Porisma

Worked solution

Average the reciprocal score and distinguish gains from guesses

Two unbiased regular tetrahedral dice have faces 1,2,3,4. The number on the face in contact with the table is recorded. Tom throws both and notes their total X. Harry guesses the total; his guess is Y. If correct, Harry wins Y2Y^2 dollars; otherwise he loses Y/2Y/2 dollars.

(i) Obtain the distribution of X and find E(1/X)E(1/X) correct to 3 decimal places. marks

(ii) Harry guesses 6 in a particular game. Find his expected gain. marks

[2006/TJC/II/24]

CJC · Review problems · Question 10

Answer

(i)distribution above; E(1/X)=0.227E(1/X)=0.227 (3 d.p.); (ii)69/16≈4.3169/16\approx4.31 dollars.

(i) Retain the complete sum grid

With independent dice, the 16 ordered outcomes are equally likely.

Die 2 / die 11234
12345
23456
34567
45678
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(i) Full distribution

xP(X=x)P(X=x)
21/161/16
32/162/16
43/163/16
54/164/16
63/163/16
72/162/16
81/161/16
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(i) Retain all seven reciprocal terms

E(1/X)=(1/2)(1/16)+(1/3)(2/16)+(1/4)(3/16)+(1/5)(4/16)+(1/6)(3/16)+(1/7)(2/16)+(1/8)(1/16)=0.227E(1/X)=(1/2)(1/16)+(1/3)(2/16)+(1/4)(3/16)+(1/5)(4/16)+(1/6)(3/16)+(1/7)(2/16)+(1/8)(1/16)=0.227 to 3 decimal places.

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(ii) Retain the gain distribution

Let W be Harry’s gain in dollars when Y=6. A correct guess wins 36; an incorrect guess loses 3.

wP(W=w)P(W=w)
363/163/16
−313/1613/16
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(ii) Expected gain

E(W)=36(3/16)+(−3)(13/16)=69/16E(W)=36(3/16)+(-3)(13/16)=69/16. Harry’s expected gain is 4.31 dollars to 2 decimal places.

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Common mistakes

E(1/X) is not 1/E(X). A wrong guess contributes a negative gain.

Solution notes

Both asks/marks, all 16 source sum-grid cells, full seven-value distribution, all reciprocal terms, two-value gain table and exact/rounded gain retained. Independent-dice source model explicit.

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