Probabilities involving dynamic sample space

[oriz123]

Sorry for the possible double post. I really need help with this...anyway let's assume we have chances of winning "something" anything...can be the lotto or whatever. We have A, B, C, D, E and each have a different chance of winning. We will also give them each a value, and the chance of winning improves linearly depending on that value. A has the value of 1,000...B has the value of 3,000...C has the value of 10,000...D has the value of 30,000 and E has the value of 100,000. So in the simplest form with just those 5, for example A would win 0.691% of the time, B would win 2.07% of the time, C 6.9% of the time, D 20.7% of the time and E 69.1% of the time(roughly...I rounded them so might not be 100% exact).

But here's where things get more complicated: what if, in each group, we had more than one? So, say we have 300 of A, 200 of B, 100 of C, 100 of D and 90 of E. Toegther we have 790 now and they are variable, they can change...in an hour it can be 250 of A, 200 of B, 110 of C, 100 of D and 85 of E for example. How do we, at each point in time and in real time, figure out the chance of each of the 790 to win?

 

[Jameson]

Re: I originally posted this in "basic" because didn't realize this thread exists

Hi oriz123,

Is this a hypothetical question or related to a real life problem? I'm just curious if you can give some more details.

In your example you say that the chance of winning linearly scales with the "value" of it. Do you mean that you have 1,000 copies of A, like 1,000 lotto tickets with the numbers of A? Or do you mean that the probability of A winning is a function of this "number"? If it's the second case then the probability of winning A becomes something like:

\(\displaystyle \frac{A}{A+B+C+D+E}\)

 

[oriz123]

Re: I originally posted this in "basic" because didn't realize this thread exists

Appreciate your quick response, let me try to explain this a little better. We're not talking about the chance of "winning A", but rather the chances of "A winning"(or rather, the chances of "little a's" that comprise of A winning. It is sort of real life but I had to change the example since in "rl" we're talking block rewards on a blockchain and didn't really want to start getting into coding. So the most similar example would be the lottery, let's try again:

The winner is chosen randomly, based on their value. Basically since it's linear, 100,000 will win 100 times more often than 1,000. Hence why in my previous example A got 0.69% of the time whereas E got it 69% of the time. Together with the inbetweens of B, C and D we reached 100%(B's value for example was 3,000 so it had a x3 chance of winning compared to A, hence the percentage was roughly 2.1%). The problem is when you have more than 1 each of A, B, C, D and E. Then they each have a smaller chance at winning, but the ratio should remain the same between them.

Ok let's pretend for the sake of the example that A bought 1,000 tickets, B bought 3,000, C bought 10,000, D bought 30,000 and E bought 100,000, and let's pretend that that improves their chances of winning linearly(we can also use dollar value instead of amount of tickets, but same principle). Now let's also assume that at each point in time, the amount of tickets they have(or the amount of money they put into them/value) can change. Say they are allowed to give some back and get a refund, or they can buy more. It constantly changes. How do you figure out the ratio of each​ of the 790 participants and their chances of winning, while still keeping it linear and retaining the same ratio between each "level"(A, B, C, D and E)?

 

[oriz123]

Re: I originally posted this in "basic" because didn't realize this thread exists

If it helps, just for the sake of the example we can change it from linear to something else. Say we have group A, group B, group C, group D and group E. ANYONE in group B has 5 times the chance of winning than anyone in group A. ANYONE in group C has 10 times the chance of anyone in group A. Anyone in D has 15 times the chance and in E 20 times the chance. The number of people in each group varies all the time(as well as the total), and we have to keep the same ratio at all times. How do we track/ensure that happening, even when the number in each group and the total number keeps changing?

 

[oriz123]

Re: I originally posted this in "basic" because didn't realize this thread exists

Hmmm..no one?

 

[Jameson]

Re: I originally posted this in "basic" because didn't realize this thread exists

I'm just not understanding your question clearly so it's hard for me to try to comment. I know you've spent time explaining but let's start from the beginning.

This quote in particular confuses me:

How do you figure out the ratio of each​ of the 790 participants and their chances of winning, while still keeping it linear and retaining the same ratio between each "level"(A, B, C, D and E)?

You ask about figuring out the ratio but also say the ratio must stay the same. These seem to contradict each other.

Let's look at this part:

Ok let's pretend for the sake of the example that A bought 1,000 tickets, B bought 3,000, C bought 10,000, D bought 30,000 and E bought 100,000, and let's pretend that that improves their chances of winning linearly(we can also use dollar value instead of amount of tickets, but same principle). Now let's also assume that at each point in time, the amount of tickets they have(or the amount of money they put into them/value) can change. Say they are allowed to give some back and get a refund, or they can buy more.

