Consequtive winning dice rolls

jacob18

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Jul 9, 2018
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At my prom for fake money I rolled 17 times in a row either a 7 or an 11. So that's 17 throws, 17 wins. Don't remember the exact amount of 7's and 11's each. But what are the odds of 17 rolls 17 wins in a row. Thank you
 
At my prom for fake money I rolled 17 times in a row either a 7 or an 11. So that's 17 throws, 17 wins. Don't remember the exact amount of 7's and 11's each. But what are the odds of 17 rolls 17 wins in a row. Thank you
What are your thoughts regarding the assignment?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

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At my prom for fake money I rolled 17 times in a row either a 7 or an 11. So that's 17 throws, 17 wins. Don't remember the exact amount of 7's and 11's each. But what are the odds of 17 rolls 17 wins in a row. Thank you

I presume this is not an assignment, just curiosity. But since our goal is to help people learn to do math themselves, and that's far more useful than just getting an answer, I'm not going to give an answer until you at least tell us what you know about probability, and give it a try. It isn't hard, and the answer is as small a probability as you probably expect.

But I do want to pass along a comment on the relative uselessness of the answer to this sort of question. See my blog post about surprising events, where I comment on such questions we have been asked on my other website. This also gives an example of how to do your calculation, though it's not about dice.

One more thought: The blog may suggest some additional information worth giving us, such as how many other times you rolled. But if you won every time you rolled, you might want to consider that, since this is not a regulated casino but a fun event, they might not have been using standard dice ... in which case, as they say, all bets are off.
 
When you roll a pair of dice there are 6 x 6= 36 possible combinations. To get a "7" you must roll (6, 1), (5, 2), (4, 3), (3, 4), (2, 5), or (1, 6). So the probability of getting a "7" is 6/36= 1/6. To get an "11" you must roll (6, 5) or (5, 6). So the probability of getting "11" is 2/36= 1/18. The probability of getting either a "7" or an "11" is 1/6+ 1/18= 3/18+ 1/18= 4/18= 2/9. The probability of doing that 17 times in a row is \(\displaystyle (2/9)^{17}\). That is approximately 0.000000000000786.
 
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