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Home/ Questions/Q 8611733
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T04:26:53+00:00 2026-06-12T04:26:53+00:00

Ok so Im a bit lost on this. So looking at a this 0

  • 0

Ok so Im a bit lost on this. So looking at a this 0 0 0 0 and looking at each 0 individually with each one has a probability of 0.01 of turning from a 0 to a 1
Ok so there are 4 numbers each number can be a 0 or a 1

Probability is measured by determining how likely an event is to happen divided by the number of possible outcomes

The possible outcomes are:

0000
0001 0010 0100 1000
1100 1001 1010 0101 0011 0110
1110 0111 1011 1101
1111

So there is a total of 16 possible outcomes or 4^2 i.e. 4 number by the possible outcomes they can be 0 or 1

So the probability of the string of number to be to turn from 0000 to 1111 would be
0.01/16

The probability of the number to be from 0000 to 0100 would also be
0.01/16?

And the probability of all the number to stay the same would i.e 0000 to 0000
would still be 0.01/16

It kinda makes sense but I don’t know would all of them be the same probability?

Or am I doing this wrong
so each number has a 0.01 chance to change so for the string to go from 0000 to 1111 would be 0.01 *4 / 16 or 0.0025

And the chance to change from 0000 to 0100
would be 0.01 * 1 /16

And to change from 0000 to 0000 would be
0.01*0 /16

Thanks for any help with this

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  1. Editorial Team
    Editorial Team
    2026-06-12T04:26:54+00:00Added an answer on June 12, 2026 at 4:26 am

    There are two things here. One, the number of possible outcomes (16) and the probability of each outcome.

    Since the probability of a single bit being flipped to 1 is skewed towards it being 0, then the distribution of outcome probabilities is not even. If there was a 50% chance of a 1 or a 0 in each spot, then the probability of each of the 16 outcomes would be 1/16. That is not the case here.

    The approach I’d take is to group the numbers into buckets. Of the 16 outcomes, 1 has zero 1’s, 4 have one 1, 6 have two 1’s, 4 have three 1’s, and 1 has four 1’s.

    The probability of four 0’s is 99%^4. The probablity of one 1 is 1% x 99%^3. The probability of 2 is 1%^2 x 99%^2. etc. Calculate each of the probabilities , then divide by the size of the bucket. Add them up, and they should equal 100% (sanity check).

    I checked it in a spreadsheet, and the results seem good:

    Outcome Probability
    0000    0.96059601
    0001    0.00970299
    0010    0.00970299
    0100    0.00970299
    1000    0.00970299
    0011    0.00009801
    0101    0.00009801
    1001    0.00009801
    0110    0.00009801
    1010    0.00009801
    1100    0.00009801
    0111    0.00000099
    1011    0.00000099
    1101    0.00000099
    1110    0.00000099
    1111    0.00000001
    
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