## 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**

January 10th, 2019 at 5:05:31 PM
permalink

Extra Credit- Does not matter which direction the ants are going

Beware, I work for the dark side.... We have cookies

January 10th, 2019 at 5:32:51 PM
permalink

Wrong solution, but I don't know why.

In a 105 minute period Amy visits the River Dragon machine 35 times, Betty 21, and Cathy 15. Service is offered 71 times for the period. The probability of who is offering service for a visit is Amy 35/71, Betty 21/71. Cathy 15/71.

“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia

January 10th, 2019 at 6:01:19 PM
permalink

Going to give this a shot before reading the other answers...

The way I did this was come up with a lowest common denominator of minutes... so let A = 3, B=5, and C=7. The LCD is 105. Meaning that if they all started at exactly the same time, after 105 minutes, they all would have come up with a finished round. Anna would have done 35 rounds, Betty would have done 21 rounds, and Cathy would have done 15 rounds. Adding those up, we get 71 rounds total. And the chances that any particular waitress would be visiting you next on her round is her percentage of that 71 total rounds.

SO... A = 35/71, B = 21/71, and C=15/71.

The chances would be as follows:

Anna: .49295774 or ~ 49%

Betty: .29577464 or ~ 30%

Cathy: .21126760 or ~ 21%

I'm probably oversimplifying this, and I look forward to seeing the other answers. But that's what I'm submitting.

SO... A = 35/71, B = 21/71, and C=15/71.

The chances would be as follows:

Anna: .49295774 or ~ 49%

Betty: .29577464 or ~ 30%

Cathy: .21126760 or ~ 21%

I'm probably oversimplifying this, and I look forward to seeing the other answers. But that's what I'm submitting.

I want to stay as close to the edge as I can without going over. Out on the edge you see all kinds of things you can't see from the center.

January 10th, 2019 at 6:09:13 PM
permalink

If it makes you feel better, BleedingChipsSlowly, I came up with the same wrong answer. :)

I want to stay as close to the edge as I can without going over. Out on the edge you see all kinds of things you can't see from the center.

January 10th, 2019 at 6:09:22 PM
permalink

Nice to see I have company. Hope, perhaps, that we are not wrong?Quote:vonnegutGoing to give this a shot before reading the other answers...

“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia

January 10th, 2019 at 6:49:25 PM
permalink

Quote:BleedingChipsSlowlyWrong solution, but I don't know why.

In a 105 minute period Amy visits the River Dragon machine 35 times, Betty 21, and Cathy 15. Service is offered 71 times for the period. The probability of who is offering service for a visit is Amy 35/71, Betty 21/71. Cathy 15/71.

You're answering the wrong question. If the question was who is the waitress for a random visit, you would be right. Why is that different that what I'm asking? Hard to explain. It's because the regular cycles. I'm still struggling with how to explain it.

Let's say you are playing in a contest with Tiger Woods to see who hits the first hole in one. Let's say that your chances per shot are 1 in 1,000,000 and Tigers are 1 in 100. Your best hope of winning is you have very good luck and Tiger has very bad luck. However, if you both hit them at exact regular cycles, you tease out some of that luck, and improve Tiger's chance of winning.

Let's say you are playing in a contest with Tiger Woods to see who hits the first hole in one. Let's say that your chances per shot are 1 in 1,000,000 and Tigers are 1 in 100. Your best hope of winning is you have very good luck and Tiger has very bad luck. However, if you both hit them at exact regular cycles, you tease out some of that luck, and improve Tiger's chance of winning.

It's not whether you win or lose; it's whether or not you had a good bet.

January 10th, 2019 at 6:52:07 PM
permalink

Quote:vonnegutGoing to give this a shot before reading the other answers...

The way I did this was come up with a lowest common denominator of minutes... so let A = 3, B=5, and C=7. The LCD is 105. Meaning that if they all started at exactly the same time, after 105 minutes, they all would have come up with a finished round. Anna would have done 35 rounds, Betty would have done 21 rounds, and Cathy would have done 15 rounds. Adding those up, we get 71 rounds total. And the chances that any particular waitress would be visiting you next on her round is her percentage of that 71 total rounds.

SO... A = 35/71, B = 21/71, and C=15/71.

