General Betting
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Probability Question
Probability Question
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Yes we can just use an equation if we can find the right equation that is. But it's good to work it out from first principles. Then you can understand where the equation is coming from with a little bit of thought.
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OK gents ..last one for me on this lets keep it simple iam a layer your a backer you come to me and say what price would you give me throwing a 6 it could be a 2 or any other number i say i will lay you 5/1 each throw so six it is your having a pound on roll it 10 times and no 6 how much have you lost ...£10 thats right heres the other side you lucky punter roll again and every throw you get a 6 how much have i lost .... thats right £50 pounds you dont need all this equation stuff.. got to go now money to be won
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it's 1/6 * (5/6 ^ 9) * 10 since where the six is thrown doesn't matter
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imagine if you change that to a football team who are priced at 5/1 and assume it is a fair price. Then tell me the probability of them winning at least once if played 10 times and the probability of them winning exactly once if played 10 times.
There is an easy way to answer the first part of this, but I don't know how to answer the second part. So, the probability of getting AT LEAST one 6: The probability on any one throw of getting a 6 is 1/6 = 0.166.. Therefore the probability of NOT getting a six on a throw is 0.833... The probability of 10 consecutive non sixes is 08.33^10, which is 0.161. Therefore there is a 0.84ish probability that you will throw at least one 6 if you make 10 attempts. |
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Cosmic
I agree with your numbers if you were going to throw the dice 36 times twice. You would expect to get 10 combos at least with one specific number in it. But what is being proposed here( I think) is that you just throw the dice twice. I still believe that in this situation, 36 things can happen, but only 2 will win you money. That is the odds are 18.00. |
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zipper, I'm not sure where that is going, although what you have said is accurate
get me a drink, what you have written is what I would think it is as well I am interested in the binomial chat earlier, but what I will do is ask a man who I know will know for sure and get back to you. Intuitively I believe this is correct, but I think there is a more complex equation when probability of repeated events is used. Did anyone do a Maths degree? Must be a one on general betting? |
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for a single 6 out of X rolls it is the probablility of
rolling a 6 on the first dice AND not rolling a six on the second AND not rolling a six on the third AND not rolling a six on the fourth AND ... AND not rolling a six on the tenth OR not rolling a six on the first AND rolling a six on the second AND not rolling a six on the third AND not rolling a six on the fourth AND ... AND not rolling a six on the tenth OR . . . OR not rolling a six on the first AND not rolling a six on the second AND not rolling a six on the third AND not rolling a six on the fourth AND ... AND rolling a six on the tenth. 10 * [1/6 * (5/6)^9] ~= 0.323 |
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Zola,
Yes that's the Binomial formula I gave Prestbury some time ago. |
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N * [p * (1-p)^(N-1)]
N = number of trials p = probability of outcome in single trial |
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^ which is the equation in the second post with M=1
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Prestbury
If you want only one 6 in 10 rolls there can only be one answer. There are 6*6*6*6*6*6*6*6*6*6 possible outcomes, = 60,466,176 It has to be 5*5*5*5*5*5*5*5*5*10 =19,531,250 for there to be one 6 in only one of the 10 rolls Therefore, odds are 3.09. Which, I believe , is the answer given below using the Excel binomial calculator. Cosmic Horizon Based on this, your calculation of 10 outcomes out of possible 36 in 2 rolls is also correct, being odds of 3.60. Sorry for doubting you. |
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Zola,
I have no idea, I think it is a long winded method of calculating the probability of "within a range of successes" (i.e between 1 and 3 sixes from 10 rolls) |
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Equi, it wasn't a question
 Your post slipped in inbetween my posts. N * [p * (1-p)^(N-1)] is the equation in the second post with M=1. As mentioned earlier Prestbury, I would be wary of just picking equations of tinternet if you are not sure of their proper derivation or applicability. Especially when probability is concerned. |
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On an unrelated note, I had a nice lunch and pint in The Plough in Prestbury yesterday. Marvelous.
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Everybody agrees then at last.
