i would call a shove with AK even if villain showed me AA because i have 6 outs and he only has 2. By my math that makes us a 75% favorite.Quote:
Originally Posted by a500lbgorilla
ez game imo.
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i would call a shove with AK even if villain showed me AA because i have 6 outs and he only has 2. By my math that makes us a 75% favorite.Quote:
Originally Posted by a500lbgorilla
ez game imo.
You keep missing the point that both of villains cards AREN'T dealt before you are dealt a card. Villain gets a card, you get a card, then he gets another card, and you get another card.
If for instance your first card is a King then the chances of villain's next card being a King is decreases. Obviously because there is one less King in the deck. The same holds if you look down and see KK in your hand. You now know there is only 2 Kings left in the deck, therefore he can only have 1 combo of KK. Likewise, because there is only 2 kings left in the deck (because you have the other two), then he can only make AK in 8 ways (4 aces * 2 kings). Reason being is because he obviously can't have your two Kings. Whereas, you had 22 he could have AK 16 ways and KK 6 ways because you know all of those cards remain in the deck.
Actually it doesn't matter what order they're dealt in, which is the point that MehFU doesn't understand, and which I might decide to refute also.Quote:
Originally Posted by XxStacksxX
I have lost count of the number of times that I have proven you wrong in this thread, and you have yet to prove anything incorrect with any argument I have given. Alas, out of the sake of how much I enjoy proving people wrong in areas of mathematics, I will provide the proof you are asking for here.Quote:
Originally Posted by MehFU
There are five piece of information that we need, and here I show the calculations for each piece of information. (Note: People observing will be interested to know that this proves that it's just as likely to be dealt a King on the first card dealt as the second, third, fourth, or even last card dealt.)
1. The chance of Villain being dealt a K on the first card is 4/52 = 0.0769230769
2. The chance of Hero not being dealt a K on the second card is (4/52)*(48/51) + (48/52)*(47/51) = 0.9230769231
3. The chance of Villain being dealt a K on the third card is (4/52)*(3/51)*(2/50) + (4/52)*(48/51)*(3/50) + (48/52)*(4/51)*(3/50) + (48/52)*(47/51)*(4/50) = 0.0769230769
4. The chance of Hero being dealt a K on the fourth card is the sum of the following:
king king king king (4/52)*(3/51)*(2/50)*(1/49)
king king blank king (4/52)*(3/51)*(48/50)*(2/49)
king blank king king (4/52)*(48/51)*(3/50)*(2/49)
king blank blank king (4/52)*(48/51)*(47/50)*(3/49)
blank king king king (48/52)*(4/51)*(3/50)*(2/49)
blank king blank king (48/52)*(4/51)*(47/50)*(3/49)
blank blank king king (48/52)*(47/51)*(4/50)*(3/49)
blank blank blank king (48/52)*(47/51)*(46/50)*(4/49)
Which is 0.0769230769, thanks to our friend Excel.
5. The chance of Hero not being dealt a K on the fourth card is (1-0.0769230769) = 0.9230769231, as proven in (4).
Now we have two cases to examine. The first case is when the cards are dealt King-blank-King-King, and the second case is when the cards are dealt King-blank-King-blank. If the probability of the first is lower than the probability of the second, then we have proven that the probabilty of Villain holding KK is reduced because our last card is a King.
The chance of the cards coming King-blank-King-King is 0.0769230769 * 0.9230769231 * 0.0769230769 * 0.0769230769 = 0.0004201534.
The chance of the cards coming King-blank-King-blank is 0.0769230769 * 0.9230769231 * 0.0769230769 * 0.9230769231 = 0.0050418403.
this thread is going nowhere, the last 50 replies have all been the same
{locked}
oh by the way, bringing up Meta in a $5NL hand history post is beyond stupid!!
Dave told me you asked if I was around. The things you do behind my back when I'm sleeping!Quote:
Originally Posted by bigspenda73