18 replies to this topic

[bonanova]

You throw m fair six-sided dice and bet that all numbers from 1 to 6 will show at least once.
What is the smallest value of m that gives you a winning bet on average?

[TimeSpaceLightForce]

Spoiler for start

m=(50% )(6^6) / 6(6P6)

[superprismatic]

Spoiler for I get

Lucky 13

[TimeSpaceLightForce]

Spoiler for low attemp

Comb(m,6)*permut(6,6)/(6^6) = avarage
m=8
Comb(8,6) =28
Permut(6,6) =720
possible w/1-6=20160
no. of ways /comb : 6^6 =46656
win chance=43.2%

[Prime]

I can get an answer, but no simple explanation. But that should do for everyday gambling needs.

Spoiler for The smallest m is...

It's relatively easy to calculate in a spreadsheet.
For m=13, the winning chance is approximately 0.51. For m=12, it's about 0.44.
The probability to see just one and the same number after n rolls is 1/6^n-1.
The probability P(k,n) to get exactly k different numbers (1 < k <= 6) after n rolls is P(k,n-1)*k/6 + P(k-1,n-1)*(7-k)/6.

Make a 6-column table for k=1 through 6. Populate the first row with 1,0,0,0,0,0. The second row with the formulas above. Then drag the formulas down.

I can't think at the moment of a simple formula instead of all that.

Edited by Prime, 24 January 2013 - 12:27 PM.

[bonanova]

I can get an answer, but no simple explanation. But that should do for everyday gambling needs.

Spoiler for The smallest m is...

It's relatively easy to calculate in a spreadsheet.For m=13, the winning chance is approximately 0.51. For m=12, it's about 0.44.The probability to see just one and the same number after n rolls is 1/6^n-1.The probability P(k,n) to get exactly k different numbers (1 < k <= 6) after n rolls is P(k,n-1)*k/6 + P(k-1,n-1)*(7-k)/6.Make a 6-column table for k=1 through 6. Populate the first row with 1,0,0,0,0,0. The second row with the formulas above. Then drag the formulas down.I can't think at the moment of a simple formula instead of all that.

Ah, but there is one. No correct answers yet.

[superprismatic]

I can get an answer, but no simple explanation. But that should do for everyday gambling needs.

Spoiler for The smallest m is...

It's relatively easy to calculate in a spreadsheet.For m=13, the winning chance is approximately 0.51. For m=12, it's about 0.44.The probability to see just one and the same number after n rolls is 1/6^n-1.The probability P(k,n) to get exactly k different numbers (1 < k <= 6) after n rolls is P(k,n-1)*k/6 + P(k-1,n-1)*(7-k)/6.Make a 6-column table for k=1 through 6. Populate the first row with 1,0,0,0,0,0. The second row with the formulas above. Then drag the formulas down.I can't think at the moment of a simple formula instead of all that.

Ah, but there is one. No correct answers yet.

Since I got my answer using Prime's method, I'll now consult Strunk and White to see what else the OP could possibly mean.  Possibly a weighted average m?  Anyway, I must look in the basement for that copy of Strunk and White which I know is down there somewhere.

[phil1882]

Spoiler for

i also get 13 by running simulations.

[plainglazed]

Spoiler for had thought...

this was an expected value problem with the expected number of rolls needed to get all six numbers equal to the sum of the expected number of rolls needed to get each new number. The average number of rolls to roll a new number being the inverse of the probability to roll a new number. So 14.7?

[bonanova]

Yes. Simple concept that cuts thru all the computational complexity.