Originally Posted by Okerchamp
so if there is 1/10000 chance that you will win. then there is only like 60% you will win it the first 10000 tries?
Correct, if your chance of winning is 1/n on one try, then the chance of winning at least once in n independent tries will be close to 1 - 1/e =~ 63.2% for large n. That's assuming that the tries are independent. We don't have 10,000 independent tries for the dice rolls. If we define a try as any 5 consecutive rolls, then these aren't independent since getting 5 in a row on say the 5th roll makes it much more likely that we get 5 in a row on the 6th roll too. There would also only be 9996 of these tries since rolls 1-4 can't produce 5 in a row. We can define a try differently so that a try ends with a non-six or 5 consecutive sixes. These tries are independent, but then the number of tries would be much less than 10,000 because the length of a try can be anything from 1 to 5, not just 1, and it won't always be the same. The average length of a try in that case would be about 1.2 rolls, and the average number of independent tries would be about 10000/1.2 or 8334 rolls, but this actual number of tries will vary. Nevertheless, this is the basis for a very accurate approximation:
1 - (1 - 1/6^5)8334
in agreement with the streak calculator.
but over lets say 1000k tries you will win it close to 100 times?
If something happens 1 time in 10,000 tries on average, then it will happen 100 times in a million tries on average. That is true even if the tries are not independent. But with the dice problem, a "try" is not the same as a roll. If we define a try as above, a try can be anywhere from 1 to 5 rolls, and on average it would be 1.2 rolls. You get 5 in a row on average once every 6^5 = 7776 tries, and this would be about 7776*1.2 = 9330 rolls. So in a million rolls, 5 in a row would occur on on average about 1000000/9330 =~ 107 times.