#1
|
|||
|
|||
How would I solve a problem like this?
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)? |
#2
|
|||
|
|||
Re: How would I solve a problem like this?
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." |
#3
|
|||
|
|||
Re: How would I solve a problem like this?
Skalansky showed 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. |
#4
|
|||
|
|||
Re: How would I solve a problem like this?
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 |
#5
|
|||
|
|||
Re: How would I solve a problem like this?
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! |
#6
|
|||
|
|||
Re: How would I solve a problem like this?
[ 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. |
|
|