w****g
发帖数: 44 | 1 1. Alice and Bob play the following game: Alice flips a fair coin N
times (N>=0). Bob flips the same coin N+1 times. Bob wins iff he gets
strictly more heads than Alice. What is the probability that Bob will win,
provide proof.
2. Solve the following differential equation:
3. Given N lightbulbs and N buttons, such that:
1. Each button toggles 1<=k
2. If button k toggles lamp j, then button j toggles lamp k
What additional conditions, if any, are required to g | g********5
发帖数: 62 | 2 1. Alice and Bob play the following game: Alice flips a fair coin N
times (N>=0). Bob flips the same coin N+1 times. Bob wins iff he gets
strictly more heads than Alice. What is the probability that Bob will win,
provide proof.
The probability Bob will win is 1/2.
Proof is easy if think of the problem in conditional probability.
N = 0, easy to see.
assume N=n, that probability is 0.5,
now for N=n+1, Let D denote difference : Bob(n+1) - Alice(n)
i.e D means the difference between Bob and Alice be | x*b
发帖数: 253 | 3 Try the following proof.
Let P ( B > A in N times ) =r1
P ( B < A in N times ) =r2
P ( B = A in N times ) =r3
Obviously r1=r2, r1+r2+r3=1, and 2r1 + r3 =1
P ( B > A in N+1 times ) = P (B>A in N+1 times | B>A in N times ) * P (B>A
in N times)+ P (B> A in N+1 times | B=A in N times) * P (B=A in N times)
= 1* r1 + 1/2 * r3 = 1/2 (2r1 + r3) =1/2
【在 g********5 的大作中提到】
: 1. Alice and Bob play the following game: Alice flips a fair coin N
: times (N>=0). Bob flips the same coin N+1 times. Bob wins iff he gets
: strictly more heads than Alice. What is the probability that Bob will win,
: provide proof.
: The probability Bob will win is 1/2.
: Proof is easy if think of the problem in conditional probability.
: N = 0, easy to see.
: assume N=n, that probability is 0.5,
: now for N=n+1, Let D denote difference : Bob(n+1) - Alice(n)
: i.e D means the difference between Bob and Alice be
| g********5
发帖数: 62 | 4 smart and short proof.
【在 x*b 的大作中提到】
: Try the following proof.
: Let P ( B > A in N times ) =r1
: P ( B < A in N times ) =r2
: P ( B = A in N times ) =r3
: Obviously r1=r2, r1+r2+r3=1, and 2r1 + r3 =1
: P ( B > A in N+1 times ) = P (B>A in N+1 times | B>A in N times ) * P (B>A
: in N times)+ P (B> A in N+1 times | B=A in N times) * P (B=A in N times)
: = 1* r1 + 1/2 * r3 = 1/2 (2r1 + r3) =1/2
| s*******u
发帖数: 35 | 5 My two cents about this problem. We denote the 'on' state by 0 and 'off'
state by 1.
Regard the states as a vector in the vector space (Z_2)^n. Each botton n
represents a vector v_n whose i-th component is 1 if the button toggles the
i-th light bulb and 0 otherwise. Then pressing the button n amounts to
adding the corresponding vector v_n to the state vector of bulbs. The
question asks when we can get to 0 vector by adding multiples of v_n's. It
is equivalent to say when any vector in (Z_2)^n
【在 w****g 的大作中提到】
: 1. Alice and Bob play the following game: Alice flips a fair coin N
: times (N>=0). Bob flips the same coin N+1 times. Bob wins iff he gets
: strictly more heads than Alice. What is the probability that Bob will win,
: provide proof.
: 2. Solve the following differential equation:
: 3. Given N lightbulbs and N buttons, such that:
: 1. Each button toggles 1<=k: 2. If button k toggles lamp j, then button j toggles lamp k
: What additional conditions, if any, are required to g
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