What is the probability of telling a secret.

KingKong433

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[FONT=.SF UI Text][FONT=.SFUIText]This post is really long, I hope you have the patience to read through it and help me.
Question: A, B, C, D, and E are playing a game. The game requires each player to throw a fair die once per turn. If they get any number between 1 and 5 inclusive, they would have to tell a secret about themselves.

Find the probability that:
(i) each of them tells only one secret by the end of their fifth turn.
(ii) A tells only at the first time, B tells only at the second time, C tells only at the third time, D tells only at the fourth time and E tells only at the fifth time.
[/FONT][/FONT]

Alright, for both of these questions, it has puzzled a lot of people, my friends, and even my teachers. So let's see what I've worked out, then I'll tell y'all why I need your help.

For question (i),
[FONT=.SFUIText]I calculated the probability of A telling a secret once in five turns, this will give us 25/7776. Reason: let's calculate the probability that A will tell only at the [/FONT][FONT=.SFUIText]first [/FONT][FONT=.SFUIText]time of the 5 turns. This will give us 5/6 * 1/6 * 1/6 * 1/6 * 1/6 = 5/7776. Since A can also tell his secret only at the 2nd turn or the 3rd, 4th, or 5th turn, thus the probability will be 5/7776 + 5/7776 + [/FONT][FONT=.SFUIText]5/7776 + [/FONT][FONT=.SFUIText]5/7776 + [/FONT][FONT=.SFUIText]5/7776 (I.e. 5/7776 * 5) = 2[/FONT][FONT=.SFUIText]5/7776[/FONT][FONT=.SFUIText]
Obviously, the probability of B, C , D, or E telling a secret once in five turns will also be the same, that is 2
[/FONT]
[FONT=.SFUIText]5/7776.
[/FONT][FONT=.SFUIText]Now, to calculate 5 turns. Then it will be 25/7776 * 2[/FONT][FONT=.SFUIText]5/7776 * 2[/FONT][FONT=.SFUIText]5/7776 * 25/7776 * 25/7776 (i.e. (25/7776)^5 ) = a very complicated number, therefore I'll just say (25/7776)^5 throughout this thread.

For question (ii),
From question (i), we already know that the probability of A telling only at the first time is 5/7776. And the probability for the other 4 people are the same. So the answer to this question should be (5/7776)^5.

At least that's what I think.

Background story: this is an exam question. Question (i) is the actual exam question. However, when I ask for the reason of how to obtain the answer (which I'll reveal later), my teacher said what the question really wants is *question (ii)*.

The answer given to question (ii) is 25/7776.

This is the explanation I obtained from my friend who is brilliant in Maths and also from my Add Math teacher:
(Somehow they straight away assumed that A will only tell at the first time, B tells at the second... just like in question (ii) )
First, calculate the probability of A telling at the first turn, that will be, as shown above, 5/7776. For B telling at the second turn, it'll also be 5/7776. Then they said since you have 5 people, you multiply 5/7776 by 5, giving you 25/7776.

My first thought on this is, this makes absolutely no sense. If you multiply by 5 because there are 5 people, then if you increase the number of people to a certain point (in this case, more than 1556 players) your probability will be more than 1 already.

But then, they told me this game cannot be played like this (another assumption they made), they told me that if you play for 5 turns, you must have at least 5 players so that each player can be assigned to one turn. So if you have 1556 players, you'll play 1556 turns. With this approach, striaghtaway multiplying 5/7776 by 5 doesn't seem wrong as the probability cannot exceed 1 (If you have 1556 players and 1556 turns, that means player A's probability of telling only at the first turn will be 5/6 * 1/6 * 1/6 * ... (1553 more 1/6s) )

Alright then, so let's take "their rule" into consideration also.

This will still not allow us to multiply 5/7776 by 5.

Even by adding "their rule" into the game, if you multiply 5/7776 by 5 what you're trying to find should be the probability of "Either A tells at the first turn or B tells at the second then or C tells at the third turn..." (note the term "or")
but what they want actually is the probability of "A tells at the first turn and B tells at the second turn and C tells at the third turn..." (note the term "and")

From my point of view, that method is wrong. Because I find that my method makes more sense. The problem here is that a lot of teachers (at least 4 teachers that I know of), upon seeing this question said that he probability is 25/7776. One even applied the binomial distribution formula straight away, not even stopping to think what it actually calculates. Besides, when I told them why I think their method is wrong, they can't accept it, yet they can't explain why they are correct; When I tell them my method, they can't accept it either, yet they can't prove me wrong. Also, the friend of mine said that the throwing the dice is an independent event, so we totally should add 5/7776 for five times (i.e. 5/7776 * 5). Personally I think he doesn't really understand what "independent" means in this context.

So, I need confirmation from other people from around the world, getting professional help. I strongly believe that what I worked out is correct (but really, I'm very open to other opinions and suggestions, just tell me what do you think, and I'll double check what went wrong).

Your time taken to read this long *** essay and energy used to help me solve this question is deeply appreciated. Thank you.[/FONT]
 
The forum is moderated. New members' first three posts won't appear until they have been approved by a moderator. No need to post the same thing, twice.

ii. If Event Q and Event R are independent, then P(Q AND R) = P(Q) * P(R). This extends to any number of events.

You have only to decide if your events are independent.
 
Last edited by a moderator:
The forum is moderated. New members' first three posts won't appear until they have been approved by a moderator. No need to post the same thing, twice.

ii. If Event Q and Event R are independent, then P(Q AND R) = P(Q) * P(R). This extends to any number of events.

You have only to decide if your events are independent.

Sorry about that, I accidentally posted it two times without knowing it.

