5 Yellow Starbursts with 1 pull

[Smackdab]

Over the lunch hour a co-worker reaches into a family size bag of Starburts and pulls out 5 pieces. ALL YELLOW!

Assuming there are 5 different colors of the candy and there is an equal amount of each color in the bag what would be the odds of pulling out 5 of the same color like this?

Not really poker related. I'm just in a weird mood today and figured one of the math whizzes could give me a quick calculation of this.

[kmvenne]

I thought Starburst only had 4 flavors, lemon, cherry, orange, and strawberry.

Assuming you want the chance of pulling 5 of any one flavor assuming you have 5 choices, it's 1*1/5*1/5*1/5*1/5 or 1 in 625. I believe the answer is 1*1/4*1/4*1/4*1/4 for Starburst though, or 1 in 256. Assuming random distribution.

[Paul2432]

[ QUOTE ]
I thought Starburst only had 4 flavors, lemon, cherry, orange, and strawberry.

Assuming you want the chance of pulling 5 of any one flavor assuming you have 5 choices, it's 1*1/5*1/5*1/5*1/5 or 1 in 625. I believe the answer is 1*1/4*1/4*1/4*1/4 for Starburst though, or 1 in 256. Assuming random distribution.

[/ QUOTE ]

This would only be correct if there were an infinite number of pieces in a bag. Otherwise the answer depends on the number of pieces in the bag.

Paul

[pokerrookie]

It is very probable that his wife doesn't like the yellow starburst, and as all husbands do, he got stuck with the dregs. Thus, probability is very high.

[carsten]

I have no idea if Starburst comes in 4 or 5 flavors, but in any case this response is not exactly correct.

First of all, the probability depends on how many pieces of each flavor you have. For example, if there are only 4 pieces of each flavor in the bag, it's impossible to pull 5 of the same flavor.

Given m flavors and n pieces of each flavor in the bag, the probability for pulling 5 pieces of any one flavor is m*(n choose 5)/(m*n choose 5), or m*n*(n-1)*(n-2)*(n-3)*(n-4)/m*n*(m*n-1)*(m*n-2)*(m*n-3)*(m*n-4).

The limit of this expression as n approaches infinity is 1/m to the fourth power, so kmvenne's response is a valid approximation as long as the sample size is much smaller than the size of the bag.

Hope this helps,

Carsten.

[Smackdab]

[ QUOTE ]
It is very probable that his wife doesn't like the yellow starburst, and as all husbands do, he got stuck with the dregs. Thus, probability is very high.


[/ QUOTE ]

DOH!

Didn't consider this option. Another one on the +EV side of my recent divorce. I now get to sample the Cherry and Strawberry flavors.

BTW.....Starburst has the following "regular" flavors:

Lemon
Orange
Cherry
AND.......... drumroll please
Strawberry

YES....4 not 5 flavors. So not as extreme to pull 5 out of a bag as I thought at first.

[SossMan]

the cherry is the best.


that is all.

[Hojglad]

Let n be the total number of starbursts in the bag. let k be the number of yellow starbursts in the bag. the probability of this happening is C(n,5)/C(k,5)

C(x,y) = x!/(y!*(x-y)!)

This all assumes that there are more than 4 yellow starbursts in the bag.