a Probability Teaser

[david185000]

Ok, my friends and I have puzzled over this for a bit and failed to come to a satisfactory answer.

There's a one hundred sided die. The chances of it landing on any number from 1 to 100 is 100 to 1.

The chances of getting a 97 are 1/100.

So if the die is rolled 100 times, what are the chances that a 97 will have come up? It cant be 100 x 1/100 because that would be 1 and therefore a certainty, which it cant be.

Any ideas?

[sciguyryan]

Ok, my friends and I have puzzled over this for a bit and failed to come to a satisfactory answer.

There's a one hundred sided die. The chances of it landing on any number from 1 to 100 is 100 to 1.

The chances of getting a 97 are 1/100.

So if the die is rolled 100 times, what are the chances that a 97 will have come up? It cant be 100 x 1/100 because that would be 1 and therefore a certainty, which it cant be.

Any ideas?

Probability = No. Of Wanted Events / No. Of Total Events

Probability = (100 * 1) / (100 * 100)

Probability = 100 / 10,000

Proability = 0.01 = 1%

Or, atleast I think it is...

Cheers,

RyanJ

[david185000]

Nah man, that wouldnt make sense. If you throw the die 1 hundred times, there isnt a 1% chance of getting a 97.

Those odds are just too small for something like this.

[The_Duck]

Statistically I think it is 1.

If you roll a dice six times you have a 1/6 chance of getting any number. Thus the chance of getting say 5 is 1 (if you roll six times)

It is the same for a 100 sided dice! If you roll it 100 times statistically you will get the number you want !

[opus]

Look at it from the other way, how much is the probability the miss the result in each throw. For the first it is 0.99, to miss it in the fist two throws 0.99*0.99 and so forth. for all the throws it 0.99^100 which is 0.36603234....
Subtrct that from 1 and you have the probabiltiy to have at least one 97 in your hundred throws.

The mistake sciguyryan made was the number of all possible results is 100^100 and the number of wanted results is more than 100. For example in two throws you can have a 97 in the first or the second and in both throws. Add them all up for the hundred throws and you will get the result from above.

[opus]

Hey The Duck, that'S only correct if you look at each throw seperatly. He searched for the total prob over all throws.

[The_Duck]

I see!

[sciguyryan]

Look at it from the other way, how much is the probability the miss the result in each throw. For the first it is 0.99, to miss it in the fist two throws 0.99*0.99 and so forth. for all the throws it 0.99^100 which is 0.36603234....
Subtrct that from 1 and you have the probabiltiy to have at least one 97 in your hundred throws.

The mistake sciguyryan made was the number of all possible results is 100^100 and the number of wanted results is more than 100. For example in two throws you can have a 97 in the first or the second and in both throws. Add them all up for the hundred throws and you will get the result from above.

Right, I see the mistake now

Cheers,

RyanJ

[Rassis]

It will be easier to find an answer for such a problem and understand the result if you run a simple simulator like the one I attach built in EXCEL.

Cheers

[Rassis]

It will be easier to find an answer for such a problem and understand the result if you run a simple simulating model like the one I attach herewith built in EXCEL.

Cheers

[sql_lall]

The results form a binomial distribution X ~ B(100, 0.01) where X is the number of times a certain value (say, 97) appears.
From this, it's easy to see that the probability the value appears at least once is the same as P(X >= 1) = 1 - P(X == 0) = 1 - ¹⁰⁰C₀ * (0.01)⁰ * (0.99)¹⁰⁰, which is just 1 - 0.99¹⁰⁰, as opus showed with logic.

[Rassis]

Correct. Simulation is just another (and easier...?) way to find an answer using nothing but logic and leaving analysis apart - specially usefull when you don´t remember the appropriate formulas and a good book on the matter is not at hand.

Cheers

[capsulecorpjx]

I have the answer.
The prob that 97 will show up in 100 rows is 1 - [the probability that 97 will NOT show up in 100 rows).

This equation is:
1 - [(99/100) ^ 100] = 0.634
Or 63.4% chance 97 will show up if you row your die 100 times.

There you go.

Ok, my friends and I have puzzled over this for a bit and failed to come to a satisfactory answer.

There's a one hundred sided die. The chances of it landing on any number from 1 to 100 is 100 to 1.

The chances of getting a 97 are 1/100.

So if the die is rolled 100 times, what are the chances that a 97 will have come up? It cant be 100 x 1/100 because that would be 1 and therefore a certainty, which it cant be.

Any ideas?

[opus]

Welcome to the forum.
If you would have rad all the reply's, you wouldhave seen this solution already.

[boku]

This would actually be impossible to do with real die because there is no way that any "hedron" shape could effectivly portray 100 equalateral sides.

[Rassis]

Boku,

That is why you use simulation on computer instead...
See my previous post in this thread with an attached Excel file.

[twanvl]

This would actually be impossible to do with real die because there is no way that any "hedron" shape could effectivly portray 100 equalateral sides.

Actually, it is possible, consider a 50 sided regular polygon, place a 'pyramid' on each side of it. Now you have a 100 sided die.
Another way is to use 2 20 sided dice, a and b and define the die c as c=(a*20+b)/4. But this is stretching the word die...

[boku]

but that is not a "hedron" shape. take a look here http://membres.lycos.fr/arjan/num100.htm