Maximizing your chances

[Gnosis]

This is directly poker related, but is a fun probablity problem:

N is a large number (say 10,000) You need to figure out how many dice you must roll, where the sum of all the dice is N (10,000) - how many dice do you need to maximize your chance of rolling N?

[Rhound50]

I'm not sure this is completely right but here is my guess the average dice roll as you approach infinte rolls is going to be 3.5. So my guess is going to be 10,000/3.5 dice. In the general form n/3.5. Just a guess.

[Kuso]

well, assuming 6-sided dice, i would agree with Rhound... but i am no probability guru.

i think the likelihood of getting exactly 10,000 would increase if you used 4-sided dice, with which the number would be n/2.5. The higher number of rolls will lessen the variance while maintaining the same EV.

[Gnosis]

[Kuso]

well, i can't really prove it, but i think that your example is not valid as 6 is within the maximum value of the die.

i think if you used 7 as an example instead of 6, you will come up with the N/3.5 answer that Rhound and I suggested.

1 die = impossible
2 dice = 6/36 (1/6) ~ 0.17
3 dice = 15/216 (5/72) ~ 0.07
...
7 dice = 1/279936 ~ really small

so by your answer, you are suggesting that 10000/6=1667 is the best answer, but this means that you will only get it 1 out of 6^1667 times. i think the N/3.5 answer is better.

i hope you can prove me wrong... just please do so in layman's terms.

[TightWad]

Kuso,

I'm not sure your example is conclusive. Gnosis is saying the best way is to use the minimum number of dice; in the case of seven, that would be 2 dice. You're saying to divide the number by 3.5; in the case of seven, that's also 2 dice. So using 7 as an example doesn't seem to tell us much.

Unless, of course, I'm missing something. That happens now and then!

-TW

[Kuso]

[Gnosis]

[Kuso]