Probabilistic Analysis of Drawing Cards and Dice Rolls for Success Conditions

rmiller70015
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I am looking for a way to think about this type of problem and this is not a coursework problem

1. Homework Statement

How many cards from a 52 card deck of ordinary playing cards would you have to draw to have either a flush or 4 of a kind

If you have a 30 6 sided dice, how many dice rolls would it take to guarantee that there are either 6 different values or 4 of the same dice

Homework Equations

The Attempt at a Solution

 
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The first one looks like it can be solved by seeing how many 3 of a kinds you can make (without making 4 of a kind), and also how many cards you can put out with 4 in the same suit without having 5 in the same suit... It looks like you can draw 16 without getting a flush and card #17 must make a (5 card) flush. That should be quite easy to do and the 4 of a kind will not be a factor. Scratch the first part of the first sentence=it appears irrelevant. ## \\ ## For the second one, you can get 5 faces 3 times. On roll #16, you either get side 6 or 4 of the same face.
 
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1st define what you mean by "guarantee".
You want a probability of 1?

You also want to be more careful about how you are specifying the conditions for a success.
You only want to know a way of thinking about the problem ... how are you currently thinking about the problem?
You want to divide the success condition into parts, and work the probability for each part separately.
Do you understand how to combine probabilities?

It is usually easier to calculate the probability that a condition does not occur.
ie. to get all six numbers appearing on N>6 dice is the same is there not being at a number absent.
So what is the probability of there being at least one number completely absent?
Is it possible for there to be 2 numbers completely absent?

With the cards you can get to p=1, but with the dice you won't since the rolls are independent.

... take an experimental approach - play with the setup to see what sort of conditions are possible. get a feel for how the odds work out, and try to use the maths to describe what you learn.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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