jasen

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?

I change my answer With The simulation Above I get this result So, I think there is no best strategy.

Ok I try again

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

No more try to answer?

from bubbled question, I need only 4 soldier !!

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

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

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

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.

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?

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?

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

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.

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.

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

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

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

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

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.

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

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 : Sub BadmintonTest() 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