simple probablility of a prop bet...

[Cap_Of_Water]

but i'm not sure if i figured it out correctly. Someone help me out pllzzz....

Side bet in a holdem game between player A and player B. Player A wins if a King, Queen, or Jack hits the flop (any three cards, just picked these randomly). Player B wins if there is no K, Q, or J on the flop. Who has the edge and what is the edge?

[tshort]

1 - Probability of no king and no queen and no jack

1 - ( (40/52) * (39/51) * (38/50) )

So prob of K, Q, or J appearing is a little over 55%

EDIT: Why are you posting the same question twice?

[Cap_Of_Water]

whats wrong with this?->

(12/52) + (12/51) + (12/50) = K,Q,or J hitting ~70.5%

[tshort]

[ QUOTE ]
whats wrong with this?->

(12/52) + (12/51) + (12/50) = K,Q,or J hitting ~70.5%

[/ QUOTE ]

Intoduction to Probability

(12/52) This is the chance the first card dealt is a K or Q or J.

(12/51) The probability of the second card dealt being a K or Q or J is dependent on what the first card was. (12/51) is the probability the second card dealt is a K or Q or J given the first card was not a K or Q or J. What if the first card dealt was a K? The prob of the second card being a KQJ is 11/51.

So you end up with all of these cases approaching it this way:

XXX Slots for cards. Enumerate probability of desired possibilities:

Let K represent K, Q, or J in a slot. n represents not KQJ.

KKK (= 12/52) * (11/51) * (10/50)
+
KKn = (12/52) * (11/51) * (48/50)
+
KnK = (12/52) * (48/51) * (11/49)
+
nKK ...
+
Knn
+
nKn
+
nnK

This is complicated so we find the probability of no KQJ:

nnn = (48/52) * (47/51) * (46/50)

[jay_shark]

[ QUOTE ]
whats wrong with this?->

(12/52) + (12/51) + (12/50) = K,Q,or J hitting ~70.5%

[/ QUOTE ]

You may be thinking of the inclusion/exclusion formula .

P(AuBuc) = P(A)+P(B)+P(C) - P(AB)-P(AC) -P(BC) +P(ABC)

Where A is the event the first card is a k,q,j.
B is the event the second card is a k,q,j
C is the event the third card is a k,q,j .

So we have 12/52*3 - 12/52*11/49*3 + 12/52*11/51*10/50 = 0.552941176

You may want to look up Venn diagrams to get a visualization of what's happening here .

[AaronBrown]

[ QUOTE ]
whats wrong with this?->

(12/52) + (12/51) + (12/50) = K,Q,or J hitting ~70.5%

[/ QUOTE ]
This, of course, is why people take the prop bet. They have an intuitive feel that there's a bit less than one chance in four of a card being K, Q or J; so they figure it's a bit less than three in four of getting a K, Q or J in three cards. It seems like a good bet to pay off 2:1 if that doesn't happen. A lot of prop bets exploit this false reasoning.

The logic works if you get paid double when two K's, Q's or J's show up; and triple when the flop is all K's, Q's and J's (in that case, it's 3 x 12 / 52 = 0.692). Since the prop bet only pays single for those cases, it's to the advantage of they guy offering it. Of course, you don't need any math to know the prop bet works to the advantage of the guy offering it.

[Sevenfold]

[ QUOTE ]
whats wrong with this?->

(12/52) + (12/51) + (12/50) = K,Q,or J hitting ~70.5%

[/ QUOTE ]

In laymans terms, what are the odds of me flipping at least one heads in two tosses?
50% + 50%=100%

You have to work backwards. The odds of not hitting a head (50%) times N flips.

Here it would be 50%(.5) x 50%(.5) = 25% (.25)

So you would hit a head 75% of the time in two flips.