Mathematics marathon

[Ordos_Koala]

Orodos you did calculate the first number correctly, but in the future please try to put your answers in spoiler tags such as not to give away information for others still working on the problem.

sry, next time, I will

whauk's problem:

Spoiler

ratio for 2 people is 365:365^2=1:365
so now for three peple, we have again 365 right answers (for each day, i won't count leap-years), but now there are 365^3 possiblities
So from what we see, the formula is 365:365^x
( =1:365^(x-1) in my opinion this one looks better, but it's not so obvious)

for my problem, look at my earlier posts

btw for the next time, could someone pls help me how to write superscripts?

 

[whauk]

whauk's problem:

Spoiler

ratio for 2 people is 365:365^2=1:365
so now for three peple, we have again 365 right answers (for each day, i won't count leap-years), but now there are 365^3 possiblities
So from what we see, the formula is 365:365^x
( =1:365^(x-1) in my opinion this one looks better, but it's not so obvious)

Spoiler

your approach is totally wrong.
take a look into http://en.wikipedia.org/wiki/Birthday_problem

 

[pr*****t]

but im unable to understand any mistake in Ordos_Koala's solution of birthday probability problem....... can u explain it to me where is he wrong?

 

[MTGjumper]

He's answering the wrong question. He's answering "given n people, what is the probability that they all share a birthday?". Whauk's question is "given n people, what is the probability that a birthday is shared by x people?" (often x=2)

 

[pr*****t]

He's answering the wrong question. He's answering "given n people, what is the probability that they all share a birthday?". Whauk's question is "given n people, what is the probability that a birthday is shared by x people?" (often x=2)

ok thanx ... i shud have read the question again

 

[Ordos_Koala]

ok, sorry I didn't understand you properly for the first time...

 

[Kynit]

Hm... (Whaulk's problem)

Spoiler

There are xCn ways to choose x people out of a group of n. This gives you xCn possibilities of finding a matching birthday.

Finding a matching birthday for x people:
365 / 365^x
or
1 / 365 ^ x-1

Therefore, the chance of finding a matching birthday is xCn / 365^x-1

Does this sound right? I'm not too familiar with the choose function.

 

[qqwref]

Spoiler

I find it interesting that the Rubik's 4x4x4's actually have 4 times as many combinations as a logo-less 4x4x4, yet the Rubik's 4x4x4's also have 4 solutions instead of just 1. This means that the ratio of solved states to scrambled states is the same for a 4x4x4 with a logo and a 4x4x4 without a logo. Of course this does not present any issue from the WCA standpoint, as the regulations clearly state that one logo is allowed. I still find it interesting that someone solving a 4x4x4 with a logo and someone solving a 4x4x4 without a logo are technically not solving the exact same kind of puzzle. For some reason I had never really thought about it this way until someone asked me this question in an e-mail. Cool stuff!

Spoiler

This is less amazing than it might seem. It is precisely *because* the white centers of a normal 4x4 are indistinguishable that the position (ignoring the logo) and the placement of the logo (on which of the 4 white centers) are independent. The position itself can give you no information on the logo placement, and the logo placement itself can give you no information on the position. So solving a logoed 4x4 is not actually any different (as long as you don't care where the logo goes, of course), because the logo neither helps nor hinders a solution. This also means that any set of positions on a normal 4x4 is bijectively related to a set of four times as many positions on a logoed 4x4 - this is true for the number of solved states as well as for the number of positions in total.

Incidentally, the same exact thing is true for many other supercube-like concepts, such as a 3x3 with the orientation of one or more centers marked, or any higher-order supercube. The position and the center placements are independent, so the number of solved positions and the number of total positions will always be multiplied by the same number when you add extra markings.

You have a cube, with side "a", which is expressed in whole centimeters bigger than 2. We painted whole cube with yellow color and then we have sliced it into cubes with side of 1 centimeter. We have then sorted these little cubes into four piles: first pile has only cubes with one yellow side, in the second one were cubes with two painted sides and in the third were cubes with three painted sides. In last pile were cubes with no yellow paint at all. Now tell me how big was the original cube, when you know, that one of these claims is true:
a) Cubes in the first and fourth pile are in ratio 4:9.
b) In the first pile are three-times more cubes than in the second pile.

I solved it quickly, it isn't hard at all, but I want you to prove your answer, not just guess.

