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"We throw 3 dice one by one. What is the probability that we obtain 3 points in strictly increasing order"?

Isn't the answer just $1/6$ as there are $3!$ possible permutations and only 1 permutation in which the 3 dice will be in strictly increasing order. However, the answer in my book is $5/54$.

Similarly, for the question of "what is the probability of 10 people being seated in strictly increasing order of age at a table?" Wouldn't the solution be $20/10!$ as direction of the increasing order has not been specified. I think this answer is correct whilst my answer for the previous question is wrong even though I am using the same logic and I can't see why this is, when they are effectively very similar questions.

Thank You

Jojo
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  • Hint: what three events are you permuting for the dice case? It is not true that there is only one way to get an increasing sequence. You could throw ${1,2,3}$,${1,2,4}$,${4,5,6}$ or many other cases. – lulu Sep 28 '15 at 13:14

4 Answers4

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For each selection of $3$ out of $6$ numbers, there is exactly one way to arrange them in order, so there are $\binom63=20$ different strictly ordered outcomes, which yields a probability of $20/6^3=5/54$.

For the seating question, your answer is correct if this is a round table and the we can start anywhere and move in either direction. The difference to the first case is that you're placing exactly those $10$ people, whereas in the case of the dice you can select some elements out of a larger set and then arrange them in order.

joriki
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For a strictly increasing sequence, you need three different numbers to show, and each such outcome has just one of strictly increasing sequence, hence

$Pr = \dfrac{\binom{6}{3}}{6^3}= \dfrac{20}{216}= \dfrac{5}{54}$

PS

I believe your answer for the 2nd question is correct for a round table. For seating in a row (assuming either direction) it would be $\dfrac{2}{10!}$

true blue anil
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  • Then how does the factor $20$ arise? – joriki Sep 28 '15 at 13:42
  • My bad, amended. – true blue anil Sep 28 '15 at 13:53
  • I don't understand what "numbered" in boldface refers to. If the numbering has any significance, I would have expected it to mean that the increasing order of age would have to be consistent with the numbers, and that would yield $1$ possibility, not $20$. What other significance could a numbering have? – joriki Sep 28 '15 at 13:56
  • Well, it does seem a bit odd, but the youngest could be allotted any of 10 numbered seats, and the others arranged clockwise or anticlockwise with respect to the youngest to yield $\frac{20}{10!}$ which, of course is the same numerically as $\frac{2}{9!}$ for an unnumbered table. What I had meant was that the exact expression $\frac{20}{10!}$ as written by OP would arise for a numbered round table, but I get your point, and am editing it. – true blue anil Sep 28 '15 at 14:19
  • @trueblueanil So my answer for the second question is incorrect as I said $20/10!$? Could you explain how you get $2/10!$ please? – Jojo Sep 28 '15 at 14:50
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    @Jojo: If they are seated in a row, there are only two arrangements in strictly ascending order, (left to right and right to left) against 10! possible arrangements – true blue anil Sep 28 '15 at 14:56
  • Thats a bit weird to me. Does it mean that if you arrange a set of integers in decreasing order it sould be called 'increasing from the right'? – Alex Sep 28 '15 at 15:48
  • @Alex: What you say is true for a set of integers by convention, but is there such a convention for people, too ? Seems like an Alice In Wonderland situation. If two directions are allowed on a round table, then why not in a row ? I concede that yours might be a more common view. Turning out to be an interesting question ! – true blue anil Sep 28 '15 at 16:10
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The possible arrangements that make that happen, since the order counts, are $123, 124, 125, 126, 134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256, 345, 346, 356, 456$ , so there's $20$ of them. How many arrangements are possible? Well, $6^3$ is the right answer so it's $216$. $$\frac{20}{216} = \frac{5}{54}$$

true blue anil
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AnalysisStudent0414
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For 2) it is $\frac{1}{10!}$ assuming all people are of different age

Alex
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