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I have three identical coins, each marked 1 and 2 on their sides.

If I flip coin 1 and coin 2, the probability coin 1 is >= coin 2 is obviously 3/4. Same for flipping coin 1 and coin 3.

If I now flip coins 1,2,3, it seems the probability that coin 1 is >= coin 2 and coin 1 >= coin 3 should just be (3/4)^2 = 9/16, since the the two events (coin 1 >= coin 2, coin 1 >= coin 3) are independent.

I can see by simply writing out the possible results this is wrong - the correct probability is 10/16 = 5/8.

I can't wrap my head around why this is: am I wrong to think the two events are independent?

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    Clearly these are dependent events! knowing that coin $1$ beats coin $2$ is evidence that coin $1$ came up two. Not proof. just evidence. good enough to increase the probability that coin $1$ also beats coin $3$. – 2017-02-03
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    They are independent given coin 1 so you can get the result as sum of two products 1/2*1+1/2*(1-3/4) – 2017-02-03
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    @lulu Thanks, that makes perfect sense. I need to practice my understanding of "event". – 2017-02-06

1 Answers 1

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Observation:

Let $X_1,X_2,X_3$ be the random variables measuring the coin results, independently and identically having a uniform distribution over $\{1,2\}$.

$$\begin{align} \mathsf P(X_1 {\geq} X_2) &= \mathsf P((X_1{=}2\cap X_2{\leq}2)\cup(X_1{=}1\cap X_2{=}1)) \\[1ex] &= \mathsf P(X_1{=}2)\,\mathsf P(X_2{\leq}2)~+~\mathsf P(X_1{=}1)\,\mathsf P(X_2{=}1) \\[1ex] &= \tfrac 12\cdot 1+\tfrac 12\cdot \tfrac 12 \\[1ex] &= \tfrac 34 \\[2ex] \mathsf P(X_1 {\geq} X_2) & = \tfrac 34 \\[2ex] \mathsf P(X_1 {\geq} X_2\cap X_1{\geq}X_3) & = \mathsf P((X_1{=}2\cap X_2{\leq}2\cap X_3{\leq}2)\cup(X_1{=}1\cap X_2{=}1\cap X_3{=}1)) \\[1ex] &= \mathsf P(X_1{=}2)\,\mathsf P(X_2{\leq}2)\,\mathsf P(X_3{\leq}2)+\mathsf P(X_1{=}1)\,\mathsf P(X_2{=}1)\,\mathsf P(X_3{=}1) \\[1ex] & =\tfrac 12\cdot 1\cdot 1~+~\tfrac 12\cdot \tfrac 12\cdot\tfrac 12 \\[1ex] & = \tfrac 58 \end{align}$$

Conclusion:

Although the three random variables are mutually independent, it is apparent that the events $X_1{\geq}X_2$ and $X_1{\geq}X_3$ are not independent.   A brief contemplation should reveal that there is an obvious source of dependency.

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    Thank you, along with the comments above this helped. – 2017-02-06