, and vice versa. Thus there must be the same number of outcomes for each.

 

[Hoosier16]

I see what you're saying, but my brain is still struggling with simple. I realize I am probably wrong, but I'm not convinced they're compliments.

Take a look at my 2 X 1 solution and explain why that is wrong. My last probability class was 35 years ago but I don't see it. To cross, you must reach one of the two islands and then the opposite shore. To get to an island, the prob of making it is 3/4. The prob of reaching the other side from either island is 5/8 (1-(3/4*1/2)). The total prob of crossing should be 3/4 * 5/8, or 15/32. Where am I going wrong?

 

[BroadwayG]

The walker succeeding and the boat failing are equal. The walker succeeding and the boat succeeding are compliments.

A simple check is to look at a 2 X 1 configuration. The answer isn't 50%, it's 15/32 (3/4 to get to the 1st set of islands by 5/8 to get to the other side).

The answer is 50%. I don't know which one of the 16 successful options your formula doesn't account for.

Spoiler

---1 2O3O4 5---1234500000000010001000011001000010100110001110100001001Y0101001011Y0110001101Y01110Y01111Y100001000110010Y10011Y1010010101Y10110Y10111Y1100011001Y11010Y11011Y1110011101Y11110Y11111Y

 

[JayMan]

I see what you're saying, but my brain is still struggling with simple. I realize I am probably wrong, but I'm not convinced they're compliments.

Take a look at my 2 X 1 solution and explain why that is wrong. My last probability class was 35 years ago but I don't see it. To cross, you must reach one of the two islands and then the opposite shore. To get to an island, the prob of making it is 3/4. The prob of reaching the other side from either island is 5/8 (1-(3/4*1/2)). The total prob of crossing should be 3/4 * 5/8, or 15/32. Where am I going wrong?

Look at your 2x1 problem with the 32 possible outcomes (each which equal probability).

If 0 bridge is left - you can't cross (0/1);

If 1 bridge is left - you can't cross (0/5);

If 2 bridges are left - you can cross on 2 scenarios (straight lines) and can't on the other 8 (2/10);

If 3 bridges are left - you can cross on 8 and can't on 2 (if the 2 north bridges or the 2 south bridges are blown) (8/10);

If 4 bridges are left - you can cross (5/5);

All 5 bridges left - you can cross (1/1).

... 16/32.

 

[jhib]

I see what you're saying, but my brain is still struggling with simple. I realize I am probably wrong, but I'm not convinced they're compliments.

Take a look at my 2 X 1 solution and explain why that is wrong. My last probability class was 35 years ago but I don't see it. To cross, you must reach one of the two islands and then the opposite shore. To get to an island, the prob of making it is 3/4. The prob of reaching the other side from either island is 5/8 (1-(3/4*1/2)). The total prob of crossing should be 3/4 * 5/8, or 15/32. Where am I going wrong?

There's another one that sounds good. Reminds me of why I had a love/hate relationship with statistics back in the day.

I'm pretty sure the way you did it removes the extra probability of getting across that's given when both initial bridges are open at the same time (you can get to both islands in your first step).

Out of the 3/4 chance you gave to get to an island:

-- 1/2 chance of only being able to get to one island (without crossing a 2nd bridge). Once you're on that island, you have a 5/8 chance of crossing the rest of the way. 1/2 * 5/8 = 5/16 = 10/32

-- 1/4 chance of being able to get to either island in that one step. Once you can get to either island, there's a 6/8 chance of being able to cross (the only way it doesn't work is if both bridges to the other side are out) 1/4 * 6/8 = 6/32

10/32 + 6/32 = 16/32

 

[Hoosier16]

The walker succeeding and the boat failing are equal. The walker succeeding and the boat succeeding are compliments.

A simple check is to look at a 2 X 1 configuration. The answer isn't 50%, it's 15/32 (3/4 to get to the 1st set of islands by 5/8 to get to the other side).

The answer is 50%. I don't know which one of the 16 successful options your formula doesn't account for.

Spoiler

---1 2O3O4 5---1234500000000010001000011001000010100110001110100001001Y0101001011Y0110001101Y01110Y01111Y100001000110010Y10011Y1010010101Y10110Y10111Y1100011001Y11010Y11011Y1110011101Y11110Y11111Y

Thanks for posting that. I knew that was what I was going to have to do.

I'm pretty sure the way you did it removes the extra probability of getting across that's given when both initial bridges are open at the same time (you can get to both islands in your first step).

That's the one. Breaking it down into two steps seems logical, but clearly doesn't work. It's going to bother me until I figure out the math.

 

[BroadwayG]

The math would be 2/4 to one island, 1/4 to either island, and 1/4 no island.

Then 5/8, 3/4, 0 to cross again, respectively.

10/32 + 6/32 + 0 = 16/32

 

[Ignoratio Elenchi]

I see what you're saying, but my brain is still struggling with simple. I realize I am probably wrong, but I'm not convinced they're compliments.

Take a look at my 2 X 1 solution and explain why that is wrong. My last probability class was 35 years ago but I don't see it. To cross, you must reach one of the two islands and then the opposite shore. To get to an island, the prob of making it is 3/4. The prob of reaching the other side from either island is 5/8 (1-(3/4*1/2)). The total prob of crossing should be 3/4 * 5/8, or 15/32. Where am I going wrong?

There's another one that sounds good. Reminds me of why I had a love/hate relationship with statistics back in the day.

I'm pretty sure the way you did it removes the extra probability of getting across that's given when both initial bridges are open at the same time (you can get to both islands in your first step).

Right, I think what's happening is that 5/8 is effectively "If I'm already on an island, what's the probability I can cross to the south shore?" So like imagine you have it like this:

*****|   |A - B|   |*****

5/8 is what you come up with if you calculate, "If I'm on A, what's the probability I can cross to the south shore by either going down, or by going right and then down." But that leaves out the case where A was the "wrong choice," and you can cross back to the north shore, and then cross directly down the right side through B:

******|    |A    B     |******

From A you can't cross to the south shore. But you can backtrack to the north and then cross through B, which isn't being accounted for in the calculation.