Skinny's Casino - Solution!

[Patient0]

Puzzle can be found here
-Lepton

Here I go:

i put aside the house-cubes combinations that don't have an inbetween number
(like x,x x,x+1 - all these combinations are considered as auto-loss for the player).

Now, let's say the house-cubes rolled 1 and 20 (aka: 1,20).
there is a 18/20 chance to win.

1/20 change for a specific number in a single roll of a single cube:

(1=)1/20 * (20=)1/20 * (all nubers between 1-20=)18/20 =

= 1*1*18/20*20*20 = 18/8000 - chance of winning.

now, the house-cubes rolled 2,20:
again we have 1/20 * 1/20 * 17/20 = 17/8000

.
.
.

we end the combinations with 18,20:
1/20 * 1/20 * 1/20 = 1/8000

Let's generalize this situations:
We have 2 possible combination of 18 inbetween-numbers: (1,20) and (20,1)
We have 4 possible combinations of 17 inbetween numbers: (1,19) , (19,1) , (2,20) , (20,2)
The final is 36 combinations of 1 inbetween number. (1,3)(3,1) , (2,4) , (4,2)... (18,20) , (20,18)

so we get:
[(2*18 + 4*17 + 6*16 + 8*15 + ... + 36*1) / 8000] * 100% =
[(36 + 64 + 96 + 120 + 140 + 120 + 96 + 64 + 36) / 8000] * 100% =
[(780/8000)] * 100% = 9.75%

You get a 9.75% chance of winning each round.
:-P

[This message has been edited by Lepton (edited 12-04-2003 03:45 PM).]

[Bicho the Inhaler]

Note to Lepton: please start a thread in Grey Labyrinth puzzles for discussion of Skinny's Casino

The probability that the house dice have n numbers between them is 2(19 - n)/20^2, if n is between 1 and 18 inclusive (we don't care if they have 0 between them, and they can't have more than 18). The probability of winning when they have n numbers between them is just n/20. Thus the total probability of winning is the sum of 2(19 - n)/20^2 * n/20 from n = 1 to n = 18, i.e., the sum of

2/20^3 * n(19 - n)

Noting that n(19 - n) = (19 - n)(19 - (19 - n)), we can sum from 1 to 9 and multiply by 2, so we have

= 2 * 2/20^3 * (18 + 34 + 48 + 60 + 70 + 78 + 84 + 88 + 90); factor out another 2:
= 2 * 2 * 2/20^3 * (9 + 17 + 24 + 30 + 35 + 39 + 42 + 44 + 45)
= 1/1000 * 285 = 0.285 = 28.5 %.

Patient0, we seem to use the same line of thinking, but I think your arithmetic needs some work

[kevinatilusa]

Another way of looking at it.

To win, you need all three numbers to be different (prob 19/20*18/20). If this happens, your number will be the middle one 1/3 of the time, so the total is (19/20)(18/20)(1/3)=0.285

[mathgrant]

I never keep up with the GL puzzles. I wasn't aware this was added. I searched VSP for Skinny's Casino to figure out what this meant.

[Bicho the Inhaler]

Sure, or you could do that
*Weeps*

[Patient0]

Well, it was my first Combination problem.
and I'm still working on my arithmetic skills :-)

kevinatilusa, your solution is very ellegant, well done!

p.s.
Can you guys publish some more chess problems, I love those.

[Lepton]

Mathgrant, I thought I had moved that entire thread to the GLOC forum. Sorry.

Bicho, duly noted.

Kevin, my solution is similar to yours. Would it be worthwhile to make it more rigourous?

Patient, I can do chess problems.

-Minotaur in disguise

[This message has been edited by Lepton (edited 10-20-2003 11:48 PM).]

[hcummins]

I get the same answer as the .285 only I did it the easy way by using excel
In column A and B I put all 400 combinations that the house could roll.

Specifically Column A contained 1 down twenty times, 2 down 20 times, etc. Column B had 1 through 20 replicated twenty times.

