How would I solve a problem like this?

[donkeyradish]

This problem is simple to describe but I don't know how to approach the calculation.

How many rolls of a die would I need to make so that the likelihood of rolling each of 1,2,3,4,5,6 at least once, is better than 50%?

A more tricky variant of this sort of problem would be, how many games of hold'em would you need to play to have a 99% liklihood of having at being dealt at least one of each of the 169 hand types (13 types of pair, 78 offsuit and 78 suited hands)?

[elitegimp]

bump because I think this is a good question but I don't know the answer.

I'm guessing there is no closed form answer, but for each of your two cases you could "guess and check."

[irchans]

Skalansky showed (http://groups-beta.google.com/group/rec.gambling.poker/msg/a345d095a74ce41b?dmode=source) that you need about n*log(n/log(2)) tries to have a 50% chace of gettting every one of n equally likely alternatives.

For example,

1) you need about 1326*log(1326/log(2)) ~= 10020 hands to see all of the 1326 possible holdem hands.

2) you need about 6*log(6/ln(2)) = 13 rolls to have a 50% chance of each possible roll (1,2,3,4,5, and 6) of a single die.

[gaming_mouse]

Let's calculate the chance that some number has not been rolled after n tries.

E1 = event that a 1 has not been rolled
E2 = same for 2
etc

We want P(E1 or E2 or ... E6)

Now P(E1) = (5/6)^n
Same for the other events.

So the first term in inclusion exclusion is:

6*(5/6)^n

The second term is:

(6 choose 2)*(4/6)^n

The complete expansion is:

(6 choose 1)*(5/6)^n + (6 choose 2)*(4/6)^n + (6 choose 3)*(3/6)^n + (6 choose 4)*(2/6)^n + (6 choose 5)*(1/6)^n

You can then try different values of n until you get an answer of .5. There may be a way to solve equations of this form, but if there is I don't remember it. pzhon or Bruce might.

HTH,
gm

[donkeyradish]

Thanks!

Actually reading that thread I realised I may even have asked the wrong question

I said how many rolls of the die give me a 50% probability of rolling all possible values.

What I really wanted to know, was how many times should I expect to have to roll it to achieve the same outcome.

The answer to that seems to be a much simpler calculation, 6*H(6), where H(6) means the sum of all the reciprocals of the mumbers from 1 to 6. This comes to 14.7

Still, I can't completely figure out in my mind why these two questions are different!

[elitegimp]

[ QUOTE ]
Thanks!

Actually reading that thread I realised I may even have asked the wrong question

I said how many rolls of the die give me a 50% probability of rolling all possible values.

What I really wanted to know, was how many times should I expect to have to roll it to achieve the same outcome.

The answer to that seems to be a much simpler calculation, 6*H(6), where H(6) means the sum of all the reciprocals of the mumbers from 1 to 6. This comes to 14.7

Still, I can't completely figure out in my mind why these two questions are different!

[/ QUOTE ]

the expected value is the mean, the probability = 50% is the median. The mean and the median are guaranteed to be the same in a symmetrical distribution, but otherwise no claim about their relationship can be made.