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11132008

#1 (permalink)

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A man walks into a bar... (probability question)
I don't really need help with this, I just wanted to throw out a problem to see who could solve it, and this seemed like the best place to put it.
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A man walks into a bar and claims that he has never yet lost money in a game of dice. The bartender clearly doesn't believe this man and so he challenges him. The bartender doesn't have a pair of dice on him, but he pulls out a perfectlyweighted quarter (50% chance of heads, 50% chance of tails).
"One cent is riding on this. Call it in the air."
The bartender flips the coin.
"Heads."
The coins shows heads.
Bartender: "Double or nothing."
The bartender flips again.
"Heads."
The coin shows heads again.
Bartender: "Double or nothing."
The bartender keeps flipping the coin, and the man keeps calling heads and winning. After 20 flips, the man stops the bartender.
"You've flipped the coin 20 times landed heads every single time. Let's try something different for the next DoubleorNothing. I'll let you keep flipping the coin as many times as you like. If at any point you have managed to land twice as many tails as you have landed heads, I'll call off all the money you owe me. The 20 you've already flipped will still count against you."
The bartender accepts the challenge.
What is the probability that the bartender will lose the bet (never stop flipping the coin)?
Last edited by sentythee; 11132008 at 09:15 PM.



11132008

#2 (permalink)

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lol...
i would say the bartender will never lose. So 0.
If you multiply infinite times of 1/2 probability you'll just get 0. I was never good at probabililty at school so thats just a guess xD



11132008

#3 (permalink)

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Theoretically, one can continue to flip the coin until that happens.
Probability wise, I think the chances of that is quite low as it would mean:
2 tails: 1 head (the closer this approaches infinity, the more accurate the requirement would be).
However, I have not taken any statistics class, so if there is some weird solution, don't kill me for being wrong.



11132008

#4 (permalink)

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50 %



11142008

#5 (permalink)

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Quote:
Originally Posted by TogSawks
lol...
i would say the bartender will never lose. So 0.
If you multiply infinite times of 1/2 probability you'll just get 0. I was never good at probabililty at school so thats just a guess xD

I DON'T understand, but it's impossible for the probability to be 0.
Law of large numbers: As the number of flips goes to infinity, the ratio of heads to tails will go to 1:1. Saying that the chance of getting a ratio of 1:2 at some point infinitely far away is 100% would contradict that law.
@Skillz: What would you say if the initial number of heads was 100? If the answer is 50% starting with 20 heads, it's impossible for the answer to be 50% with 100 heads.
Quote:
Originally Posted by Xpecial
However, I have not taken any statistics class, so if there is some weird solution, don't kill me for being wrong.

All you need is (love) algebra



11142008

#6 (permalink)

YOUUUUUUUUUUUUUUUUUU
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Unless that coin is actually improportional due to weight balances, it's 50%.
Quote:
Originally Posted by sentythee
@Skillz: What would you say if the initial number of heads was 100? If the answer is 50% starting with 20 heads, it's impossible for the answer to be 50% with 100 heads.

Theory and reality are two different environments. Even though you may have a probability that situation B may occur with situation A, that doesn't mean that probability will account for what will happen.
Last edited by k0n; 11142008 at 10:07 AM.



11142008

#7 (permalink)

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Alright, I think I understand why people are saying 50%. And to clear that up:
Quote:
Originally Posted by sentythee
"You've flipped the coin 20 times landed heads every single time. Let's try something different for the next DoubleorNothing. I'll let you keep flipping the coin as many times as you like. If at any point you have managed to land twice as many tails as you have landed heads, I'll call off all the money you owe me. The 20 you've already flipped will still count against you."

>>



11142008

#8 (permalink)

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So aren't we looking for a probability of landing the next 40 coins tails



11142008

#9 (permalink)

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No, the bartender can also flip 5 heads and 50 tails to win (for a total of 25 heads and 50 tails). He can also flip 100 heads and 240 tails to win (120 heads, 240 tails total).



11142008

#10 (permalink)

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Doesn't it multiply after every try?
Like so
1st flip :
1/2% heads  1/2% tail

2nd flip 1/2 and 1/21/2 and 1/2

31/2l 1/21/21/21/2l1/21/2l1/2
so on and so on..
let's say for 3 heads in a row chances would be (1/2*1/2*1/2)= 0.125 or 12.5% chance
flipping heads 20 times = 0.0000009536743 or 0.00009536743% ( lol)
I dunno about the problem, I was just trying to remember how to do probability problems, it's been really long time :0
Also, the chance of him getting heads infinite times will never be 0%, there will always be a chance, even if it's incredibly small



11142008

#11 (permalink)

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in relation to probability of 0, anything tending towards infinity can be 0 or a really large number.
1/infinity is infact 0



11142008

#12 (permalink)

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I thought that infinity is not a number, therefore cannot be used in algebraic equations. So you can't divide by infinity, right?



11142008

#13 (permalink)

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In limits, using infinity is allowed to determine end behaviour.
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11142008

#14 (permalink)

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so i don't really see how you can use algebra to solve this.
all i can say is probability was bartender winning is 100% at infinity, unless the OP gives another logical solution.



11142008

#15 (permalink)

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Well, obviously he doesn't just mean Algebra 1, or concepts specifically taught in those classes.
Probability is in Algebra btw.



11142008

#16 (permalink)

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._.
I'd say 100% due to luck? Since theres in a 50% chance of getting either heads or tails, you need to get lucky enough to get more tails than heads, so if you keep flipping EVENTUALLY you'll get 2:1?
If you want to be a realistic bastard though, just say the bartender will die before he can flip 2:1 since you'd have to flip that coin for an incredibly long time unless he has LUCK HAX.



11142008

#17 (permalink)

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if LK doesnt work on trickster it won't work on maths either hahahaha.
Maths will always have some retarded solution to anything, so luck is out of the question. However, I can agree that if i was an realistic bastard, the bartender would have already died.



11152008

#18 (permalink)

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I already posted why it was impossible for the answer to be 0%.
More in depth:
Assumptions: As the number of flips approaches infinite, the ratio of heads to tails will approach 1:1 (the law of large numbers says the the ratio of different outcomes will approach the ratio of the probability of each outcome as the number of outcomes approaches infinite).
The chance that the bartender loses is 0% for any initial number of heads (this is what you're saying)
k is the initial number of heads (in this problem, 20). I'm going to make k the variable here because, according to you, the bartender can't lose this bet regardless of what k is (as long as it's a positive integer).
As k approaches infinite, you're saying the the ratio of heads to tails at some point infinitely far away will be 2:1. This violates the law of large numbers, so one of the assumptions must be wrong.
Also, keep in mind that the chance of getting twice as many tails as heads decreases every time the bartender flips a head.
I'll give the answer to individual people, but I won't post it here.
Last edited by sentythee; 11152008 at 10:52 AM.



11152008

#19 (permalink)

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Law of large numbers what the hell?



11152008

#20 (permalink)

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That's theory for you, the more you think about it, the more your head hurts.
Love these awesome Law names
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