Porisma

Worked solution

Count first-winning throws and distinguish payout from net gain

A player wins by balancing a beam with three alternating items: ball–cube–ball or cube–ball–cube, as shown. An unbiased die determines each item: scores 1 or 2 give a ball; otherwise a cube. The player has at most 5 throws. An item of the same type as the last accepted item is rejected.

Let X be the number of throws required to win. Show that P(X=4)=2/9P(X=4)=2/9. marks

Find the probability of winning. marks

The entry fee is a dollars. A win on throw 3 pays back 2 a dollars, on throw 4 pays back a/2 dollars, and on throw 5 pays back a/3 dollars. Find a if the owner expects a profit of 1.20 dollars per game. marks

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The initial beam is unbalanced. A winning beam has the prize on the left and either ball–cube–ball or cube–ball–cube stacked on the right. Balls are circles; cubes are squares.
The initial beam is unbalanced. A winning beam has the prize on the left and either ball–cube–ball or cube–ball–cube stacked on the right. Balls are circles; cubes are squares.
The initial beam is unbalanced. A winning beam has the prize on the left and either ball–cube–ball or cube–ball–cube stacked on the right. Balls are circles; cubes are squares.

[2006/NJC/II/29(OR)]

CJC · Review problems · Question 12

Answer

P(X=4)=2/9P(X=4)=2/9; P(win)=50/81P(\text{win})=50/81; a≈3.10a\approx3.10 dollars (2 d.p.).

Read the source stacks and rejection rule

Let B denote a ball, with probability 2/6, and C a cube, with probability 4/6. Independent die throws are used. BCB or CBC are the three accepted items; repetitions of the most recently accepted type are rejected but still use a throw. The source phrase “if it is the last drawn accepted item” means the same type as that item. On losing by the five-throw limit, X has no finite winning value; probabilities below concern wins on throws 3,4,5.

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Retain all four first-win-on-four paths and products

P(X=4)=P(BCCB)+P(BBCB)+P(CCBC)+P(CBBC)P(X=4)=P(BCCB)+P(BBCB)+P(CCBC)+P(CBBC).

P(X=4)=(2/6)(4/6)(4/6)(2/6)+(2/6)(2/6)(4/6)(2/6)+(4/6)(4/6)(2/6)(4/6)+(4/6)(2/6)(2/6)(4/6)=2/9.\begin{aligned}&P(X=4)\\ & =(2/6)(4/6)(4/6)(2/6)\\ &{}+(2/6)(2/6)(4/6)(2/6)\\ &{}+(4/6)(4/6)(2/6)(4/6)\\ &{}+(4/6)(2/6)(2/6)(4/6)\\ & =2/9.\end{aligned}

Each path has exactly one rejected repeated item before its third accepted item.

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Retain both three-throw paths

P(X=3)=P(BCB)+P(CBC)=(2/6)(4/6)(2/6)+(4/6)(2/6)(4/6)=2/9P(X=3)=P(BCB)+P(CBC)=(2/6)(4/6)(2/6)+(4/6)(2/6)(4/6)=2/9.

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Retain all six five-throw paths and products

P(X=5)=P(BCCCB)+P(BBBCB)+P(BBCCB)+P(CBBBC)+P(CCCBC)+P(CCBBC)P(X=5)=P(BCCCB)+P(BBBCB)+P(BBCCB)+P(CBBBC)+P(CCCBC)+P(CCBBC).

Writing p=2/6 and q=4/6, their six source products in the same order are pq3p+p3qp+p2q2p+qp3q+q3pq+q2p2q=14/81pq^3p+p^3qp+p^2q^2p+qp^3q+q^3pq+q^2p^2q=14/81. These represent the source’s explicit factors 2/6 and 4/6, grouped into powers for repeated items.

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Winning probability

P(win)=P(X=3)+P(X=4)+P(X=5)=2/9+2/9+14/81=50/81P(\text{win})=P(X=3)+P(X=4)+P(X=5)=2/9+2/9+14/81=50/81.

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Retain the full player net-gain table

Let Y denote the player’s net gain after subtracting the entry fee.

yEventP(Y=y)P(Y=y)
awin on throw 32/92/9
−a/2-a/2win on throw 42/92/9
−2a/3-2a/3win on throw 514/8114/81
−ano win1−2/9−2/9−14/81=31/811-2/9-2/9-14/81=31/81
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Retain the owner-profit equation and rounded fee

An expected owner profit of 1.20 dollars gives E(Y)=−1.20E(Y)=-1.20. Thus 2a/9−a/9−28a/243−31a/81=−1.202a/9-a/9-28a/243-31a/81=-1.20. Solving gives a≈3.10a\approx3.10 dollars to 2 decimal places, as in the source. The unrounded solution is 729/235≈3.10213729/235\approx3.10213; charging the rounded cent amount makes the expected profit approximately 1.20 dollars.

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Common mistakes

The source prize amounts are payouts, so subtract the entry fee before constructing net gains. Stop counting a path as soon as it wins.

Solution notes

All asks/marks, initial unbalanced beam and both winning-stack panels reconstructed accessibly, complete four/ two/ six winning path lists and products, winning sum, full net-gain table and fee equation retained. Source rejection wording made explicit; losing outcome does not have a finite winning X. Exact fee supplements the source rounded answer.

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