Freethought & Rationalism Archive

Friar Bellows · 07-17-2002, 06:53 PM

Step right up, test your mathematical reasoning skills.

Seven seperate events are about to occur inside a seven day period, from Sunday to Saturday (inclusive). Within this one-week period, it is completely random when a particular event will occur -- they could each happen at any time with an equal probability.

What is the likelihood that one event will occur on each day, i.e. one event on Sunday, one event on Monday, one on Tuesday, ... , and one on Saturday? Show your reasoning.

tronvillain · 07-17-2002, 08:11 PM

Well, I make the calculation to be something like this:

[1/7(6/7)^6][1/6(5/6)^5][1/5(4/5)^4][(1/4(3/4)^3][1/3(2/3)^2][1/2(1/2)^1][1/1]=0.0576

In other words, there is about a six percent chance that one event will occur on each day.

What is my reasoning you ask? Well, the probability of an given event happening on the first day rather than any of the others is 1/7, and the chance of a given event not happening on the first day is 6/7. Multiply one of the first by six of the second and you get the probability of a single event happening on the first day. Now, repeat the calculation for six days and six events, and so on, then multiply it all together to get the total probability.

Friar Bellows · 07-17-2002, 09:20 PM

Quote:

Originally posted by tronvillain:
<strong>[1/7(6/7)^6][1/6(5/6)^5][1/5(4/5)^4][(1/4(3/4)^3][1/3(2/3)^2][1/2(1/2)^1][1/1]=0.0576</strong>

You must have accidently pressed the wrong keys on your calculator somewhere, because:

[1/7(6/7)^6][1/6(5/6)^5][1/5(4/5)^4][1/4(3/4)^3][1/3(2/3)^2][1/2(1/2)^1][1/1] = 0.00000121426567890...

which is close to one in a million (and certainly not 0.0576). To see it without the pain of reentering all of it on your calculator, note what happens when you multiply, for example, (6/7)^6 and (5/6)^5. All but one of the 6's cancel out. It turns out that your entire expression reduces to:

(1/7)^7

Wow, that simplifies it.

I will not comment yet on your reasoning. But your answer is roughly one in a million.

Are you still happy with your answer?

Undercurrent · 07-17-2002, 10:14 PM

First we count the total number of ways that the events could be assigned to days. First we pick a day for the first event, which we can do in 7 ways, then for the second event, which we can in 7 ways also, &c... This means that there are 7^7 = 823543 possible assignments of all seven events to the seven days.

We now count how may ways there are to assign the seven events to the seven days that have the property that each event is on a different day. Start by picking a day for the first event, of which there are seven to choose. Then for the second event, there are only six possible days to choose (since it can't be only the same day as the first event), and so on. So there are 7! = 5040 ways to allocate events to days with the restriction that each event occur on a different day.

So the probability is 7!/7^7 = 5040/823543 = 720/117649 ~= 0.00612

m.

Friar Bellows · 07-18-2002, 12:39 AM

Michael is correct and his explanation is spot on. I think tronvillain missed out the fact that there are many (equivalent) permutations of "seven events in seven different, consecutive days". In fact, as Michael points out, there are 7! of them, i.e. 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040.

I was once reading a book, I think it was Rudy Rucker's Infinity and the Mind, where the two functions, x! and x^x, were compared. It was quite a coincidence when the same night saw me working on the problem of this thread (which I found on a web site), the answer of which is equal to the ratio of these same two functions, for x=7. Cool.

tronvillain · 07-18-2002, 10:59 AM

Ah well.

Mr. Sammi · 07-18-2002, 11:09 AM

However, the probability of an event occurring has not been factored in.

What has been factored in is some highly insignificant events falling on 7 consecutive days.

My answer is infinity. The reasons are clear.

Sammi Na Boodie ()

Abacus · 07-18-2002, 12:18 PM

Quote:

Originally posted by Sammi:
[QB]My answer is infinity. The reasons are clear.
[QB]

The probability is infinity? That makes no sense at all. Please clarify.

Abacus · 07-18-2002, 01:48 PM

Okay fellas. Let's see if you can figure this one out:

What's the probability of 1 event occuring on Sunday, 3 events occuring on Monday, 2 events occuring on Friday, and 1 event occuring on Saturday?

tronvillain · 07-18-2002, 06:28 PM

Random Number Generator:

Quote:

The probability is infinity? That makes no sense at all. Please clarify.