So here you seem to be saying that the chance of winning is proportional to the number of "tickets" bought. A couple of questions:

1. Each "ticket" is equivalent correct?
2. Is this list of groups complete? By that I mean does one of the groups have to win or are these simply 5 groups that could possibly win?

I suggest we start with understanding the initial probabilities then move onto any changes. If you can clarify some more, I'll try to help but the above points aren't clear to me.

Also I think writing these as ratios or probabilities is fine, just slightly different. For example if A:B is 1:2, that's a ratio but we could equivalently write that A = 1/3 and B=2/3 in terms of probability.

 

[oriz123]

Re: I originally posted this in "basic" because didn't realize this thread exists

Alright thanks for your reply let's try one more time: Just for the sake of this example forget what I said before lol. Say you have let's call it a competition not a lottery, the amount you win depends on how many "tickets" you have. The chance improves linearly. So, say me and 320 other people have 1,000 of those. Another group with 300 people have 3,000. Another group with 200 people have 10,000. Another group with 100 people have 30,000. And the last group with 50 people have 100,000. I'm trying to figure out a way to calculate the chance of each of those 971 people winning something.

If we for example used the example I have given before, say instead of having 971 people total we only had 5 people, 1 in each group. Then it would look something like this roughly(I rounded it up)

Those holding 1,000 will win 0.691% of the time, 3,000 would win 2.07% of the time, 10,000 6.9% of the time, 30,000 20.7% of the time and 100,000 69.1% of the time. If you add it all up you get, again, roughly 100%. But where it gets complicated is when we have different numbers of people in each group, and we need to figure out the chance of each individual, while keeping the above relationship similar.

 

[Jameson]

Re: I originally posted this in "basic" because didn't realize this thread exists

Alright thanks for your reply let's try one more time: Just for the sake of this example forget what I said before lol. Say you have let's call it a competition not a lottery, the amount you win depends on how many "tickets" you have. The chance improves linearly. So, say me and 320 other people have 1,000 of those. Another group with 300 people have 3,000. Another group with 200 people have 10,000. Another group with 100 people have 30,000. And the last group with 50 people have 100,000. I'm trying to figure out a way to calculate the chance of each of those 971 people winning something.

If we for example used the example I have given before, say instead of having 971 people total we only had 5 people, 1 in each group. Then it would look something like this roughly(I rounded it up)

Those holding 1,000 will win 0.691% of the time, 3,000 would win 2.07% of the time, 10,000 6.9% of the time, 30,000 20.7% of the time and 100,000 69.1% of the time. If you add it all up you get, again, roughly 100%. But where it gets complicated is when we have different numbers of people in each group, and we need to figure out the chance of each individual, while keeping the above relationship similar.

Ok I think I understand slightly better.

Let's say there are 3 groups to make this simpler and that if you are in these groups you have the following number of tickets no matter what.

Group A: 1,000 tickets
Group B: 10,000 tickets
Group C: 25,000 tickets

Now we can write the total number of tickets as follows:
\(\displaystyle \text{Total #}=1,000 \times (\text{# people in Group A})+10,000 \times (\text{# people in Group B})+25,000 \times (\text{# people in Group C})\)

That's the total number of tickets held by all people across all groups. Now for each group you simply take that number of people times tickets and divide by the total above. That will tell you that chance of the group winning. To refine it down to a person within the group, divide again by the number of people.

For example: \(\displaystyle \frac{1,000 \times (\text{# people in Group A})}{\text{Total #}}\)

To generalize this for coding you need to let the ticket number for each group be variable as well as the number of people in that group, but the logic remains the same. Is this what you are looking for?

 

[oriz123]

Perfect thank you! now what do you do if it's not exactly linear but close? say the relationship is like this: Say group B has 3 times the tickets, but they get 3.15 times the chances instead of 3. Say group C has 10 times, but they get 11 times the chance. What do I change to adjust to that as well?

 

[oriz123]

Here's the issue I'm running into now: if trying to add an extra percent(bonus), like an extra 10% for group C basically meaning 11 times right? But if I just add that to it then the total comes out more than 100%. How do we get around that?

 

[HOI]

You are talking about a "weighted average".

"Say group B has 3 times the tickets, but they get 3.15 times the chances instead of 3."
Then use 3.5 in place of 3.

"Say group C has 10 times, but they get 11 times the chance."
Then use 11 in place of 10.

 

[oriz123]

Thanks for your help guys!