MHB Warped Coin Probability: Flipping 5 Times - Find More Heads than Tails

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The warped coin has a 0.5 probability of landing on Heads, 0.4 for Tails, and 0.1 for landing on its Edge. When flipped five times, the challenge is to determine the probability of getting more Heads than Tails. The correct solutions were provided by members anemone and MarkFL. The calculations involve analyzing the outcomes of the flips, considering the unique probabilities of the warped coin. Ultimately, the discussion centers on the mathematical approach to solving this probability problem.
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A warped coin has probability of 0.5 of landing Heads, probability of 0.4 of landing Tails, and probability 0.1 of landing on its Edge. It is flipped 5 times. What is the probability that more Heads occur than Tails?
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Congratulations to the following members for their correct solutions:

1) anemone
2) MarkFL

Solution (from anemone):
The probability that more Heads occur than Tails

$\small=\text{P(5 Heads)}+\text{P(4 Heads 0 Tail 1 Edge)}+\text{P(4 Heads 1 Tail 0 Edge)}+\text{P(3 Heads 0 Tail 2 Edges)}+\text{P(3 Heads 2 Tails 0 Edges)}+\text{P(3 Heads 1 Tail 1 Edge)}$

$\;\;\small+\text{P(2 Heads 0 Tail 3 Edges)}+\text{P(2 Heads 1 Tail 2 Edge)}+\text{P(1 Heads 0 Tail 4 Edge)}$

$=0.5^5+\dfrac{0.5^4\cdot0.1\cdot5!}{4!}+ \dfrac{0.5^4\cdot0.4\cdot5!}{4!}+\dfrac{0.5^3\cdot0.1^2\cdot5!}{3!\cdot2!}+ \dfrac{0.5^3\cdot0.4^2\cdot5!}{3!\cdot2!}+\dfrac{0.5^3\cdot0.1\cdot0.4\cdot5!}{3!}+\dfrac{0.5^2 \cdot 0.1^3\cdot5!}{2!\cdot3!}+\dfrac{0.5^2\cdot0.4\cdot0.1^2\cdot5!}{2!\cdot2!}+\dfrac{0.5^1\cdot0.1^4 \cdot 5!}{4!}$

$=0.03125+0.03125+0.125+0.0125+0.2+0.1+0.0025+0.03+0.00025$

$=0.53275$
 
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