Spoiler

There are clearly \( a^3 \) cubes in total. Consider the bigger cube like an a*a*a Rubik's cube, with corners, edges, and centers (as well as hidden interior pieces). The first pile contains the centers, for a total of \( 6(a-2)^2 \) pieces; the second pile contains the edges, for a total of \( 12(a-2) \) pieces; the third pile contains the corners, for a total of 8 pieces; and the last pile contains the hidden pieces, for a total of \( (a-2)^3 \) pieces.

a) The problem here is: \( 9*6(a-2)^2 = 4*(a-2)^3 \), or \( 54 = 4*(a-2) \), but there is no integer solution for a, because 54 isn't divisible by 4.
b) The problem here is: \( 6(a-2)^2 = 3*12(a-2) \), or \( 6(a-2) = 36 \), which means a=8.

 

[Ordos_Koala]

Spoiler

There are clearly \( a^3 \) cubes in total. Consider the bigger cube like an a*a*a Rubik's cube, with corners, edges, and centers (as well as hidden interior pieces). The first pile contains the centers, for a total of \( 6(a-2)^2 \) pieces; the second pile contains the edges, for a total of \( 12(a-2) \) pieces; the third pile contains the corners, for a total of 8 pieces; and the last pile contains the hidden pieces, for a total of \( (a-2)^3 \) pieces.

a) The problem here is: \( 9*6(a-2)^2 = 4*(a-2)^3 \), or \( 54 = 4*(a-2) \), but there is no integer solution for a, because 54 isn't divisible by 4.
b) The problem here is: \( 6(a-2)^2 = 3*12(a-2) \), or \( 6(a-2) = 36 \), which means a=8.

nicely done

Kynit
Hm... (Whaulk's problem)

Spoiler

There are xCn ways to choose x people out of a group of n. This gives you xCn possibilities of finding a matching birthday.

Finding a matching birthday for x people:
365 / 365^x
or
1 / 365 ^ x-1

Therefore, the chance of finding a matching birthday is xCn / 365^x-1

Does this sound right? I'm not too familiar with the choose function.

read posts before, this isn't what he asked (I thoght he did too )

whauk's problem v.2

Spoiler

it's much easier just to count chances that they DON'T have birthday on same day... so for two people it's:
365*364*363*...*(365-n+1)/365^n
it means the first one can be born whenever in year, but the second one can't be born at the same day like the one before him and we go on like this (it's how many possiblities are right for that condition) slash all possiblities including the wrong ones

for three people it's similar:
above in this fraction will be again 365 beacause the first one can be born at any day again, so can be the second, because that condition is broken only when 3 people are born at the same day. So if these two would be born at same day, the third one can't be, so now it's only 364, but it's again same for the fourth, because these three people can make only one day inappropriate... below in fraction, it is same as for 2 people sharing birthday

365*365*364*364*363*.../365^n
(above are n elements)

for any higher number of people for the same birthday there will be more 365s above, but I hope you understand what I mean. I realize there is probably nice equasion with some factorials and no dots, but I don't know much about factorials, so I'd fail somewhere

of course at the end you have to calculate the ratio for that they ARE born at the same day: 1-x=y
where x is probability that they won't (that's what we have counted earilier) and y the number we are seeking

 

[Hiero]

The following problem is from an elementary mental math test we had here in Texas last year. Each problem should take about 5-10 seconds. I know the solution to this but have a hard time seeing how an elementary student would see the answer so quickly, but maybe there is another pattern that I am missing.

"The next number in the sequence 0, 7, 26, 63,...is___________"

Give the next number and the pattern you used. Try not to google the answer.

 

[MTGjumper]

124. I thought it was fairly obvious (hopefully I'm not shooting myself in the foot here). How old is an elementary student?

(Out of interest, could anyone point me in the direction of a question and its solution which concerns a question where there doesn't appear to be enough information to answer it, but with additional information such as "the second brother has a ginger beard" there is a solution. There used to be a similar problem on Shelley's site.)

 

[Hiero]

124. I thought it was fairly obvious (hopefully I'm not shooting myself in the foot here). How old is an elementary student?

(Out of interest, could anyone point me in the direction of a question and its solution which concerns a question where there doesn't appear to be enough information to answer it, but with additional information such as "the second brother has a ginger beard" there is a solution. There used to be a similar problem on Shelley's site.)

It is 124. What was your reasoning? This would be for 9-11 years old.

 

[Ordos_Koala]

if you know that murderer has a ginger beard, who killed this guy? sry, but I don't really get your point...

 

[Rinfiyks]

Had this optional brain teaser at the end of a question sheet for theory of programming:

Spoiler

There are three gods, Angel, Demon and Chaos. Angel will always answer truthfully, Demon will always lie, and Chaos will answer your questions randomly. The gods are standing before you, but you do not know which is which. Your task is to determine the identity of the gods, by asking them three questions. To make matters worse, the gods will answer your questions only with 'da' or 'ja', which mean true and false in their own language, but you don't know which is which.

