The Odds Must Be Insane!!!

icracknuts · #1 03-20-2006, 03:31 AM

Anybody know what the odds of holding hole cards and the board showing up all cards that match your hole cards?

E.X. Hole cards J10 The board JJJ1010

I just happened to me

UATrewqaz · #2 03-20-2006, 06:59 AM

Odds of being dealt a non-pair = .94

Total possible boards = 50 choose 3 = 19,600

Number board appearing with a full house of your two non-paired cards = 6

6 / 19,600 = .00030612244897959183673469387755102

.94 * .000306 = 0.00028764

In other words once every 3,476 hands, approx.

drbst · #3 03-20-2006, 08:34 AM

No cigar.

Total possible boards is 50 choose 5.

UATrewqaz · #4 03-20-2006, 08:53 AM

Correct, I realized this later.

50 choose 5 = 2,118,760

6 / 2,118,760 = 2.83e-6

.94 * 2.8318450414393324397288980346995e-6 = 7.538e-12

Or rather once every 132,657,917,777 hands

correct?

drbst · #5 03-20-2006, 09:03 AM

Right.

Phil153 · #6 03-20-2006, 10:44 AM

[ QUOTE ]
.94 * 2.8318450414393324397288980346995e-6 = 7.538e-12

[/ QUOTE ]
OK, what am I missing?

Tom1975 · #7 03-20-2006, 11:27 AM

You took a number with an order of magnitude of 10^-6, multiplied it by .94 and wound up with a number of order of magnitude 10^-12.

mostsmooth · #8 03-20-2006, 11:31 AM

[ QUOTE ]
I just happened to me

[/ QUOTE ]
100%

Kerth · #9 03-20-2006, 11:35 AM

Yet another reason to avoid RTG casinos.

UATrewqaz · #10 03-20-2006, 11:36 PM

My only friends in those calculations were my brain and MS calculator, so cut me some slack, I'm sure I made some errors here and there, what are the correct numbers btw?

I see that multiplication should be 2.6619343389529724933451641526171e-6 (not e-12)