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In lecture we learned that rolling two indistinguishable die and how the chance of doubles is the same probability as rolling distinguishable die. But I'm having difficulty with a certain component of the lecture.

For the indistinguishable 2 6-sided dice we have $\{(1,1)..(6,6)\} \cup \{(1,2), (1,3)....(5,6)\}$ 21 total outcomes and a probability space of total 21. Then we learn that the probability, q, of not getting doubles is twice as likely as getting doubles, p.

So $q=2p$

But I got confused on the above step. How is it double since 6/21 for doubles vs. 15/21 for non doubles?

Thanks

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    $\begingroup$ Do not forget... the probability of occurrence can be given as the ratio of favorable outcomes divided by total outcomes only if each of the outcomes are equally likely to occur. That is not the case here. There are two outcomes to a lottery, either you win or you lose but you don't win with probability $\frac{1}{2}$. $\endgroup$ – JMoravitz Nov 8 at 15:38
  • $\begingroup$ Now... as for the statement that was actually said, I worry you may be confusing things. Regardless of whether or not you roll two distinguishable dice or two indistinguishable dice, the probability of having rolled doubles is $\frac{6}{36}=\frac{1}{6}$ (and is not $\frac{6}{21}$). What is probably intended to have been said, is that the probability of having rolled a specific double (such as $(1,1)$) is half as likely as having rolled a specific pair of different numbers (such as $(1,2)=(2,1)$ since these are considered the same in the indistinguishable case) $\endgroup$ – JMoravitz Nov 8 at 15:43
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    $\begingroup$ If the indistinguishableness of the dice confuses you, don't worry you aren't alone. Imagine that you see color normally and the two dice are green and pink, obviously different and distinct and you roll the dice and you expect certain probabilities for certain outcomes. You hand the dice to your colorblind friend who can't tell the difference between them and he rolls the dice. The probabilities don't suddenly change just because he can't see color, the only things that change is his ability to distinguish between certain results. $\endgroup$ – JMoravitz Nov 8 at 15:46
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As @JMoravitz pointed in the comments what happen here is that the probably to get $(1,1)$ is not the same to get $(1,2)$ when the dice are not distinguishable. When you write $6/21$ or $15/21$ you are assuming that every event have the same probability to occur, but this is not the case.

If the dice are not distinguishable then, as you knows, you have $21$ different outcomes, however they doesn't have the same probability to occur. If the dice are distinguishable then there are $36$ different events and this time each event have the same probability to occur.

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