How about this. If you flip just one time, and get heads, have you learned anything from the problem? You do not know for sure what type of coin you are holding still. The monty hall explanation only applies if you actually garner information from the results. Since a heads can be indicative of any of the ten coins on the table, you have not changed the problem. Maybe you should read further into the monty hall problem, as it has good explanations of why the odds do or do not change. Thanks.

Lets restate the problem so we know where we're going. We have 10 coins, 9 of which are fair, and 1 is a 2-headed coin. You randomly pick a coin and toss it. It comes up head. You claim that you gained no information from the toss. Actually, you just gained some information, just that it isn't a lot of information.

After you pick up the coin, but before the toss, you have no information whatsoever, so the probability that your coin is biased is 1/10. After one toss and the coin comes up head, the chance of it being a biased coin just increased a bit. Instead of giving calculations, let me present two examples.

To argue that you gain information from 1 toss, let's look at the opposite side. Assume that you randomly pick a coin and it comes up tail upon the flip. The probability of the coin being 2-headed goes from (1/10) before the toss to 0. You just gained some information from 1 toss.

Suppose you have ten 1000-sided dice. 9 of which have sides numbered from 1-1000. The last die has all sides labelled with 314. You randomly pick a die and toss it once. The number 314 comes up. What die do you think you got?

How is it that people are under the assumption that the number of iterations will change the outcome? If we have not changed the initial conditions of the problem, then the chance of flipping a coin heads on the 1st or the Nth iteration remain constant. (you do not get decreasing odds from multiple iterations if you do not change the data set) If this was true, then for each iteration, you would say that your odds are constantly changing (55/45, 45/55. 35/65, etc... after each flip, which is not how statistics work)

I think there is a fundamental difference in the thinking of the two camps here. The first camp assumes a priori that p, chance of a head, is fixed at 1/2. Consequently, regardless of the sequence of heads, the chance of the next toss is still 1/2.

psychic_mind and others are taking the bayesian point of view, which is that p is unknown. They then estimate p given the data. In that approach, given the 20 heads, the probability of head is very near 0.

Given the original wording, I'd go with the second approach. Perhaps this is a rewording of the original problem that makes it clearer. Assume that psychic_mind approaches you and proposes a game. He'll take a coin out of his pocket and flip it. If you can correctly guess the coin's orientation, he gives you 1 dollar. If you're incorrect, you give him a dollar.

You figure head is equally likely as tail, so you guess tail 20 times. The coin come up head 20 times, and you're down 20 dollars. Would you guess tail on the 21 try?

I tried simulating that strategy with 10 jars and five reads. My first attempt failed. Here's what I had...

1 6
2 5
3 4
4 3
5 2
6 1
7 10
8 9
9 8
10 7

Jar  Init

I would expect a similar outcome if Jar #1 started at 51 and scaled down to 1, then filled in the remainder. If you encounter all H or all L in your first pass of 50, you don't have the option of trading an L for an H. You might easily deduce that you could swap Jars 1 and 6 in the above case (even though you don't know the contents of Jar #6), but the rules don't allow it. With only four prisoners, the first jar would still contain an H value in an L jar.

Good catch. Back to the drawing board.

I think 4 prisoners can do it, we merge the first 4 rounds from my previous answer into 2:

Prisoner 1: starts swapping H for L, beginning with 1 vs. 100. Here's the process:

Lojar = 1, hijar=100;

loop

if lojar and hijar both contain H, decrement hijar and loop

if lojar and hijar both contain L, increment lojar and loop

if lojar contains H and hijar contains L, (swap contents, increment lojar, decrement hijar, and loop)

if lojar contains L and hijar contains H, increment lojar, decrement hijar and loop

P1 does this until 50 jars are opened.

Prisoner 2 finishes the process, starting at the other end! That is, lojar = 50, hijar=51;

loop

if lojar and hijar both contain H, increment hijar and loop

if lojar and hijar both contain L, decrement lojar and loop

if lojar contains H and hijar contains L, (swap contents, decrement lojar, increment hijar, and loop)

if lojar contains L and hijar contains H, decrement lojar, increment hijar and loop

At this point, all the L numbers are in L jars, all Hs in H jars.

Prisoner 3 puts the Ls in their proper places

Prisoner 4 puts the Hs in their proper places

Bravo, CaptainEd. I agree that 4 is pretty much the minimum number of turns required for guarantee a sorted array. This is is solved way too fast, I should make them harder next time.

