ELi5 Probability odds help. Im not math genius 15% chance 3 times but if I hit on any of those chances I get the whole thing

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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?

In: 2

20 Answers

Anonymous 0 Comments

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Anonymous 0 Comments

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%.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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%.

Anonymous 0 Comments

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%.

Anonymous 0 Comments

Start from this:

You have a formula for “what is the chance of a thing happening *every* time”, and that’s to multiply the chances of each trial together.

A chance of a coin flip being heads both times in two flips is 1/4, because each flip had a 1/2 chance, and 1/4 is 1/2 of 1/2.

A chance of a coin flip being heads 3 times in 3 flips is 1/8, because that’s 1/2 of 1/2 of 1/2.

So if you have this rule, then you can invert the odds of it to figure out your chances of FAILING to have a thing happen every time.

If your odds of getting 3 heads in 3 flips is 1/8, from the above, then your odds of getting anything OTHER than 3 heads in those 3 flips is the remaining 7/8. (which is 1 – 1/8).

So you do that here with your example. If success means you just have to hit 15% once in 3 tries, that’s the inverse of saying “to fail you have to hit 85% every time for 3 tries.”

And the odds of hitting 85% three times in a row is 85% of 85% of 85%.

Which is 61.4125%. That’s your chance of NOT hitting 15% 3 times in a row.

So the inverse of that is your chance of hitting 15% at least once out of 3 tries: 100% – 61.4124% = 38.5875%.

Anonymous 0 Comments

You have two ways of calculating this

1. You can try and calculate the probability that you win the first time *or* you lose the first and win the second *or* you lose the first two and win the third.
2. You can calculate the probability that you won’t win any of them, and subtract that from the whole. (do you see why this works?)

The first approach is complicated. It requires figuring out the probability of three different scenarios and working out how the hell to combine them.

The second approach is simple: You just need to work out the probability of a single scenario: something happening three times in a row (where *something* is “not winning”)

On each draw, there’s an 85% (i.e. 0.85) probability of not winning. To work out the probability you lose all three, you just take 0.85 × 0.85 × 0.85. That’s the probability that you lose all three, so the probability that you win at least one of them is 1 – (probability of losing all three)

Anonymous 0 Comments

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Anonymous 0 Comments

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%.

Anonymous 0 Comments

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.