Timeline for Two dice problem
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Jul 10, 2013 at 14:42 | vote | accept | Ivana | ||
| Jul 9, 2013 at 22:29 | answer | added | whuber♦ | timeline score: 8 | |
| Jul 9, 2013 at 9:15 | comment | added | Glen_b | In essence we're just exploiting the symmetry of the bivariate cdf. But as I said earlier, likely there's a simpler way. Indeed, I can see ways to exploit the symmetry further, but I expect there are ways to do it even more easily. | |
| Jul 9, 2013 at 9:03 | comment | added | Glen_b | With that in mind, the bivariate distribution of the winner's score against the losers looks like this, where probability of interest is the proportion of the distribution above the pink line. The complement - the proportion below it, relates to my previous comment (where you don't differentiate winner from loser). You may not be able to compute it algebraically, but many programs will compute the bivariate cdf for you. | |
| Jul 9, 2013 at 8:50 | comment | added | Glen_b | @COOLSerdash If you treat the pair of scores as bivariate normal, could you work out $P(\text{both players score} < 100.5)$? | |
| Jul 9, 2013 at 8:29 | comment | added | COOLSerdash | @Glen_b Thanks for your comments. I have managed to calculate the probability that a given player will collect more than 100 point in 100 throws, the expectation and variance of the score of a single throw and the covariance of the per-round score of the two participants. I just don't know how to proceed from here. | |
| Jul 9, 2013 at 5:32 | comment | added | Ivana | I know I'm supposed to avoid $n^{th}$ convolution. I also know how to work out that one person will win more than 100 points in 100 throws. I have trouble when it comes to two players and distribution of value differences. | |
| Jul 9, 2013 at 0:16 | answer | added | user25658 | timeline score: 1 | |
| Jul 9, 2013 at 0:14 | comment | added | user25658 | I'm posting some simulation code to validate your answers. | |
| Jul 8, 2013 at 23:05 | comment | added | Glen_b | @whuber I have the same answer as you. | |
| Jul 8, 2013 at 22:45 | comment | added | Glen_b | It's not too difficult to also work out the covariance of the per-round score of the two participants, and hence of the total score. From there, it's not difficult to compute the relevant probability - but there is probably an easier way to do it than that. | |
| Jul 8, 2013 at 22:38 | comment | added | Glen_b | Ivana - the point with the CLT is that you avoid the $n^{th}$ order convolution. Can you work out the probability that a given person (without knowing they won or lost overall) collects > 100 points in 100 throws? More basically, but more to the point, can you work out the (approximate) distribution of their total score? Failing that, can you work out the mean and variance of the distribution of the number of points they gain from a single throw? | |
| Jul 8, 2013 at 21:40 | comment | added | Ivana | Yes, I am familiar with CLT. I assumed I would have to get convolution for the difference of values, but I don't know how since distribution for die is discrete. | |
| Jul 8, 2013 at 20:37 | comment | added | whuber♦ | The answer--if I'm correct this time ;-)--will be around 0.713. | |
| Jul 8, 2013 at 20:34 | comment | added | whuber♦ | @COOL Yes, that's realistic and it's related to this problem, but it's not the answer. You have accurately computed the chance that a given player will collect more than 100 points in 100 throws. The question asks for the chance that the winner will collect more than 100 points. | |
| Jul 8, 2013 at 20:33 | comment | added | COOLSerdash | @whuber I'm very unsure about this, but I'm getting an estimate of $0.3998819$. Is that realistic? | |
| Jul 8, 2013 at 20:22 | comment | added | whuber♦ | I was wrong anyway. :-(. How accurate an estimate do you need? Are you familiar with the Central Limit Theorem? | |
| Jul 8, 2013 at 19:51 | comment | added | Ivana | I was hoping for help with method more than the final number... | |
| Jul 8, 2013 at 19:34 | review | First posts | |||
| Jul 8, 2013 at 20:17 | |||||
| Jul 8, 2013 at 19:17 | history | edited | Ivana |
edited tags
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| Jul 8, 2013 at 19:17 | history | edited | Peter Flom |
Added self-study tag, since it is clear that this is such
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| Jul 8, 2013 at 19:15 | comment | added | Peter Flom | Is this a homework problem? Sure sounds like it, so, it should have the self-study tag | |
| Jul 8, 2013 at 19:14 | history | asked | Ivana | CC BY-SA 3.0 |
