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Forum: Probability 12-10-2005, 07:48 AM
Replies: 2
Views: 97
Posted By SwissPoker
2 boxes problem

You have 2 boxes with 2 pigeonholes each. Box A has 1 gold coin in each pigeonhole. Box B has 1 silver coin in one pigeonhole and 1 gold coin in the other. Now, you are blindfolded and chose a box...
Forum: Probability 12-03-2005, 11:00 AM
Replies: 1
Views: 85
Posted By SwissPoker
5 people choose a digit from 0-9

What is the probability that at least 2 people choose the same digit?

Can I use the backdoor approach like ...

1- [9/10*8/10*7/10*6/10] = 69.76%

... or is this the inclusion-exclusion...
Forum: Probability 11-24-2005, 10:31 AM
Replies: 1
Views: 84
Posted By SwissPoker
How do I get P from distinct EV

I need a formula to determine P in case I have a,b, and c:

a = -20
b = +80
c = +54

EV = (P)(a)+(1-P)(b) = c

How do I determine P? What is the formula?
Thanks for any help.
Forum: Probability 10-22-2005, 07:03 PM
Replies: 3
Views: 96
Posted By SwissPoker
Re: I don\'t get the Bernoulli Equation right

Thanks for all the answers.
Forum: Probability 10-22-2005, 11:49 AM
Replies: 3
Views: 96
Posted By SwissPoker
I don\'t get the Bernoulli Equation right

I understand this ...

If I throw a dice 11 times, what is the probability that 8 times I throw a six?

C(11,8) * (1/6)^8 * (5/6)^3 = 0.00006 = 0.006%

Now, I thought I can apply this formula...
Forum: Probability 05-15-2005, 02:11 PM
Replies: 4
Views: 87
Posted By SwissPoker
Re: More 5Draw Probability Questions

Thank you very much for your answers.

I still wonder how you calculated the kickers.

For 3 ranks *42* possible kickers: 47 cards - 2 aces - the 3 discards?
For 4 ranks *41* possible kickers:...
Forum: Probability 05-14-2005, 09:18 PM
Replies: 4
Views: 87
Posted By SwissPoker
More 5Draw Probability Questions

1. I was wondering how Mike Caro arrives at 2592 combinations to Aces-up in 5draw (no joker) when drawing 3 to 2 aces?

I tried to figure it like this:

pair combinations (3 ranks with 3 cards):...
Forum: Probability 05-14-2005, 06:39 PM
Replies: 2
Views: 82
Posted By SwissPoker
Re: Probability of making a full house

Thank you very much for your help.

I corrected the calculation ...

Total of combinations: C(47,3) = 16'215

9 ranks with 4 cards left:
9 * C(4,3) = 9 * 4 = 36
9 * C(4,2) * 2[aces] = 9 * 6 *...
Forum: Probability 05-14-2005, 11:20 AM
Replies: 2
Views: 82
Posted By SwissPoker
Probability of making a full house

I was wondering how Mike Caro arrives at 97.3 to 1 to a full house in 5draw (no joker) when drawing 3 to 2 aces?

I tried to figure it like that:

Since I keep two aces 50 cards are left. Thus...
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