The chances would be as follows:

Anna: .49295774 or ~ 49%

Betty: .29577464 or ~ 30%

Cathy: .21126760 or ~ 21%

I'm probably oversimplifying this, and I look forward to seeing the other answers. But that's what I'm submitting.

Nice try. You had the same answer as BleedingChipsSlowly. See my response to him for my attempt to explain why it's wrong, but it probably isn't the best explanation. Maybe Don can put it better.

It's not whether you win or lose; it's whether or not you had a good bet.

January 10th, 2019 at 7:26:12 PM
permalink

Quote:BleedingChipsSlowlyWrong solution, but I don't know why.

In a 105 minute period Amy visits the River Dragon machine 35 times, Betty 21, and Cathy 15. Service is offered 71 times for the period. The probability of who is offering service for a visit is Amy 35/71, Betty 21/71. Cathy 15/71.

I thought this at one point as well, but...

Suppose instead that all three are going the same speed. By your reasoning, each has an equal chance of being first. However, if Amy is just ahead of Betty, who is just ahead of Cathy, then Amy will be the first to reach the machine far more often than the other two.

I also know I'm supposed to wait 24 hours, but since somebody already posted the correct answer, I figure it's safe now...and I just realized this is pretty much just like a problem I solved some time ago

Let A, B, and C be the number of minutes before Amy, Betty, and Cathy will get to the machine. A is in the interval [0,3], B is in [0,5], and C is in [0,7].

Divide the interval [0,7] into three pieces; 0-3, 3-5, and 5-7.

Amy is always in 0-3

Betty is in 0-3 3/5 of the time and 3-5 2/5 of the time

Cathy is in 0-3 3/7 of the time, 3-5 2/7 of the time, and 5-7 2/7 of the time

3/5 x 3/7 = 9/35 of the time, all three are in 0-3; each of the three is equally likely to get to the machine first

3/5 x 4/7 = 12/35 of the time, A and B are in 0-3, and C is in either 3-5 or 5-7; Amy and Betty are equally likely to be first

2/5 x 3/7 = 6/35 of the time, A and C are in 0-3, and B is in 3-5; Amy and Cathy are equally likely to be first

2/5 x 4/7 = 8/35 of the time, A is in 0-3, B is in 3-5, and C is in 3-5 or 5-7; Amy will get there first

P(Amy getting there first) = 9/35 x 1/3 + 12/35 x 1/2 + 6/35 x 1/2 + 8/35 = (18 + 36 + 18 + 48) / 210 = 4 / 7

P(Betty getting there first) = 9/35 x 1/3 + 12/35 x 1/2 = (18 + 36) / 210 = 9 / 35

P(Cathy getting there first) = 9/35 x 1/3 + 6/35 x 1/2 = (18 + 18) / 210 = 6 / 35

The earlier question involved A being in the interval [0,1], B in [0,2], and C in [0,3]

Last edited by: ThatDonGuy on Jan 10, 2019

January 10th, 2019 at 7:38:32 PM
permalink

I'd be interested in knowing what the house edge is set at for various games at the Wizard Casino.

Have you tried 22 tonight? I said 22.

January 10th, 2019 at 8:20:05 PM
permalink

Quote:ThatDonGuyI also know I'm supposed to wait 24 hours, but since somebody already posted the correct answer, here's how I get it (at least, the probability that Amy gets there first):

No triple integral needed, although I did use one (where, as it turns out, one wasn't necessary)

Let A, B, and C be the points in the interval (0,1) where they are located, with 0 being where the machine is, and they are moving from 1 towards 0.

A's speed is 1/3, B's is 1/5, and C's is 1/7

A will get to 0 before B if 3A < 5B -> 3/5 A < B

A will get to 0 before C if 3A < 7C -> 3/7 A < C

At any given point A, the probability that A will get to 0 before B and C is (1 - 3/5 A)(1 - 3/7 A) = 1 - 36/35 A + 9/35 A^2

The integral over A from 0 to 1 of (9/35 A^2 - 36/35 A + 1) dA = 3/35 - 18/35 + 1 = 4/7.

That's how I did it too.

It's not whether you win or lose; it's whether or not you had a good bet.