General Betting
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11 Oct 10
10:50
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06 Feb 06
Does anyone have a link the probability equation for repeated events. For example, the probability of rolling a 6 on a die if you roll it 10 times.
Thanks in advance
Thanks in advance
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I have found this..
((N!)/(N-M)!M!)*(P TO THE POWER OF M)*((1-P) TO THE POWER OF (N-M)) where M is the exact number of successes, N is the number of trials, P is the individual probability. To give a bit of context, I am trying to calculate when a money back special on a bad beat multiple becomes value when compared to exchange prices. Its not for arbing purposes but just for my own knowledge. I need a better mathematician than me to help if you can |
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Any number on a dice is 5/1 against so 10 polls = 50/1 to do the double good luck ....
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Zipper, what does that mean? It seems dubious mathematically.
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Baz a dice 6 munbers to name any one number is 5/1 against so you say number 6 roll the dice there are 5 numbers who can beat you hence 5/1 against .... ok each throw is 5/1 against so 5 times 10 = 50 to get the double
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It's all ok until the final part. The chance of the 1st rolls alone both being 6 is 35-1. You have 9 chances of any 2 in a row? In fairness I may be misunderstanding the question.
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1st 2 rolls that should be
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Baz i thought you were talking about one dice ... two dice is 35/1 against each throw .... god iam hard work .
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I think it's one dice thrown 10 times? The 1st 2 individual rolls would be 35-1 to both be 6 etc. You have 9 chances of getting 2 in a row. Feels like a 7/2 shot roughly.
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Baz 7/2 to throw a double //// you work for one of the big 3
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I am actually wanting to know the probability of rolling any 6 with 10 rolls. This is where you move away from basic probability. If there were two rolls then the probability of rolling a six would be roughly 1/6 + 1/6. If three rolls then roughly 1/6 1/6 +1/6. I say roughly because with each roll the odds diminish away from the calulations given here. If you roll it six times the probability obviously can't be 6* 1/6 = 1 because that states that you will definitely roll a 6 which obviously might not happen. I can't remember the formula to calculate the probability as its very long, and was wondering if anyone knew it or could find it for me. The complicated formula above I think is it but was looking for someone to confirm
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Prestbury,
Do you mean the probability of rolling 1 or more sixes from 10 rolls of a dice? |
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Yes equimine, exactly
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Just find the probablity of NOT rolling it, then subtract this from 1. Hence 1-(5^10/6^10) whatever that works out at.
Gets more complicated though if throwing a particular number more than once is required. Binomial theorem or something? Dunno, I only did CSE maths. |
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Agree with Cosmic.
1 - (5/6)^10 1 -( .83^10) equals .33 Therefore probability of throwing 1 or more sixes from 10 throws is 67%. |
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Sorry shouldn't try and do two things at once
1 - (5/6)^10 1 - (.83 ^10) equals .16 Therefore probability of 1 or more sixes from 10 throws is 84% |
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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. Thank you thogh cosmic for that. Its a long time since I studied probability, but it was at university and I remember a very long equation that hurt my head a little. Maybe I am wrong about this, but you two sound like you have a good grasp of this. Any ideas?
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You need one 6(1/6) and 9 non-6s (5/6). There are 10 different ways in which this can be achieved (6 first, 6 second, 6 third, ...)
The equation you need is therefore 10 x 1/6 x (5/6)^9 = 32.3% |
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Equimine, that is what I had, using 1.2*10 to establish chance of no 6. That gives a 16.15% chance of no 6 being thrown.
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I can't fathom out any need to move away from probability at all.
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Prestbury,
You need a Binomial Distribution. If you use, or have access to Excel use the inbuilt one in statistical formulas. Number would be 1 Trials would be 10 Probability would be 0.167 Use "false" in cumulative The answer is 32.25% |
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Basically as follows
Roll die 2 times,there are 6*6 outcomes = 36 5*5 outcomes will not contain a 6, =25 So 11 will contain a 6, thus probability of there being at least one 6 =36/11= 3.27 Roll a die 3 times, there are 6*6*6 outcomes = 216 5*5*5 outcomes will not contain a 6, = 125 So 91 will contain a 6, thus probability ot there being at least one 6 = 216/91 = 2.37 etc etc |
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I don't know the equation. Binomial? Asking for one 6 EXACTLY is different from asking for at least one six since at lest one six will include all the permutations for 2 6's etc..
