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Annoying Math Problem


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Came across another problem similar to this recently. My friend was playing roulette and was betting on four numbers only on each spin. In 10 spins, one of his four numbers came up 6 of the ten times.Using what I think is the correct formula, it was(4 / 37) * (4 / 37) * (4 / 37) * (4 / 37) * (4 / 37) * (4 / 37) * (33 / 37) * (33 / 37) * (33 / 37) * (33 / 37) * 10 = 1.0101793 × 10-5 likely for this to happen? This is almost 100,000-1, yes?
Not quite, but you're close. You should replace the "10" by the number of ways in which the winning 6 tries could have been distributed over the 10 tries. That number is(10*9*8*7*6*5)/(6*5*4*3*2*1) = 210Explanation: say i want to put the letters A-F in 10 boxes. The boxes in which the letters end up correspond to the winning tries. I could put the A in 10 different boxes, then the B in 9 remaining boxes, etc. This gives 10*9*8*7*6*5 possibilities. However, in the example above I don't care about the ordering: putting A in the first box and B in the second box would give me the same two winning tries as doing it the other way around. So I have overcounted every possibility by the number of ways of reordering the six letters, which (by the same reasoning) is 6*5*4*3*2*1.Long story short: your answer should be 21 times as large.
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Not quite, but you're close. You should replace the "10" by the number of ways in which the winning 6 tries could have been distributed over the 10 tries. That number is(10*9*8*7*6*5)/(6*5*4*3*2*1) = 210Explanation: say i want to put the letters A-F in 10 boxes. The boxes in which the letters end up correspond to the winning tries. I could put the A in 10 different boxes, then the B in 9 remaining boxes, etc. This gives 10*9*8*7*6*5 possibilities. However, in the example above I don't care about the ordering: putting A in the first box and B in the second box would give me the same two winning tries as doing it the other way around. So I have overcounted every possibility by the number of ways of reordering the six letters, which (by the same reasoning) is 6*5*4*3*2*1.Long story short: your answer should be 21 times as large.
ah yes, instead of 10 it should be 10C6?like, in the first example, it wasn't 4, but 4C3 which just happens to equal 4? I get it now, ty
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Good thing its Einsteins and not Schroediners cat..the box is open so there are no quantum issues any longer....in this universe anyway.
This needed more love.I think we need to integrate pascal's triangle in here somewhere.
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I haven't read the other replies yet. I noticed that this thread already has two pages, so no doubt the correct answer has already been supplied.Nevertheless, here it is:1) Your odds of winning any one tournament is 50:1 or 1/51 = 0.0196...2) Hence your odds of losing any one tournament is 50/51 = 0.980...3) The outcome in any one tournament is independent of the outcome in the other three tournamentsTherefore, the final equation is:(1/51)*(1/51)*(1/51)*(50/51)*4 = 200/6,765,201So you will win exactly 3 out of a block of 4 tournaments 200 out of every 6.77 million attempts (or approximately 1 in every 33,826 times).Good luck!
This is soooo wrong
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This is soooo wrong
Well, when you put it that way......nope, it's still right.
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I'm well aware of that option, but ...<--- Feels that the two best words of the english language are "DE" and "FAULT"
This is the greatest answer ever to the "posts per page" response EVER. DE FAULT DE FAULT!!!!
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Well, when you put it that way......nope, it's still right.
I assume the bit he meant was wrong was the 'independent events' part.To save you another post in this thread Merbs, I'm just going to post what you would say.
Well, when you put it that way......nope, it's still right.
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