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I have the following problem for which I'm not sure my solution is correct:

A dice is constructed in such a way that 1 dot occurs twice more often than the rest of the points. The probabilities for the rest of the dots are mutually equal. The dice is thrown 2 times.

Calculate the probability that the dots on the second dice are more than the dots on the first one.

My solution:

Let x be the probability for 1, and y the probability for anything else.

$$ \left\{ \begin{array}{c} x=2y\\ x+5y=1 \end{array} \right. $$

I get that $x=\frac{2}{7}$ and $y=\frac{1}{7}$. I have four different scenarios for the dots - $(1, 1), (1, i), (i, j), (i, 1)$, where $2 \le i \le 6$ and $2 \le j \le 6$. I have denoted those cases $H_1, H_2, H_3 $ and $H_4$ respectively. For the probability of the desired event I'm using the formula for total probability:

$$P(A)=\sum_{i=1}^4P(H_i)P(A|H_i)=\frac{2}{7}\frac{2}{7}0+\frac{2}{7}\frac{5}{7}1+\frac{5}{7}\frac{5}{7}(\frac{10}{49})+\frac{5}{7}\frac{2}{7}0=\frac{740}{49^2} \approx 0.30$$

Now, is this correct and are there other ways to solve this problem?

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  • $\begingroup$ So it's a 7-sided die?But is there even a 3D object with 7 sides so that the probability of any side coming 'up' is the same for all 7 sides? Maybe they mean it's 6-sided, two of the ones are 1, and you have 4 'other' sides? $\endgroup$
    – Bram28
    Aug 29, 2017 at 13:03
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    $\begingroup$ No, it's a 6-sided die, with faces numbered 1 to 6. The die is weighted so that one side comes up more often than any of the others. $\endgroup$
    – Especially Lime
    Aug 29, 2017 at 13:06
  • $\begingroup$ @EspeciallyLime Oh, I get it ... the 'occurs' means how often it comes up, not how many sides. THanks! $\endgroup$
    – Bram28
    Aug 29, 2017 at 13:07

2 Answers 2

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I don't think the term $\frac57\frac57\frac{10}{49}$ is right. $\frac57\frac57$ is the probability that both are greater than $1$, but if that happens the probability that $i<j$ is quite a bit bigger than $\frac{10}{49}$.

The right way to approach this is that the probability of the second die showing more dots is equal to the probability of the first die showing more dots, so the probability you want is just half the probability that the two dice show different values. So work out the probability they show the same value, subtract from $1$, and halve.

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  • $\begingroup$ It should be noted that this only works because the probability distributions of the two dice are the same (in this case because it is the same die). If OP were asked to find the probability the biased die rolled a higher score than a fair die, he would have to sum conditional probabilities. $\endgroup$
    – Tez LaCoyle
    Aug 29, 2017 at 13:09
  • $\begingroup$ Actually if you want to know the probability that the biased die rolls a higher score than a normal die, that is exactly $6/7$ of the probability that a normal die rolls higher than another normal die, because you can think of the result of a biased die as the linear combination of $\frac67X+\frac17Y$, where $X$ is a normal die and $Y\equiv 1$ (and rolling a $1$ always loses). $\endgroup$
    – Especially Lime
    Aug 29, 2017 at 13:18
  • $\begingroup$ True - I was thinking more in terms of a general case where you can't apply the symmetry argument, but the example I gave was clearly not that difficult to solve either. $\endgroup$
    – Tez LaCoyle
    Aug 29, 2017 at 13:37
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The $\frac{10}{49}$ term should be $\frac{10}{25}$: out of the 25 cases where you get two non-1's, 5 cases are where $i=j$, leaving half of the remaining 20 where $i<j$

The fact that you ended up with such a weird fraction for your final answer should have been a clue that something is wrong, as clearly you should end up with something of the form $\frac{x}{49}$

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