# Add probabilities expressed as decimals

This is going to sound like a dumb question. But what method do I use to add probabilities expressed as decimals . For fractions it is easy add the numerators and denominators then simplify

1/2 + 1/2 = 2/4

In excel if I have 0.5 + 0.5 obviously this equals 1, but when dealing with probabilities I should get back to 0.5

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Sorry, this makes no sense. What do you mean by "when dealing with probabilities I should get back to $0.5$"? – Gerry Myerson Sep 21 '12 at 12:37
What do you exactly mean? Usual fractions add up by the rule $\frac ac+\frac bc=\frac{a+b}c$. – Berci Sep 21 '12 at 12:38
Sorry if I have made myself unclear. If i wish to add together the probabilities 0.5 and 0.5, how do I do it on a calculator or excel ? – Scott Sep 21 '12 at 12:43
What do you mean by "get back to 0.5"? Also, $\frac{1}{2} + \frac{1}{2} \ne \frac{2}{4}$... – Epictetus Sep 21 '12 at 12:48
What's unclear is the sense in which adding probabilities is different from adding any other kind of number. You must be using the word "adding" to mean something differing from what everyone else means. What exactly do you mean? – Gerry Myerson Sep 21 '12 at 12:51

$\frac 12 + \frac 12 = 1$, not $\frac 24$. This is the same as $0.5+0.5=1$. Why do you think the answer should be $\frac 12$ or $0.5$?

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Not only is the operation you want not "addition" in any conventional sense, it is not well-defined at all.

If we apply it to $\frac13$ and $\frac 23$ we get $\frac{1+2}{3+3}=\frac36=\frac12$.

But if we apply it to $\frac26$ and $\frac 23$ we get $\frac{2+2}{6+3}=\frac49$.

But $\frac13$ is equal to $\frac26$. Surely even if your operation is not conventional addition, one has to require that it gives equal results when applied to equal operands.

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Reading your answer, I finally understand OP's question (I think). If something happened once in three tries yesterday, and twice in five tries today, then it has happened $(1+2)/(3+5)=$ three-eighths of the time. But as you point out, if we only know it happened 40 percent of the time today, we don't know whether that's twice in five tries, or 200 times in 500 tries, and we can't combine the two days' results without that knowledge. – Gerry Myerson Sep 23 '12 at 12:51
@Gerry: Yes, that is probably the intuitive motivation. But then, of course, the OP is wrong when he wants to simplify $\frac{a+c}{b+d}$ after adding the numbers on each floor. – Henning Makholm Sep 23 '12 at 13:18

I know what this person was getting at, though probably a bit late for them. Example - if you toss two coins at the same time and you want to know what is the chance that at least one head is tossed, how do you display this in decimal probability. If you add the decimal probability of each toss up (0.5 + 0.5) this does not work as you know it cannot be 100%.

Ans: number of coin tosses = 2, probability of a head being tossed per coin = 0.5

probability of at least one head being tossed = 1 - (0.5/2) = 0.75 (75%)

so addition does not come into it, the equation is x = 1 - (y/z) where x is the overall probability, y is the probability of a single event and z is the number of events.

so if you are trying to add a probability of 0.5 to another probability of 0.5 the equation would be: 1 - (0.5/2) = 0.75

Hope that makes sense.

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The probability ought to be $1-(1-y)^z$ - where $(1-y)$ is the probability of an event not happening, $(1-y)^z$ is the probability of it not happening $z$ times in a row, and $1$ minus that is the probability of it happening at least once. – Milo Brandt Oct 19 '14 at 16:28