Do they get the "one night of preperation" like in the previous ones?

Yup, they do get the obligatory one night of preparation.

This topic is inspired by previous prisoners on a death row problems.

There are 10 prisoners. The warden implements a simple game. He will put 100 slips, numbered from 1-100, randomly into 100 identical jar. Each prisoner will be able to open 50 jars and shuffle their contents. The prisoners will each do this individually, and in sequential order. The waiting room and the shuffling room are separate, so those prisoners waiting their turns can't see the jars being opened.

Each prisoner does not have to open the 50 jar sequentially. Each prisoner can remove the 50 slips from the jar and place them back in any order he desired. However, at the end, each jar must contain exactly 1 slip and no modification can be made to the jars' appearances, placement, orientation, etc.

After open and closing the 50 jars, the prisoner will be moved to a new room and have no communication with the ones who have yet to have their turn.

What is the minimum number of turns it takes to guarantee a sorted array of jars? Describe the strategy.

ARRGGH! I was gonna say something like this! I thought it was original though

You could find a way of measuring mass. Drop something from the roof, and use either a bathroom scale, or a viscous substance or water to determine the mass of the impact as the object strikes.

The further through the water the object moves (say a rubber ball, which floats) the more its mass will have increased.

(is it mass that increases?)

I think you're onto something here. if you measure the force of the impact, the you can estimate the velocity at which it travels upon impact if you know the item's mass.

It is the momentum ( momentum = velocity * mass ) that increases with speed. The mass of an object under classical physics is constant. Under relativity, every gets a bit heavier as it travels, but the mass increases for non near-light-speed are negligible.

Yeah I was thinking that or the watch.

Just thought of some more

Estimate the rate of ascent of a balloon with a few trials in your home. Let it go near the building on a windless day. The rate of ascent of the balloon should be constant if the building is not too high. Time how long it takes for the balloon to clear the building top. Multiply time by rate.

Synchronize the two watches. Let the friend stand at the bottom of the building and clap really hard at a previously specified time. Stand on the building top and record the time when you heard the clap. Multiply the time difference by the speed of sound.

use a triangle with a 90 and some other two angles. the greater the angle nearest your eye, the less you'll have to walk. follow the same procedures as before. take your stride lenght x count+height number and put it in to the equation:

tangent (angle nearest your eye) = building height / ((stride length x count)+ eye height)

take a piece of cardboard, make a triangle with 90,45,45 angles. hold the triangle so that it is at eye level and you are looking up the hypotenuse. walk back till you just see the top of the building in line with the hypotenuse. count the number of strides it takes to walk to the base of the building. measure your stride. multiply stride length by number of strides, add the height of your eyeball. thats the building height

Very nice, Geoff. I'm very impressed by your creativity. Since you mentioned food, here's another way

Open your fridge and take out a nice piece of chocolate cake. Go to the building's superintendent and say "I'll give you this nice chocolate cake if you tell me this building's height".

One approach is to consider all possible ways of having six kids. It is very rare for a family not to satisfy these condition in 6 kids or less. There are 64 different permutations, and those who are familiar with logic tables or probability should be able to construct such a table quickly. Once you have such a table, go down row by row and circle the 'stopping point', or the point at which the family satisfies the condition. For instance, suppose the row is this- S D D D S D- you would circle the 5 kid, since that is the point where the family satisfies the condition and stops.

Once done, look at each column and count how many stopping points you circled in that column. From there, we should have a reasonable estimate of what is the best guess.

There is a high-rise building in front of your house. You want to measure the building's height. Using common household items, list out the strategies to measure the building's height. (And no, common household items don't include laser measurement tool or a coil of rope as long as the building's height, though I guess watches with atomic accuracy are allowed).

I hope this will be more of an exercise in creativity. Members of this board have certainly surprised me with their novel and creative solutions before, and I certainly expect to see creative solutions for this one also.

Correct me if I'm wrong

Ok, the best bet will actually be 4, and here is why.

Probability of having 2 sons and 2 daughters out of any 4 children = .375

Probability of having 2 sons and 2 duaghters out of any 5 children = .625

however, the possibilities for 2 sons and 2 daughters in 4 children also exist in the possibility for 2 sons and 2 daughters in 5 children, so...

