Like. I know they are different and that one is less likely. But could someone explain this in a way that I can explain it to my partner? I know it is true but cannot remember anything about why and how to explain my point.
In: 6
Because you don’t add that sort of probability together. The total probabilities of all the possible out comes must equal 100%.
If you added 7% ten times, then you would also be implying that the total probabilities
of all possible outcomes need to sum to 1000%
Also, it might be better to ask what way should you add them, rather than picking a specific way and asking why not.
For example, if you add 1/2 + 1/2, why do you get 2/2 (add just the numerator) instead of 1/4 (add just the denominator), or 2/4 (add numerator and denominator)
I’ll focus just on the ten 7% chances and then compare to a single 70% chance.
Having something happening after ten times is the opposite of avoiding something happening ten times.
If you have one go, the chance of something happening is 7%, and the chance of avoiding it is 93%. The chance of avoiding it with two goes, means avoiding it the first AND second time, which is 93% x 93%. So this is 86.49% chance of avoiding it after two goes. This means after two goes, the chance of something happening is 13.51%. Crucially, the chance of something happening with two attempts has gone up because you could have either have the something happen on the first *or* the second go.
With more and more attempts, the odds of avoiding the thing happening goes down. Therefore, the odds of it happening goes up. After ten attempts, your odds of something happening is about 52%. So you have a 52% chance, rather than a 70% chance from the single go.
It turns out that you would need 16-17 attempts for a 7% chance to be equal to the single 70% chance.
You can carry this logic on. If you had infinitely many goes at something happening, with each attempt having a probability of 7%, you would expect the odds of the something happening to be 100%. (Likewise, if you had no attempts, your chance of the something happening would be 0%.)
imagine a lottery where u have 10 tickets each with 7% chance of winning
what are the chances of u not winning with any of these tickets?
it would be 0.93*0.93*0.93*0.93*0.93*0.93*0.93*0.93*0.93*0.93=48%
thus the chance of u winning atleast one of them is 52%. This is much lower than 70% but there is also a possibility of u winning this lottery multiple times.
Let’s say you have two coins, and you want to know the chances of getting a “heads” somewhere, if you flip both of them.
Well, that’s Two 50% chances, right? But you know you don’t have a 100% chance of getting a heads – you might just flip tails both times.
The Probability of a 70% chance event occuring is 70%. Easy peasy
The probability of a 7% chance occuring at least once in 10 try’s is 100% – [The chance that it doesn’t]
The chance you try the 7% event 10 times and never “hit” is 93%^10=48% ; which means you have a 52% chance of hitting
So in other words: A Single 70% chance is 70%, but Ten 7% chances is only 52% which is very different.
Take an extreme example of that question. Whats the difference between a 100% chance and two 50% chance? Would you rather have a 100% chance of getting a million dollars, or two 50% chances of a million dollars? The first case you get 1million dollars no matter what. The second case You have 25% chance of no money, 50% chance of 1 million and 25% chance of 2 million.
It depends on the situation. If you’re talking about OR probabilities (i.e. P(A or B)) and the events are mutually exclusive, then then we’d just add the individual probabilities and get 0.07+0.07… 0.07=10*0.07=0.7. But if the events aren’t mutually exclusive (i.e. more than one of them can happen) we have to subtract the compound probability. That is, the general rule for OR probabilities is P(A or B) = P(A)+P(B)-P(A and B). (If you wanna know why that’s the formula, Google “venn diagrams for or probability”; I’d post some diagrams here, but I’m on mobile).
On the other hand, if we’re talking about AND probabilities, they don’t combine by addition. If the events are independent (i.e. knowing that one event occurred or didn’t doesn’t effect the probability of the other events), then P(A and B) = P(A)•P(B). More generally, P(A and B)=P(A)•P(B|A)=P(B)P(A|B).
Because probabilities work with multiplications, not sums.
If ten 7% chances were the same as a 70% chance, then 2 coin flips (two 50% chances) would be 100% heads AND 100% tails which is obviously not the case.
I would consider if you had 20 times 7% chances.
That adds up to 140%
So that might show it doesn’t work that way.
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