# First to roll a six dice game (Probability)

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#1
Hi, Came across this Maths question:

In a game four players take it in turns to roll a normal unbiased six-sided die.
The winner of the game is the first person to roll a 6.
Find the probability that
a) The first player wins
b) The last player wins

I am wondering if my method/answers are right. I got 216/671 for the first answer and 125/671 for the last.
I did this by forming a geometric sum where a= 1/6 for the first player and r = 5/6 all to the power of 4 (4 players) is this thinking correct? I can explain my reasoning more if helpful! Many thanks!
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3 weeks ago
#2
(Original post by tande33)
Hi, Came across this Maths question:

In a game four players take it in turns to roll a normal unbiased six-sided die.
The winner of the game is the first person to roll a 6.
Find the probability that
a) The first player wins
b) The last player wins

I am wondering if my method/answers are right. I got 216/671 for the first answer and 125/671 for the last.
I did this by forming a geometric sum where a= 1/6 for the first player and r = 5/6 all to the power of 4 (4 players) is this thinking correct? I can explain my reasoning more if helpful! Many thanks!
The first player wins if the first roll is a 6. The probability of this is just 1/6.
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3 weeks ago
#3
the fact that you said "came across" suggests that this isn't formal so I am wondering if I am permitted to give my ideas for a conclusive answer. first I will give hints because they are allowed.
so the first one is simple. this is just the probability of an outcome (in this case the probability of rolling a 6) which is pretty simple so I won't touch that.
the last person winning is a little different. it is just asking what are the odds the 4th roll will be a 6. you can use a probability tree or use simple logic.
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#4
(Original post by β0b)
the fact that you said "came across" suggests that this isn't formal so I am wondering if I am permitted to give my ideas for a conclusive answer. first I will give hints because they are allowed.
so the first one is simple. this is just the probability of an outcome (in this case the probability of rolling a 6) which is pretty simple so I won't touch that.
the last person winning is a little different. it is just asking what are the odds the 4th roll will be a 6. you can use a probability tree or use simple logic.
yeah no this is not a question I have been set or hw or anything, this is for fun!
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3 weeks ago
#5

You could also “sense check” your approach by calculating the probability for the 2nd and 3rd players (which should be easy once you have the 1st player) - and then adding up the probabilities for all players to check they equal one.
1
#6
I have been practising combinatorics and probability in general, mainly for fun, but also they (combinatorics) aren't really on the a level, from what I have seen so far. Does anyone know of some good resources I could look at?
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#7
(Original post by RDKGames)
The first player wins if the first roll is a 6. The probability of this is just 1/6.
I think the question is asking more about the first player rather than the first roll!
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#8
Here is another question, bit more stuck on this one. I can do part a, but wondering if there is a way to simplify the rest of the question, rather than brute forcing!
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3 weeks ago
#9
(Original post by tande33)
I think the question is asking more about the first player rather than the first roll!
I think so too, the way it is phrased. If no one wins in the first round, then go round again.
1
3 weeks ago
#10
(Original post by tande33)
Here is another question, bit more stuck on this one. I can do part a, but wondering if there is a way to simplify the rest of the question, rather than brute forcing!
There are 1 line answers for each of (a), (b), (c), although for (c) at least brute force is going to be fairly painless.

Incidentally, this looks very much like an Oxford MAT question, in which case there are threads specifically dedicated to the Oxford MAT in the Maths Exams subforum.
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#11
(Original post by mqb2766)
Quite a lot on combinations/counting in problem solving, for instance
https://kheavan.files.wordpress.com/...0471789011.pdf
Has a decent chapter and questions.
Oh thanks, I will have a look
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#12
(Original post by DFranklin)
There are 1 line answers for each of (a), (b), (c), although for (c) at least brute force is going to be fairly painless.

Incidentally, this looks very much like an Oxford MAT question, in which case there are threads specifically dedicated to the Oxford MAT in the Maths Exams subforum.
so for b do you think systematic counting is a valid method
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3 weeks ago
#13
(Original post by tande33)
so for b do you think systematic counting is a valid method
It's *valid*, but if by systematic counting you mean enumerating every possibility it's going to take rather a long time. Not the recommended approach.
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3 weeks ago
#14
(Original post by tande33)
I am wondering if my method/answers are right. I got 216/671 for the first answer and 125/671 for the last.
I did this by forming a geometric sum where a= 1/6 for the first player and r = 5/6 all to the power of 4 (4 players) is this thinking correct? I can explain my reasoning more if helpful! Many thanks!
Agreed on the answers. There's actually a very short way of doing this. If you define A, B, C, D as the events that each player wins, it's easy to form expressions for p(B), p(C), p(D) in terms of p(A). Then use the fact that p(A)+p(B)+p(C)+p(D) = 1 to find p(A) and thence p(B), p(C), p(D).
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#15
(Original post by DFranklin)
Agreed on the answers. There's actually a very short way of doing this. If you define A, B, C, D as the events that each player wins, it's easy to form expressions for p(B), p(C), p(D) in terms of p(A). Then use the fact that p(A)+p(B)+p(C)+p(D) = 1 to find p(A) and thence p(B), p(C), p(D).
Is it that each chance is 5/6 times smaller as you go from player 1 to 2 to 3 to 4?
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3 weeks ago
#16
(Original post by tande33)
Is it that each chance is 5/6 times smaller as you go from player 1 to 2 to 3 to 4?
Yes. There's a 5/6 chance that A doesn't win on the first roll, after which B is in exactly the same position that A was in terms of chance of winning. And so on down the line.
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#17
(Original post by DFranklin)
Yes. There's a 5/6 chance that A doesn't win on the first roll, after which B is in exactly the same position that A was in terms of chance of winning. And so on down the line.
I am thinking about the grid of counter problem still, will get back to you once I think I have a method! Thanks for all the help btw
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3 weeks ago
#18
For the grid and counters, I get the following.
Spoiler:
Show

1. 16!/(4!12!) - Just a straightforward combination of how 16 slots can be filled with 4 white counters.
2. 4^4 (4 ways a single white counter can appear in row one x 4 ways a single counter can appear in row 2 x ...)
3. 4*3*2*1 = 12
(4 ways a single counter can appea in the first row , then 3 ways one can appear in the second row and two ways one can appear in the third row and one way for the last row.)
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3 weeks ago
#19
(Original post by HedgePig)
For the grid and counters, I get the following.
Spoiler:
Show

1. 16!/(4!12!) - Just a straightforward combination of how 16 slots can be filled with 4 white counters.
2. 4^4 (4 ways a single white counter can appear in row one x 4 ways a single counter can appear in row 2 x ...)
3. 4*3*2*1 = 12
(4 ways a single counter can appea in the first row , then 3 ways one can appear in the second row and two ways one can appear in the third row and one way for the last row.)
Essentially fine, but 4*3*2*1 is not 12.
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#20
I know this is wrong, but what is the error with the thinking:

Total number of different outcomes is 16 C 4

Chance of them all being in different rows: 12/16 * 8/16 * 4 /16

Hence total number of out comes that work = total times chance.

I am pretty sure the error is with my 'chance calculation' am I double counting with that method?
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