Clarifications:
Each question can be directed only at one god, but you may ask two (or three) of your questions to the same god, if you so wish.
The gods, being gods, are all-knowing and perfect logicians.
There's no "trick", like asking them to answer paradoxes.
The three gods all speak the same language.

 

[cmhardw]

I was watching Deal or No Deal a couple days ago, and an interesting probability situation came up. It's very similar to another problem I've asked on here before.

The lady playing had gotten down to only 2 cases, the one with $750,000 and the one with $1,000,000 dollars. Before she was to open her case, Howie gave her the option to switch cases and open the one still on stage rather than the one she picked at the beginning of the game. She had been having such good luck throughout the game that she decided not to switch cases.

Did she make the right decision? Or did it not matter one way or the other whether she swapped cases or not?

 

[Keroma12]

I was watching Deal or No Deal a couple days ago, and an interesting probability situation came up. It's very similar to another problem I've asked on here before.

The lady playing had gotten down to only 2 cases, the one with $750,000 and the one with $1,000,000 dollars. Before she was to open her case, Howie gave her the option to switch cases and open the one still on stage rather than the one she picked at the beginning of the game. She had been having such good luck throughout the game that she decided not to switch cases.

Did she make the right decision? Or did it not matter one way or the other whether she swapped cases or not?

Spoiler

Wrong decision. When there are n cases, and you pick one, there is a 1/n chance you are right, and so a (n-1)/n chance you are wrong and it is in another case. Since it wasn't in the cases already opened, there is a (n-1)/n chance she should have swapped. I don't watch this show, but I assume n > 2

 

[qqwref]

Spoiler

Since nobody told the lady what the contents of any given box was until after she opened it, we could have switched the $750,000 and $1,000,000 boxes before the game started, and had the exact same probability of ending up in this situation at the end. So switching is irrelevant, because each box is equally likely to be the $750,000 or the $1,000,000, given that we end up with these two boxes left.

 

[cmhardw]

Spoiler

Since nobody told the lady what the contents of any given box was until after she opened it, we could have switched the $750,000 and $1,000,000 boxes before the game started, and had the exact same probability of ending up in this situation at the end. So switching is irrelevant, because each box is equally likely to be the $750,000 or the $1,000,000, given that we end up with these two boxes left.

Spoiler

Michael, my hat's off. I initially thought of this as a Monty Hall style problem, but I can see the flaw in that now because of the same probability of ending up with the same two cases at the end even if they were swapped.

I'm imagining a different version of Deal or No Deal where there are 26 cases, one case contains $1,000,000 and the other 25 are empty. Imagine in this game if you pick a case at the beginning, then proceed to open 24 empty cases. At this point Howie asks if you would like to swap cases or not. Since you are down to two cases, one empty and one the $1,000,000 I wonder if this works the same way.

My impression is to try to calculate it.

The probability of winning the $1,000,000 case assuming you do not switch would be 1/26 since you open the case that you initially picked.

The probability of winning the $1,000,000 assuming that you do switch is:
\( \frac{25}{26} * \frac{24}{25} * \frac{23}{24} * \ldots * \frac{2}{3} * \frac{1}{2} = \frac{1}{26} \)

Wow, so it does not matter whether you switch or not even in this variation, the probability of winning the million is still 1/26. Very neat result! I definitely got it wrong when watching the show. I thought she should have switched. Cool, you learn something new every day.

 

[qqwref]

Spoiler

I'm imagining a different version of Deal or No Deal where there are 26 cases, one case contains $1,000,000 and the other 25 are empty. Imagine in this game if you pick a case at the beginning, then proceed to open 24 empty cases. At this point Howie asks if you would like to swap cases or not. Since you are down to two cases, one empty and one the $1,000,000 I wonder if this works the same way.

Spoiler

Once you've managed to open 24 empty cases, you are indeed less likely to have chosen the million dollars than in the beginning - in 25 of 26 possibilities the million dollar case will be in the last of the 25 non-chosen ones, but in only 1 possibility will you have chosen the million dollars. Of course, you're a lot more likely to have opened 24 empty cases if you started with the million, but once you've done it the probability of the past event isn't relevant anymore.

The difference between this version and the version you saw on the real show was that the $750,000 and $1,000,000 cases function identically, so neither is biased against by the act of removing every other case. In this example, the $1,000,000 case and empty case you are left with are NOT identical, because there were more cases just like the empty one at the start.

 

[LewisJ]

It is 124. What was your reasoning? This would be for 9-11 years old.

The differences between terms are 7, 19, 37; the differences between these are 12 and 18; the difference between 37 and the next increased amount is thus 24, so the next increased amount is 61, which added to 63 gets 124. This is not a 5-10 second problem for 9-11 year olds, but it is certainly solvable within a minute or two by a determined individual in that age group.