Since I don't know the equation then we have to do it the long way round. If any other number apart from 6 = X, then the permutations are: 6xxxxxxxxx x6xxxxxxxx etc with the 6 in a new position each time making up 10 permutations. Each permutation will be 1/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6* Therefore the probabilty of throwing one 6 exactly should be 10 times this value? |
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Jus' used excel to calculate the answer. Probability comes out at 0.323 if my reasoning is correct. Might not be though. It's astonishingly easy to get confused with maths!
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Ah! My answer is the same as Equimines! So looks like my reasoning is correct.
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Nope you're both wrong I'm afraid.
My answer is the correct one. Remember we are only looking for at least one 6. |
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The equation is as follows I believe
Binomial = Trials C Successes (in this case 10 and 1) * Probability of Success ^ Number of Successes (0.167 ^1) * Probability of Failure ^ Number of Failures (.833^9) 10 * ((.167 * (.833^9))) = .3225 32.25% |
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The easiest way to approch these things is always to establish the convers first as I have done,
Then simple deduction gets your answer. |
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The more recent request is for one 6 exactly.
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I don't think so Baz but if it is then that's really easy
Example, roll a die 2 times, = 36 outcomes. To get only one 6, you must either get in on the first or the second, so only 2 outcomes win. Therefore probability of there being only one 6 are 36/2 = 18.00 |
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Probabilities have to be less than 1. You must be referring to odds? Odds of 18 is probability of 0.056. If only thrown twice then it's 2*(1/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6)= 0.064602233.
Or you can use binomial which is probably the way I'm doing it but put in its general form. So your answer is wrong. |
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Equimine
If you're saying what I think you are, that is if you roll a die 10 times, you will give me odds of over 3 to 1 that I can't get one 6 up, then that's what I call a real yummy value bet. Just think about your maths a bit. As I say before the odds of me getting at least one 6 up in 3 rolls is 2.37 You guys are overcomplicating a relatively simple situation. |
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No that's wrong. If only thrown twice the permutations are:
XX 6X X6 66 So we want either 6X or X6 . Both will be equal probabilities of 5/6*1/6. So probability = 2*(5/6*1/6) = 0.277777778 I think! |
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First of all, thanks for all the answers. One six exactly can be done by going the long way around, yes, but I was looking for the equation as what I need isn't do roll a die. The binomial equation I think is what I need for at least one six. If you both (equimine and cosmic) got the same result then that looks good. I'll use that and see what I get.
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Cosmic
If you roll a die twice there must be 36 outcomes. If you want to get one specific number only once, then it has to occur either on the first throw or the second, and not both. That is 2 outcomes out of 36. Odds are 18.00. |
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FAFH,
With all due respect, before you start lecturing on the intricacies of statistical probability, most people realise it is advisable to know the question you are trying to answer. Prestbury wrote as follows, "I want it to show me the price to throw one 6 exactly from 10 rolls" You are not even correct about the odds of getting at least one six which is 84% (1.19) |
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FINE AS FROG, I know where you are coming from but this is the crux of what I am asking. The odds of something happening is simplified in the dice example, but 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. Maybe I am wrong as I have said, but the method you are using does not give a true probability even though intuitively you expect it to.
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The outcomes are:
11 12 13 14 15 16 21 22 23 24 25 26 31 32 33 34 35 36 41 42 43 44 45 46 51 52 53 54 55 56 61 62 63 64 65 66 Not there's 36 if you count them all. How many have one 6 exactly? Well count them. It's 10. So the probabiity is 10/36 = 5/18 = 0.277777778 The exact same answer I got before! |
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If you look at the equation I have posted in my second post, this is more like the complexity of equation that I believe I need
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good to get the brains working though, eh?
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