Question

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

Answer 1

`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`

Answer 2

`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`