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Well, now that I reconsider the problem:

I agree that there are 27 permutations altogether. To be more specific, if we represent the 3 different colors with a, b and c, we would have:

aaa

aab

aac

aba

abb

abc

aca

acb

acc

baa

bab

bac

bba

bbb

bbc

bca

bcb

bcc

caa

cab

cac

cba

cbb

cbc

cca

ccb

ccc

As we can see, any combinations of 2 of the letters (out of 3), are repeated 3 times. For example, we have 3 x ?aa ?ab and ?ac, also 3 x b?a etc. Therefore with 9 tries at maximum, we can have a guaranteed win (2 colors in the correct place, thus 2 black pegs).

Thanks Olinguito! So my understanding is that in the end, you do not subtract the winning sequence, right? It is just 27-6.

Can you please explain "Hence there are 27 − 6 − 1 = 20 permutations giving at most one black peg"?

**anna_gg**- Replies: 5

We are playing a mastermind game with 3 colors and 3 positions (holes). There is no white key peg (indicating the existence of a correct color code peg placed in the wrong position) - just black pegs, for each code peg from the guess that is correct in both color and position.

What is the minimum number of tries, to get at least 2 black pegs?

Duplicates are allowed.

For length 3, it would be YYY, YYN, YNY, NYY, NYN (5 options).

For length 4: YYYY, YYYN, YYNY, YNYY, YNYN, NYYY, NYYN and NYNY (8 options) and so on (13, 21 etc), thus for length 15 it would be 1597.

Thank you guys for your assistance!!

Hmmm... Fibonacci numbers...

(I am neither a physicist nor a respectable mathematician, thus not contemptuous about proof...)

**anna_gg**- Replies: 14

Hi everyone!

A computer data center is controlled by an electrical fuse board consisting of 15 fuses in one row.

The power in the data center goes off if any two consecutive fuses are in the OFF position (for example, if Y is ON and N is OFF, the power goes off if we have YYNNYYYYYYYYYYY but not if we have YNYNYNYNYNYYYYY). In how many different ways can we arrange the fuses, so as to always have power?

bobbym wrote:

How did you get 7 / 16 by multiplying?

The probability for both parts NOT to produce a knot is 3/4 for each (since the probability to produce a knot is 1/4). Therefore the total probability NOT to have a knot is 3/4*3/4 = 9/16, thus giving us probability to have a knot 1-9/16 = 7/16.

It wasn't me who solved it, I just unraveled anonimnystefy's solution

Got it. You don't add the two sections, you multiply them

anonimnystefy wrote:

No, both probabilities are 1/4.

So, how do you get 7/16? Am confused

anonimnystefy wrote:

Well, I looked at the two parts separately, because they are virtually independent. For one, there are 8 possibilities for one and 16 for the other. Then I just combine them.

So you mean that in the right part there are only 3/16 possibilities that a knot is produced? Because at the left part I see there are 2/8, that is 4/16.

anonimnystefy wrote:

Um, have you seen my answer?

Yes, but I do not understand how you calculate it

**anna_gg**- Replies: 13

A shoelace is lying on the floor, and attached you can see its shadow. If I pull it, what is the probability that it will produce a knot?

bobbym wrote:

Hi;

Correct! Thank you!!!

So, to have equal probability for both cases, we have to substitute 0.3 with 1-sqrt(0.50) = 1-0.707 = 0.292

**anna_gg**- Replies: 3

Get two random numbers between 0 and 1. Subtract the smaller from the bigger. What is the probability that the result is <0.3? Is it greater than the probability to be > or equal to 0.3?

For what number (instead of 0.3) the probability for both cases is the same?

**anna_gg**- Replies: 3

Person A left Town X at 10:18 am. He walked at a constant speed and arrived at town Z at 1:30 pm. On the same day, Person B left town Z at 9 am. Person B walked the same route in the opposite direction at a constant speed. Person B arrived at town X at 11:40 am. The road crosses a wide river. By coincidence, both arrived at the bridge on opposite sides of the river at the same instant. Person A left the bridge 1 minute later than Person B. At what time did they arrive at the bridge?

bobbym wrote:

Hi anna_gg;

Thanks. If YOU don't understand it, then I won't even try! Let alone that I have no clue where to find this book

**anna_gg**- Replies: 3

Draw 4 points, not belonging (altogether) on the same straight line or same circle. How many straight lines and circles are equidistant from these 4 points? Just to clarify, the definition of the distance of a point P from a circle with center C, is the straight PA, where A is the node of the straight PC with the circle circumference. Please explain the reasoning!

bobbym wrote:

Do not worry, it gets worse.

A shortcut consists of playing spot the pattern and going on from there proving it later by induction or whatever.

From some small problems of say 3 Hits and 4 Misses, I could guess at this formula:

This was followed by a large amount of empirical testing so that at least there is a chance it is correct.

Now for the last question. Since we have a formula we can just sum it:

Are you sure about the last part? Because my calculation (based on your calculations, actually!!!) results to

anna_gg wrote:

bobbym wrote:Hi;

The contest is a continuing weekly for total points so it will never be over until I have the most points or die trying.

They want me to submit solutions but it is much easier for me to do the problems than to explain what I did. I will try to explain my thoughts as I was doing it.

We call a hit H and a miss M.

The probabilty of H hits and M misses is always the same no matter how they are ordered. That is to say {H, M, H, T, H} has the same probability of occuring as {M,H, H, H, M}. No matter what arrangement of H,H,H,M,M,M,H,M... that produces 50 H's and 50 M's. This is because the numerators and denominators of the fractions produced are the same just in different order.

We only need to know how many arrangements are there.

The first two do not need to be figured. They are always {H,M} or {M,H}. So we are really only looking for 49 H's and 49 M's.

If this is not extremely confusing we can move on to the second question.

Well, it is quite a bit

So what is this big number in the denominator of the second fraction? Where does it come from?

bobbym wrote:

Hi;

The contest is a continuing weekly for total points so it will never be over until I have the most points or die trying.

They want me to submit solutions but it is much easier for me to do the problems than to explain what I did. I will try to explain my thoughts as I was doing it.

We call a hit H and a miss M.

The probabilty of H hits and M misses is always the same no matter how they are ordered. That is to say {H, M, H, T, H} has the same probability of occuring as {M,H, H, H, M}. No matter what arrangement of H,H,H,M,M,M,H,M... that produces 50 H's and 50 M's. This is because the numerators and denominators of the fractions produced are the same just in different order.

We only need to know how many arrangements are there.

The first two do not need to be figured. They are always {H,M} or {M,H}. So we are really only looking for 49 H's and 49 M's.

If this is not extremely confusing we can move on to the second question.

Well, it is quite a bit

How did the contest go? Can you now explain your solution?