So I have a math probablity question.
Lets say there is a pile of money that I can win. I get 15 random numbers assigned to me (i get no say in the matter) and then a number out of 100 is drawn at random. If I hit on that 15% chance I win the money.
Now lets say I get three shots. Get 15 random numbers, number gets drawn out of 100 if I win I win game stops, If i lose we do it again. Get 15 random numbers, number gets drawn out of 100 if I win I win if I lose we do it again. Get 15 random numbers, number gets drawn out of 100 if I win I win if I lose I lose.
So basically I have a 15% chance of winning three times.. what is my overall chances of winning the money?
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You need to look at the “complement”. The only way to *not* win is to lose all three times. There’s an 85% chance of losing on each attempt. 1-0.85^(3) ≈ 0.39 = 39% chance of winning.
You need to look at the “complement”. The only way to *not* win is to lose all three times. There’s an 85% chance of losing on each attempt. 1-0.85^(3) ≈ 0.39 = 39% chance of winning.
Your odds of winning the first time are .15. Your odds of losing the first time and winning the second time are .85 * .15. Your odds of losing the first and second times but winning the third time are .85 * .85 * .15. Add those up and your odds of winning are about .3858, or 38.58%.
Your odds of winning the first time are .15. Your odds of losing the first time and winning the second time are .85 * .15. Your odds of losing the first and second times but winning the third time are .85 * .85 * .15. Add those up and your odds of winning are about .3858, or 38.58%.
From your phrasing, it looks like the number you drew goes back in after each draw, so each trial is independent (whether you win on one trial doesn’t affect whether you win on another).
In general, the formula for the probability that BOTH of two events happen is:
P(X and Y) = P(X) * P(Y|X)
where P(Y|X) means “the chance that Y happened if you know that X happened”.
In this case, because the events are independent, knowing something about X tells you nothing about Y. That means the probability of Y doesn’t change if you know X happened, so P(Y|X) is just P(Y). In other words, **because these events are independent**:
P(X and Y) = P(X) * P(Y)
and the same goes for all three events together:
P(X and Y and Z) = P(X) * P(Y) * P(Z)
But in this case, we want to know the chance you win on **one** of your draws. There [are formulas](https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle) we could use for this, but it’s easier if we think about the problem a different way.
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*Winning* on *one* of your draws is the same thing as *not* losing on *all* of your draws. So instead of thinking about it as P(X or Y or Z) (which has a somewhat ugly formula), we can think of it as one minus P(not X and not Y and not Z) (since the chance something doesn’t happen is one minus the chance it does happen).
And that lets us use our formula from above:
P(not X and not Y and not Z) = P(not X) * P(not Y) * P(not Z)
P(not X) is the chance you lose on the first draw, or 0.85. Same for P(not Y) and P(not Z), so:
P(not X and not Y and not Z) = 0.85 * 0.85 * 0.85
which is 0.614125, or a bit more than 61%. This is the chance you **lose** all three draws, so the chance you *don’t* lose all three draws (that is, you win one of them) is 1 – 0.614125 = 0.385875 = about 39%.
From your phrasing, it looks like the number you drew goes back in after each draw, so each trial is independent (whether you win on one trial doesn’t affect whether you win on another).
In general, the formula for the probability that BOTH of two events happen is:
P(X and Y) = P(X) * P(Y|X)
where P(Y|X) means “the chance that Y happened if you know that X happened”.
In this case, because the events are independent, knowing something about X tells you nothing about Y. That means the probability of Y doesn’t change if you know X happened, so P(Y|X) is just P(Y). In other words, **because these events are independent**:
P(X and Y) = P(X) * P(Y)
and the same goes for all three events together:
P(X and Y and Z) = P(X) * P(Y) * P(Z)
But in this case, we want to know the chance you win on **one** of your draws. There [are formulas](https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle) we could use for this, but it’s easier if we think about the problem a different way.
—-
*Winning* on *one* of your draws is the same thing as *not* losing on *all* of your draws. So instead of thinking about it as P(X or Y or Z) (which has a somewhat ugly formula), we can think of it as one minus P(not X and not Y and not Z) (since the chance something doesn’t happen is one minus the chance it does happen).
And that lets us use our formula from above:
P(not X and not Y and not Z) = P(not X) * P(not Y) * P(not Z)
P(not X) is the chance you lose on the first draw, or 0.85. Same for P(not Y) and P(not Z), so:
P(not X and not Y and not Z) = 0.85 * 0.85 * 0.85
which is 0.614125, or a bit more than 61%. This is the chance you **lose** all three draws, so the chance you *don’t* lose all three draws (that is, you win one of them) is 1 – 0.614125 = 0.385875 = about 39%.
Chance of losing in the first run is 85/100, in the second run it is 84/99, in the third run it is 83/98. So the chance of you losing is (85 * 84 * 83)/(100 * 99 * 98) = approx 0.61. The chance of winning is 1 – 0.61 = 0.39, or 39%.
Chance of losing in the first run is 85/100, in the second run it is 84/99, in the third run it is 83/98. So the chance of you losing is (85 * 84 * 83)/(100 * 99 * 98) = approx 0.61. The chance of winning is 1 – 0.61 = 0.39, or 39%.
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