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Part a) Let A be the number that the left-hand die shows. Let B be the number that the right-hand die shows. Both dice are fair and therefore the probability that B is greater than A is equal to the probability that B is less than A (i.e. P(B > A) = P(B < A)). Furthermore, all possibilities can be summarized by B > A, B < A, and B = A. Therefore

P(B > A) + P(B < A) + P(A = B) = 1

Note that P(A = B) is 1/6 because whatever number B shows, A will show the same number one out of six times. Therefore equation 1 becomes:

P(B > A) + P(B < A) + 1/6 = 1

Since P(B > A) = P(B < A), we can arrive at our solution with:

2P(B > A) = 1 − 1/6 = 5/6

Therefore

P(B > A) = 5/12

The probability that the right-hand die shows a larger number than the left-hand die is 5/12

Part b)

Given that the left hand die is rolled first and it shows 5 (i.e. A = 5). What is P(B > A|A = 5)? The answer is 1/6 because to be greater than A = 5, die B must show 6 (which occurs one out of six times for a six sided die).

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    $\begingroup$ What is your question exactly? $\endgroup$ – Tara B Feb 28 '13 at 0:27
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For a, as long as both dice follow the same distribution, normal, uniform, or whatever, the probability of each die being higher is the same. This gives you $P(B \gt A) = \frac 12(1-P(A=B))$. Usually the normal distribution is a continuous one so the chance of equality is zero.

For b $P(B \gt A|A=5)=P(B\gt 5)$. Again, is the distribution of $B$ continuous?

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  • $\begingroup$ Dice with continuous result? Come on, you must be pulling the OP's leg! $\endgroup$ – Did Feb 28 '13 at 17:18
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For two independent identically distributed random variables $X$ and $Y$ that have a continuous distribution (such as the normal), the situation is even simpler. Since $\Pr(X=Y)=0$, we get that $\Pr(X\gt Y)=\frac{1}{2}$.

Analogues of the second question can be solved in the continuous context, usually by an integration. One nice thing is that we do not have to distinguish between $\lt$ and $\le$. This is because if $X$ has continuous distribution, then, for example, $\Pr(X=5)=0$.

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  • $\begingroup$ Which part(s) of the question is this addressing? $\endgroup$ – Did Feb 28 '13 at 17:18
  • $\begingroup$ Part a) for sure. Also part b), interpreted as asking how one computes $\Pr(X\gt Y|Y=5)$ in the case $X$ and $Y$ are independent and have known continuous distributions, possibly normal. $\endgroup$ – André Nicolas Feb 28 '13 at 17:34
  • $\begingroup$ Seriously? Both parts of the question are concerned with the results of dice. $\endgroup$ – Did Feb 28 '13 at 17:44
  • $\begingroup$ We do not appear to have the same view of the intent of the question. Perhaps the OP can clarify. $\endgroup$ – André Nicolas Feb 28 '13 at 17:50

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