BrainDen.com - Brain Teasers # jasen

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## Posts posted by jasen

1. ### Bloody Roulette game

In a bloody Roulette game, usually a gun with even bullet's holes is used,
then the gun is fired alternately to the head of 2 player.
This is a fair game, because if the bullet is in odd hole so the first shooter will win,
but if the bullet is in even hole so the second shooter will win.

Now, How if we use 2 guns ?
first gun has 1 more hole then second gun.
(such as 1st gun 6 hole and 2nd gun 5 holes)
first player hold first gun and shoot first.
both player shoot alternately.

Is this a fair game or not?
If not who has bigger chance to win?

2. ### Pick a red card, randomly?

Option Explicit

Dim Cards As Variant
Dim SumRed As Variant
Dim i As Integer
Dim j As Single
Dim SumBlack As Integer
Dim YouWin As Single

Sub SuffleCards()
Dim i As Integer
Dim PosCard1 As Integer
Dim PosCard2 As Integer
Dim TempCard As Integer
Randomize
For i = 1 To 50
PosCard1 = Int((52 * Rnd))
PosCard2 = Int((52 * Rnd))
If PosCard1 <> PosCard2 Then
TempCard = Cards(PosCard1)
Cards(PosCard1) = Cards(PosCard2)
Cards(PosCard2) = TempCard
End If
Next i
End Sub

Sub StopCardGame()
Dim xx As Integer
ReDim SumRed(52)
Const NumOfGames = 1000000
For xx = 1 To 10
YouWin = 0
Randomize
Cards = Array(1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0)
For j = 1 To NumOfGames
SuffleCards
SumRed(0) = Cards(0)
For i = 1 To 50
SumRed(i) = SumRed(i - 1) + Cards(i)
SumBlack = i - SumRed(i)
If SumBlack = 26 Or SumBlack = SumRed(i) + xx Then
If Cards(i + 1) = 1 Then YouWin = YouWin + 1
GoTo gameOver
End If
Next i
If Cards(51) = 1 Then
YouWin = YouWin + 1
GoTo gameOver
End If
gameOver:
Next j
Selection.TypeText Text:=vbCrLf + win + str(YouWin) + " times of " + str(NumOfGames) + " games with " + str(xx) + " more black"
Next xx
End Sub

With The simulation Above I get this result

win 500096 times of  1000000 games with  1 more black
win 500356 times of  1000000 games with  2 more black
win 500055 times of  1000000 games with  3 more black
win 500070 times of  1000000 games with  4 more black
win 499398 times of  1000000 games with  5 more black
win 500050 times of  1000000 games with  6 more black
win 499553 times of  1000000 games with  7 more black
win 499117 times of  1000000 games with  8 more black
win 499168 times of  1000000 games with  9 more black
win 500105 times of  1000000 games with  10 more black

So, I think there is no best strategy.

3. ### Pick a red card, randomly?

in my computer simulation

the best strategy is
1. anytime you find 2 black more then red (such as 4B 2R) then you stop the dealer
2. or all black has appear then stop the dealer
3. but if until last card no condition above appear then bet for last card.

and you will win with probability 50.030%. (very-very small)

I've tried many combination, like 1 black more, or 3 black more, or combination of them, but those do not better then solution above.

4. ### Maths Teaser

Ok I try again

Use base 3 numbering, from 0 to 22220 to the barrels

Example:  barrel no 10201 means :
Soldier 1 and 5 drink at time 0
Soldier 3 drink at time 24

Barrel no 22010 means :
Soldier 4 drink at time 0
Soldier 1,2 drink at time 24

Now to identify which barrel poisonous is like this :
if soldier 3 died first (form time 0 to 24)
then soldier 1 and 5 died (from time 24 to 48)
Means : barel no 20102 (or 173 (base 10)) poisionous

5. ### shooting game

Hidden Content

K-man is right

I was thought shooting skill when open eyes is not important, but I was wrong.

I have created VBA code to test this, and this agree with K-man answer

Sub AccuracyTest()
Dim win As Single
Dim cnt As Double
Dim Shot As Integer
Const openEyes = 37 'percentage if open eyes
Const divFactor = 5  'divison factor if closed eyes
Const NumOfGames = 1000000
Const NumOfTries = 6

'test open eyes
win = 0
For cnt = 1 To NumOfGames
Shot = Int(100 * Rnd)
If Shot < openEyes Then win = win + 1
Next cnt
Selection.TypeText Text:=vbCrLf + str(win / NumOfGames)

'test closed eyes
win = 0
For cnt = 1 To NumOfGames
For j = 1 To NumOfTries
Shot = Int(100 * divFactor * Rnd)
If Shot < openEyes Then
win = win + 1
GoTo finish
End If
Next j
finish:
Next cnt
Selection.TypeText Text:=str(win / NumOfGames)
End Sub

7. ### Maths Teaser

Correction

Then each group divided into 5 more smaller group

I mean 4 more smaller group, not 5.

8. ### Maths Teaser

from bubbled question, I need only 4 soldier !!

