ewjones080
ewjones080
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January 14th, 2014 at 2:17:28 AM permalink
I just wanted to make sure I was right. And there's really two parts to this question. It pertains to a couple questions I posed to some fellow dealers..

Part one: A man is betting 4 numbers straight up on roulette and not really hitting. So I ask my fellow dealer, "What's the probability this player doesn't hit in three spins?" .. I say in any given spin he won't hit in (34/38).. So in a 3 spin sequence he won't hit (34/38)^3 .. My coworker says its in the vein of 34/38, bringing up the fact the last spin doesn't have any bearing on the next. I say well my question is slightly different, cause I'm talking about a 3 spin trial, not three 1 spin trials.

Part Two: After break, on a dead craps game, I pose this street game: You and I will flip a coin four times and if you get exactly two heads, you win, but any other result, I'll win. I'll even put up three to your two. Would you take the bet? My three coworkers had a surprising answer. No, because they thought they should've gotten 5:1 odds. They reasoned there's only five outcomes (so they REALLY meant 4:1) which are zero, one, two, three, or four heads. Thus, there's a one in five chance they win. At least two posited that HTTT wasn't really different from TTTH. Of course I posited that it WAS different. In actuality you need to look at ALL possibilities--which should be 16--of flipping 4 coins. Six of these contain exactly two heads. Everyone still questioned me. I tried to explain that not counting the four different versions of a single head was the same as counting just one version of a seven on dice.. "No, because two six sided dice is different than four coins" was the rebuttle.... Needless to say, I wasn't getting through. I even mentioned that this game was like 4 rolls of a two sided dice v. Two rolls of two six sided dice. So who's right?
ewjones080
ewjones080
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January 14th, 2014 at 3:49:56 AM permalink
Also, I came up with a little formula (which I don't think is special or unique) which is, when flipping a coin N times, where N is an even whole number, the probablilty that I get exactly 50% heads is: (N c N/2) / (2^N) .. For what it's worth..
ewjones080
ewjones080
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January 14th, 2014 at 5:54:46 AM permalink
Bump.. I wanna here some input!!..
Wizard
Administrator
Wizard
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January 14th, 2014 at 6:53:36 AM permalink
Quote: ewjones080

Bump.. I wanna here some input!!..



Official warning -- rule 9 violation. Next time it will be a suspension.
It's not whether you win or lose; it's whether or not you had a good bet.
Ibeatyouraces
Ibeatyouraces
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January 14th, 2014 at 7:02:09 AM permalink
deleted
"And that's the bottom lineeeee, cuz Stone Cold said so!"
ewjones080
ewjones080
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January 14th, 2014 at 7:09:48 AM permalink
Quote: Wizard

Official warning -- rule 9 violation. Next time it will be a suspension.



I apologize and WAS worried I might be doing something frowned upon.. I just figured if it left the main page I wouldn't get any replies..
1BB
1BB
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January 14th, 2014 at 7:13:27 AM permalink
Quote: ewjones080

Quote: Wizard

Official warning -- rule 9 violation. Next time it will be a suspension.



I apologize and WAS worried I might be doing something frowned upon.. I just figured if it left the main page I wouldn't get any replies..



You forgot to tell a joke.
Many people, especially ignorant people, want to punish you for speaking the truth. - Mahatma Ghandi
thecesspit
thecesspit
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January 14th, 2014 at 7:35:08 AM permalink
Quote: ewjones080

I just wanted to make sure I was right. And there's really two parts to this question. It pertains to a couple questions I posed to some fellow dealers..

Part one: A man is betting 4 numbers straight up on roulette and not really hitting. So I ask my fellow dealer, "What's the probability this player doesn't hit in three spins?" .. I say in any given spin he won't hit in (34/38).. So in a 3 spin sequence he won't hit (34/38)^3 .. My coworker says its in the vein of 34/38, bringing up the fact the last spin doesn't have any bearing on the next. I say well my question is slightly different, cause I'm talking about a 3 spin trial, not three 1 spin trials.



You are correct. Your co-worker is wrong. Think about it if you went for 10,000 spins. He's saying it'll be 34/38 that he won't hit on of his four numbers? The way you did it is precisely because each spin is independent... whats the chance of missing on the first spin? The second spin? The third spin? Multiply them all for the chance of not hitting on spin 1 AND spin 2 AND spin 3.
Quote:


Part Two: After break, on a dead craps game, I pose this street game: You and I will flip a coin four times and if you get exactly two heads, you win, but any other result, I'll win. I'll even put up three to your two. Would you take the bet? My three coworkers had a surprising answer. No, because they thought they should've gotten 5:1 odds. They reasoned there's only five outcomes (so they REALLY meant 4:1) which are zero, one, two, three, or four heads. Thus, there's a one in five chance they win. At least two posited that HTTT wasn't really different from TTTH. Of course I posited that it WAS different. In actuality you need to look at ALL possibilities--which should be 16--of flipping 4 coins. Six of these contain exactly two heads. Everyone still questioned me. I tried to explain that not counting the four different versions of a single head was the same as counting just one version of a seven on dice.. "No, because two six sided dice is different than four coins" was the rebuttle.... Needless to say, I wasn't getting through. I even mentioned that this game was like 4 rolls of a two sided dice v. Two rolls of two six sided dice. So who's right?



There's 16 results, not 5, as you correctly state. HHHH is rarer than HHTT. Easy answer, tell them they flip, and pay YOU 3:1 if you get HHTT.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
BleedingChipsSlowly
BleedingChipsSlowly
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January 14th, 2014 at 7:37:12 AM permalink
For the coin flip your analysis is spot on. Four flips has 2**4=16 possible results. Six results win the proposed bet (HHTT,HTHT,HTTH,THHT,THTH,TTHH) for 3:5 odds. You offer 3:2 for a win. Big Red is a better bet.
BleedingChipsSlowly
BleedingChipsSlowly
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January 14th, 2014 at 7:57:33 AM permalink
You are right about the probability of not hitting four numbers played for three roulette spins: (34/38)**3 = 71.6%. Whether or not the spins are consecutive does not matter.Changing which numbers are played does not matter, either.

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