In column C I wrote a formula that determined the number of successful rolls the participant could have against any roll the house could have.

The formula was as follows: =IF(ABS(A1-B1)>=2,ABS(A1-B1)-1,0)

I then summed the column and the successful combinations were 2,280. Knowing that there are 8000 possible combinations i.e., 20^3 leads me to the conclusion that the house needs to give approximately $3.50 to a $1.00 bet to make this game "fair". This of course doesn't incude the free drinks, and opulent surroundings of a gaming establishment.

[Bicho the Inhaler]

hcummins, for the easy way, see reply 2

[Patient0]

I got another combination question, it's very old but for those of you who haven't heard, here goes:

You have 100 lamps in a room. all are turned off.
You also have 100 children. all good clean nice children.
the first boy comes inside the room and turns on all the lamps.
boy number 2 comes in and CHANGES the statue of every second lamp (if the lamp was on - he turned it off and black vise).
Now, unexpectedly, boy number 3 comes in and changes the statue of every third lamp.

How many lamps will be turned on after the 100th boy will finish his job?
(Including explenation, ofcourse)

[Patient0]

arrggg! i ment "explanation".
(damn microsoft keyboard...)

[a little bird]

Yes, but did you mean "state" ?

[hcummins]

bicho, yes indeed your answer is much easier. I am a financial guy and my first inclination is to put it on a spreadsheet. It only took a few minutes, and while it was easy for me it was not nearly as elegant as your solution. For me to replicate your answer would have caused me to think outside my box. Thanks for the insight.

It kind of reminds me of a question that was given to me a few years ago. It went like this: If you had a string around the earth at the equator (exactly 25,000 miles) then added twelve inches to it how far would the new circle be above the old one. Once I worked the problem I discovered that when looking at the delta between concentric circle radiuses, it makes now difference how big the original circle is: the difference is a constant and is dependent only on how much you add to the circumference.

H.

[DLove28]

10 lamps. i'm a spreadsheet geek too. i had to use one before i saw the pattern. Patient0, a switch would have to be turned on odd number of times in order for the light to remain on. since each Lamp Number switch is turned by only the Boy Number that is a factor of it, only lamps with on odd number of factors are turned an odd number of times. (that is lamp number 3 is only turned by boys 1 and 3 leaving it off. lamp #25 is turned by boys 1, 5, and 25 leaving it on). So that means only squared numbers (which have an odd # of factors) would be left on. So 10 lamps - 1,4,9,16,25,36,49,64,81,and 100

[Patient0]

DLove28, you're correct!
now, accordint to tradition, it's your turn to present us with a new question.

[fritfrat]

all these riddles are way out of my league, but I do wonder if such a problem could be solved:

how many sides would be on your die to give you the best chances of winning? and obviously, what (consecutive, no doubled) integers would they be?

[Lepton]

A n-sided die (with an infinite number of sides) will give you a 1/3 chance of winning, which is the best you can do. Knowing Skinny, I'm sure that he could sell you one. Unlabeled, of course.

[DLove28]

well i'm more for solving than giving but since you threw down the gauntlet, here goes. this one got me going since i'm more mathematically inclined than not.


What do all six of the items in group A have in common that the items in group B do not?

Group A: Astronomical, Raze, Ape, Loom, Listening

Group B: Lack, Portable, Rocket, Usual, Atom

[Snail]

DLove28, I suppose six is just a typo and you really mean five. Unless that's part of the puzzle. You might have better luck finding an answer if you posted this in [url=http://www.greylabyrinth.com/cgi-bin/forumdisplay.cgi?action=topics&forum=<!--2-->Visitor+Submitted+Puzzles&number=5]Visitor Submitted Puzzles[/url].

