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I have ordered sets. Each index in the set is random and independent.

Here is a sample of possible sets:

$X_1 < X_2 < X_3 < X_4 < X_5 < X_6 < X_7 < X_8 < X_9 < X_{10}$:

$B_1[{X1,X2,X3,X4,X5,X6,X7,X8,X9,X10}]$

$B_2[{X2,X3,X4,X5,X6,X7,X8,X9,X10,X1}]$

$B_3[{X3,X4,X5,X6,X7,X8,X9,X10,X1,X2}]$

$B_4[{X4,X5,X6,X7,X8,X9,X10,X1,X2,X3}]$

$B_5[{X5,X6,X7,X8,X9,X10,X1,X2,X3,X4}]$

$B_6[{X6,X7,X8,X9,X10,X1,X2,X3,X4,X5}]$

$B_7[{X7,X8,X9,X10,X1,X2,X3,X4,X5,X6}]$

$B_8[{X8,X9,X10,X1,X2,X3,X4,X5,X6,X7}]$

$B_9[{X9,X10,X1,X2,X3,X4,X5,X6,X7,X8}]$

$B_{10}[{X10,X2,X3,X4,X5,X6,X7,X8,X9,X1}]$

The probability of getting X10 on first try is 1/10.

I want to find the probability of getting X10 (after choosing the first index of a sequence $B_i$ ) by choosing the next highest value after the first index.

Here is what I did.

If I choose X1, Then my probability to choosing a higher value and X10 is 1/9; so on;

X2, my probability of getting a higher value and X10 is 1/8 (excluding X1 since if I get X1 I can choose again)

X3,1/7

X4,1/6

X5,1/5

X6,1/4

X7,1/3

X8,1/2

X9,1

X10,0

Here is where I am unsure.

Since these events are disjoint: 1+1/9+1/8+1/7+1/6+1/5+1/4+1/3+1/2 = 2.8 Err probability is more than one.

What can I do?

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The events you are describing in your list are conditional probabilities. You cannot directly add them. If we let $D_1$ and $D_2$ be your first draw and second draw (ignoring draws that are lower than $D_1$), the what you have calculated is $P(D_2=10|(D_1=i\cap D_2>D_1))$. The disjoint events you are looking for is $P((D_1=i \cap D_2=10)|D_2>D_1)$

The best way to think of this is as pairs of draws $(D_1,D_2)$. We want to restrict our sample space to just the pairs where $D_2>D_1$. Then, we want to look at the number of pairs where $D_2=10$.

How do we get this probability? -- let's use Bayes rule:

$P((D_1=i \cap D_2=10)|D_2>D_1) = P(D_1=i|D_2>D_1)P(D_2=10|(D_1=i\cap D_2>D_1)) =P(D_1=i|D_2>D_1)\frac{1}{10-i}$,

This uses the result you calculated in your post. The only part we are missing is the first probability:$P(D_1=i|D_2>D_1)$. Given that the second draw is greater than the first, how many of these pairs involve $D_1=i$?

The number of such pairs is $10-i$. The total number of such pairs is $\sum_{i=1}^{10} (10-i)=100-\sum_{i=1}^{10} i = 100 - \frac{10\times 11}{2} = 45$.

Therefore, your probabilities are:

$P((D_1=i \cap D_2=10)|D_2>D_1) = \frac{10-i}{45}\times \frac{1}{10-i} = \frac{1}{45} \approx 0.022$

Also, in your post you have included the case where $D_1=10$, even though its impossible to satisfy your conditions if you draw 10 first, since you will never get to a second draw.

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  • $\begingroup$ $\frac{1}{45}= .022 \neq .22$ $\endgroup$ – RohanLatinSindhi Sep 12 '14 at 18:02
  • $\begingroup$ @RohanLatinSindhi LOL...oh boy, its one of those days. Thanks for pointing that out. $\endgroup$ – user76844 Sep 12 '14 at 18:19

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