As Sammi says, the reason are clear: he has nothing coherent to say and simply writes what sounds good to him at the time.

7[1/7(6/7)^6]6[(1/6)^3(5/6)^3]3[(1/2)^2(1/2)^1]1[1/1]=0.00239

So, there is about a 0.24% chance of it happening that way, assuming I haven't made another error.

*sigh* Never mind. Apparently I need to brush up on these.

[ July 19, 2002: Message edited by: tronvillain ]</p>

07-17-2002, 06:53 PM	#1
Friar Bellows Veteran Member Join Date: Apr 2001 Location: arse-end of the world Posts: 2,305	Brain Teaser Step right up, test your mathematical reasoning skills. Seven seperate events are about to occur inside a seven day period, from Sunday to Saturday (inclusive). Within this one-week period, it is completely random when a particular event will occur -- they could each happen at any time with an equal probability. What is the likelihood that one event will occur on each day, i.e. one event on Sunday, one event on Monday, one on Tuesday, ... , and one on Saturday? Show your reasoning.

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07-17-2002, 08:11 PM	#2
tronvillain Veteran Member Join Date: Oct 2000 Location: Alberta, Canada Posts: 5,658	Well, I make the calculation to be something like this: [1/7(6/7)^6][1/6(5/6)^5][1/5(4/5)^4][(1/4(3/4)^3][1/3(2/3)^2][1/2(1/2)^1][1/1]=0.0576 In other words, there is about a six percent chance that one event will occur on each day. What is my reasoning you ask? Well, the probability of an given event happening on the first day rather than any of the others is 1/7, and the chance of a given event not happening on the first day is 6/7. Multiply one of the first by six of the second and you get the probability of a single event happening on the first day. Now, repeat the calculation for six days and six events, and so on, then multiply it all together to get the total probability.

07-17-2002, 10:14 PM	#4
Undercurrent Veteran Member Join Date: Jan 2001 Location: Santa Fe, NM Posts: 2,362	First we count the total number of ways that the events could be assigned to days. First we pick a day for the first event, which we can do in 7 ways, then for the second event, which we can in 7 ways also, &c... This means that there are 7^7 = 823543 possible assignments of all seven events to the seven days. We now count how may ways there are to assign the seven events to the seven days that have the property that each event is on a different day. Start by picking a day for the first event, of which there are seven to choose. Then for the second event, there are only six possible days to choose (since it can't be only the same day as the first event), and so on. So there are 7! = 5040 ways to allocate events to days with the restriction that each event occur on a different day. So the probability is 7!/7^7 = 5040/823543 = 720/117649 ~= 0.00612 m.

07-18-2002, 12:39 AM	#5
Friar Bellows Veteran Member Join Date: Apr 2001 Location: arse-end of the world Posts: 2,305	Michael is correct and his explanation is spot on. I think tronvillain missed out the fact that there are many (equivalent) permutations of "seven events in seven different, consecutive days". In fact, as Michael points out, there are 7! of them, i.e. 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040. I was once reading a book, I think it was Rudy Rucker's Infinity and the Mind, where the two functions, x! and x^x, were compared. It was quite a coincidence when the same night saw me working on the problem of this thread (which I found on a web site), the answer of which is equal to the ratio of these same two functions, for x=7. Cool.

07-18-2002, 10:59 AM	#6
tronvillain Veteran Member Join Date: Oct 2000 Location: Alberta, Canada Posts: 5,658	Ah well.

07-18-2002, 11:09 AM	#7
Mr. Sammi Banned Join Date: Jun 2002 Location: Montr�al Posts: 367	However, the probability of an event occurring has not been factored in. What has been factored in is some highly insignificant events falling on 7 consecutive days. My answer is infinity. The reasons are clear. Sammi Na Boodie ()

07-18-2002, 01:48 PM	#9
Abacus Veteran Member Join Date: Mar 2002 Location: South Dakota Posts: 2,214	Okay fellas. Let's see if you can figure this one out: What's the probability of 1 event occuring on Sunday, 3 events occuring on Monday, 2 events occuring on Friday, and 1 event occuring on Saturday?

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