「大富翁骰戲」的概率分析(Probability analysis of the "Monopoly Express", or "Don't go to Jail")

「大富翁骰戲」的概率分析

(未完成)

Probability Analysis of the "Monopoly Express", or "Don't Go To Jail"

(Not completed)

—方潤 (推廣弈棋委員會)

by Fong Fu Yun (FCC)

(2006/11/05, 未完成版 / not completed)

(「大富翁骰戲」一名由筆者自譯 / The Chinese name of this game means"Monopoly Dice Game", created by the author)

規則簡介 / Basic Rules

骰子分成三種﹕入獄骰、地組骰和建屋骰。

Dice could be divided into three types: Go to Jail Dice, Groups Dice, and the House and Hotel Die.

遊戲開始時,先擲入獄骰(共三粒)分先後,擲得最少警察圖案的先行,然後順時針方向進行。

At the start, roll all three "Go to Jail Dice", the player who rolls fewest policemen icon play first, and then go clockwise.

每一玩者的回合中,先擲入獄骰和地組骰(共七粒),擲進骰盅之內。

Each turn starts with rolling all the dice except "House and Hotel Die" into the bowl.

1. 入獄骰 / Go to Jail Dice

每逢擲到警察「入獄」圖案,就要儲起,直到三粒入獄骰都儲起了,就要出局,那回合的分數全數取消。

擲到箭頭「起點」圖案,每次可以得到 $200。

Keep the dice if the Police icon ("Go to Jail") face up, if you collect all 3 dice with Police icon, your turn is over and all money get in that turn lose.

If you roll the Arrow icon ("Go"), get $200 on each roll.

2. 地組骰 / Groups Dice

擲骰後,可以自由決定保留哪粒骰,以便集齊一組。餘下的骰可以繼續重擲。但決定保留後,骰子不可重新再擲或改變骰面。

擲到問號圖案的「機會」,可以放到任何一組中,作「百搭」用。每一組只可以用一次「機會」。

If you roll the Question mark ("Chance"), you may use them to help complete any group. Only one Chance can be used per group.

儲齊一組地(火車站和公用事業不算)之後,可以加入建屋骰同擲。

After collected a complete colour group (not railways or utilities), you may add the "House and Hotel Die" to the bowl and roll with other dice.

3. 建屋骰 / House and Hotel Die

擲到左起第一個圖案,可以得到一間屋﹔

擲到第二個圖案,損失一間屋(如有的話)﹔

擲到第三個圖案,如果已經有四間屋,可以建酒店(得到 $5000)﹔

擲到第四個圖案(出獄卡),可以把已儲起的「入獄骰」放回骰盅。

Start from the left,

if you roll the first icon, you may take a house;

If you roll the second icon, return a house (if you have any);

If you roll the third icon, and you have four houses already, you get a hotel (score $5000);

If you roll the fourth icon ("Get out of Jail free"), you may return one "Go to Jail" die to the bowl (if you have any).

玩者可以一直重擲,直至三粒入獄骰都儲起為止。(出局,該局所有分數取消)

玩者可隨時決定停止重擲,計分完成回合。

Player could roll until all 3 dice with Police icon collected. (Turn is over and all money get in that turn lose)

Player could decide stop rolling at anytime, count the score and end your turn.

計分時,

每一組集齊的地組,可以得到上圖所列的總分數﹔

未集齊的組,可以得到骰面分數加起來最多一組(只可選一組)的分數﹔

每一次擲到「起點」,得 $200﹔

每一間屋,加 $1000﹔

如果有四間屋,再擲到「酒店」,加 $5000。

When scoring,

For each completed group, you may get the full score indicated on the chart above;

For incompleted groups, you may choose one group with highest value, add the value indicated on each die to your score (only one group could be counted);

For each "Go" you rolled, add $200;

For each house collected, add $1000;

If you have four houses and roll a "Hotel", add $5000.

骰面分佈 / Faces of Dice

各骰骰面分佈如下圖所示﹕(不依任何骰面順序)

Faces of different dice are described in following table: (faces not described by any sequences)

概率分析 / Probability analysis

第一擲可以集齊一組的機率﹕ P(一擲全組出現) - P(一擲三粒「入獄」的機會)

= P(一擲全組出現) * P(三粒「入獄」沒有同時出現的機會)

= P(骰面 1 出現機會)*...P(骰面 n 出現機會) * [1-(2/6)^3]

Probability for collecting one complete group in first roll:

P(complete group face up) - P(three "Go to Jail" face up)

= P(complete group face up) * P(three "Go to Jail" not appear coherently)

= P(face 1 appear)*...P(face n appear) - (2/6)^3

(三位有效數字 / corr. to 3 sig. fig.)

入獄出局的機會﹕

Probability for Go in Jail:

在該擲出局的可能情況 / Probable outcomes when Go to Jail in certain roll﹕

(數字﹕每一次擲骰擲到的「入獄」數字﹔n﹕擲不到「出獄」﹔y﹕擲到「出獄」)

(numbers: the "Go to Jail" faced up in each roll; n: "Get out of Jail free" not faced up; y: "Get out of Jail free" faced up)

每一粒骰擲到「入獄」的機會是 2/6,擲不到的機會是 4/6。

多粒骰同時擲到的「入獄」的機會是 (2/6)^n,同時擲不到的機會是 (4/6)^n。

用建屋骰擲到「出獄」的機會是 1/6,擲不到的機會是 5/6。

每一擲出局的機率,以每一種情況出現的機率加總。

例如有使用建屋骰時,第二擲出局的機率為﹕

The probability for getting Out in each roll, is the sum of probabilities for each of the outcomes.

For example, when "House and Hotel Die" is used, the probability for getting Out in second roll is:

P = (4/6)^3*(2/6)^3*5/6 i.e. P(03n)

+(4/6)^2*2/6*5/6*(2/6)^2*5/6 P(1n2n)

+(4/6)^2*2/6*1/6*(2/6)^3*5/6 P(1y3n)

+4/6*(2/6)^2*5/6*2/6*5/6 P(2n1n)

+4/6*(2/6)^2*1/6*(2/6)^2*5/6 P(2y2n)

+(2/6)^3*1/6*2/6*1/6 P(3y1n)

= 0.039971...

在每一擲中不出局的機率,以上一擲未出局的機率減去此一擲出局的機率計算。

The probability of not getting Out in each roll, is the remainder of "probability for not getting out in the roll before" subtracted by the "probability for getting Out in this roll".

在每一擲中的出局機率 / Probability for getting Out in each roll﹕

(三位有效數字 / corr. to 3 sig. fig.)

由於未有時間推算擲骰次數和出局概率的變化關係方程式,故暫且利用所得的少數資料以外推法估計多次擲骰後的出局情況。(如讀者有興趣協助解決,請聯絡本人。)

Because I've not enough time to construct equations explaining the relationship between number of rolls and the probability of Go to Jail, so extrapolation is used to prediction the probability change for more rolls. (If you are interested to solve this problem, please contact me.)

建酒店而不入獄的機率﹕(暫缺)

Probability for getting hotel and not go to Jail: (not completed yet)

(版權聲明)

(Announcement: Copyrights reserved.

English version of the part "Basic Rules" based on the Rules provided by Hasbro)