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SoulMonster

Help with probability question

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The reason it's confusing is that your question doesn't really make sense.

You've said there is a 30% chance that each ball is red and a 70% chance that each ball is blue.

But - the probability is different for each ball if you're pulling them out of the bag one-by-one.
When you remove one of the balls from the pile (red or blue), the odds of picking a blue or red ball shift. It's not a fixed question.

Consider the first ball you pull, assuming you had 14 red and 33 blue balls of 47 total:

Red = 14/47 = 29.7872%
Blue = 33/47 = 70.21%

You pull a blue ball out of the bag. There are now 14 red and 32 blue of 46 total.

This changes the equation on the next ball:

Red = 14/46 = 30.4347%,
Blue = 32/46 = 69.5652%

So - the odds have will change with the selection of each ball and move closer to 50-50 as you go on.

The only way your question makes sense is if you are asking what the distribution would be if you were to empty the bag all at once. In which case, the answer is as simple as you're instinctively thinking (14).

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7 hours ago, SoulMonster said:

This is similar to die rolls. The probability of rolling a 6 (or any other number) when throwing one die, is 1/6 (because only one of the six sides have 6 pits). If you throw two dice, the probability of both being 6, is 1/36 (1/6 x 1/6). In this way I can calculate the probability of, say, 27 balls being red and the rest being blue, but I am not interested in calculating the probability of a specific outcome, I am interested in knowing what the average outcome will be, e.g. assuming the probabilities are Poisson distributed, what the maximum is. So, let's say I did this 1000 times, and counted the number of blue balls each time, what would the average number be?

If you are interested in the details:

We have an internal R&D project going where we order synthetic DNA sequences. When we order them we ask the subcontractor who does the actual synthesis to insert some randomness in the sequences. Usually you would order something like 5'-ATGGCTAGTCGC-3' and so on, which is a perfectly defined sequence with no room for deviation. But in this project we want the sequences to contain random changes (or mutations), so instead we order something like 5'-ATG11111111TCGC-3' where each 1 in the sequence means that there should be a 70 % of no change and a 30 % chance of a different nucleotide being inserted. We then get millions of these molecules and want to know what the average number of mutations per sequence is.

It's different to die rolls because you're removing a ball from the bag each time. 

If you're pulling a ball from the bag and putting it back in before pulling another one, that's more like the dice roll scenario.

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3 hours ago, highvoltage said:

It's different to die rolls because you're removing a ball from the bag each time. 

If you're pulling a ball from the bag and putting it back in before pulling another one, that's more like the dice roll scenario.

Not when they are independent? 

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36 minutes ago, SoulMonster said:

Not when they are independent? 

You mean measuring the results of numerous attempts with 47 balls in the bag each time?

It looks like there may be software that does what you're looking for (based on my limited understanding of your requirements): http://www.bioinfo.de/isb/2006060024/

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1 hour ago, highvoltage said:

You mean measuring the results of numerous attempts with 47 balls in the bag each time?

I mean, the result of a die roll is not affected by the outcome of a previous die roll. Similarly, the result of one ball (being red or blue), is not affected by the colors of the other balls. 

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5 hours ago, SoulMonster said:

I mean, the result of a die roll is not affected by the outcome of a previous die roll. Similarly, the result of one ball (being red or blue), is not affected by the colors of the other balls. 

Then the ball example is based on a flawed premise. It doesn't help provide real-world context to your problem. The chance of a ball being red or blue has to be affected by the other balls and the total balls in the bag.

Unless you put the previous ball back into the bag before you draw another out, it is not like the dice roll.
The die has 6 possible states which remain constant. You need the same constant with balls in the bag.

If you do indeed put the ball back into the bag (and there are 47 balls in total for each "ball draw"), then it's 14 and 33 balls respectively. Don't get "chance" and "randomness" confused. Remember the real world almost never lines up with calculated probability. 

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15 hours ago, SoulMonster said:

I mean, the result of a die roll is not affected by the outcome of a previous die roll. Similarly, the result of one ball (being red or blue), is not affected by the colors of the other balls. 

Correct.  As long as you put the ball back in before you pull the next ball. 

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