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A magician holds one six-sided die in his left hand and two in his right. What is the probability the number on the dice in his left hand is greater than the sum of the dice in his right?

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closed as off-topic by user99914, NCh, Jyrki Lahtonen, PSPACEhard, Deepesh Meena Sep 11 '18 at 3:28

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Some ideas: count the appropiate events

$$\begin{align}&Left\;\; Hand&Right\;\; Hand\\{}\\ &3&(1,1)\\{}\\ &4&(1,1),\;(1,1),\,(2,1)\\{}\\ &5&(1,1),\,(1,2),\,(2,1),\,(2,2),\,(1,3),\,(3,1)\end{align}$$

and I leave to you to fill the last line and do the final mathematics.

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  • $\begingroup$ 1,1 must be 1,2 and Thankz.. $\endgroup$ – Ujjwal Agrawal Jul 4 '15 at 21:19
  • $\begingroup$ Why would $\;(1,1)\;$ be $\;(1,2)\;$ ? It fits exactly with what the other answer tells you: for example, with a $\;3\;$ in the left hand you win only if the right hand's dice show $\;(1,1)\;$ , which is $\;1/36\;$ , and etc. $\endgroup$ – Timbuc Jul 5 '15 at 7:05
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Bag one contains six elements $$\{1,2,3,4,5,6 \}$$ Bag two is the multiset of all possible sums of two dice rolls, of which there are 36, $$\{2,3^2,4^3,5^4,6^5,7^6,8^5,9^4,10^3,11^2,12\}$$

Considering the first bag, a $1$ or a $2$ wins $0/36$th of the time (never). A $3$ wins $1/36$th of the time. A $4$ wins $3/36$ of the time.

How many combinations are there of one element from each bag? How many of these combinations win?

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