Porisma

Worked solution

Separate disjoint prizes and subtract the entry fee once

A biased tetrahedral die is thrown onto a table; X is the number on the face in contact with the table. Its distribution is

xP(X=x)P(X=x)
−11/81/8
01/21/2
21/81/8
31/41/4

(i) X1X_1 and X2X_2 are independent observations of X, and Y=X1+X2Y=X_1+X_2. Show that P(Y=2)=3/16P(Y=2)=3/16.

(ii) A player pays 2 dollars to throw two such dice. The player receives 16 dollars if the maximum score is −1, or 3 dollars if the sum is prime, and receives nothing otherwise. Find the expected gain.

[2017/HCI/II/6]

CJC · Discussion · Question 8

Answer

(i) 3/163/16; (ii) expected gain −0.25 dollars, an expected loss of 0.25 dollars.

(i) Retain both kinds of ordered pair

The sum 2 occurs with (2,0),(0,2),(3,−1),(−1,3). Independence gives P(Y=2)=2P(X1=2,X2=0)+2P(X1=3,X2=−1)=2(1/8)(1/2)+2(1/8)(1/4)=3/16P(Y=2)=2P(X_1=2,X_2=0)+2P(X_1=3,X_2=-1)=2(1/8)(1/2)+2(1/8)(1/4)=3/16.

How did this part go?

(ii) Maximum −1 requires two scores of −1

The source uses the same independent-dice model in the game. Since −1 is the smallest possible score, P(max⁡(X1,X2)=−1)=P(X1=−1,X2=−1)=(1/8)2=1/64P(\max(X_1,X_2)=-1)=P(X_1=-1,X_2=-1)=(1/8)^2=1/64.

How did this part go?

(ii) Retain every possible prime-sum calculation

The only possible prime sums are 2,3,5. From part (i), P(Y=2)=3/16P(Y=2)=3/16.

P(Y=3)=2P(X1=0)P(X2=3)=2(1/2)(1/4)=1/4P(Y=3)=2P(X_1=0)P(X_2=3)=2(1/2)(1/4)=1/4.

P(Y=5)=2P(X1=3)P(X2=2)=2(1/8)(1/4)=1/16P(Y=5)=2P(X_1=3)P(X_2=2)=2(1/8)(1/4)=1/16.

The maximum −1 event has sum −2, so it is disjoint from the prime-sum prize.

How did this part go?

(ii) Expected payout minus the fixed cost

The expected gain in dollars is 16(1/64)+3(3/16+1/4+1/16)−2=−0.2516(1/64)+3(3/16+1/4+1/16)-2=-0.25. Thus the player has an expected loss of 0.25 dollars per game.

How did this part go?

Common mistakes

The prizes are payouts. Subtract the two-dollar entry fee once to obtain gain;1 and negative integers are not prime.

Solution notes

Full printed probability table, both asks, all ordered-pair factors, maximum event, all three prime-sum probabilities and full expected-gain calculation retained. Independence carried from part (i) into the source game model explicitly; no marks printed.

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