Problem with combinations: Find fraction of total w/ more A’s than B’s

Peter Duncan

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Joined
Jan 11, 2018
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Problem with combinations: Find fraction of total w/ more A’s than B’s

Hi,

I would like to know which formula to use for the following combinations, please.

A or B can be chosen x times to generate a series of length x. A and B can be chosen as often as you like. Order does not matter. The question for each series is: For what fraction of the total are there more A’s than B’s?
Examples: If x=2, possibilities are A-A, A-B, B-A, B-B so how often are there more A’s than B’s? That is 1 out of 4=0.25.
If x=4, possibilities are A-A-A-A, A-A-A-B, A-A-B-A, A-A-B-B, A-B-A-A, A-B-A-B, A-B-B-A, A-B-B-B, B-A-A-A, B-A-A-B, B-A-B-A, B-A-B-B, B-B-A-A, B-B-A-B, B-B-B-A, B-B-B-B so how often are there more A’s than B’s? That is 5 out of 16=0.3125.
I can do these by hand, but it becomes more difficult when x=62.
Note that I am not giving any examples where x is an odd number, because that answer would always be 0.5.
A formula would be sooooo helpful!
Thank you, Peter
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,627
Hi,

I would like to know which formula to use for the following combinations, please.

A or B can be chosen x times to generate a series of length x. A and B can be chosen as often as you like. Order does not matter. The question for each series is: For what fraction of the total are there more A’s than B’s?
Examples: If x=2, possibilities are A-A, A-B, B-A, B-B so how often are there more A’s than B’s? That is 1 out of 4=0.25.
If x=4, possibilities are A-A-A-A, A-A-A-B, A-A-B-A, A-A-B-B, A-B-A-A, A-B-A-B, A-B-B-A, A-B-B-B, B-A-A-A, B-A-A-B, B-A-B-A, B-A-B-B, B-B-A-A, B-B-A-B, B-B-B-A, B-B-B-B so how often are there more A’s than B’s? That is 5 out of 16=0.3125.
I can do these by hand, but it becomes more difficult when x=62.
Note that I am not giving any examples where x is an odd number, because that answer would always be 0.5.
A formula would be sooooo helpful!
Thank you, Peter
I think in effect you are asking for the probability that, if you toss a coin x times, there will be more heads than tails.

I would find the probability that there are exactly the same number of heads and tails (A's and B's), subtract that from 1 to find the probability that there are more of one than the other, and divide by 2 (since for every case with more A's, there is a case with more B's).

Does that make sense?

In terms of combinations, in how many ways can you have exactly x/2 of each? Or, if it helps you think more clearly, if x = 2n, in how many ways can there be exactly n of each?
 
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