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Assuming that I use a standard die, what is the probability that I roll a 2 before I roll two odd numbers? The odd numbers do not have to be distinct. For example, 1,6,4,2 wins and 3,3 loses.

heyhuehei
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2 Answers2

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First observe that the numbers $4,6$ do not affect the outcome at all here, so it is equivalent to consider a $4$ sided die with sides $1,2,3,5$. The fact that it does not matter which odd number shows up means that we could consider a $4$ sided die with sides: $2$, odd, odd, odd.

The probability that I roll $2$ odd numbers before rolling a $2$ is then $$\frac34 * \frac34=\frac9{16}$$

Hence your probability is $1-\dfrac9{16}=\dfrac7{16}$.

Mathmo123
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  • Thanks, I understand this now. Can you also help me with http://math.stackexchange.com/questions/928177/what-is-the-probability-that-i-roll-a-2-i-roll-a-second-odd-number/928191 ? – heyhuehei Sep 12 '14 at 00:09
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    That link is taking me to this question! – Mathmo123 Sep 12 '14 at 00:10
  • Oops haha, http://math.stackexchange.com/questions/919928/how-can-this-technique-be-applied-to-a-different-problem – heyhuehei Sep 12 '14 at 00:11
  • Just keep in mind that, while we don't care which odd number shows up, it's critical that the odd-number faces are distinguishable. Otherwise the statistics are rather different. – Carl Witthoft Sep 12 '14 at 15:53
  • Can you clarify what you mean by distinguishable? Is there a problem with this answer? – Mathmo123 Sep 12 '14 at 15:59
  • @Mathmo123 No, the answer is correct in our physical world, as experiments have shown. It's just that there are some cases in quantum physics where particles (analogous to the three "odd" faces) are indistinguishable so the statistical properties change. It's related to the fact that there are 6 permutations of (1,2,3) but only one permutation of (1,1,1). – Carl Witthoft Sep 12 '14 at 16:37
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It is reasonably clear that this probability exists. Call it $p$, the probability of winning. We use the method of conditioning on the result of the first toss.

If the first toss is a $2$, we have won..

If the first toss is a $4$ or a $6$, the probability we ultimately win remains at $p$.

If the first toss is odd, then we want the probability of tossing a $2$ before the second odd. This is $\frac{1}{4}$, since there are $3$ odd and only one $2$. Thus $$p=\frac{1}{6}+\frac{2}{6}p+\frac{3}{6}\cdot\frac{1}{4}.$$ Solve for $p$. We get $p=\frac{7}{16}$.

André Nicolas
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