The 240 barrels of wine is divided into 4 group (each group with 60 barrels), Let say group A to Group D

Each soldier drink a spoon of all wine from 1 group. Soldier A drink from group A, and so on.

Then each group divided into 4 smaller group (each group with 15 barrels), Let say group AA to group DD

6 hour latter. Each soldier drink a spoon of all wine from 1 group in each bigger group. Soldier A drink from group AA, and so on.

Then each group divided into 5 more smaller group (each group with 4 or 3 barrels), Let say group AAA to group EEE

6 hour latter. Each soldier drink a spoon of all wine from 1 group in each bigger group. Soldier A drink from group AAA, and so on.

Then each group divided into 5 more smaller group (each group with only 1 barrel), Let say group AAAA to group EEEE

6 hour latter. Each soldier drink a spoon of wine from 1 group in each bigger group. Soldier A drink from group AAAA, and so on

from the time each soldier died the King can identify which barrel is poisonous

9. ### Maths Teaser

How many bottles of wine will be served In the party ?

If only 1, the King just need 1 soldier.

if the soldier poisoned, so the wine that the soldier take is poisonous, serve other 999 bottles

if the soldier is not poisoned, so serve 1 bottles.

10. ### Dice Game

There is an advantage to student A, where he can win in the first roll.

If we do not allow student A (12) win in the first roll, so the probability is 50:50

11. ### The Android unlock pattern

i'd say its

A good start.

Someone has create a program to list all the possibility and there are only 1/3 possibilities of your answer.

12. ### Random bullets

I want to solidify the idea of using expected speeds to establish the accuracy of the (1/2)(3/4) ... formula

 It gives an exact result. As does the formula. So if there's any difference, no matter how small, it's wrong. It must agree exactly with the formula in order to be correct. That criterion is not available to us when we compare the formula to the result of a finite number of random cases.

 It gives the correct result for the four-bullet case, as shown here:

Hidden Content

Very clear explaination Bonanova. and the formula is a beautiful math equation.

You said it is a conjecture, I wonder why nobody has proved it mathematically.

13. ### shooting game

I'm going with:

Hidden Content

Right !

Now I modifiy the question

One day on a shooting game, you are given two options
1. You only have 1 chance shooting a target with open eyes.
2. You are given 6 chances shooting the target with closed eyes.
You realize that your shooting ability is only 1/5 with closed eyes than open eyes.

What option do you take to maximize your chance to win?

14. ### shooting game

One day on a shooting game, you are given two choices.
1. You only have 1 chance shooting a target with open eyes.
2. You are given 3 chances shooting the target with closed eyes.
You realize that your shooting ability is only 1/3 with closed eyes than open eyes.

What choice do you take to maximize your probability to win?

15. ### The Science & Faith Paradox

Yes, based on the fact that Human can never find "theory of everything" scientifically. Only religion can.

16. ### Need an Alibi

2 Alibi

1. Chalk dust is the key. When I meet the Professor, I'm clean, no chalk dust anymore on my body, so I'm not the killer.

2. If I'm the killer, I will not call the police. I will go immediately, so nobody knows the killer.

17. ### The Android unlock pattern

The Android unlock pattern

The Android unlock pattern has 9 dots on the screen organized in a 3×3 matrix. To unlock the phone a so called pattern has to be drawn on the screen, which means connecting certain points in a certain order. So how many valid patterns are there? For this let’s first observe the rules of pattern drawing:

at minimum 4 dots have to be used
at maximum 9 dots can be used
one dot can be used only once
the order in which the dots are connected matters (thus making it a directed graph)
dots are connected with a straight line meaning that all points on the path of the line get connected
The last rule introduces some conditional connection paths. When connecting two points with a straight line it is valid only if there is no unused point in the way. For example: you cannot connect points 1 and 3 unless point 2 is already is used. So by default drawing a line from 1 to 3 will result in the pattern 1→2→3. However if point 2 is used the transition 1→3 becomes valid, such as in 2→1→3, making a previously invalid transition valid (1→3).

Also what some people seem to omit is that connecting points in a slight diagonal is possible (especially on Android 4, since the dots became smaller) such as 2→7.  18. ### Random bullets

Error! I left out a case: corrected cases are:

Hidden Content

20000 samples of this yields 0.371. This is close enough to bonanova's statement of 3/8 being spot on, that I'm looking forward EAGERLY to the simple analytical explanation :-)

CaptainEd, before creating the code above, I also thinking like you, but then I reliaze the question is not that hard.

Look at again at prop 1 and 2

1. Every second, a gun shoots a bullet along a straight line.
2. Each bullet has a random speed between 0 and 1   ----> I guest bonanova means 0 m/s to 1m/s (not 0 to invinity)

which means that if  V3> v2 > v1 then second bullet will hit first bullet, before third bullet fired.

19. ### Random bullets

And for 4 bullets are

P of all anhilated =  0.374813
P of all anhilated =  0.374194
P of all anhilated =  0.375763

so I conclude the probability is 37.5 percent

20. ### Random bullets

Sorry, I was misunderstanding the question.

