## Another probability problem

### Another probability problem

I found this problem elsewhere on the internet. Suppose you have a six-sided die that's weighted such that there's a probability of k/21 for the die to land on k (for 1 <= k <= 6). On average, how many rolls will it take before you roll each number at least once?

### Re: Another probability problem

I'm doubtful, but should it be 21 ?

Number of such rolls will be determined by the "lowest probability side" which is "1". We need 21 rolls on average to get "1" once. All other sides will have greater expectation in this series.

Number of such rolls will be determined by the "lowest probability side" which is "1". We need 21 rolls on average to get "1" once. All other sides will have greater expectation in this series.

### Re: Another probability problem

It has to be greater than 21, since it's harder (less probable) to roll both a 1 and a bunch of other stuff in a given number of rolls than just to roll a 1. If you have a coin with a 1/21 chance of landing heads and a 20/21 chance of landing tails, the expected number of flips to get both a heads and a tails is equal to 21.05 rather than just 21. (IIRC the answer to the problem with the die is 26 point something.)

### Re: Another probability problem

Could you post how to do the calculations for problems like this?

### Re: Another probability problem

The calculations would look like this :

(1+21)*20/21+(1+21/20)*1/21=21.05

the first part assumes that the first throw will be a tails then you expect to have to flip 21 more times to get the heads notice that is the probability of heads inverted then multiply it by the probability that this will occur. The second part is similar only it assumes that a heads is throw first then you invert the probability of a tails and multiply it by the probability that this will occur.

Next if we look at a regular six sided dice we get the following

1+6/5+6/4+6/3+6/2+6=14.7 rolls to get one to six on a regular six sided dice

I'm not sure how to figure out the problem given though because there are 6!=720 possibilities for the order that you could get 1 to 6 and order does matter in the problem because of the varying probabilities unlike the regular six sided dice example and it has way more possibilities than the coin example.

(1+21)*20/21+(1+21/20)*1/21=21.05

the first part assumes that the first throw will be a tails then you expect to have to flip 21 more times to get the heads notice that is the probability of heads inverted then multiply it by the probability that this will occur. The second part is similar only it assumes that a heads is throw first then you invert the probability of a tails and multiply it by the probability that this will occur.

Next if we look at a regular six sided dice we get the following

1+6/5+6/4+6/3+6/2+6=14.7 rolls to get one to six on a regular six sided dice

I'm not sure how to figure out the problem given though because there are 6!=720 possibilities for the order that you could get 1 to 6 and order does matter in the problem because of the varying probabilities unlike the regular six sided dice example and it has way more possibilities than the coin example.

### Re: Another probability problem

Here's a hint: Let X be the number of the first roll at which you've rolled all six numbers, and for each k between 1 and 6, let X_k be the number of the first roll at which you've rolled the number k. A little thought shows that X = max{X_1, X_2, ..., X_6}. You can then use the maximum-minimums identity and the fact that the expected value operator is linear to find the answer.