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The Magic Cafe Forum Index » » Magical equations » » "Odds" question. (0 Likes) Printer Friendly Version

DR STEVE HOSKINS
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Hi, A question to all the clever people who hang out in this forum!

If a casino dice / die is rolled three times what are the odds of the Mentalist correctly predicting all those three numbers?

- Now maths is not my strong point but pure logic says it must be a one in eighteen chance??? but I'm probaly wrong, right?

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landmark
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Steven, you need to multiply, not add, to get the probability of a number of independent events all happening. So, 1/6 x 1/6 x 1/6 = 1/216 for getting the roll of a single die correctly. For two dice it's a little trickier, because to roll a total of eight, for example, there are five different ways (2 and 6, 3 and 5, 4 and 4, 5 and 3, 6 and 2), out of 36 possible outcomes. So for the sum of eight on three different rolls the probability is 5/36 x 5/36 x 5/36 . Adjust accordingly for other sums.
DR STEVE HOSKINS
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Hi Landmark,
Thanks for the help, - but I'm even more confused!

If I have a die, I give that die to the spectator to roll three times, we only count the uppermost number on the die, (1-6).
So lets say it lands on a three, six and finaly a two, - I correctly name all those three numbers, what are the odds/chances of me correctly predicting that?

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S2000magician
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Quote:
On 2011-03-23 06:52, DR STEVE HOSKINS wrote:
If I have a die, I give that die to the spectator to roll three times, we only count the uppermost number on the die, (1-6).
So lets say it lands on a three, six and finaly a two, - I correctly name all those three numbers, what are the odds/chances of me correctly predicting that?

If your prediction includes the order in which they rolled the numbers - You will roll a three first, followed by a six, and, finally, a two - the probability is 1/216; the odds against doing that are 215:1.

If your prediction merely lists the numbers without specifying the order - You will roll a two, a three, and a six - the probability is 1/36; the odds against doing that are 35:1.
DR STEVE HOSKINS
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Ahh ok, so that's my answer; the odds of me being able to correctly predict three differnt spectators who will each have one roll of the die is 215 to 1.
Thank you both for taking the time to help me with that.

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S2000magician
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My pleasure.

(By the way, technically, 215:1 are the odds against a correct prediction; the odds in favor of a correct prediction are 1:215. "Odds against" are what are normally quoted (e.g., on William Hill).)
Decomposed
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Wow, those are heavy odds.
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