[gopower]

Close but no cigar for you!
The house can roll between 2 and 40. But you win only between 2 and 20 (house cannot roll a 1. How many ways can the house roll between 2 and 20? 2 = (1,1)
3 =(1,2;2,1) 4 =(2,2;3,1,1,3) and so on, so the ways to roll are 1 way for the number 2 through 19 ways for the number 20. The sum of 1 through 19 = 19(20)/2
or 190 ways. You can win with only 19 ways (2-20) so the chances of winning are 1 in 10 or 10%. Check it out!

[gopower]

You know. Sometimes even I can make a misteak! How many ways can 2 - 20 sided dice be rolled? 400 different ways! How many result in a sum between 1 and 20?
only 2-20 are possible, so 190 ways (from previous reply). Since all your rolls will fall between 2 and 20, you win 190/400. or 47.5% I like this answer better. People will like to play this game because they will win often, but the house will win in the long run! Yes. This is the answer...

[Bicho the Inhaler]

gopower, nowhere in the problem are the numbers on the dice added. You win by getting a number between the two numbers on the house dice.

[/dev/joe]

DLove, got your puzzle with the words.
The answer is all the words in the first group form new words if you add the letter G at the beginning.

[gopower]

Third tries the charm thanks to Bicho the Inhaler who correctly corrected my misunderstanding. That withstanding,there are 400 possible combinations that the house can roll. 20 of them are pairs ( zero chance of winning) 38 rolls are one number apart (still no chance of winning) 36 rolls are two numbers apart -giving one chance to win and so on up to 18 chances to win if the numbers are nineteen apart (1,20, 20,1) There are 18(19)/2 winning opportunities and 400 different ways to roll or 171/400. That yields 42.75% chance of winning. How's that Inhaler?

[Kutti]

Gopower, your argumentation seems to be right but the result of 42.75% is not. I think, the reason is, that all the 171 winning opportunities are not of the same probability. If you weight them with their probabilities (36/400, …, 2/400), the total chance of winning comes out to be 28.5%, the same result as in the elegant solution of kevinatilusa.

[HappyFunBall]

I'm kicking myself for doing this one the hard way, and not seeing Kevin's very slick solution. However, a few nitpicks with the puzzle itself. "Between" is a little ambiguous. It could mean inclusively or exclusively. Everyone so far has interpreted it as exclusively, but if you consider it inclusively, your chance of winning comes out to 153/400 (38.25%), which is quite a bit higher. And folks, the singular of "dice" is "die". Sorry to be such a spelling nazi, but it's a pet peeve .

[Samadhi]

How is x "between" x and x?

[Beartalon]

Samadhi - play Tic-Tac-Toe, and let x win.

[HappyFunBall]

> How is x "between" x and x?

The same way that a train that runs between New York and Washington actually stops in both cities. The use of "between" in English is vague like that, and whenever it's used in a problem like this, it should be noted which meaning is intended. And if you'd like a more mathematical example, consider the closed interval [3,3]. 3 is certainly contained in it. It's not much of a jump from the continuous case to the discrete.

Also, just for kicks, here's the win probability for an arbitrary n-sided die:

exclusive: (n^2 - 3n + 2)/3n^2
inclusive: (n^2 + 3n - 1)/3n^2

[Sugaar]

i've got gopower's same result (47.5%) though my reasoning is less mathematical and more logical.

reasoning: for house's die #1 drawing "1" and #2 drawing also "1" (P=1/400), you've got a 95% (19/20) chance of winning; for die #1 drawing "1" and die #2 drawing "2" you've got 90%, and so on until the house's draw is 1-20, where you've got no possibility at all (0%). so the media for die #1 drawing "1" and die #2 drawing anything (P=1/20) is 47.5%. When you try with the possibilities at house's die #1 drawing "2" and #2 drawing all the 20 possibilities, you find out that you've got the same chance. so you've got the pattern and don't follow (you can of course) with die #1 drawing "3" and so on. definitely you've got a 47.5%.

as gopower says the house's win is in the long run but you can still be lucky enough to win once or even several times, provided that the dice are not tricked.

