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10-14-2005, 05:38 PM
I have been trying to compute the probability of throwing one triple of say sixes in a throw of 4 dice.

I figure that a triple of sixes with 3 dice is 1 (event) out of 216 (outcomes)--0.004629

I know that adding one dice to the throw, making 4 dice total, must increase the probability of throwing at least 3 sixes (a triple). However, when I do the computation, I come up with 4 out of 1296--0.003086

I am asking someone to show me the error in my ways so I can figure this out properly.

10-14-2005, 05:50 PM
I'm not sure what you're doing, but here are two ways to approach this:

The chance of rolling at least 1 tripple 6 is:
1/216+215/216*(1/216+215/216*(1/216+215/216*(1/216)))

Alternatively, the chance of not rolling any tripple sixes will be
(215/216)^4
so the chance of rolling at least one, must be
1-(215/216)^4

yellowjack
10-14-2005, 05:59 PM
P(rolling a 6) = 1/6
P(not rolling a 6) = 5/6
(1/6)^3 * (5/6) = 5/1296

For dealing with combinations of outcomes, you are correct with the 1296 (C(6,1)^4) possible outcomes. However there are C(4,3)*C(5,1) ways of getting triple 6s with a "non-6 face" for your 4th die.

The probability of triple 6s with 4 dice is:
C(4,3)*C(5,1) / 1296 = 20/1296 = 0.01543

I hope someone can confirm this answer or tell me why this is wrong.

AaronBrown
10-14-2005, 06:14 PM
Yellowjack correctly computed the probability of throwing exactly three sixes out of four dice. You asked for "at least" so you have to add one.

There is one way to get four sixes out of four dice.

There are 20 ways to get three sixes. The non-six can be any of five values (1 to 5) and on any of the four dice. 4x5 = 20.

21/1,296 is the probability of getting at least four sixes. This is 3.5 times the probability of getting three sixes on three dice.

yellowjack
10-14-2005, 10:04 PM
[ QUOTE ]
There are 20 ways to get three sixes. The non-six can be any of five values (1 to 5) and on any of the four dice. 4x5 = 20.

[/ QUOTE ]

I think you were rewriting your post and forgot to change this =o

10-15-2005, 02:41 AM
First of all, thanks to all for the replies.

Ok. So YellowJack and AaronBrown are saying that the non-six showing die has 5 non-six faces which then is used to multiply my original combination computation of 4 to get 20 plus the 1 event of all 4 dice showing six. (What a long sentence :-)

So I end up with 21 over 1296 or about 3.5 times my previous computation. So for 5 dice the math would be C(5,3)=10 x 5 x5=250+1=251 then divided by 7776.

As for where this started; I was playing Zonk against someone and was using a strategy of not taking all my dice/points on the first roll. That way I left more dice for my next roll with a higher chance of getting a triple and a higher score than if I took all of my dice/points and then did or did not roll again.

When pressed by my opponent, I was unable to do the math to convince him, and that sparked a trip to the Internet, library and book store to brush up on probability. Usually that does it, but not this time--I think probability is Latin for "makes one cry." No, I don't plan to confront my opponent with this; he wouldn't understand it anyways. I'll just keep on winning.

wickss
10-15-2005, 12:07 PM
[ QUOTE ]
The chance of rolling at least 1 tripple 6 is:
1/216+215/216*(1/216+215/216*(1/216+215/216*(1/216)))

[/ QUOTE ]
I don't follow. This could be simplified as 1*(1*(1*(1/216))) or 1/216 which is the probibility for 3 dice.

wickss
10-15-2005, 12:12 PM
The way I figure is this.
Chances on first die: 1/6
Chances on second die: 1/6
Chances on either third or forth die: 2/6

1/6 * 1/6 * 2/6= 1/108

Someone tell me if this is wrong.

AaronBrown
10-15-2005, 02:58 PM
There are two problems with your calculation.

First of all, it assumes you have to get sixes on the first two dice, then at least one of the last two. This gives you approximately half the correct answer, because you could miss your six on one of the first two dice or one of the third and fourth.

Second, the chance of getting a six on at least one of the last two dice is 11/36, not 2/6. You have to subtract 1/36 for the chance of getting sixes on both the last two dice. Another way to do this is to take 1 - (5/6)*(5/6) = 11/36, which is one minus the probability of missing six on the third die and the fourth die.

1/6 * 1/6 * 11/36 = 11/1,296. 2 times this is 22/1,296. You have to subtract 1/1,296 because we're double counting the case when you get four sixes, so you get the correct answer, 21/1,296.

There's more than one way to skin a cat.