General Betting
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Apologies if this has been discussed ad infinitum before ...I couldn't see any threads on the subject and a Google search didn't turn up any relevant results ...
Basically, I would like it if one or more of the more mathematically adept members of the forum could confirm or correct my calculations regarding Deal or No Deal probabilities ...and yes I am well aware that this has the potential for huge thread drift ... not a problem for me ...
What I am interested in establishing is the probability of the player taking out one or other, or both, of the top two boxes (ie the £250,000 and the £100,000) at the various deal points of the game.
The way I see it , it is easy enough to calculate the probability of the player taking out any specified box ( for the sake of argument the specified box will be the one containing £250,000)
You work out the probability of the player NOT picking the box, and subtract it from 1.
So the probability that they will NOT pick the box containing £250,000 in the first round of five selections, is 21/22*20/21*19/20*18/19*17/18, which comes to 17/22 . That means that the probability that the player WILL find the box in the first round is 5/22 . which is 22.7% ...so the odds against finding the big one are just a hair short of 7/2.
Am I right so far ...or a meringue?
Next question is...what is the probability of taking out at least one of the top two ...(ie either the 250K or the 100K) in the first round of five selections.
My thinking is that the chances of NOT hitting either of the top two are 20/22*19/21*18/20*17/19*16/18 ... which by my calculations works out at .588 ...so the probability that the player WILL hit one or other of the top 2 in the first round is .411 ... say 6/4 near as dammit.
Am I right so far ?
Now the question which slightly bugs me is this ...what is the probability that the player will take out BOTH of the top two in the first round?
Intuitively, I am thinking that it is a straight multiplication of the individual odds ...ie .227*.227 ...which works out around .05 ...or in odds terms 20/1 ...is this in fact the case or am I overlooking or miscalculating something ?
If my approach is in fact correct, then I can easily work out the relevant probabilities for 17 box, 14 box, 11 box, 8 box and 5 box.
All advice from experts gratefully received.
Basically, I would like it if one or more of the more mathematically adept members of the forum could confirm or correct my calculations regarding Deal or No Deal probabilities ...and yes I am well aware that this has the potential for huge thread drift ... not a problem for me ...
What I am interested in establishing is the probability of the player taking out one or other, or both, of the top two boxes (ie the £250,000 and the £100,000) at the various deal points of the game.
The way I see it , it is easy enough to calculate the probability of the player taking out any specified box ( for the sake of argument the specified box will be the one containing £250,000)
You work out the probability of the player NOT picking the box, and subtract it from 1.
So the probability that they will NOT pick the box containing £250,000 in the first round of five selections, is 21/22*20/21*19/20*18/19*17/18, which comes to 17/22 . That means that the probability that the player WILL find the box in the first round is 5/22 . which is 22.7% ...so the odds against finding the big one are just a hair short of 7/2.
Am I right so far ...or a meringue?
Next question is...what is the probability of taking out at least one of the top two ...(ie either the 250K or the 100K) in the first round of five selections.
My thinking is that the chances of NOT hitting either of the top two are 20/22*19/21*18/20*17/19*16/18 ... which by my calculations works out at .588 ...so the probability that the player WILL hit one or other of the top 2 in the first round is .411 ... say 6/4 near as dammit.
Am I right so far ?
Now the question which slightly bugs me is this ...what is the probability that the player will take out BOTH of the top two in the first round?
Intuitively, I am thinking that it is a straight multiplication of the individual odds ...ie .227*.227 ...which works out around .05 ...or in odds terms 20/1 ...is this in fact the case or am I overlooking or miscalculating something ?
If my approach is in fact correct, then I can easily work out the relevant probabilities for 17 box, 14 box, 11 box, 8 box and 5 box.
All advice from experts gratefully received.
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You're right so far. An easier way of working these out is to use the nCr formula ( = from n items, how many combinations to choose r of them) - http://mathworld.wolfram.com/Combination.html
So there are 22 boxes in DOND. In the first round, you have to pick 5 boxes. Probability of picking the top box = 1c1 x 21c4 / 22c5 = 1x5985/26334 = 0.22727 Probability of picking ONLY one of the top 2 boxes = 2c1 x 20c4 / 22c5 = 2x4845/26334 = 0.36797 Probability of picking BOTH of the top 2 boxes = 2c2 x 20c3 / 22c5 = 1x1140/26334 = 0.04329 Probability of picking NONE of the top 2 boxes = 20c5 / 22c5 = 15504/26334 = 0.58874 Probability of picking ONE OR MORE of the top 2 boxes = 1 - p(NONE) = 1-0.58874 = 0.41126 |
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Not disputing your maths Mr Magoo, but I would still like a K.I.S.S. step by step explanation of why the probability of picking ONLY one of the top two is 0.367, and why the probability of picking BOTH is 0.4329, which differs markedly from my admittedly tentative assumption that it might be 0.05.
