I roll a die and flip three coins at the same time, and repeat this multiple times. What is the probability that I flip three heads twice in a row before I roll two 6's twice in a row?
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Solution:
The probability, in two (2) sequential events, that a set of three coins presents three heads twice before a die presents two (2) sixes, is equal to their respective ratios when calculated as odds against in the (fully populated) sample space.
The three coins and six-sided die have (8*6) 48 unique arrangements.
The two (2) events have (48^2) 2304 unique pairs of these arrangements.
Of the (2304) pairs of arrangements ONLY the pairs that have either two sixes or two three-head coin sets are of interest. The other 2205 pairs are irrelevant.
Sixty-four (8^2) of these pairs will have two (2) sixes and thirty-six (6^2) of these pairs will have two (2) sets of three heads. Note that one (1) pair will have both a pair of sixes and a pair of three-head sets that is shared by the (64) and (36) pairs. It’s necessary to subtract (1) from the (36) pairs of three-head sets, because this simultaneous appearance fails the requirement that the two sets of three-heads appear before the two sixes. (This does not affect the success of the die pair.)
The new sample space is now populated with 99 elements with the ratio of (64) pairs of die with a six to (36-1) pairs of three-head coin sets. When sampled randomly the odds against the coins are 64:35. This gives a probability of 35.356%
So the probability that a set of 3 coins has three heads twice in sequence before a die has two sixes in sequence is 35.356%
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Descriptive Arrays
The two arrays list the pairs of index numbers (from the array in post #10) that gives success to the Die or the Coins.
To use: choose a number on the left and then one of the corresponding numbers on the right.
For example (7, 3): Number 7 corresponds to (6-TTH) and the 3 corresponds to (6-HTH).
\(\begin{array}{|c|c|} \hline \text {Die} & \text { successes }\\ \hline \text {First selection} & \text{Second selection}\\ \hline 1 & \;\;\;2,3,4,5,6,7,8 \\ 2 & 1,2,3,4,5,6,7,8 \\ 3 & 1,2,3,4,5,6,7,8 \\ 4 & 1,2,3,4,5,6,7,8 \\ 5 & 1,2,3,4,5,6,7,8 \\ 6 & 1,2,3,4,5,6,7,8 \\ 7 & 1,2,3,4,5,6,7,8 \\ 8 & 1,2,3,4,5,6,7,8 \\ \hline \end{array} \text { } \begin{array}{|c|c|} \hline \text {Coin} & \text {successes }\\ \hline \text {First selection} & \text{Second selection }\\ \hline 1 & 1, 9, 17, 25, 33, 41 \\ 9 & 1, 9, 17, 25, 33, 41 \\ 17 & 1, 9, 17, 25, 33, 41 \\ 25 & 1, 9, 17, 25, 33, 41 \\ 33 & 1, 9, 17, 25, 33, 41 \\ 41 & 1, 9, 17, 25, 33, 41 \\ & \\ & \\ \hline \end{array} \)
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The title says this is tricky. There is an illusion when solving this: The die has six sides and eight selections, and the three coins have eight arraignments and six selections. Because of this cross match, I inverted the logic in the first solution –subtracting one from the die sets instead of the coin sets. I also doubled counted the success selections.
My quantum state of dumbness opened a portal and transported a shovelful of BS into my solution. This is definitely Spooky Dumbness at a Distance. The spookiest part is I probably would not have noticed this for months, or ever, if not for the parallel solution I constructed at your behest. Then the contamination became instantly obvious.
Now that I’ve extracted the bovine excrement, I’ll use it to fertilize my quantum flower garden. Flower equations thrive on this.
GA