Probability of NOT having 2 sons and 2 daughters in 4 children, but then acheiving this in 5 = .625-.375 =.25

If we take this out to 6 total children, it becomes more likely that with 6 kids, there will be 2 sons and 2 daughters, but the odds of not getting them in 4 or 5 and then getting them with # 6 will be even less than .25

I'd write out the statistics that are used for this, but the truth is I don't remember the formulas, so I just did it longhand, basically wrote out every possibility outcome for 4 and 5 total children.

The best bet you could make on this one is 4 total kids, and the odds of you winning are 3 in 8

OBoyOBoyle got it. Well done.

Mr. and Mrs. Smith are friends whom you haven't seem for a long time. The last time you saw them was at their wedding ceremony, where they confided that they wanted to have 2 daughters and 2 sons, and they would have as many kids as needed until they have a minimum of 2 daughters and 2 sons. (For instance, suppose they have 2 daughters in a row, then would then give birth as many times as necessary until they get 2 sons, at which point they will immediately stop having kids. Likewise, suppose their first three kids are son, daughter, and son. They would then try again until their number of daughters reach 2. ) Assume that the probability of getting a son is .5.

You haven't seen the couple for years since the wedding. One day, you happened to meet Mr. Smith, who told you that his wife and he succeeded in getting the family they wanted. He then offered you a simple game. If you can correctly guess his total number of children, he would give you 100 dollars. You can only guess once. What is the best guess?

I wish I had an interesting story on how I figured this out, but basically I spent a half-hour figuring the odds for every W/L scenario in Excel.

n number of games::

0: <0.01%

1: 0.01%

2: 0.19%

3: 1.69%

4: 8.10%

5: 21.37%

6: 31.05%

7: 24.58%

8: 10.40%

9: 2.33%

10: .26%

11: .01%

...approximately 68.64%.

if a random game is picked out of the 11, A's average chance of winning is 6.1/11 = .5545, so B's average chance of winning is 4.9/11 or .4455 or so

say that 'a' is .5545 and 'b' is .4455

for A to win the competition, A needs to win 6,7,8,9,10 or 11 games

(a^6)(b^5) + (a^7)(b^4) + (a^8)(b^3) + (a^9)(b^2) + (a^10)(b^1) + (a^11) should give the right answer if I'm right (hell I'm probably wrong, maybe getting the average first was a bad idea). Also I might need to add the binomial coefficients from the 11th row of Pascal's triangle starting from the sixth and moving right, but I'll just try with coefficients of 1 and see if it's the correct answer. Probability isn't my strong suit

anyway I calculated the above as:

0.0056667911

or half of 1%.

Yeah that's got to be wrong. I bet the binomial coefficients have to be put on (but not the coeffiecients for .5 but the coefficients for .5545 I think). Oh well I tried

This approach of randomly drawing a game from the 11 and calculating its chance of success (.55) is a novel one. I didn't consider it. If we assume that we can do repeat random draws and then use the binomial theorem as unreality suggested, the chance of winning the tournament for A is about 64%. It's a bit off from the true percentage, but it's way closer than previous attempts.

The reason the answer is off because this approach isn't a true binomial experiment. While the first random draw from the pool of 11 games has a chance of success = .55, subsequent repetition of this exercise draw from an increasingly smaller pool of games, and thus the chance of success changes.

I haven't figured out an exact solution to this problem. But I have an approximate solution that is accurate to within 2 decimal digits, and increases in accuracy as the number of game increases. Hint = normal approximation.

if they play 10 times at each game, player A will win game 1 9 times, game 2 9 times, and so on. They will have then played 110 games, of which player A will have won 61. Odds are 61/110 or just a bit under 55 1/2% that player B will come in second.

61 is actually the average number of wins for A over 10 repetition of the tournament. Dividing that over the number of games does not give the expect change of winning a single competition. The chance of A winning a single tournament is actually higher than that.

1 lift.

We take advantage of the fact that all three labels are wrong. Lift once on the GRAY label to see its color. Since the labelling is wrong, the box must contain 2 blacks or 2 whites. Seeing the color of one will determine both.

Since the other two boxes now contain two configurations, but the labelling is wrong. It's easy to figure out what box contain what.

Saw this on another site.

Players A and B are taking part in a competition consisting of 11 different games. Each of the 11 games is always won outright by either A or B (i.e. a draw is not possible). 1 point is scored for winning a game, 0 points for a loss. So the overall competition has to be won outright by either A or B. Whoever has a higher score at the end of 11 games wins the competition.

