4 coin problem..

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?

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..

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

Quote: ewjones080

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

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

deleted

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..

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.

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.

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.

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.

Perhaps you can get your buddies to bet on whether or not only one head or one tail will come up on four coin flips. That's an even odds event, but I think most people will have a gut feeling it is not.

Quote: thecesspit

There's 16 results, not 5, as you correctly state. HHHH is rarer than HHTT.

Not true, HHTT has exactly the same probability as HHHH, they are both 1 of the 16 possible combinations. It is true that HHHH is rarer than a two head result.