Compound probability (but not quite) question...

[OMRB]

Let's I have 2 possible outcomes. For example: winning a game or losing a game.

Winning the game can happen by any of 3 different ways, for example: scoring more points, looking better, or having more fans show up.

And let's say I estimate I have a 25% chance of scoring more points, 20% chance of looking better, and 10% chance of having more fans show up.

Again, any one of these 3 things wins the game for me. How do I calculate my percentages of winning the game?

Thank you!

 

[pka]

Winning the game can happen by any of 3 different ways, for example: scoring more points, looking better, or having more fans show up.
And let's say I estimate I have a 25% chance of scoring more points, 20% chance of looking better, and 10% chance of having more fans show up.

It is already figured for you: \(\displaystyle .25+.20+.10=55\%\) is the chance of winning and \(\displaystyle 45\%\) is the chance of loosing.

 

[Dr.Peterson]

You really have 4 events, and want the probability of the event W1 or W2 or W3.

In order to find that, you'll need some additional information. Can you assume the three winning outcomes are independent? If so, then you can find the probabilities of the various "ands" (intersections), and use inclusion-exclusion to find the "or". (I'm assuming they aren't mutually exclusive.)

Or, you could find the probability that none of the three winning things happens, and subtract that from 1.

 

[OMRB]

You really have 4 events, and want the probability of the event W1 or W2 or W3.

In order to find that, you'll need some additional information. Can you assume the three winning outcomes are independent? If so, then you can find the probabilities of the various "ands" (intersections), and use inclusion-exclusion to find the "or". (I'm assuming they aren't mutually exclusive.)

Or, you could find the probability that none of the three winning things happens, and subtract that from 1.

Hmm, thanks. They are not independent. Just one could happen, 2 could happen, or all 3 could happen. In each case (1, 2, or all 3), I win. So I'm not attempting to find the probability of W1 OR W2 OR W3. I'm trying to calculate the probability of ANY combination of those 3 things happening.

pka's response doesn't make sense to me because, assuming we moved the percentages to 33%, 33%, and 34%, his answer says I have a 100% chance of winning. That can't be true, right? If we have 3 separate things, each with just a mere 33.33% of happening, we can't say there is 100% chance of one of them happening...

I guess another way to look at it is with dice. Let's say I have three different dice that I'm going to roll at the same time. To win I need to roll a 6 on any individual dice. If a 6 shows up on any one of the 3 dice, I win, doesn't matter what happens with the other 2 dice. If 6 doesn't show up on any of the three dice, I lose. Each individual dice gives me a 16.7% chance of a 1. What are my chances of winning? It seems like there must be a formula for this that isn't just 16.7 + 16.7 + 16.7.

 

[OMRB]

It is already figured for you: \(\displaystyle .25+.20+.10=55\%\) is the chance of winning and \(\displaystyle 45\%\) is the chance of loosing.

Thanks, pka. See my other response for why this doesn't seem to make sense. It's as if we had a game of dice where I get to throw 6 dice at once. To win, all that needs to happen is for a 1 to show up on any of the 6 dice. Based on your answer, I'd have a 100% chance of winning (as each individual dice has a 16.7% chance of being a 1, and we'd just add that 6 times to get 100%), but we know that can't be correct.

 

[pka]

Thanks, pka. See my other response for why this doesn't seem to make sense. It's as if we had a game of dice where I get to throw 6 dice at once. To win, all that needs to happen is for a 1 to show up on any of the 6 dice. Based on your answer, I'd have a 100% chance of winning (as each individual dice has a 16.7% chance of being a 1, and we'd just add that 6 times to get 100%), but we know that can't be correct.

It is not like a dice game. I was fully aware that the ways to win were not disjoint (may over lap). But any one of the three wins for you.
So 45% loses. This 55% wins. If you don't loose you win.

 

[OMRB]

It is not like a dice game. I was fully aware that the ways to win were not disjoint (may over lap). But any one of the three wins for you.
So 45% loses. This 55% wins. If you don't loose you win.

Interesting, thanks.

So I guess the dice game example is a bit different but I'm still interested in the probability there... If I am throwing 6 dice at the same time, and I can win my make-believe game by rolling a 1 on ANY dice, what are my odds of winning and how do you calculate that? Thanks again!

 

[Dr.Peterson]

Hmm, thanks. They are not independent. Just one could happen, 2 could happen, or all 3 could happen. In each case (1, 2, or all 3), I win. So I'm not attempting to find the probability of W1 OR W2 OR W3. I'm trying to calculate the probability of ANY combination of those 3 things happening.

