1. Probability averaging?

I might answer this myself as I go along but anyway here goes.

Say two people roll a dice, the highest number wins you would expect
each person to win 50% of the time. But what if one person was luckier
and won 55% of the time. You could say he was 5% luckier than average.

How extend that to 3 people, you would expect each person to win 33%
of the time. How say one persoon was luckier as in the first example, and
also the same %age luckier. Would that be 5% luckier, ie 33+5=38%?

I think this is too high, I think the answer is he would be 3.3% luckier for
it
to be the same amount of 'luckierness' than in the first example.

Also I think I am into problems here with sample size possibly.

Going back to the first example another way is to say 55% is 10%
luckier than normal (5/50 times 100).Thats how I get the 3.3% value for
the second example.

So for four players it would be 2.5%, 5 players=2%, 10 players=10% etc...

You see I want to compare how lucky a person is in poker 'showdowns'
there would be no problem if it was just 2 players but sometimes there are 3
or maybe 4, the max possible (but rather unlilkely) being 10.

Problem X
------------

But lets take 10 players as an example, say one player wins 20% of the time,
which is twice as lucky as expect. Now if he got the same amount of luck
in 2 player games what would the figures be? It can't be twice because
twice 50% is 100% and that don't seem right.
-----------------

I think I am into sample size problems here too, I never did statistics
unfortunately.

Well I pretty much know I am into sample size stuff because I know that
over say a million game even winning 5% more than average would be
impossible (at least statistically anyway).
I guess I need to look at something like standard deviation (yuk) or
something like that?

OK back to problem X.
I can't even guess the right answer so I am going to make it easier and say
it is on a sample of 100 to see if that helps.

So in the 10 player games normal is 10 in 100 but he wins 20 in 100

So in a 2 player game normal is 50 in 100 but he wins ?? in 100 with the
same amount
of good luck.

What is the correct value of ??
One guess would be 60 but that seems to low, another guess is 100 but that
seems
way to high. 75 feels about right but I can't explain why.

OK I will have a crack at Problem X.
I think a better way of thinking is how often would he win 20 in 100 which
would have
a value 'x' which means it would happen once in every x number of trials so
then I
would need to work out the value which happened once in every 'x' trials for
two
players, and this would be the answer!??

Trouble is I am not sure how to do it (yet). I think I could do a simulation
though.
On my computer :O)
I would just get the maximum value which occured in 'x' trials. I better
still obtain
an average value over repeated trials??

I expect there is a (well known?) equation for this, which I might be able
to work
out if I spent few years thinking about it!!!

2. Originally Posted by esbo
I might answer this myself as I go along but anyway here goes.

Say two people roll a dice, the highest number wins you would expect
each person to win 50% of the time. But what if one person was luckier
and won 55% of the time. You could say he was 5% luckier than average.

How extend that to 3 people, you would expect each person to win 33%
of the time. How say one persoon was luckier as in the first example, and
also the same %age luckier. Would that be 5% luckier, ie 33+5=38%?

I think this is too high, I think the answer is he would be 3.3% luckier for
it
to be the same amount of 'luckierness' than in the first example.

Also I think I am into problems here with sample size possibly.

Going back to the first example another way is to say 55% is 10%
luckier than normal (5/50 times 100).Thats how I get the 3.3% value for
the second example.

So for four players it would be 2.5%, 5 players=2%, 10 players=10% etc...

You see I want to compare how lucky a person is in poker 'showdowns'
there would be no problem if it was just 2 players but sometimes there are 3
or maybe 4, the max possible (but rather unlilkely) being 10.

Problem X
------------

But lets take 10 players as an example, say one player wins 20% of the time,
which is twice as lucky as expect. Now if he got the same amount of luck
in 2 player games what would the figures be? It can't be twice because
twice 50% is 100% and that don't seem right.
-----------------

I think I am into sample size problems here too, I never did statistics
unfortunately.

Well I pretty much know I am into sample size stuff because I know that
over say a million game even winning 5% more than average would be
impossible (at least statistically anyway).
I guess I need to look at something like standard deviation (yuk) or
something like that?

OK back to problem X.
I can't even guess the right answer so I am going to make it easier and say
it is on a sample of 100 to see if that helps.

So in the 10 player games normal is 10 in 100 but he wins 20 in 100

So in a 2 player game normal is 50 in 100 but he wins ?? in 100 with the
same amount
of good luck.

What is the correct value of ??
One guess would be 60 but that seems to low, another guess is 100 but that
seems
way to high. 75 feels about right but I can't explain why.

