In one version of British pool, you have 6 yellow balls, 6 red balls and one black. They are arranged as follows:
R
Y R
R B Y
Y R Y R
R Y R Y Y
I usually set them up by putting them at random into the triangle and then moving the appropriate ones to get the correct arrangement.
One time, when I looked to see what I needed to move, I realised that I’d thrown them into the triangle in the correct arrangment already. So, my question is, what are the odds of this happening?
I’ve tried to calculate it, but I keep getting confused in that if I have a red in one place, it forces a yellow to appear elsewhere, and the black could be anywhere.
Assuming that you meant (as is shown in your diagram) 7 red balls and 7 yellow balls, your chances of randomly arranging the fifteen balls properly is 0.00194 %, or 1 in 51480.
Thanks Donut. Yes, I did mean 7 of each; don’t know how I didn’t spot that.
1 in 51480…wow! No wonder I’ve only done it once. Could you tell me how you went about calculating it? I’m just one of these people who prefers to know the “why” rather than just the “what”
What is it with all the probability questions lately?
There are 15 spaces. Label them in such a way that the black ball goes in space 15 and then the red balls go in spaces 1 to 7 and the yellows balls in 8 to 14, or vice versa. Label the red balls R[sub]1[/sub] to R[sub]7[/sub] and label the yellow balls similarly.
How many correct arrangements are there? The red balls can be arranged in spaces 1 to 7 in 7! = 5040 ways and similarly for the yellow balls. This gives 7![sup]2[/sup] = 25,401,600 ways of placing the balls correctly with reds in 1 to 7 and the yellows in 8 to 14 ( note we do not have to worry about the black here. If the red and yellow balls are all placed correctly then the black ball must be as well). There are as many arrangements again with the yellows in spaces 1 to 7 and the reds in 8 to 14. Thus there are 50,803,200 correct arrangements of the 15 balls.
How many arrangements are there of the balls in total? This is simply 15! = 1.308[symbol]´[/symbol]10[sup]12[/sup].
Finally, the probability is one divided by the other, or 1 in 25,740. This is twice Donut’s answer. This is because I have allowed red and yellow to be reversed in your diagram. If this is not allowed, Donut’s answer is the correct one.
There’s several ways he could have arrived at it. As a check, I get the same figure thusly:
There are 15! equally likely arrangements of balls, however two groups of 7 are identical so 15! / (7! 7!) = 51480. One of these is your desired arrangement.
If you’re not a probability ace, a simpler way is just to go one by one and figure the probability of each, then multiply them. Since we can go in any order, start with the black. There’s a 1/15 chance that you’ll get it right. Now for the first red ball. There are seven reds to chose from for that slot out of 14 total, so the chance of getting that one right is 7/14. Now for the second red ball (let’s leave the yellow for the end), there are six out of 13, so 6/13. Continuing down the list gives the series:
At this point, all you have are red balls for the red slots, so the chance of getting them right is 1. As Jabba said, if you can switch red and yellow, it’s cut in half.
Thanks guys. I understand the explanations; I think I was having problems because I didn’t simplify the problem enough in my head. I still had the idea of the position in the triangle, rather than just thinking slots they could fit in (if this makes any sense to you!). Also, I didn’t stop to think that if the other balls were in the correct position, then the black must also be.
Jabba: It was actually the other probability questions that got me thinking about this one. It’s not just a theoretical question for me: I actually did do this and noticed it because in the thousands of times I’ve set the balls up, that was the only time they hadn’t needed any rearranging.
I don’t think swapping red for yellow is allowed in formal games, but in informal ones it’s OK.