It sounds like you are misunderstanding "independent". If they are independent, then any combination can occur. It's if they are mutually exclusive (which pka denies assuming, but I think he has done so) that combinations would be impossible. Independence means that whether A happens doesn't affect the probability that B happens -- implying that A and B can happen either together or separately.

You are looking for the probability of W1 or W2 or W3 -- that means any one or more of them happening, which is exactly what you say you want. ("Or" in math is inclusive, and doesn't mean "only this or that", but "this, or that, or both".)

The trouble is that you can't just assume independence. It is conceivable (if your hypothetical situation is real) that, say, you might score more points because there were many fans encouraging you, so that W1 is more likely when W3 occurs.

But if you do assume independence, then the probability is

P(W1 or W2 or W3) = 1 - P(neither W1 nor W2 nor W3) = 1 - (1-0.25)(1-0.20)(1-0.10) = 1 - (0.75)(0.80)(0.90) = 1 - 0.54 = 0.46​

This is a little less than the sum of the three probabilities, because it takes into account their overlap.

 

[pka]

So I guess the dice game example is a bit different but I'm still interested in the probability there... If I am throwing 6 dice at the same time, and I can win my make-believe game by rolling a 1 on ANY dice, what are my odds of winning and how do you calculate that?

If you are interested in rolling six dice then LOOK AT THIS.
There are from 6 to 36 possible sums and they can be gotten in \(\displaystyle 6^6=46656\) possible ways.
In that expansion the term \(\displaystyle 4221x^{20}\), tell us that there are \(\displaystyle 4221\) ways to roll a sum of \(\displaystyle 20\).
So the probability of getting a sum of \(\displaystyle 15\) on six dice is \(\displaystyle \frac{1666}{6^6}\). Can you use the expansion to see why?
To use a different number of dice, say \(\displaystyle K\), then expand \(\displaystyle (x+x^2+x^3+x^4+x^5+x^6)^K\)

 

[OMRB]

It sounds like you are misunderstanding "independent". If they are independent, then any combination can occur. It's if they are mutually exclusive (which pka denies assuming, but I think he has done so) that combinations would be impossible. Independence means that whether A happens doesn't affect the probability that B happens -- implying that A and B can happen either together or separately.

You are looking for the probability of W1 or W2 or W3 -- that means any one or more of them happening, which is exactly what you say you want. ("Or" in math is inclusive, and doesn't mean "only this or that", but "this, or that, or both".)

The trouble is that you can't just assume independence. It is conceivable (if your hypothetical situation is real) that, say, you might score more points because there were many fans encouraging you, so that W1 is more likely when W3 occurs.

But if you do assume independence, then the probability is

P(W1 or W2 or W3) = 1 - P(neither W1 nor W2 nor W3) = 1 - (1-0.25)(1-0.20)(1-0.10) = 1 - (0.75)(0.80)(0.90) = 1 - 0.54 = 0.46​

This is a little less than the sum of the three probabilities, because it takes into account their overlap.

Thanks! This is exactly what I was looking for. My hunch was that the actual probability had to be a bit less than the sum. Is there a term for this calculation/formula?

Your point about not being able to assume independence is also interesting, and makes sense for my example. I probably could have come up with a better example where there was true independence. For example, the probability of a jury giving a not guilty verdict when their 2 options are a) the suspect didn't do it (let's estimate 20% chance they'll say this) and b) the suspect did do it but was insane (let's estimate 80% chance they'll say this). If I'm doing the math correctly, the chances of a not guilty verdict are 84% (1 - (.80)(.20)) Right?

 

[Dr.Peterson]

The important point is that you can't assume independence, but must determine it based on the specific events. So an arbitrary example can totally misstate the problem. In the typical case, you would need far more information, something along the lines of a Venn diagram showing the probability of every possible combination.

The basic term for what I did is to use complements. This is a standard technique for "at least one" type problems.

In your new example, we definitely do not have independence -- your possible verdicts are in fact mutually exclusive, meaning that they can't both be chosen (by the same person). So there, the probability of a not-guilty verdict is .80 + .20 = 1!

It is common for students initially to confuse these two concepts. But in fact independence and mutual exclusivity are (almost) mutually exclusive -- the only way events can be both independent and mutually exclusive is if they are both impossible (probability zero).