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4 people spinning a slot machine at same time, each do 10 spins. During The spins there's a special symbol to collect. Probability for collecting the symbol is 0.45 each spin.

What's the probability that at the end of the 10 spins that 1 person will have collected the most OR that 2 to 4 people collected the exact same

I've calculated this as 8/15 is this correct? Got this by getting all the combinations of a win or loss and didnt take probabilty into account because everyone has the same probability? (keep in mind if people get equal that = a win)

E.g (below are all 15 possible combinations and 8 include "A" in them)

  1. A Wins

  2. B Wins

  3. C Wins

  4. D Wins

  5. A & B Wins

  6. A & C Wins

  7. A & D Wins

  8. B & C Wins

  9. B & D Wins

  10. C & D Wins

  11. A & B & C Wins

  12. A & B & D Wins

13 A & C & D Wins

  1. B & C & D Wins

  2. A & B & C & D Wins

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    This method is flawed, because the probabilities of the 15 events are not the same: 1–4 are the same as each other, but not the same as 5–10, etc. A brute force calculation shows that the answer is about $0.3393$. – 2017-01-29
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    Are you able to explain how you got this ? I'm quite interested – 2017-01-29
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    Oh I literally had a computer compute all possibilities. I had imagined that this was in the context of a class where you were looking for a more high-level approach...? – 2017-01-29
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    Trying to solve this on excel basically. – 2017-01-29
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    Ok: condition on the number of wins that $A$ gets. If $A$ wins exactly $n$ times, then calculate the probability that $B$, $C$, and $D$ each win at most $n$ times (those three probabilities will be the same, since spins are independent, I assume), and multiply by the probability that $A$ wins $n$ times. (All these probabilities are easy to calculate, using a binomial distribution.) Then sum from $n=0$ to $n=10$. – 2017-01-29

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