A common dice game (sometimes called “You blew it”)consists of: Any number of pl

Question

A common dice game (sometimes called “You blew it”)consists of:

Any number of players

Each in turn rolls 5 dice

A “1” scores 100 points, a “5” 50points, others none

After each roll the player can score and keep any of the“1 or 5” and roll the remaining die or dice.

If any roll does not contain a 1 or 5, the player“busts” and the total score for that turn becomes zero.Play moves to the next player.

The player can stop or continue any time.

If all 5 dice are scored the player can continues with another 5dice roll

Scores accumulate and an arbitrary total (say 500) is thegoal.

(a) If on my first roll I get 5,3,1,5,1 what is my score ifI stop? If I continue and roll the 3 again:

(b) What is the probability of a 1 or 5?

(c) What is my expected score if I get a 1 or 5? (onenumber)

(d) What is the probability of losing all my points?

(e) What is the total mathematical expectation ofcontinuing? Given the choice of stopping or rolling the 3 again,which is the better strategy? Why?

(f)   If I hold the 1, 1 and roll the other threedice, what is my probability of busting?

(g) Of not busting?

(h) Why is this a better strategy?

(i)    What is the chance (expressed as 1 in ?)of busting on the first roll?

(j)    If I hold 1,1,1 should I stop or roll theremaining two dice again? Discuss the options.

Explanation / Answer

(a) Score is 300 (b) P(1or5) = 2/5 = 0.4 (c) E(X) = 100*1/5 + 50*1/5 = 30 (d) P(losing all points) = 3/5 = 0.6 (e) Total mathematical expectation = 400(0.2) + 300(0.2) + 0(0.6) =150      From the probability, stopping is thebetter choice. The chance to loss all the point is 0.6 and expectedscore is lower than this. (f)P(bust) = 0.63 = 0.216 (g)P(not bust) = 1-0.216 = 0.784 (h)Throwing another more time is the better choice, the choice toget anoth 1 or 5 is higher (i)P(bust in first roll) = 0.6^5 = 0.07776 (j)P(bust) = 0.62 = 0.36    P(one is '1' another is none) = 0.6*0.2*2 = 0.24    P(one is'5' another is none) = 0.6*0.2*2 = 0.24    P(one is '1' and another is '5') = 0.2 * 0.2 * 2=0.08    P(both is '1') = 0.04    P(both is '5') = 0.04 E(X) = 0 + 0.24(400) + 0.24(350) + 0.08(450) + 0.04(500) +0.04(400) = 252 Not trowing again is the better choice, since the expected value ofthrowing again is lower than 300.

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