[機統] 投擲硬幣的問題

[tokyo291 (工口工口)]

題目為

A fair coin is independently flipped n times, k times by A and n-k times

by B. Show that the probability that A and B flip the same number of heads

is equal to the probability that there are a total of k heads.

前半部的機率為

t=min(k,n-k)

  t   k                    n-k
 Σ  C *0.5^(a)*0.5^(k-a)*C *0.5^(a)*0.5^(n-k-a)
 a=0  a                    a

後半部的機率為
        n
P(X=k)=C *0.5^(k)*0.5^(n-k)
        k

然後兩式都有0.5^n可以消掉,也就是說要證明

  t   k  n-k  n
 Σ  C *C   =C         t=min(k,n-k)
 a=0  a  a    k

有代數字下去作t=k 和t=n-k的情況等號都會成立

有試過用數學歸納法,結果卡住QQ

請問該用何種技巧呢?

--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 218.175.102.167
→ doom8199    :(1+x)^k * (1+x)^(n-k) = (1+x)^n
推 LPH66       :其實這題不必動到算式..只要把 A 所扔的結果給反過來
→ LPH66       :就可以直接證明有一一對應  於是機率相等
→ tokyo291    :不好意思請問把A扔的結果反過來是指@@?
推 yanchenglin :P(A)=P(B)=0.5 => 1-P(A)=0.5 利用這個特性吧@@
→ tokyo291    :那個A B在此是指兩個人投擲@@
※ 編輯: tokyo291        來自: 218.175.96.193       (04/06 00:24)
推 recorriendo :P(totoal k heads)=Σ[P(a get n-k head)P(b get k h
→ recorriendo :ead) (k summed from 0 to n), and then note that
→ recorriendo :P(a get n-k heads)=P(a get k heads)
→ recorriendo :二樓應該是這個意思 (抱歉英文錯很多)