The odds are 3.09. Bingo. |
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What Zola's Back Heel said is exactly the same as what I did. So it's def about 32%
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Just to clarify this. Let's imagine we throw the die (dice) 3 times rather than 10 times. We want to know the probability it will land on 6 EXACTLY once. So the 6 will occur either with the 1st throw, 2nd throw or 3rd throw. Let X = any of the numbers from 1-5. So we have to have either:
1. 6XX = (1/6)*(5/6)*(5/6) = 25/216 2. X6X = (5/6)*(1/6)*(5/6) = 25/216 3. XX6 = (5/6)*(5/6)*(1/6) =25/216 They each come to the same answer for the same reason as 2*3 = 3*2. We add all of them up and we get 75/216 or 0.347222222 in decimal. We can work out the general equation. 1/6 = the probability of throwing a 6 with one throw. Call that P So in "1" above this turns from (1/6)*(5/6)*(5/6) to p*(1-P)*(1-P) The number of times we multiply (1-P) with itself will be the total number of throws minus 1. Let the total number of throws = N. So it becomes P*(1-P)^(N-1) "2" and "3" will be exactly the same equation. So in fact we multiply p*(1-P)^(N-1) by the number of throws = N*(p*(1-P)^(N-1)) So prestbury road this is the exact same equation as you mentioned in your 2nd post where M = 1. You pasted in the equation: ((N!)/(N-M)!M!)*(P TO THE POWER OF M)*((1-P) TO THE POWER OF (N-M)) But M=1 so we get ((N!)/(N-1)!*1!)*(P^1)*((1-P)^(N-1)) which is ((N!)/(N-1)!*P*(1-P)^(N-1) N! means N*(N-1)*(N-2)*(N-3) . . .etc. So for example 5! means 5*4*3*2*1 so N!/(N!-1) must equal N since all numbers less than N simply cancel each other out on the top and bottom of the fraction. Thus we get N*(P^1)*((1-P)^(N-1)) In other words the equation we worked out from first priciples. This is something I never knew before! Could come in useful sometime! |
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Holy crap! Thats a lot of text
 I solved it using excel. Probability rounded to 4 decimal places: Sixes P 1 0.1615 2 0.3230 3 0.2907 4 0.1550 5 0.0543 6 0.0130 7 0.0022 8 0.0002 9 0.0000 10 0.0000 |
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You only need Combin(), the rest is arithmetic.
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oops copied & pasted wrong, let me correct that...
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Sixes P
0 0.1615 1 0.3230 2 0.2907 3 0.1550 4 0.0543 5 0.0130 6 0.0022 7 0.0002 8 0.0000 9 0.0000 10 0.0000 |
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FINE AS FROG HAIR Joined: 12 Mar 07
Replies: 653 11 Oct 10 15:12 Everybody agrees then at last. The odds are 3.09. Bingo. I'll lay at that price? |
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^
in a 0% commission bet. |
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Or if somoene else wants to back at 3.10....
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.
http://www.sendspace.com/file/ipndmb for the spreadsheet (link at bottom of page) |
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Can't say I can make head or tails of the numbers you posted Investor.
It's pretty clear that the odds of only getting one six up in 10 rolls of the dice is 3.09. Remember we're only talking about one six exactly. |
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Yes FAFH, I agree (disregarding the rounding error).
1/0.323=3.0959... |
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Investor/FAFH, you're both agreeing with each other. the 0.323011166 for 1/10 sixes is 3.096 expressed as decimal odds.
But this totally fails to answer the OP anyway, who wanted to know the probability of at least one six, not only one six. In which case you need 1-0.161505... = 0.8384944171 At least 1 is easy to calculate, it's 1-(5/6)^10 because it's the complement of 10 rolls without a six. If you want to know at least 2, at least 3 etc, then the formula to use is the cumulative binomial distribution, which has been covered elsewhere in this thread but can easily be done in excel with Binomdist() function. |
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ror
It's a bit confusing what the OP wants. I thought at first like you say, that is at least one six. Then it was changed to only one six. Who knows and who cares by now. It's all gone beyond meaning in a reality sense. |
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ror:
prestbury road Joined: 06 Feb 06 Replies: 191 11 Oct 10 12:43 Sorry, I think I have mislead you all a bit, I want it to show me the price to throw one 6 exactly from 10 rolls. My bad, sorry about that. |
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but can easily be done in excel
the death of thought |