If we hold 87, what is the chance that we flop a gutshot (4 out) straight draw? Include the times we also flop a pair or flush or flush draw, but don't include the times we flop an 8 out straight draw or flop a straight.
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If we hold 87, what is the chance that we flop a gutshot (4 out) straight draw? Include the times we also flop a pair or flush or flush draw, but don't include the times we flop an 8 out straight draw or flop a straight.
in the following blocks of combinations take for example JTx. x cannot be 9 else its not a gutshot , nor can it be a 6 as the 6TJ combinations will be counted in the T6x block therefore for each subblock theres 8 ineligible cards.
6 blocks of combinations TJx,T6x,9Jx,46x,45x,59x.
each block can be made up as
(
JxT TxJ (4/50*41/49*4/48)*2= 1312/117600
+
JTx TJx (4/50*4/49*40/48)*2= 1280/117600
+
xJT xTJ (42/50*4/49*4/48)*2= 1344/117600
)
*
6 blocks of combinations
= (3936)*6/117600
=23616/117600
=20.08%
EDIT: as linaker has just pointed out double gutshots are not allowed either so they will have to deducted from my total. easiest way is to rework with 2 blocks with 8 ineligible cards and 4 blocks with 12 ineligible cards.
in the following blocks of combinations take for example JTx. x cannot be 9 else its not a gutshot , nor can it be a 6 as the 6TJ combinations will be counted in the T6x block therefore for each subblock theres 8 ineligible cards.
2blocks of combinations TJx,45x,
each block can be made up as
(
JxT TxJ (4/50*41/49*4/48)*2= 1312/117600
+
JTx TJx (4/50*4/49*40/48)*2= 1280/117600
+
xJT xTJ (42/50*4/49*4/48)*2= 1344/117600
)
+
4 blocks of combinations T6x,9Jx,46x,59x.
Tx6 6xT (4/50*37/49*4/48)*2= 1184/117600
+
6Tx T6x (4/50*4/49*36/48)*2= 1152/117600
+
xT6 x6T (38/50*4/49*4/48)*2= 1216/117600
)
= (3936*2+3552*4)/117600
=22080/117600
Hope you get better soon Nish. You forgot to eliminate the 8 out SDs tho.
Cards to flop a gutshot:
a) JTx
b) J9x
c) T6x
d) 95x
e) 64x
f) 54x
In each case, 1 value of x will make a straight. ie 9 with JT makes JT987.
In cases b) to e), 1 value of x will make a double gutshot. ie 5 with J9 makes J 987 5, so we have an 8 out SD
In cases b) to e), 1 value of x will make an OESD. ie 6 with J9 makes J 9876.
Flops containing JTx, where x does not make a straight (ie the flop is not a combination of JT9):
(4*4*44)*(6) = 4224
(44 because the four nines are excluded).
Times six because there's six combinations to get a gutshot:
4224*6 = 25344
But there are 2 additional cards in cases b to e that make an 8 out SD. Number of J96 or J95 flops is
4 * 4 *8 *6 = 768
So 768 *4 = 3072 flops contain an OESD, so the number of flops that only contain a 4 out SD are:
25344 - 3072 = 22272
The chance of making a 4 out gutshot is
22272/117 600 = 18.9%
This is wrong. If you think about it, you are saying that an AhJT flop happens more frequently than a JTAh flop.Quote:
Originally Posted by Keith_MM
In the first block you are working out the number of flops, given that the first card is a J and the third is a T. x can't be a 9 or a 6, so there are 42 possible cards for x. But we know the first card is one of the 4 jacks and the third card is one of the four Ts, so there are in fact only 40 possible cards for x. The same is true for the other blocks, since you have fixed the other two cards in each case.
In fact, you are double counting. In the JxT example you are including the T at position 3 in x, so counting it twice. This is why your xJT result > JxT > JTx.
LOOK AT THE LITTLE BASTARDS GO AHAHAHAHAHAHAHAHAHHAAHAHAHAHAHAHAHA
spoonitdrunk at 5pm?
My "guess":
[C(4,1)*C(4,1)*C(40,1)*4+C(4,1)*C(4,1)*C(36,1)*2]/C(50,3)
(16*40*4+16*36*2)/19600=18.94%
This comes from the following flops:
1) J9x -- no T, no 6
2) JTx -- no 9, no 6
3) 6Tx -- no 9, no 5, no 4
4) 59x -- no 6, no T, no J
5) 45x -- no 6, no 9
6) 46x -- no 5, no 9
18.29
I'm surprised more people didn't comment on how surprised they were by this, but 18.9% is the winnar.
Why would I be surprised by 18.9%?Quote:
Originally Posted by spoonitnow