Probability Challenge

[GoSox]

I am sure there is someone on this forum who appreciates these kinds of questions, both happened to me the other night in the same hour. (NL HE)

1) Easier - Probability of getting AK 4 out of five hands.
Harder - Probability of getting 4 flops so awful to make nothing out of it. (Axx, 3-suited, 3 to a high straight)

2) Harder - Probability that a flop will contain two suited cards and that three of the remaining players who saw the flop will hold Ax Kx Qx all of that same suit.

For what it's worth I had Kx and we all hit the flush on the turn.

thanks.

[Paul2432]

Well your second problem is impossible to answer because it depends on the probability of a player not folding Ax, Kx, or Qx before the flop. If you assume that a player will always call preflop, then it depends on the number of players.

Paul

[BruceZ]

[ QUOTE ]
1) Easier - Probability of getting AK 4 out of five hands.

[/ QUOTE ]

P(4 or 5 AK suited or offsuit) =

(16/1326)^5 + 5*(16/1326)^4 * (1310/1326) = 1 in 9,526,575.


[ QUOTE ]
Harder - Probability of getting 4 flops so awful to make nothing out of it. (Axx, 3-suited, 3 to a high straight)

[/ QUOTE ]

Assuming AK offsuit:

P(Axx) = 1 - 47*46*45 / (50*49*48) = 17.27%

P(3 suited matching A or K w/no A) = (12*11*10 + 11*10*9) / (50*49*48) = 1.96%

P(QJT not all same suit) = (4*4*4-2) / C(50,3) = 0.32%

[ 1 - (17.27% + 1.96% + 0.32%)]^4 = 41.9%.


[ QUOTE ]
2) Harder - Probability that a flop will contain two suited cards and that three of the remaining players who saw the flop will hold Ax Kx Qx all of that same suit.

[/ QUOTE ]

Assuming 9 players, and these hands always see flop. You can scale by actual probability that they see flop:

C(9,3)*4*10*9*8 / [C(52,2)*C(50,2)*C(48,2)] * 7*6*39 /(46*45*44) = 1 in 421,140.

[Ed Miller]

3) Harder - Probability that BruceZ's numbers will make GoSox feel better about dropping 40 bets to a bunch of nincompoops.

[UMTerp]

I think your arithmetic is wrong in #1. I keep getting 1 in ~47,000,000.

[BruceZ]

[ QUOTE ]
I think your arithmetic is wrong in #1. I keep getting 1 in ~47,000,000.

[/ QUOTE ]

It's not wrong. Another way to do it is to sum the binomial distribution from 0 to 3 for 5 trials and subtract this from 1. In Excel, this would be

=1-BINOMDIST (3,5,16/1326,TRUE) = 1.0497E-07 = 1 in 9,526,575 exactly the same.


I think we might also want to include Kxx as a "non-awful" hand. This changes the answer to the second part of 1 to 18.2%. Also, I wasn't sure what he meant by "3 to a high straight", so I just took QJT, but perhaps he meant any 3 for the purpose of bluffing. I redid the calculation to include any 3 to a straight from A high down to Q high (with no aces). That made the answer 14.6% for the case where we like Kxx, and 32.8% otherwise.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
2) Harder - Probability that a flop will contain two suited cards and that three of the remaining players who saw the flop will hold Ax Kx Qx all of that same suit.

[/ QUOTE ]

Assuming 9 players, and these hands always see flop. You can scale by actual probability that they see flop:

C(9,3)*4*10*9*8 / [C(52,2)*C(50,2)*C(48,2)] * 7*6*39 /(46*45*44) = 1 in 421,140.

[/ QUOTE ]

This should be:

C(9,3)*4*30*18*8 / [C(52,2)*C(50,2)*C(48,2)] * 7*6*39*3 /(46*45*44) = 1 in 23,397.

See why?

[GoSox]

Thanks all !!