Player A's win odds for all 11 games are known (see below), and it follows that Player B's win odds for any game n is [ 1 - P(A wins game n) ].

P(A wins game 1) : 0.9

P(A2) : 0.9

P(A3) : 0.9

P(A4) : 0.9

P(A5) : 0.7

P(A6) : 0.6

P(A7) : 0.5

P(A8) : 0.2

P(A9) : 0.2

P(A10): 0.2

P(A11): 0.1

What is the probability of A winning the overall competition (having a higher total score than B)?

This isn't a puzzle per se, but I just saw this movie scene and it's bothering me about whether it's possible in reality.

In case you haven't seen Transporter 3, at the end of the movie, the protagonist encounters a scenario where his car, an Audi A8 W12, plunges into a lake and sinks to the bottom. The protagonist then nonchalantly swims to his trunk and takes out four balloon-shaped canvas sheets. He attaches those canvas to his tires, one canvas per tire, and then releases the air from the tires into the canvas, thus inflating them like balloons under water. The four canvas balloons then lift his car from the bottom of the lake to the surface.

What you do think about this scenario? I feel that it's unlikely, but perhaps someone here can positively convince me so or otherwise.

This is not a trick question. This is a real math problem so don't say that a bus has no legs.

There are 7 girls in a bus.

Each girl has 7 backpacks.

In each backpack, there are 7 big cats.

For every big cat there are 7 little cats.

The bus driver is not on the bus.

Question: How many legs are there in the bus?

Number of legs = 7 * 2 + 7*7*7*4 + 7*7*7*7*4 = 10990 legs

I guess is that the bus driver is a smart man and got off the bus earlier. There's no way any sane man would want to be cramped into a bus with 2700 cats.

I realize that in my first post I was talking about numbers instead of names, but it is easier to deal with numbers and I assume that prisoners with perfect memory would also know everyone else's name and order alphabetically, which they could translate into numbers for purposes of sorting. An interesting twist on this problem would be if the prisoners did not know each others' names, but I won't go there now.

Interestingly, I think CaptainEd's sort and search approach is the optimal solution for the case where the prisoners don't know each others' names. The reason HoustonHokie's strategy gives superior performance is because we are taking advantage of extra information- the fact that each prisoner knows the other 99's names. The sort-and-search approach makes does not use of that piece of information, and thus equivalent to the strategy where the prisoners don't know each other's name.

Some interesting extension. Suppose that each prisoner has limited memory and can only remember the names of 49 other people. Let's say that the night before the game, the prisoners randomly divide into two groups, each with 50 members. Each prisoner then memorizes the names of all members of his group. What is the optimal strategy for this situation and what is the expected number of prisoners saved. I expect it to be bigger than 99.45 ( no one knows each others' names) but less than 99.5 ( each prisoner knows all others' names ).

737.7559108 inches

[pi * (1)^2] + [pi * (1.01)^2] + .... + [pi * (2)^2]

If you're integrating the length of the tape as the sum of successive concentric circles, shouldn't you use the fomula for the circumfence 2*pi*r as opposed to the area of the circle pi * r²?

A reel is 2 inches in diameter. A thin tape (1/50 of an inch in thickness) is wrapped round the reel until the end of the tape is reached. The diameter of the reel and tape combined is now 4 inches. How long is the tape?

The area that the tape makes in a cross section is

pi (2)² - pi (1)² = 3 pi

The length of the tape is then 3pi*50 = 150 pi = 471 in

HoustonHokie's approach is one that I didn't consider. Very nice. However, I think the expected number for his approach will be a tiny wee bit smaller, though. My approach is along the line of CaptainEd's, though with a slight difference.

We can modify's CaptainEd's approach a bit. Let the first guy sort the first 50. Let the second prisoner spend 6 searches on the first sorted 50 to find his name. He then can spend the remaining searches to sort jar 51-94. That way, we can bump his chance of living to about 94%. The 3rd guy will need to do a binary search on jar 1-50, another on 51-94, and a simple search on the last 6, with guaranteed 100% survival. So the expected number of prisoners saved is (.51 + .94 + 98 ) = 99.45

Some clarification to the original post.

Each prisoner, in turn, is allowed to open and shuffle the contents of 50 jars. At the end of each prisoner's turn, he identifies one jar as the one containing his name. The jar is not removed from play, but the warden determines whether the choice is correct or not at the end of every turn, and consequently, whether that prisoner will live or die.