## badreferences 3 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?

@UnkleRhaukus @TuringTest @KingGeorge @experimentX @estudier

2. TuringTest

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

3. TuringTest

I'll be curious to hear what others say though

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

5. Libniz

I think this is geometric distribution?

Well, no I don't think so.

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

@mukushla

I do not think this solution works.

10. sauravshakya

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

11. sauravshakya

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

on the right track, but not it.

13. sauravshakya

Did I missed something?

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.

@saifoo.khan This would interest you, methinks.

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

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

18. sauravshakya

I got |dw:1349578761373:dw|

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

20. sauravshakya

What is r ?

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

23. sauravshakya

WELL GOOD NIGHT

24. saifoo.khan

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

25. sauravshakya

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

26. gregohb

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)

@gregohb That sets $$r=0$$.

28. UnkleRhaukus

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

29. gregohb

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.