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#1
08-02-2002, 06:14 AM
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Don\'t Use Formula For This

Using simple fractions and simple logic, tell me how to calculate the chances of gettiing, (when tossing five dice, five times), exactly one three exactly one toss (the other tosses result in 0,2,3,4,or 5 threes).
#2
08-02-2002, 08:06 AM
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Re: Don\'t Use Formula For This

For a given throw to have exactly 1 three, any of the 5 dice can have the three, and the other 4 dice can have any of 5 remaining numbers, so there are 5*5^4 ways for the throw to have exactly one 3. The total number of ways to throw 5 dice is 6^5, so the probability on one throw is 5*5^4/6^5 = 40%.

Now there is a 60% probability of NOT getting exactly 1 three on any throw, so the probability of getting it on the first throw and not on the next 4 is .4(.6)^4 = 5.2%. But since we can get it on any of the 5 throws we multiply this by 5 to get 26%.
#3
08-02-2002, 12:54 PM
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Looks good to me. *NM*

#4
08-02-2002, 03:31 PM
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Re: Don\'t Use Formula For This

!3,!3,!3,!3,3 = (1- 1/6)^4 * 1/6 = 625/7776 * 5 ~= .4

n,n,n,n,y = (1 - 3125/7776)^4 * 3125/7776 = x *5 ~=

25.7%

#5
08-02-2002, 05:58 PM
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Re: Don\'t Use Formula For This

* Using simple fractions and simple logic, tell me how to calculate the chances of gettiing, (when tossing five dice, five times), exactly one three exactly one toss (the other tosses result in 0,2,3,4,or 5 threes). * ... call it p

... when tossing 25 dice, one time ... call it p1

... when tossing one dice, 25 times ... call it p2

... when tossing one dice, 25 times ... gettiing exactly one three in the 7th toss ... call it p3

p3= (5/6)^24*(1/6)

p2=25*p3

p1=p2

p=p1

p= (5/6)^24*(1/6)*25= 5,24 %

#6
08-02-2002, 06:26 PM
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Mat ... please delete above post !

Jack has lost his mind or he can't read !?

brad and Bruce got it right.

#7
08-02-2002, 07:16 PM
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ACK--work in progress above, errors

Thaty wasn't supposed to be posted, I was still working on it and se an error already.

6^5 = total number of combinations given one toss. For each of the 5 slots the 3 can fall into, the other slots together hold 5^4 combos. So we have 5(5^4)/6^5 giving the fraction of ways a 3 will fall on one toss.

The total combos given 5 tosses are 6^5^5.

That's how far I am now but I keep getting disconnected and its time for poker

#8
08-02-2002, 07:19 PM
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my isp is whacked now, thought I had posted but no

can't post til tomorrow isp keeps disc. and it led me to believe I had posted when I hadn't ...got Message Posted screen but it was like a ghost post, so my post referring to DS's was actually intended to refer to my own (invisible post)
#9
08-02-2002, 10:29 PM
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Stick To Fractions!

Here's how I was hoping you would do it.

Throw five dice. The chances the first one is a three and the rest aren't, is 1/6 times 5/6 to the fourth power. 08038. But that three could also be on one of the other four dice. Total chances for exactly one three is .402

Do above five times. Chances that the first throw of five dice contains exactly one three and the other four throws, don't is .402 times .598 to the fourth power. .514 Since the throw with exactly one three could also be the 2nd 3rd 4th or 5th as well, the total chances are five times as much or .257.
#10
08-03-2002, 02:03 AM
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Re: Stick To Fractions!

That is nice and compact and I like it.

It took me maybe 5 minutes to really follow your explanation through and see where you got the numbers (like where you got .598 had me baffled briefly, until it suddenly struck me that .598 and .402 added up to 1.0).

I am basically new to counting problems, but I will practice a few examples with variations on this theme over the next few days.

Thank you.

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