I have VBA create this code, I think this time I'm right

Sub RandomBulletTest()
Dim Anhilated As Integer
Dim Bullets As Variant
Dim AllAnhilated As Single
Dim TotalBullets As Integer
Bullets = Array(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -2)

For Cnt = 1 To 1000000
''random inisial speed between 0 to 1
For i = 0 To 9
Bullets(i) = Rnd
Next i
TotalBullets = 10
' value -2 for anhilated bullets
ulang:
For i = 0 To TotalBullets
If Bullets(i) < Bullets(i + 1) Then
TotalBullets = TotalBullets - 2
For j = i To TotalBullets
Bullets(j) = Bullets(j + 2)
Next j
Bullets(TotalBullets + 1) = -2
Bullets(TotalBullets + 2) = -2
GoTo ulang
End If
Next i
finish:
'check how many bullet anhilated
For i = 0 To 9
If Bullets(i) < 0 Then
Anhilated = Anhilated + 1
End If
Next i

'test wheather all bullet anhilated or not
If Anhilated = 10 Then
AllAnhilated = AllAnhilated + 1
End If
Anhilated = 0
Next Cnt
Selection.TypeText Text:=" P of all anhilated = " + str(AllAnhilated / 1000000)
End Sub

--------------------------

3 times running the code the results are

P of all anhilated =  0.245612
P of all anhilated =  0.245541
P of all anhilated =  0.246631

so I conclude the probability is 24.6 percent

21. ### Dice Game

Thank You bubbled,  Now I know my Mistake.

In my code I didn't set newValue to 0 before asign it to lastValue

After using the modified code, I get same result as Markov chain proof

Three times running the modified code, I get this result :

0.537232
0.5372334
0.5372318

22. ### Dice Game

It doesn't look iike you allow for a roll where player 1 can win and player 2 can't (the first roll).

Your procedure for a game should be this:

While Winner = 0
LastValue1 = NewValue1
LastValue2 = NewValue2
NewValue1 = Int((6 * Rnd) + 1)
NewValue2 = Int((6 * Rnd) + 1)
If NewValue1 = 6 And NewValue2 = 6 Then Winner = 1
If NewValue1 + NewValue2 = 7 And LastValue1 + LastValue2 = 7 Then Winner = 2
Wend

I have checked back my code, and I think nothing wrong there. Student A still can win even at first roll.

23. ### Half-balls

I think just sort all the half sphere, then merge the lightest with the heavyest as 1 pair, Keep the procedure 4 times.

24. ### Random bullets

Simulation by VBA :

Sub RandomBulletTest()
Dim Anhilated As Integer
Dim Bullets As Variant
Dim AllAnhilated As Single
Bullets = Array(0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

For Cnt = 1 To 500000
''random inisial speed between 0 to 1
For i = 0 To 9
Bullets(i) = Rnd
Next i

' i for front bullet and j for bullet behind i
' if speed j > speed i then both bullets anhilated
' i assign -2 for anhilated bullets
For i = 0 To 9
For j = i + 1 To 9
If Bullets(i) >= 0 Then
If Bullets(j) >= 0 And Bullets(j) > Bullets(i) Then
Bullets(j) = -2
Bullets(i) = -2
j = 10
End If
End If
Next j
Next i

'check how many bullet anhilated
For i = 0 To 9
If Bullets(i) < 0 Then
Anhilated = Anhilated + 1
End If
Next i

'test wheather all bullet anhilated or not
If Anhilated = 10 Then
AllAnhilated = AllAnhilated + 1
End If
Anhilated = 0
Next Cnt
Selection.TypeText Text:="      P of all anhilated = " + str(AllAnhilated / 5000000)
End Sub

---------------------------------

P of all anhilated =  .0349778
P of all anhilated =  .0349992
P of all anhilated =  .0350404

Sory I thought I was wrong, The code should be like this :

Dim JohnWin As Single
Dim JulieWin As Single
Dim P5of9 As Single
Dim P4of7 As Single
Randomize
For i = 1 To 500000
For j = 1 To 9
NewValue = Int((2 * Rnd) + 1)
If NewValue = 1 Then JohnWin = JohnWin + 1
If NewValue = 2 Then JulieWin = JulieWin + 1
Next j
If JulieWin = 5 Then P5of9 = P5of9 + 1
JulieWin = 0
Next i
Selection.TypeText Text:=" percentage 5of9 = " + str(P5of9 / 5000000)

For i = 1 To 500000
For j = 1 To 7
NewValue = Int((2 * Rnd) + 1)
If NewValue = 1 Then JohnWin = JohnWin + 1
If NewValue = 2 Then JulieWin = JulieWin + 1
Next j
If JulieWin = 4 Then P4of7 = P4of7 + 1
JulieWin = 0
Next i
Selection.TypeText Text:="      percentage 4of7 = " + str(P4of7 / 5000000)
End Sub

---------------------

after 3 times I get this :
percentage 5of9 =  .0245682      percentage 4of7 =  .0274882
percentage 5of9 =  .0246054      percentage 4of7 =  .0273542
percentage 5of9 =  .0245194      percentage 4of7 =  .027356

So it more likely that julie will win 4 out of 7 games than 5 out of 9

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