The event where A gets to tell a secret on his first turn is independent right, same as how B gets to tell his secret at the second turn. So that means the probability for question (ii) should be ([FONT=.SFUIText]5/7776)^5[/FONT]
 
[FONT=.SF UI Text][FONT=.SFUIText]This post is really long, I hope you have the patience to read through it and help me.
Question: A, B, C, D, and E are playing a game. The game requires each player to [/FONT][/FONT]
throw a fair die once per turn. If they get any number between 1 and 5 inclusive[FONT=.SF UI Text][FONT=.SFUIText], they would have to tell a secret about themselves.

Find the probability that:
(i) each of them tells only one secret by the end of their fifth turn.
(ii) A tells only at the first time, B tells only at the second time, C tells only at the third time, D tells only at the fourth time and E tells only at the fifth time.
[/FONT][/FONT]

Alright, for both of these questions, it has puzzled a lot of people, my friends, and even my teachers. So let's see what I've worked out, then I'll tell y'all why I need your help.

For question (i),
[FONT=.SFUIText]I calculated the probability of A telling a secret once in five turns, this will give us 25/7776. Reason: let's calculate the probability that A will tell only at the [/FONT][FONT=.SFUIText]first [/FONT][FONT=.SFUIText]time of the 5 turns. This will give us 5/6 * 1/6 * 1/6 * 1/6 * 1/6 = 5/7776. Since A can also tell his secret only at the 2nd turn or the 3rd, 4th, or 5th turn, thus the probability will be 5/7776 + 5/7776 + [/FONT][FONT=.SFUIText]5/7776 + [/FONT][FONT=.SFUIText]5/7776 + [/FONT][FONT=.SFUIText]5/7776 (I.e. 5/7776 * 5) = 2[/FONT][FONT=.SFUIText]5/7776[/FONT][FONT=.SFUIText]
Obviously, the probability of B, C , D, or E telling a secret once in five turns will also be the same, that is 2
[/FONT]
[FONT=.SFUIText]5/7776.
[/FONT][FONT=.SFUIText]Now, to calculate 5 turns. Then it will be 25/7776 * 2[/FONT][FONT=.SFUIText]5/7776 * 2[/FONT][FONT=.SFUIText]5/7776 * 25/7776 * 25/7776 (i.e. (25/7776)^5 ) = a very complicated number, therefore I'll just say (25/7776)^5 throughout this thread.

For question (ii),
From question (i), we already know that the probability of A telling only at the first time is 5/7776. And the probability for the other 4 people are the same. So the answer to this question should be (5/7776)^5.

At least that's what I think.

Background story: this is an exam question. Question (i) is the actual exam question. However, when I ask for the reason of how to obtain the answer (which I'll reveal later), my teacher said what the question really wants is *question (ii)*.

The answer given to question (ii) is 25/7776.

This is the explanation I obtained from my friend who is brilliant in Maths and also from my Add Math teacher:
(Somehow they straight away assumed that A will only tell at the first time, B tells at the second... just like in question (ii) )
First, calculate the probability of A telling at the first turn, that will be, as shown above, 5/7776. For B telling at the second turn, it'll also be 5/7776. Then they said since you have 5 people, you multiply 5/7776 by 5, giving you 25/7776.

My first thought on this is, this makes absolutely no sense. If you multiply by 5 because there are 5 people, then if you increase the number of people to a certain point (in this case, more than 1556 players) your probability will be more than 1 already.

But then, they told me this game cannot be played like this (another assumption they made), they told me that if you play for 5 turns, you must have at least 5 players so that each player can be assigned to one turn. So if you have 1556 players, you'll play 1556 turns. With this approach, striaghtaway multiplying 5/7776 by 5 doesn't seem wrong as the probability cannot exceed 1 (If you have 1556 players and 1556 turns, that means player A's probability of telling only at the first turn will be 5/6 * 1/6 * 1/6 * ... (1553 more 1/6s) )

Alright then, so let's take "their rule" into consideration also.

This will still not allow us to multiply 5/7776 by 5.

Even by adding "their rule" into the game, if you multiply 5/7776 by 5 what you're trying to find should be the probability of "Either A tells at the first turn or B tells at the second then or C tells at the third turn..." (note the term "or")
but what they want actually is the probability of "A tells at the first turn and B tells at the second turn and C tells at the third turn..." (note the term "and")

From my point of view, that method is wrong. Because I find that my method makes more sense. The problem here is that a lot of teachers (at least 4 teachers that I know of), upon seeing this question said that he probability is 25/7776. One even applied the binomial distribution formula straight away, not even stopping to think what it actually calculates. Besides, when I told them why I think their method is wrong, they can't accept it, yet they can't explain why they are correct; When I tell them my method, they can't accept it either, yet they can't prove me wrong. Also, the friend of mine said that the throwing the dice is an independent event, so we totally should add 5/7776 for five times (i.e. 5/7776 * 5). Personally I think he doesn't really understand what "independent" means in this context.

So, I need confirmation from other people from around the world, getting professional help. I strongly believe that what I worked out is correct (but really, I'm very open to other opinions and suggestions, just tell me what do you think, and I'll double check what went wrong).

Your time taken to read this long *** essay and energy used to help me solve this question is deeply appreciated. Thank you.[/FONT]
I have not seen a fair dice with five faces yet!
 
I have not seen a fair dice with five faces yet!
The OP doesn't mention a five sided dice.
It seems pretty clear - Each player throws a fair (six sided ) die once per turn. If they throw 1 to 5 inclusive, they have to tell a secret about themselves. (If they throw a 6 they do nothing.)
 
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