

Hi all,
Quick probability question. Let's say that I have 48 balls in a bag, 40 red and 8 blue. I take 3 out at once. What is the probability that among the 3 balls will be 1 blue?
What if it's 42 red and 6 blue? Again, what is the probability that among 3 balls picked at once will be 1 blue?
Thanks in advance!


Dan Blum
United States Wilmington Massachusetts

Do you want the probability that EXACTLY one will be blue or AT LEAST one will be blue?




tool wrote: Do you want the probability that EXACTLY one will be blue or AT LEAST one will be blue?
Sorry, should have clarified that. I meant the probability that at least one will be blue.


Dan Blum
United States Wilmington Massachusetts

The Hound wrote: tool wrote: Do you want the probability that EXACTLY one will be blue or AT LEAST one will be blue? Sorry, should have clarified that. I meant the probability that at least one will be blue.
Well, that's the easy case. The simplest way to calculate that is to calculate the probability that all three are red and then subtract it from 1:
1  40/48 * 39/47 * 38/46 = 43% (approximately)
For the 42/6 case it's
1  42/48 * 41/47 * 40/46 = 34% (approximately)


Russ Williams
Poland Wrocław Dolny Śląsk

The Hound wrote:
Hi all,
Quick probability question. Let's say that I have 48 balls in a bag, 40 red and 8 blue. I take 3 out at once. What is the probability that among the 3 balls will be 1 blue?
P(first=blue) = 8/48 + P(first=red) * P(second=blue) = 40/48 * 8/47 + P(first=red) * P(second=red) * P(third=blue) = 40/48 * 39/47 * 8/46 = 8/48 + 40/48 * 8/47 + 40/48 * 39/47 * 8/46 = 0.428769657724329
Or more simply (and to sanitycheck the calculation) 1  P(all 3 are red) = 1  (40/48 * 39/47 * 38/46) = 0.428769657724329
Quote: What if it's 42 red and 6 blue? Again, what is the probability that among 3 balls picked at once will be 1 blue?
I trust I have taught you how to build a fire well enough to leave this one for you.




This gives me tremendous perspective on the problem I'm working on  I want it to be very rare for the player to draw a blue ball (or card, as it were). Looks like I'll have to both shave down the blues a bit and add to the total pool...
Thanks everyone for the help!


Stephen Eckman
United States Oviedo Florida

Also wanted to point you to this thread for future reference:
Post Probability Questions Here


Jeremy Lennert
United States California

The Hound wrote: This gives me tremendous perspective on the problem I'm working on  I want it to be very rare for the player to draw a blue ball (or card, as it were). Looks like I'll have to both shave down the blues a bit and add to the total pool... Or reduce the number of balls the player draws at a time. If drawing a blue ball or not is the ONLY thing that matters, then you're probably better off having players draw only 1 ball at a time, rather than 3, as that would make the probabilities really easy to calculate, and will also tend to make the game play faster and reduce the total number of balls you need. (Of course, players may have a different psychological reaction to drawing 3 balls rather than 1 even if the odds are exactly the same...)
You might want to come at the math from the other direction: figure out how likely you want it to be for the player to draw a blue ball, and then calculate how many blue balls you should have based on that. (Rather than picking a number of balls and then calculating the odds.)
Also keep in mind that as the number of balls becomes bigger, it becomes less likely that players are doing a good job of mixing them up after every draw. The theoretical distribution doesn't mean much if players keep drawing the same 10% of the balls over and over because they draw from the top and then put them back on top.



