概率問題-急

Question

小明與小英玩擲硬幣遊戲, 遊戲是最先擲到"公"那方勝, 擲硬幣次序為小明先擲, 請問小明能勝出的概率是? 請用算式輔助解答.

Answer 1

0.5 + 0.5* 0.5* 0.5+ 0.5*0.5*0.5*0.5*0.5 + ............
= 0.5^1+ 0.5^3 + 0.5^5 + 0.5^7 +..............
= 0.5 /(1- 0.5^2)
= 2/3

2006-11-19 04:05:35 補充：
第一個0.5 係小明直接擲公的機會0.5 * 0.5 * 0.5 = 小明第一次中字, 小英都要擲字, 小明先至擲到公0.5 * 0.5 * 0.5 = 小明第一次中字, 小英都要擲字, 小明第２次再擲中字，小英第２次都係再擲中字, 小明第３次先擲中公如此類推再用 GP sum to infinty 加埋晒佢

2006-11-19 05:36:46 補充：
更正:第一個0.5 係小明直接擲公的機會0.5 * 0.5 * 0.5 = 小明第一次中字, 小英都要擲字, 小明先至擲到公0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 小明第一次中字, 小英都要擲字, 小明第２次再擲中字，小英第２次都係再擲中字, 小明第３次先擲中公如此類推再用 GP sum to infinty 加埋晒佢

Answer 2

H=公
T=字

They will make two throws, one each by Ming and Ying.

The outcomes are:
HT 小明
HH 小明
TH 小英
TT undecided

The first two cases represent 小明's win (0.5), and the third for 小英 (0.25).
The fourth case is a tie, so the same repeats.
小明's probability of winning is thus:

0.5+0.25(0.5+0.25(0.5+0.25(...))))
=0.5(1+1/4+1/16+1/64+....)
=0.5(4/3)
=2/3

Similarly, the probability for 小英's success is:
0.25+0.25(0.25+0.25(0.25+....)))
=0.25(1+1/4+1+1/16+...)
=0.25(4/3)
=1/3

Check: 2/3+1/3=1 OK

Note: (1+1/n+1/n^2+1/n^3,,,)=n/(n-1)

It is a very interesting problem!

Answer 3

你又吾講佢擲幾次 =-=
因為得兩面
點擲落去都係50%=1/2