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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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badreferencesBest ResponseYou've already chosen the best response.1
@UnkleRhaukus @TuringTest @KingGeorge @experimentX @estudier
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
I am terrible at probability, but everyone always asks me about it.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
I'll be curious to hear what others say though
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
So am I. Took me like 3 days to solve this. But I do have the solution.
 one year ago

LibnizBest ResponseYou've already chosen the best response.0
I think this is geometric distribution?
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
Well, no I don't think so.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
@Mariantheduke You might be interested. This problem came from the physics dept.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
I do not think this solution works.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
1/n * (n)/n + 1/n *(n1)/n + 1/n * (n2)/n+...
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
1/n^2 (n+n1+n2+n3+...nn)
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
on the right track, but not it.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Did I missed something?
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
The answer is\[\left(1\frac1n\right)^{nr}\]where \(r\) is the roll that has just occurred, and \(n\) the number of sides the dice has.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
@saifoo.khan This would interest you, methinks.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
I'm sorry, that's the probability for the first player to lose. Wait a second.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
The probability that the first player wins is\[1\left(1\frac1n\right)^n\]where \(r=0\).
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
I got dw:1349578761373:dw
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
I have it in good confidence that my answer is correct. I don't know what assumption you made that was wrong, though.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
The roll just made.
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
Well, goodnight, Sir. I will see you another day perhaps.
 one year ago

saifoo.khanBest ResponseYou've already chosen the best response.0
@badreferences : No man! I can't do probability. :S
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
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
 one year ago

gregohbBest ResponseYou've already chosen the best response.0
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)
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
@gregohb That sets \(r=0\).
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
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
 one year ago

gregohbBest ResponseYou've already chosen the best response.0
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 + ... (n1)/n) = (n1)*n/(2n^2) = (n1)/(2n) = a The probablity of the other guy to not lose is 1a. So For 1st guy to lose and second guy to win (since these are indepedendent events) is a*(1a). 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.
 one year ago
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