Daniel J Isom
United States Smyrna Delaware
Send me a geekmail if you'd like to buy a copy!
I have to return some videotapes.

I suck at figuring out probabilities so I'm asking the brighter minds of Chit Chat to help me. I'll even make the example about burritos to keep your interest.
I have 10 burritos to give away. There are 200 tickets to buy. 200 people all buy 1 ticket. What is the probability for an individual ticket holder to win one of the 10 burritos?

Billy McBoatface
United States Lexington Massachusetts
KGS is the #1 web site for playing go over the internet. Visit now!
Yes, I really am that awesome.

Seems straightforward to me. 5%: 10/200 × 100% = 5%.

Mike Norris
United States Sacramento California
...and JPop too!!!!
Raina rocks!!!!!

disom wrote: What is the probability for an individual ticket holder to win one of the 10 burritos?
With Rudy around???
Not gonna happen...ever....

David Bush
United States Radiant Virginia

Can you win more than one burrito, or are you limited to one?
If you are limited to one, the odds are simple, 1 in 20 = 0.05 If it's allowed to win multiple burritos, the simplest way to calculate the odds is to determine how likely it is you will win no burritos, subtract that from one, and that gets you the odds you will win AT LEAST one burrito. 1 (199/200)^10 = 0.0488899 (EDITED)
EXACTLY ONE odds are more complicated.

maf man
United States Waunakee (madison area) WI

disom wrote: I have 10 burritos to give away. I'm lost right there

Daniel J Isom
United States Smyrna Delaware
Send me a geekmail if you'd like to buy a copy!
I have to return some videotapes.

wmshub wrote: Seems straightforward to me. 5%: 10/200 × 100% = 5%.
That's what I thought but I was thinking it was more complicated than that.

Billy McBoatface
United States Lexington Massachusetts
KGS is the #1 web site for playing go over the internet. Visit now!
Yes, I really am that awesome.

disom wrote: wmshub wrote: Seems straightforward to me. 5%: 10/200 × 100% = 5%. That's what I thought but I was thinking it was more complicated than that. It depends. If, after winning, you throw out the winning ticket and draw from the rest, then it is the simple 5%. That is what I assumed when I gave my answer.
If you put the winning ticket back, then draw the ticket for the next burrito, then you can win more than once and you need to look at David Bush's answer.

Bobby F.
United States Henrico Virginia
Whatever you do will be insignificent. But it is very important that you do it.

twixter wrote: Can you win more than one burrito, or are you limited to one? If you are limited to one, the odds are simple, 1 in 20 = 0.05 If it's allowed to win multiple burritos, the simplest way to calculate the odds is to determine how likely it is you will win no burritos, subtract that from one, and that gets you the odds you will win AT LEAST one burrito. 1 (199/200)^10 = 0.0488899 (EDITED)
EXACTLY ONE odds are more complicated. Twixter I might be crazy but I don't think this is correct.
If you can win only 1 burrito and your ticket is not replaced your odds are: 5.1162% (1 in 200 or 1 in 199 or 1 in 198... etc)
If you can win any number of burritos, your odds of winning at least 1 burrito are: 5% (1 in 200 * 10)
Edit: I was crazy. Figured out my logic flaw. The odds that this thread will increase burrito sales... very high.

Daniel J Isom
United States Smyrna Delaware
Send me a geekmail if you'd like to buy a copy!
I have to return some videotapes.

wmshub wrote: disom wrote: wmshub wrote: Seems straightforward to me. 5%: 10/200 × 100% = 5%. That's what I thought but I was thinking it was more complicated than that. It depends. If, after winning, you throw out the winning ticket and draw from the rest, then it is the simple 5%. That is what I assumed when I gave my answer. If you put the winning ticket back, then draw the ticket for the next burrito, then you can win more than once and you need to look at David Bush's answer.
It's the former situation. Thanks again for the help.

Daniel J Isom
United States Smyrna Delaware
Send me a geekmail if you'd like to buy a copy!
I have to return some videotapes.

Quaseymoto wrote:
If you can win only 1 burrito and your ticket is not replaced your odds are: 5.1162% (1 in 200 or 1 in 199 or 1 in 198... etc)
This is exactly what I was thinking.

United States Plainwell Michigan
Mi cabasa esta muy verde
Et in Vantasia ego

disom wrote: I suck at figuring out probabilities so I'm asking the brighter minds of Chit Chat to help me. I'll even make the example about burritos to keep your interest.
I have 10 burritos to give away. There are 200 tickets to buy. 200 people all buy 1 ticket. What is the probability for an individual ticket holder to win one of the 10 burritos?
Uh, hmm.
Van's Law States each person will have a 50/50 chance to win one of the 10 burritos.
They will.
or.
They won't.

John Holder
United States Centennial Colorado

disom wrote: wmshub wrote: Seems straightforward to me. 5%: 10/200 × 100% = 5%. That's what I thought but I was thinking it was more complicated than that.
It is more complicated! The probability is determined by figuring out, at each draw, what the odds of losing are, and then multiplying them all together and subtracting from 1 to get the odds of winning a burrito (multiply that by 100 to get the percent chance).
So, for the first draw, your odds of not getting a burrito are 199/200. Second draw, 198/199...
199 198 197 196 195 194 193 192 191 190 1  (  *  *  *  *  *  *  *  *  *  ) 200 199 198 197 196 195 194 193 192 191
Simplifies to:
1  ( 190/200 ) = 0.05
So yeah, 5% chance of winning...

The Seal of Approval
Austria Vienna

I have another math problem:
11 people are on a bus. At the first stop, 7 people leave the bus, and 3 people enter the bus At the second stop, 10 people leave the bus At the third stop, 5 people enter the bus
In the end, how many people are on the bus?

Steve Vondra
United States Charlottesville Virginia

The way I heard the "bus problem"
Asperamanca wrote: I have another math problem:
11 people are on a bus. At the first stop, 7 people leave the bus, and 3 people enter the bus At the second stop, 10 people leave the bus At the third stop, 5 people enter the bus
In the end, how many people are on the bus? How many more miles until the end of the route?

Mike Norris
United States Sacramento California
...and JPop too!!!!
Raina rocks!!!!!

Asperamanca wrote: I have another math problem:
11 people are on a bus. At the first stop, 7 people leave the bus, and 3 people enter the bus At the second stop, 10 people leave the bus At the third stop, 5 people enter the bus
In the end, how many people are on the bus?
It takes 4 people to make 1 Chipotle burrito. (That is pretty accurate actually.)
How many people does it take to make 1 burrito short of infinity?

Brian Bankler
United States San Antonio Texas
"Keep Summer Safe!"

demondude777 wrote: How many people does it take to make 1 burrito short of infinity?
Infinity would like you to know he is "Average plus" in height, and a generally good guy.

James Newton
United Kingdom Wiltshire
In the interest of giving credit where credit is due, my avatar is a scan of a handdrawn caricature by cartoonist Jim Naylor which was done at my company's 20th anniversary dinner.

cooler king wrote: The way I heard the "bus problem" Asperamanca wrote: I have another math problem:
11 people are on a bus. At the first stop, 7 people leave the bus, and 3 people enter the bus At the second stop, 10 people leave the bus At the third stop, 5 people enter the bus
In the end, how many people are on the bus? How many more miles until the end of the route? Clearly way too long  who wants to spend more time with all those negative people?

Mike Norris
United States Sacramento California
...and JPop too!!!!
Raina rocks!!!!!

Bankler wrote: demondude777 wrote: How many people does it take to make 1 burrito short of infinity?
Infinity would like you to know he is "Average plus" in height, and a generally good guy.
You sir....got it right in one try!


