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  #1  
Old 11-29-2007, 02:56 AM
imfatandugly imfatandugly is offline
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Default probability of n ties in a race to three (roshambo)

Too lazy to figure this out. I was playing rock, paper scissors with a friend and had an improbable amount of ties. Given that two people randomly pick paper, rock, scissors what is the probability that there will be n ties before someone wins twice? How about n or less?

thanks.
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Old 11-29-2007, 03:35 AM
pzhon pzhon is offline
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Default Re: probability of n ties in a race to three (roshambo)

The number of ties before a decisive throw follows a geometric distribution with mean 1/2. There is a 50% chance that a best-of-three-decisive-throws match will last 2 decisive throws, and a 50% chance that it lasts 3. So, the distribution is 50% of a convolution of 2 geometric distributions of mean 1/2 plus 50% of a convolution of 3 geometric distributions of mean 1/2.

P(n) = 1/2 (n+1 C 1) (1/3)^n (2/3)^(2) + 1/2 (n+2 C 2) (1/3)^n (2/3)^3.
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Old 11-29-2007, 09:24 PM
Siegmund Siegmund is offline
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Default Re: probability of n ties in a race to three (roshambo)

You can replace the phrase "convolution of n geometric distributions" with "negative binomial distribution" (and "with mean 1/2" by "with probability of success 2/3") to make it look a little less scary.

But it's the same calculation.
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Old 11-29-2007, 11:42 PM
imfatandugly imfatandugly is offline
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Default Re: probability of n ties in a race to three (roshambo)

thanks, sometimes I think phzon forgets that he isn't talking to grad students. Is there an easy way to see this? Link to derivation...

phzon...when you look at problems like this are you like "duh.... so obviously geometric distribution, and so obvious that it is equally like to end in two decisive throws as three."

Thanks though, your input is always appreciated.
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Old 11-30-2007, 11:37 AM
rufus rufus is offline
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Default Re: probability of n ties in a race to three (roshambo)

[ QUOTE ]
thanks, sometimes I think phzon forgets that he isn't talking to grad students. Is there an easy way to see this? Link to derivation...

phzon...when you look at problems like this are you like "duh.... so obviously geometric distribution, and so obvious that it is equally like to end in two decisive throws as three."

[/ QUOTE ]

You can think of this sort of question in terms of transitions. On a throw-by-throw level, each result (p1 wins, p2 wins, tie) has a chance of 1/3 of occuring, so you could work it out like a 3-dimensional pascal's triangle.
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  #6  
Old 11-30-2007, 09:46 PM
pzhon pzhon is offline
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Default Re: probability of n ties in a race to three (roshambo)

[ QUOTE ]
thanks, sometimes I think phzon forgets that he isn't talking to grad students.


[/ QUOTE ]
I hope it doesn't take a graduate student to evaluate (n+2) choose 2, the most complicated part of the formula I gave.

When I give an answer, I often expect an interested reader to take some time to think about what I said. The amount of time it takes depends on the reader's interest and background. If you expect to get everything at the first reading, you will be disappointed in mathematics.

Your complaints made me regret helping you again. I'm not inclined to spoonfeed you, and if I were, what I would need to say to make everything clear to you would depend on your background, which I don't know.

[ QUOTE ]
when you look at problems like this are you like "duh.... so obviously geometric distribution, and so obvious that it is equally like to end in two decisive throws as three."

[/ QUOTE ]
No. Here is a rough transcript of my thought processes: "Is there anything ifu could have meant other than best of 3 decisive results? I don't think so. After 2 decisions, there is a 50% chance it's over, and a 50% chance it will take one more. The number of ties before each decisive result is geometric. What is the convolution of geometric distributions called again? "Hypergeometric" would make sense, but I think that's something else. Yes, Wikipedia says hypergeometric is something else. Oh well. It's easy to read off the explicit formula from basic combinatorics, but that's prone to producing errors. Let me put the formula into Mathematica to check that the total probability is 1. Yup. Check a few values for plausibility. Hmm, the denominators for P(0) and P(1) are equal to 27, but that pattern can't continue. The expected value is 5/4. That makes sense, as that is the average of 2/2 and 3/2. Post."
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  #7  
Old 11-30-2007, 10:20 PM
imfatandugly imfatandugly is offline
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Default Re: probability of n ties in a race to three (roshambo)

[ QUOTE ]
[ QUOTE ]
thanks, sometimes I think phzon forgets that he isn't talking to grad students.


[/ QUOTE ]
I hope it doesn't take a graduate student to evaluate (n+2) choose 2, the most complicated part of the formula I gave.

When I give an answer, I often expect an interested reader to take some time to think about what I said. The amount of time it takes depends on the reader's interest and background. If you expect to get everything at the first reading, you will be disappointed in mathematics.

Your complaints made me regret helping you again. I'm not inclined to spoonfeed you, and if I were, what I would need to say to make everything clear to you would depend on your background, which I don't know.

[ QUOTE ]
when you look at problems like this are you like "duh.... so obviously geometric distribution, and so obvious that it is equally like to end in two decisive throws as three."

[/ QUOTE ]
No. Here is a rough transcript of my thought processes: "Is there anything ifu could have meant other than best of 3 decisive results? I don't think so. After 2 decisions, there is a 50% chance it's over, and a 50% chance it will take one more. The number of ties before each decisive result is geometric. What is the convolution of geometric distributions called again? "Hypergeometric" would make sense, but I think that's something else. Yes, Wikipedia says hypergeometric is something else. Oh well. It's easy to read off the explicit formula from basic combinatorics, but that's prone to producing errors. Let me put the formula into Mathematica to check that the total probability is 1. Yup. Check a few values for plausibility. Hmm, the denominators for P(0) and P(1) are equal to 27, but that pattern can't continue. The expected value is 5/4. That makes sense, as that is the average of 2/2 and 3/2. Post."

[/ QUOTE ]

Obviously it wasn't the binomial that was giving trouble. It was the "distribution is 50% of a convolution of 2 geometric distributions ". Yeah I looked it up and I get what it means.
Your help is always appreciated, like i've said a dozen times. Just sometimes your answers seem to come out of thin air, which is frustrating because I don't want to just know the answer, I want to know why it is the answer. Like you said it's hard to make things clearer if you don't the background of the person your responding to.
I should of thought of that. I guess i'm just used to answers with proofs or lines of reasoning (makes the answers seem like they haven't fallen out of the sky). I'm sure when you talk to whoever in real life if you spout out an answer and they ask you to explain your reasoning you don't get offended.....
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