OK I will have a crack at Problem X.
I think a better way of thinking is how often would he win 20 in 100 which
would have
a value 'x' which means it would happen once in every x number of trials so
then I
would need to work out the value which happened once in every 'x' trials for
two
players, and this would be the answer!??

Trouble is I am not sure how to do it (yet). I think I could do a simulation
though.
On my computer :O)
I would just get the maximum value which occured in 'x' trials. I better
still obtain
an average value over repeated trials??

I expect there is a (well known?) equation for this, which I might be able
to work
out if I spent few years thinking about it!!!

you already asked this. please do not double post. it is against the rules. someone will see your question and answer it as soon as they are able to

3. Originally Posted by Jhevon
you already asked this. please do not double post. it is against the rules. someone will see your question and answer it as soon as they are able to
Yes I did.
I was not sure which forum I was supposed to ask the question in because I don't know the level of difficulty of the question, if I did I know that it is likely I would also know the answer too!!
I guess I could have asked which forum to post the question in but that would
seem to involve posting the question too but then I would not know which was the appropiate forum to post that question in too so I guess I would be back to square one.

I am sure you have an easy solution to my dilemma and I look forward to hearing it to ensure I do not repeat the crime

4. Originally Posted by esbo
Yes I did.
I was not sure which forum I was supposed to ask the question in because I don't know the level of difficulty of the question, if I did I know that it is likely I would also know the answer too!!
I guess I could have asked which forum to post the question in but that would
seem to involve posting the question too but then I would not know which was the appropiate forum to post that question in too so I guess I would be back to square one.

I am sure you have an easy solution to my dilemma and I look forward to hearing it to ensure I do not repeat the crime
post in a reasonable forum. your question is about probability, so post in one of the probability forums if you're not absolutely sure. if the thread is not in the appropriate place, the moderators have the power to, and will, move it to the appropriate forum.

...or you can always post in urgent homework help, homework help, or miscellaneous if you have no clue what topic something comes under. (of course you shouldn't post in a homework forum if it's not homework, but that goes without saying)

probability is not my strong suit either, but to attempt to give you some ideas about your problem X. you said that being twice as lucky would not make sense for the player because twice 50% is 100%. you must remember, the 50% here is not luck, it is what is expected if there is no luck. the "luck percentage" would be the percentage in excess of 50%. that is what would be doubled. this percentage would not exceed to the point that the said player wins 100% of the time unless the game is rigged

but maybe that's not the sense in which you were speaking

5. Originally Posted by Jhevon
post in a reasonable forum. your question is about probability, so post in one of the probability forums if you're not absolutely sure. if the thread is not in the appropriate place, the moderators have the power to, and will, move it to the appropriate forum.

...or you can always post in urgent homework help, homework help, or miscellaneous if you have no clue what topic something comes under. (of course you shouldn't post in a homework forum if it's not homework, but that goes without saying)

probability is not my strong suit either, but to attempt to give you some ideas about your problem X. you said that being twice as lucky would not make sense for the player because twice 50% is 100%. you must remember, the 50% here is not luck, it is what is expected if there is no luck. the "luck percentage" would be the percentage in excess of 50%. that is what would be doubled. this percentage would not exceed to the point that the said player wins 100% of the time unless the game is rigged

but maybe that's not the sense in which you were speaking
Well obviously getting 50% is not luck and is to be expected, but that has got me thinking a bit...The probability of getting 50% is, I think, 100% because you would expect it to happen. So maybe I need to rethink my figures.
So if I was going down that probably erroneous line of thought I should be thinking about 200%!!
But anyway I think it is complicated.
For example I know rolliing 2 sixes on a dice is 36-1, but 2 sixes in 3 rolls is a lot harder.
I think it is..well the ways to do it are 66Q, 6Q6 and Q66, so three ways.

Each of which is.... 6-1, X 6-1 X 5/6. ( Q is not a 6 and hence chance is 5/6)
so that would be 1/3 X 6-1, X 6-1 X 5/6 = 10-1.

So the chances of, in a ten player hand of hiting 20 10-1 shots is....rather complicated. But if you only have to get the expected 10 10-1 shots I guess that is, 10,000,000,000, divided by the number of possible permutation of doing that. I originally thought that would also be 10,000,000,000 to give you the answers of 100%, but I know the perm is not 10,000,000,000 so I thought I had gone wrong, but then I realised the chances are not 100% they are something else, well at least I hope so.................

6. So possibly I could work from there but that is for a specific sample size and the sample sizes would be different........

7. Actually I think I could do it now, even if the sample size varied.
Because I would know the probability I wanted, and it would be a case of
workng backwards, either by trial and error 'guessing' or some more horrible maths