## Poll

No votes (0%) | |||

2 votes (13.33%) | |||

2 votes (13.33%) | |||

1 vote (6.66%) | |||

No votes (0%) | |||

3 votes (20%) | |||

No votes (0%) | |||

9 votes (60%) | |||

5 votes (33.33%) | |||

5 votes (33.33%) |

**15 members have voted**

All three are at work, but you have no idea when each started or where they are in their rotations. You arrive and plop down at a River Dragons machine, because the large jackpot is over $4,950. What is the probability each one of them arrives to you first to offer a nice libation?

Like last time, I'd like to ask Don to give this a 24-hour waiting period to give someone else a chance. As usual, please put answers and solutions in spoiler tags.

The question for the poll is what is the probability Amy is first?

Since I ‘assume’ that one of the poll choices is the correct answer, using no math, I picked 1/2 simply because it feels right... at least from the options provided.

For what it’s worth, I used the same type of picking an answer on many of my SAT questions. I did rather well.

The sine waves don't completely repeat until 105 minutes, but there will be co-visits to the same machine at 15 minute, 21 minute, and 35 minute intervals. You could have sat down anywhere within that time frame. So am I looking for times that end with co-visits, or only solo visits? If co-visits count, do I give Amy full credit, or 1/2 credit?

Edit: I am assuming they all clock in at hour 0 and leave the cocktail station at the same time.

Quote:beachbumbabsHow are you counting Amy arriving at the same time as one of the others? Is Amy there first, though tied, or does that disqualify that visit/round from her?

Time is divided into infinitely small units, so ties are not an issue. Much like the probability of picking pi from from the uniform distribution from 0 to 10 -- the probability of a tie is zero -- but it could happen.

p.s. Just so it's clear, let's assume the River Dragons machine is just a point on the floor.

Quote:WizardTime is divided into infinitely small units, so ties are not an issue. Much like the probability of picking pi from from the uniform distribution from 0 to 10 -- the probability of a tie is zero -- but it could happen.

Please see the spoiler I added as you were responding. I'm calculating in whole minutes.

Quote:beachbumbabsPlease see the spoiler I added as you were responding. I'm calculating in whole minutes.

Again, there can be no ties, so your answer is to a different question.

To help visualize it, assume the casino is like a big circle and you are at a random point on it. The three waitresses are like clock hands. What is the probability the clock hand that makes a 3-minute revolution reaches your point first?

Quote:WizardAgain, there can be no ties, so your answer is to a different question.

To help visualize it, assume the casino is like a big circle and you are at a random point on it. The three waitresses are like clock hands. What is the probability the clock hand that makes a 3-minute revolution reaches your point first?

I think the clock analogy is a good one for demonstrating there CAN be ties. The minute hand and the hour hand will be tied 12 times in 12 hours. The minute hand and second hand will be tied every just-over-a-minute. As will the second hand and the hour hand. And all 3 will be together 12 times in every 12 hours. What if I'm seated at exactly 12 on your clock face?

First stab with simple math... Might come back and check later.

We split into 4 cases. 1st Case - all 3 waitresses are coming in 3 minutes. 2nd Case - 1st and second waitress are coming in 3 minutes. 3rd Case - 1st and 3rd waitress are coming in 3 minutes. 4th Case - only 1st waitress is coming in 3 minutes. Since waitresses coming in 3 minutes all have the same probability of showing up first, this leads to probabilities of 1/3, 1/2, 1/2, and 1. Then, we calculate the probabilities of each case. 1st - (3/5)(3/7) = 9/35, 2nd - (3/5)(4/7) = 12/35, 3rd - (2/5)(3/7) = 6/35, 4th - (2/5)(4/7) = 8/35. Summing up, we get 3/35 + 6/35 + 3/35 + 8/35 = 20/35 = 4/7.

I guess I only did Amy... The same cases can give the probabilities for the other two. Betty can win in case 1 and 2, yielding 3/35+6/35 = 9/35. Cathy can win in case 1 and 3 yielding 3/35 + 3/35 = 6/35. The total probability is 4/7 + 9/35 + 6/35 = 35/35, so that checks out at least.

For every 1 revolution for Amy, Betty only gets 21/35ths of the way around, and Cathy only 15/35ths. At best, the two slowpokes would cover 1 and 1/35ths of the stations every time Amy made a round, so 34 out of 35 times Amy will get there first.