GAME: MindTrap

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Originally Posted by edball

Even if that's true, 42% is still greater than 1/3 (33%). https://forums.windrivers.com/images.../2005/03/1.gif

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You are so like, totally right!

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Originally Posted by Deity

Who? :confused:
:D

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La Dee Da....we're not really playing right now......um uh...maybe we'll play later....https://forums.windrivers.com/images.../2005/03/1.gif

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Originally Posted by Mindwarp

You are so like, totally right!

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Well, I'm sure you are also, so let me congratulate you in advance for your 3 points. https://forums.windrivers.com/images.../2005/03/1.gif

I meant to say less than 3/6 chance.

Quote:

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Originally Posted by Mindwarp

I meant to say less than 3/6 chance.

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I know, I was just messing with you for getting the right answer when mine was wrong. https://forums.windrivers.com/images.../2005/03/1.gif

Quote:

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Originally Posted by Mindwarp

His chances of winning are less than a third.

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I think you give edball pretty good odds there. :D

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Originally Posted by Mindwarp

His chances of winning are less than a third.

If he rolls all three dice at the same time the probability of each die not landing on six is 5/6. Multiplied together there is a 125/216 chance that the man will NOT roll a 6 on any die. The inverse of that is a 91/216 chance that he does roll at least one 6 which is 42.1296296%, etc, which is less than a third.

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I still don't see why you would multiply instead of add.

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Originally Posted by Deity

I think you give edball pretty good odds there. :D

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So, you didn't fall for my devious ploy ? https://forums.windrivers.com/images.../2005/03/1.gif

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Originally Posted by edball

So, you didn't fall for my dubious ploy ? https://forums.windrivers.com/images.../2005/03/1.gif

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Nope. ;)

Because what I'm really doing is looking at the total number of outcomes. There are 136 outcomes when three dice are rolled which is when the multiplication comes in. Each die can land on 1,2,3,4,5 or 6 on each roll. There are 125 outcomes where no 1 is present. The other 91 outcomes out of the 136 possible are the man's chances of winning.

That wasn't very clear. Start with two dice. If the first one lands on 1, the other can still be 1-6. 6 possibilities. If it lands on 2, the other can still be 1-6. 6 more possibilities. 12 total. You can continue that way until you discover that the total number of possibilities for two dice is 36 (6 for each number on the the first die). Then, you add a third die into the equation. Think of the possibilities as numbers (11, 12, etc). If you roll 1 on the third die it changes the first combination to 111, the second to 121, the third to 131, multplying again the number of possibilities by 6. I don't think this is a better explanation, but I tried. :)

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Originally Posted by a d e p t

A swindler once approached an honest man with a die. He handed him the die and told him about the bet. The die had six sides. If the man rolled a ONE, he wins, and gets back twice the amount of his bet. If not, the swindler would keep the bet.

"But...my chances are only one out of six," retorted the man.

"True," grinned the swindler, "But I'll give you three tries to get a one."

The man considered. Three tries, with each try having a 1/6 chance of winning. So his chances of winning is 1/2. Why not give it a try?

Is the bet really fair? If not, what are the chances of the man winning?

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Fair = no
first roll = 1 of 6
sec. roll = 1of 5
thrid roll = 1 of 4
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_________3 of 15
1 out of 5 chance

I don't see multiple dice mentioned... I took it as 1 die - singular, to be rolled 3 times.

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Originally Posted by NooNoo

still 1/6 as the odds do not diminish because each throw starts and ends with a 1/6 chance. If the number was somehow knocked off the die each time and could not be rolled again, only then the odds would improve.

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I agree with the above if a normal dice is used. However, the guy is a swindler, so the dice would be loaded or not square, meaning the dice would probably never land on a 1 if he had 100 throws.

My guess is that the odds are non-existant because the die is loaded!

Adept?