Hi, two questions for the math guys:

what is the possibility of hitting 5 6's in a row with a dice?

and

if I roll the dice 10.000 times, what is the chance that I will hit 5 6's in a row once?

Thanks

If you start right now and say, "I will roll 5 sixes in a row on a 6-sided fair die" the probability is 1/6^5 or 1/6 * 1/6 * 1/6 * 1/6 * 1/6 = .000129.

The second question is a bit harder and I believe it requires a streak calculator to solve. BruceZ has dedicated quite some time to correcting problems with a variety of inaccurate streak calculators and has made his own.


edit: In a real quick simulation that I did not double check I found that 65.17% of the time (out of 10,000 trials) if you rolled a die 10,000 times you would find a string of 5 or more "sixes" in a row.

This one is accurate:

http://www.pulcinientertainment.com/...tor-enter.html

and it gives 65.8% for at least one run of at least 5 sixes in a row out of 10,000 rolls. To get that, enter 10000, 5, and 0.16666667 in the first 3 boxes.

That's based on the recursive equation:

p(n) = p(n-1) + [1 - p(n-6)]*(5/6)*(1/6)⁵ for n > 6

p(n) = 0 for n < 5

p(5) = (1/6)⁵

p(6) = (5/6)*(1/6)⁵

where p(n) is the probability of having gotten a run of at least 5 in a row in n rolls, so we want p(10,000).

It is also possible to approximate this very closely by hand, but the above gives the exact answer.

You can download a fancier streak calculator here that runs in Excel and also gives the probabilities for multiple streaks.

Thanks!!

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?

but over lets say 1000k tries you will win it close to 100 times?

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 =~ 65.8%

in agreement with the streak calculator.


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.

I believe that percentage approaches 1 - 1/e as N gets large. That is, for any P=1/N then in N trials where N is large, the chance to hit is 1 - 1/e. Which is ~63.2%


EDIT: type too slow for Bruce.