Any Stats experts out there ...

[Bingo]

question for you.

What are the actual odds/chances of the following happening.

- six guys asked to pick a price to the penny without knowing the choice of the other five
- for simplicity lets assume the band of choices is 1.00 (100 potential answers for each individual).

Chances of ANY two guys picking the exact same price to the penny?

[Buff]

C'mon man that is like high school math... or maybe even jr. high! Sheesh! you can't remember how to figure that out?

I can't remember either

I think it is less than 1% though.

[Bobblehead]

Since probablity always adds up to 1 then:

the first guy picks a number
The odds that the second guy picks a different number is 99/100
The odds that the third guy picks a different number is 98/100
The odds that the fourth guy picks a different number is 97/100
The odds that the fifth guy picks a different number is 96/100

So the odds that they all pick different numbers is .99x.98x.97x.96 = 0.9034

So it is 90.34% chance they will all pick different numbers, or 9.66% chance that at least 2 will pick the same number.

[oddsrule]

I'm certainly no stats expert, but if there's no reason that any possible prices are more likely than any others wouldn't it simply be 1/(100 * 100)*number of different people?

So for dude 1 it would be 1/(100 * 100)*5
dude 2 1/(100 * 100)*4

and so on

leaving a final result of 0.15% likelihood of any 2 people picking the same number.

but looking at this it doesn't make sense, and it'll bug me all day, so I'll check back later when I figure it out

[Bingo]

So I have two choices? 10% and 0.15%

Can someone break this tie?

[Bingo]

Worked it out in the office ... Bobblehead is right. 9.66% chance.

Thanks guys

[Bend it like Bourgeois]

Originally Posted by Bingo

Worked it out in the office ... Bobblehead is right. 9.66% chance.

Thanks guys

I'm not a stats guy, but that can't be right. 10% of the time sounds crazy high.

first guy picks a number

second guy has a 1 in 100 chance of getting same
third guy has a 2 in 100 chance
fourth guy a 3 in 100 chance
fifth guy a 4 in 100 chance
sixth guy a 5 in 100 chance

15 chances in 500, or 3% I think.

Edit.

hmmm. Or if you don't count the order of choosing, is it 5 chances in 100 times 6 guys?

this is why I'm not a stats expert, but if someone can explain the answer that'd be cool.

[Mike Oxlong]

10% has to be WAY too high.

If the first guy basically has a 1% chance of picking that number (1/100)
How can the second guy have a 10% chance of picking the same number?

It has to be something like 1/100 x 1/100 x 1/100 etc.....

It would be a very small percentage.

I am brutal with math though. I could be way off.....

[Shawnski]

Originally Posted by Bobblehead

Since probablity always adds up to 1 then:

the first guy picks a number
The odds that the second guy picks a different number is 99/100
The odds that the third guy picks a different number is 98/100
The odds that the fourth guy picks a different number is 97/100
The odds that the fifth guy picks a different number is 96/100

So the odds that they all pick different numbers is .99x.98x.97x.96 = 0.9034

So it is 90.34% chance they will all pick different numbers, or 9.66% chance that at least 2 will pick the same number.

I follow that logic, but you have five people... not six. Thus would it not be .99 x .98 x .97 x .96 x .95 = 0.858278 of picking unique numbers, thus 14.17 % chance of at least one set of duplicates.

To do a random test in excel, put "=ROUND(RAND()*100,0)" into six cells and refresh as many times to see how many times the same number comes up... Strangely enough, the very first time I did this, I got two 9's. And on my eighth refresh I got another pair. One in seven is about right.


Edit... and I want to confirm what you are looking for, Bingo. Do they need to pick a particular number (pick the RIGHT price??) IF that is the case, then you have to factor in that two people have a 1 in a hundred shot at it. So take the 14.17 and divide by 10,000. IF it is just whether two people pick the same number, the logic above stands.

[Buff]

Like I said in my reply, I don't remember how to figure this out properly but I am certain that it would be less than 1% chance that two fellers would pick the same number.

Bobblehead's way of figuring things out looks familiar though.

[FurnaceFace]

Ask Dr. Math:
http://mathforum.org/dr.math/

[Bend it like Bourgeois]

Maybe it's just me, but i'm amazed that smart people can have such differeing views. from less than 1% to 15% is a wide range

[Bingo]

14% right you are.

Basically the first two guys have a 1% chance of nailing the same number, but the next four improve the odds of getting at least one match in the total of six.

hence a number that's bigger than 1%.

You don't care if the first guy is ever matched, just that it gets match by two of the six.

Office pool around the old Nymex Natural Gas settle price. Six guys in for each contract each month and the winner gets lunch. This time around two guys picked the exact same price and won.

Just wondering the odds.

[Buff]

Originally Posted by Bend it like Bourgeois

Maybe it's just me, but i'm amazed that smart people can have such differeing views. from less than 1% to 15% is a wide range

Never said I was smart !

[Bobblehead]

Originally Posted by Shawnski

I follow that logic, but you have five people... not six. Thus would it not be .99 x .98 x .97 x .96 x .95 = 0.858278 of picking unique numbers, thus 14.17 % chance of at least one set of duplicates.

To do a random test in excel, put "=ROUND(RAND()*100,0)" into six cells and refresh as many times to see how many times the same number comes up... Strangely enough, the very first time I did this, I got two 9's. And on my eighth refresh I got another pair. One in seven is about right.


Edit... and I want to confirm what you are looking for, Bingo. Do they need to pick a particular number (pick the RIGHT price??) IF that is the case, then you have to factor in that two people have a 1 in a hundred shot at it. So take the 14.17 and divide by 10,000. IF it is just whether two people pick the same number, the logic above stands.

Oops - I missed the sixth person

[comrade]

Originally Posted by Mike Oxlong

10% has to be WAY too high.

If the first guy basically has a 1% chance of picking that number (1/100)
How can the second guy have a 10% chance of picking the same number?

It has to be something like 1/100 x 1/100 x 1/100 etc.....

It would be a very small percentage.

I am brutal with math though. I could be way off.....

Shawnski and Bobblehead are right. The problem with your math is that by using multiplication on the probabilities is like saying the word and. Whereas you'd really be using the word or since all that's required is that the first and second guy are the same or first and third or etc.

So you're argument would look like:
-first guy picks a number
-chance that second guy picks same number 1/100
-chance that third guy picks first two numbers 2/100
...
Which looks like 0.01+0.02+0.03+0.04+0.05=0.15 which is in approximate agreement with the other way. I think the discrepancy between the answers is due to my assumption that none of the earlier guys pick the same number which would impact the available picks for the later guys. I'm not entirely sure though.

[bigpete]

Math cannot be trusted. The only way to settle this is to carry out the experiment. Get six guys together and have them all pick a number between one and a hundred and see if any of their picks match. Repeat this about a thousand times to make sure the law of averages works out.

Then you will have your answer. Be sure to post what you find out.

[evman150]

Shawnski is right. It is better to approach this problem from the "one minus" side of things. So find the odds of uniqueness and subtract them from one and you get the odds of non uniqueness.

But you would get a different answer if you were looking for exactly two same answers, no more, no less. But it would probably be just below 14% anyway.

This problem is similar to the teacher who bets his students even money that there is at least one pair of kids with the same birthday among the 30 students. The normal thinking is that, hey 30 students, 365 days.....that's a bad bet for the teacher. But it's actually about 3:2 in favour of the teacher, or something like that. You can figure it out exactly if grinding through maths is what you like doing on Saturday afternoons.