If you roll a single die, what is the expected number of rolls necessary to roll every number once?

2 Answers
May 7, 2018

#14.7 " rolls"#

Explanation:

#P["all numbers thrown"] = 1 - P["1,2,3,4,5, or 6 not thrown"]#
#P["A or B or C or D or E or F"] = P[A]+P[B]+...+P[F] -#
#P[A and B] - P[A and C] .... + P[A and B and C] + ...#
#"Here this is"#
#P_1 = 6*(5/6)^n - 15*(4/6)^n + 20*(3/6)^n - 15*(2/6)^n + 6*(1/6)^n#
#P = P_1(n) - P_1(n-1)#
#= 6*(5/6)^(n-1)(5/6 - 1) - 15*(4/6)^(n-1)(4/6-1) + ...#
#= -(5/6)^(n-1)+5*(4/6)^(n-1)-10*(3/6)^(n-1)+10*(2/6)^(n-1)-5*(1/6)^(n-1)#
#"The negative of this is our probability."#

#sum n*a^(n-1) = sum (d/{da})(a^n)#
#= (d/{da}) sum a^n = (d/{da}) (1/(1-a)) = 1 / (1-a)^2#

#=> E[n] = sum n * P["all numbers thrown after n throws"]#
#= sum n*((5/6)^(n-1) - 5*(4/6)^(n-1) + ...#
#= 1/(1-5/6)^2 - 5/(1-4/6)^2+10/(1-3/6)^2-10/(1-2/6)^2+5/(1-1/6)^2#
#= 36 - 45 + 40 - 22.5 + 7.2#
#= 15.7#
#"We have to subtract one because of the begin condition P_1(0)"#
#"gives a faulty value P=1 for n=1."#
#=> P = 15.7 - 1 = 14.7#

May 8, 2018

#6/6+6/5+6/4+6/3+6/2+6/1 = 14.7#

Explanation:

Think of it like six mini-games. For each game, we roll the die until we roll a number that hasn't been rolled yet—what we'll call a "win". Then we start the next game.

Let #X# be the number of rolls needed to roll every number at least once (i.e. win all 6 mini-games), and let #X_i# be the number of rolls needed to "win" mini-game number #i# (for #i# from 1 to 6). Then each #X_i# is a Geometric random variable with distribution #"Geo"(p_i)#.

The expected value of each Geometric random variable is #1/p_i#.

For the first game, #p_1 = 6/6# since all 6 outcomes are "new". Thus, #"E"(X_1) = 6/6 = 1#.

For the second game, 5 out of the 6 outcomes are new, so #p_2=5/6#. Thus, #"E"(X_2)=6/5 = 1.2#.

For the third game, 4 of the 6 possible rolls are new, so #p_3=4/6#, meaning #"E"(X_3) = 6/4 = 1.5#.

By this point, we can see a pattern. Since the number of "winning" rolls decreases by 1 for each new game, the probability of "winning" each game goes down from #6/6# to #5/6#, then #4/6#, etc., meaning the expected number of rolls per game goes from #6/6# to #6/5#, to #6/4#, and so on, until the last game, where we expect it to take 6 rolls to get the last number.

Thus:

#"E"(X) = "E"(X_1+X_2+X_3+X_4+X_5+X_6)#

#color(white)("E"(X)) = "E"(X_1)+"E"(X_2)+...+"E"(X_5)+"E"(X_6)#

#color(white)("E"(X)) = 6/6+6/5+6/4+6/3+6/2+6/1#

#color(white)("E"(X)) = 1+1.2+1.5+2+3+6#

#color(white)("E"(X)) = 14.7#