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What are the odds of throwing $10$ six sided dice and landing all the same number. Also, how many throws would I need to do to achieve a $100$ percent success of this happening. Is this even possible in terms of mathematical probability?

David_Shmij
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Sean Winter
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    However many times one throws, there cannot be certainty of success. – André Nicolas Jul 24 '16 at 22:23
  • It seems straightforward to compute the chance of this happening in one throw. Did you work through the details? – hardmath Jul 24 '16 at 22:27
  • The answers posted so far are bare assertions that don't explain how to find the solution. $\qquad$ – Michael Hardy Jul 25 '16 at 00:26
  • It depends what you mean by throw. Will you throw all $10$ at once or one at a time? Also I disagree with Andre Nicolas. If you get all the same the that is a certain success. Just keep going until you observe it. The number of throws cannot be determined beforehand to get to this state but if you count the states as you go then observe your desired goal condition, that is your answer. Power of observation. Using computer simulation , you can verify that it is possible to achieve this special condition of all $10$ the same. – David Jul 25 '16 at 04:21
  • I know the level of certainty never hits $100$% but it will be so close that so for all intents are purposes, it is basically $100$%. Simulate $1$ billion $10$ dice rolls and what is the probability of never getting all $10$ dice the same? Practically $0$%. – David Jul 25 '16 at 04:28

3 Answers3

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To answer your first question: The probability that ten six-sided dice land with the same number is equal to the probability that nine of the six-sided dice have the same number as the tenth i.e. $1/6^9$.

Your second question does not make sense, unless you are asking how many rolls you need to make for the probability of ten six sided dice landing on the same number to happen almost surely. Then the answer is an infinite number of rolls.

smcc
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The odds are $1$ to $6^9-1$. The probability is $\frac1{6^9}$.

This is because there are $6$ choices for the face and $6^{10}$ values for $10$ dice.

After $n$ throws, the chances of getting all $10$ the same at least once is $$ 1-\left(\frac{6^9-1}{6^9}\right)^n $$ It never is exactly $1$.

However, to get a $50\%$ chance of seeing all faces the same, you would need to throw the dice $6,\!985,\!327$ times, and to get a $99\%$ chance, $46,\!409,\!503$ times.

robjohn
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For your first question: The probability of throwing $10$ six sided dice and having them all land on the same arbitrary number is = $(\frac{1}{6})^{10} \times $ ${6}\choose{1}$ =$\frac{1}{10077696}$ assuming fair six-sided dice.

Notice that if you are simply looking for answers to these kinds of questions, you can use WolframAlpha, for example try this.

This related post can help with some methodology if required.

Community
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David_Shmij
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