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 2 years 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?
 2 years 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?

This Question is Closed

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1@UnkleRhaukus @TuringTest @KingGeorge @experimentX @estudier

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I am terrible at probability, but everyone always asks me about it.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I'll be curious to hear what others say though

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1So am I. Took me like 3 days to solve this. But I do have the solution.

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0I think this is geometric distribution?

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1Well, no I don't think so.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1@Mariantheduke You might be interested. This problem came from the physics dept.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1I do not think this solution works.

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.01/n * (n)/n + 1/n *(n1)/n + 1/n * (n2)/n+...

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.01/n^2 (n+n1+n2+n3+...nn)

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1on the right track, but not it.

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0Did I missed something?

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1The 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.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1@saifoo.khan This would interest you, methinks.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1I'm sorry, that's the probability for the first player to lose. Wait a second.

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

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0I got dw:1349578761373:dw

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

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1The roll just made.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1Well, goodnight, Sir. I will see you another day perhaps.

saifoo.khan
 2 years ago
Best ResponseYou've already chosen the best response.0@badreferences : No man! I can't do probability. :S

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0when 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

gregohb
 2 years ago
Best ResponseYou've already chosen the best response.0If 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)

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1@gregohb That sets \(r=0\).

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0each 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

gregohb
 2 years ago
Best ResponseYou've already chosen the best response.0There 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.
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