Is this math right?

cavalier · #1 02-04-2003, 11:49 AM

The likelihood of being dealt a 7-stud hand WITH a 10, J, or Q is 55.29%
1 - [ (40/52) * (39/51) * (39/50) ] = 55.29%

The odds of starting an 8 player 7-stud hand with NO PLAYER having a 10, J, or Q is
(40/52)*(39/51)*(38/50)*(37/49)*(36/48)*(35/47)*(34/46)*(33/45)*(32/44)*(31/43)*(30/42)*(29/41)*(28/40)*(27/39)*(26/38)*(25/37)*(24/36)*(23/35)*(22/34)*(21/33)*(20/32)*(19/31)*(18/30)*(17/29) = 0.000147406931056658 or 1 in 6784 hands.

Let's assume you were dealt a 7-stud hand that doesn't have a 10, J or Q.

If there are 8 other players in, you'll know what 10 cards are ( your three and the other 7 on board ).

If you don't see ANY 9, J or Qs out, please let me know if the following is correct.

What is the chance that one of those other players has one in the hole.
Is it -> (30/42)*(29/41)*(28/40)*(27/39)*(26/38)*(25/37)*(24/36)*(23/35)*(22/34)*(21/33)*(20/32)*(19/31)*(18/30)*(17/29) =
0.00275107916728688 or 1 in 363?

Logic used was
1) You have seen 10 cards leaving 42 in the deck.
2) There are 12 cards which are a 10, J or Q, leaving 30 that aren't.
3) You multiply the probability of each card NOT BEING one of the 12 for the rest of the 14 cards you can't see.

Now, assume you don't have one in your hand. However, you see one on board. What are the odds of someone having one in the hole?
Are they -> (31/42)*(30/41)*(29/40)*(28/39)*(27/38)*(26/37)*(25/36)*(24/35)*(23/34)*(22/33)*(21/32)*(20/31)*(19/30)*(18/29) =
0.00501667377564078 or 1 in 199?

Thanks in advance!

Bozeman · #2 02-05-2003, 02:26 AM

It looks like in the last two you calculated the probability that no one has another one, not the probability the someone has one as you stated (which is 1-the prob. you calculated, so very likely).

Numbers are at least approximately right with this caveat.

Craig

cavalier · #3 02-05-2003, 11:20 AM

You are right... see it now.

Question #1)
What is the chance that one of those other players has one in the hole?
( 1 - 0.00275107916728688 ) or 99.72% of the time

Question #2)
Now, assume you don't have one in your hand. However, you see one on board. What are the odds of someone having one in the hole?
( 1 - 0.00501667377564078 ) or 99.50% of the time.

Like you said, both very likely.

Thanks!