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Both 0.04329
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Art Decko,
Your tentative assumption for calculating the probability of picking both of the 'big two' in the first round of five picks is incorrect (this is apparent despite my p**s poor probability skills, albeit with the benefit of reference to Mr Magoo's earlier post). To pick both of the big two in the first round, 1 of the 5 selected boxes from the original total of 22 boxes must contain the big one AND 1 of the remaining 4 selected boxes from the remaining total of 21 boxes must contain the second of the big two (£100k). Therefore, simply squaring the probability of picking the 'big one' (or any other specified box) in the first round is incorrect. If the problem were simplified to 3 boxes, including the big two, and the first round consisted of 2 picks, then the probability of picking the big one (or any other specified box) in that round is 2/3. Therefore, using exactly the same logic as you used in your original tentative assumption as to the probability of picking both of the big two in the first round, the [incorrectly assessed] probability would be (2/3 * 2/3) = 4/9. The correct probability of picking the big two in the round is (2/3 * 1/2) = 1/3. Thus, the correct probability of picking both of the big two in the first round, of 5 selections from 22 boxes, is: (5/22 * 4/21) = 10/231 = 0.04329 The probability of picking only 1 of the top two can be simplified to the probability of either of the big two being amongst the 5 selected boxes from 22, multiplied by the probability of the other of the big two boxes being amongst the unselected 17 of the remaining total of 21 boxes i.e. not being amongst the remaining 4 selected boxes: (2 * 5/22 * 17/21) = 85/231 = 0.36797 I'll leave the maths and detailed explanations to Mr Magoo and the many others with more competence than me. (MC: They aren't the same. (5/22)^2 = 0.051652 and is wrong) |
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* 0.051653
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Art Decko,
As Money Chaser pointed out, you've missed a zero from the probability of picking both boxes, it is 0.043, i.e. it is much less likely than picking only one box, as you'd expect. To break down the formulas a bit and to try and explain a bit better: A key number here is this '22c5' - the number of ways that you can choose 5 items from a selection of 22 (the order in which you pick the 5 is irrelevant). 22c5 is 26334, so there's 26334 different ways of picking boxes in the first round. Now, we need to know for each of your scenarios, how many ways are there of picking the magic boxes. Then, it's just dividing one by the other in order to get the probability. This is why you see everything being divided by 22c5 in my formulas. Let's take the case of picking ONLY ONE of the top two prize boxes. There are 2 top prize boxes, and 20 remaining boxes. 1) How many ways of choosing one top prize box from the two? 2c1 = 2 2) How many ways of choosing four of the remaining boxes? 20c4 = 4845 So, how many possible ways are there of picking ONLY ONE top prize box, and four other boxes? 2 x 4845 = 9690 Finally, what's the probability of this? 9690 / 26334 = 0.36797 All the other scenarios can be worked out in exactly the same way. |
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Thank you all, gentlemen. My "0.4329" above was a typo btw, I meant to write "0.04329".
I can see now that I was overcomplicating the issue of the probability of picking the big one in the first round ...22 boxes, 5 selections so it's just 5/22. So assuming it's still there in the next round the probability of finding it will be 3/17 ...or 0.176. The third round will be 3/14 ...or 0.214 The fourth round will be 3/11 ... or 0.272 The fifth round will be 3/8 ... or 0.375 At 5 box it becomes 3/5 .... or 0.6 ...odds on for the first time. So basically, at the start of the game, it's an even chance that player will hit the big one in the first eleven boxes. |
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sorry, cross posted there Mr Magoo ...
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Apologies for resuscitating this thread but in view of today's game I just wondered if one of you maths whizzkids could tell me how to work out the probability of the last two boxes in the game being the two biggies ... ie the £100,000 and the £250,000. I do fully realise that the probability is exactly the same as the last two boxes being the 1p and the 10p ... or indeed any two other previously nominated boxes ...
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22 boxes, and doesn't matter which way round they are, so there are 22 x 21 / 2 combinations. Therefore the probability of the final two being £100k and £250k is 1/22 x 1/21 x 2 = 1/231 = 0.433 %.
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Pick two boxes from the original 22 boxes. One of the two selected boxes must contain the £250k and the remaining selected box from the remaining total of 21 boxes must contain the £100k:
(2/22)*(1/21) = 1/231 = 0.43% |
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Ignore beaten again
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Thank you gentlemen ... it all seems so obvious once it's explained clearly and succinctly
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Watched this prog recently,the above methods of probability calculation are correct-however the player holds one of the boxes,making the target only 21 boxes to choose from.
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I know this shouldn't come as a surprise but I've never seen the dealer offer more than the mathematical expectation, has anyone?
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He does not offer a fair amount. But the value of money is not constant. If your expectancy was £50,0000,000 you might be correct to settle for accepting only £10,000,000.
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