[ChienFou]

2 and 40 have same frequency; 3 and 39 etc. So the house's edge is simply the probability of 21 being rolled. This is obviously 20/400 (well it is to me). 5%, so 47.5/52.5. Trivial. - and I still like kevin's solution, very elegant.

[pokerfaced]

Sugaar - you seem to be thinking that you need to roll higher than both dice to win, whereas the problem requires you to roll between both dice to win.

[Jingle47]

There are 400 possible ways to roll the house dice. Ther are 2 ways to roll a 1 and a 20 with a 18/20 probability of rolling in between them. OR the house dice could be rolled in such a way as to be 17 possible numbers between them(4 ways to do this)(1,19)(19,1)(2,20)(20,2). This pattern continues in such a way that the probability of rolling in between the two dice is:
(2*18 + 4*17 + 6*16 +... + 34*2 + 36*1)/8000=28.5%

IT is that simple right?

[Samadhi]

HFB: My point was that you were obviously including instances where all the dice were equal. I agree that [n, n+1] has could be construed to have a win of n & n+1 if the "between" is meant to be inclusive.

However, "between" implies two different values. You can travel between New York and Washington but traveling between New York and New York is ludicrous. Therefore, [n, n] can not result in a win.

[Siuol]

Fritfrat asked earlier. How many sides should your die contain and which
numbers should it have for you to have the best chance of winning?

For a normal 20 sided die 2280 of the 8000 possible throws is a win for
you. Which is 28.5%

Every side taken of your die decreases the posible throws by 400.
If n is any side from 1 to 10. If you take of side n or 20-(n-1) the
winning combinations is reduced by ((n-1)*(20-n))*2.

With an 18 sided die containing 2 through 19 you still have 2280 chances to
win but this time out of a total of 7200 posible throws. This gives you a 31.6%
chance. This is because 1 and 20 is never between anything.

Now if you also take of the sides 2 through 5 and 16 through 19 you end up
with 1640 chances to win out of 4000 posible throw combinations. This gives
you a 41% chance of winning.

If you take of any other numbers, your chance to win will decrease.

I believe that the 41% is the best you can get.

[mathgrant]

How about a die with 20 sides labeled 10.5? That has a 50% chance.

If that's not kosher, then 20 10's will give 45%.

[Westiex]

Quotith Patient0:

You have 100 lamps in a room. all are turned off.
You also have 100 children. all good clean nice children.
the first boy comes inside the room and turns on all the lamps.
boy number 2 comes in and CHANGES the statue of every second lamp (if the lamp was on - he turned it off and black vise).
Now, unexpectedly, boy number 3 comes in and changes the statue of every third lamp.

How many lamps will be turned on after the 100th boy will finish his job?
(Including explenation, ofcourse)

End Quotith

I believe the answer is 49

My reasoning:

All lights are on.

Second boy comes in, turns every second light off, so that there now 50 lights on, fifty lights off.

Third boy comes in, changes the state of every third light.

That means:

1st On, No change, No change (on)
2nd On, Off, No change (off)
3rd On, No change, Off (off)
4th On, Off, No change (Off)
5th On, No change, No change (On)
6th On, Off, On (On)

So out of every six lights, up till the 96th light, Three lights are on, three lights are off. 96/6 (Number of lights/pattern)= 16, 16 * 3 (lights on per pattern) = 48. Of the remaining 4, only one is on.

Thus the answer is 49, NOT 10.

[Lepton]

There are 100 boys that switch the lights, not just 3.

[sodasipper]

http://www.greylabyrinth.com/Puzzles/puzzle025.htm
Why is this still being discussed? Move on already!

[ComputerHacker]

Write a simple code using a language of your choice.

For example:

For i= 1 to 20 do
For j= 1 to 20 do
For k= 1 to 20 do
If ((k>i) and (k<j)) or((k<i) and k>j))
win=win+1
end
end
end

you will find 2281 winning possibilities out of 8000.
Hence the chances of winning is 2281/8000 = 28.5125%