Question

Probability question challenge. Two players roll an \(n\)-sided dice. The player who fails to get better rolls than previous ones loses. How likely is it for the first player to win?

Answer 1

@UnkleRhaukus @TuringTest @KingGeorge @experimentX @estudier

Answer 2

I am terrible at probability, but everyone always asks me about it.

Answer 3

I'll be curious to hear what others say though

Answer 4

So am I. Took me like 3 days to solve this. But I do have the solution.

Answer 5

I think this is geometric distribution?

Answer 6

Well, no I don't think so.

Answer 7

@Mariantheduke You might be interested. This problem came from the physics dept.

Answer 8

@mukushla

Answer 9

I do not think this solution works.

Answer 10

1/n * (n)/n + 1/n *(n-1)/n + 1/n * (n-2)/n+...

Answer 11

1/n^2 (n+n-1+n-2+n-3+...n-n)

Answer 12

on the right track, but not it.

Answer 13

Did I missed something?

Answer 14

The answer is\[\left(1-\frac1n\right)^{n-r}\]where \(r\) is the roll that has just occurred, and \(n\) the number of sides the dice has.

Answer 15

@saifoo.khan This would interest you, methinks.

Answer 16

I'm sorry, that's the probability for the first player to lose. Wait a second.

Answer 17

The probability that the first player wins is\[1-\left(1-\frac1n\right)^n\]where \(r=0\).

Answer 18

I got

Answer 19

I have it in good confidence that my answer is correct. I don't know what assumption you made that was wrong, though.

Answer 20

What is r ?

Answer 21

The roll just made.

Answer 22

Well, goodnight, Sir. I will see you another day perhaps.

Answer 23

WELL GOOD NIGHT

Answer 24

@badreferences : No man! I can't do probability. :S

Answer 25

when n=1 and n=2 your and my answer gives the same result Now, when n=3 according to u the answer is 0.7037 and according to me the answer is 0.66666
Now, when n=3 we can do it in a simple way.
1/3 *1 + 1/3 *2/3 + 1/3 *1/3 = 0.66666

Answer 26

If you know r, then its a conditional probability. the actual question was what was the overall probability of the first player winning (regardless of his roll)

Answer 27

@gregohb That sets \(r=0\).

Answer 28

each player rolles the same dice with expectation value \(\langle roll\rangle\)

the players are expected to roll the same number
there will be variation in the result , hence player one will win half of these and play two will win the other half , occasionally the players will roll the same number and in this event play one wins,

so if they play for a long time player one is more likely to win

Answer 29

There is something wrong with the original problem. If they roll one each, and stop, and first guys loses. Thats one thing. Or if they keep rolling until one loses, what if they both lose simultaneously. Or what if they keep rolling over and over again and each one wins some and loses some? Then do they keep count? Its not clear.

The prob of one guy losing on his own from one roll to the next is
(1/n)*(0 + 1/n + 2/n + ... (n-1)/n) = (n-1)*n/(2n^2) = (n-1)/(2n) = a

The probablity of the other guy to not lose is 1-a. So For 1st guy to lose and second guy to win (since these are indepedendent events) is a*(1-a).

Now if there is a series of rolls and first must continue to lose and 2nd guy continue to win, or whatever, then that needs to be expanded. but i am not clear what the problem is actually asking precisely.