Probability of Ace of Spades on fFop
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January 28th, 2018 at 11:08:42 AM
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There seems to be a lot of confusion as to the formula for solving this. What is the probability of the ace of spades (or any unique card) hitting the flop? Hole cards unknown. My gut says apx 6%. This is from ADDING the probabilities of each card as dealt. - 1/52 + 1/51 + 1/50 = apx 6%.
January 28th, 2018 at 11:28:32 AM
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I think I agreeQuote: wrc555What is the probability of the ace of spades (or any unique card) hitting the flop? Hole cards unknown.
Flop being 3 cards from 52
I would use this
[C(1,1)*C(51,2)] / C(52,3)
Wolfram Alpha
result
3/52 or
0.0576923
C(1,1) = 1 unique card and choose 1
C(51,2) = 51 other cards and choose 2
C(52,3) = 52 total cards and choose 3
Sally
Last edited by: mustangsally on Jan 28, 2018
I Heart Vi Hart
January 28th, 2018 at 11:51:15 AM
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that can't be correct the more I look at itQuote: wrc555This is from ADDING the probabilities of each card as dealt. - 1/52 + 1/51 + 1/50 = apx 6%.
a=1/52 if the As on 1st draw
for the 2nd card
b=51/52 * 1/51
3rd card
c=51/52 * 50/51 * 1/50
now add
a+b+c = 3/52
1/52 + 51/52 * 1/51 + 51/52 * 50/51 * 1/50 = 3/52
looks much better
now we agree
I Heart Vi Hart
January 28th, 2018 at 1:47:51 PM
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Of course, the answer is different if you are holding a two-card hand and you know that neither is the Ace of spades. Then there are only 50 unknown cards and its 3/50 or exactly 6%.
What is the probability of the Ace of spades being in the flop if you are in a 6-player game and all 5 other players have called your opening raise? Now we have slipped the bonds of objective mathematics and are into game theory. I would answer "<6%" because there is evidence that the other five players may have been dealt an unusually large number of high cards. So the ace of spades would have a larger probability than usual of being in one of the other 5 hands.
What is the probability of the Ace of spades being in the flop if you are in a 6-player game and all 5 other players have called your opening raise? Now we have slipped the bonds of objective mathematics and are into game theory. I would answer "<6%" because there is evidence that the other five players may have been dealt an unusually large number of high cards. So the ace of spades would have a larger probability than usual of being in one of the other 5 hands.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
January 28th, 2018 at 1:53:55 PM
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One simple approach is to consider the deck before you deal any cards. The Ace of Spades, or any other card you nominate, has an equal chance of being in 1st, 2nd... 52nd position. Three of these positions are used to create the flop, so there are three chances in 52 that the As is one of them.
January 29th, 2018 at 8:46:55 AM
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Quote: mustangsallythat can't be correct the more I look at it
a=1/52 if the As on 1st draw
for the 2nd card
b=51/52 * 1/51
3rd card
c=51/52 * 50/51 * 1/50
now add
a+b+c = 3/52
1/52 + 51/52 * 1/51 + 51/52 * 50/51 * 1/50 = 3/52
looks much better
now we agree
As stated, I do not get that to reduce to 3/52 or even close.
January 29th, 2018 at 11:44:11 AM
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Flop is meaningless... same as saying "What's the odds of drawing the Ace of Spaces when drawing 3 cards randomly from a deck?" The flop only holds meaning if we know other cards, such as player hole cards.
Answer then should be fairly clear...
P(1st Try) = 1/52
P(2nd Try) = 1/51
P(3rd Try) = 1/50
P(drawing in 3 cards) = P(1st) + P(2nd) + P(3rd) = 1/52 + 1/51 + 1/50 = .01923 + .01961 + .02 = .05884
Answer then should be fairly clear...
P(1st Try) = 1/52
P(2nd Try) = 1/51
P(3rd Try) = 1/50
P(drawing in 3 cards) = P(1st) + P(2nd) + P(3rd) = 1/52 + 1/51 + 1/50 = .01923 + .01961 + .02 = .05884
Playing it correctly means you've already won.
January 29th, 2018 at 12:07:04 PM
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There are 22,100 possible flops. 1,275 of them contain the As. (remove the As, then multiply 51x50/2 to get all of the possible 2 card combos with the As).
1 in 17.333...
1 in 17.333...
DUHHIIIIIIIII HEARD THAT!
January 29th, 2018 at 12:13:31 PM
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that is different from others that say 3/52 is the answerQuote: RomesP(drawing in 3 cards) = P(1st) + P(2nd) + P(3rd) = 1/52 + 1/51 + 1/50 = .01923 + .01961 + .02 = .05884
they are close
which is the correct answer?
I think 3/52 is correct
because adding 3 values that are NOT independent is not
following some basic probability rules.
let us go the other way
prob 1st card is NOT As
51/52
2nd card is NOT As
50/51
3rd card is NOT As
49/50
51/52 * 50/51 * 49/50
we multiply because we want all of these events to occur
= 49/52 = a
Wolfram Alpha
1-a = probability As is in the 1st 3 cards drawn
3/52
Sally
I Heart Vi Hart
January 29th, 2018 at 12:46:23 PM
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Thanks Sally. I think your thinking is correct and the answer is apx 5.75% or 3/52 but I don't follow your last line. I don't think that reduces to 49/52.
"51/52 * 50/51 * 49/50
we multiply because we want all of these events to occur
= 49/52 = a
"51/52 * 50/51 * 49/50
we multiply because we want all of these events to occur
= 49/52 = a
January 29th, 2018 at 12:47:17 PM
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Sally, I was able to understand your post!!! ...and appreciate your view point =).Quote: mustangsallythat is different from others that say 3/52 is the answer
they are close
which is the correct answer?
I think 3/52 is correct
because adding 3 values that are NOT independent is not
following some basic probability rules....
I'm torn... in your example you say they are independent events (NOT drawing the As each time), but you say in my example they're not independent events... even though they're the mirror opposite of one another, correct?
Playing it correctly means you've already won.
January 29th, 2018 at 12:51:58 PM
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I believe the differences you guys are getting is due to rounding.
Again 22,100 ÷ 1,275 = 17.333333333333333... or 0.0576923077%
Again 22,100 ÷ 1,275 = 17.333333333333333... or 0.0576923077%
DUHHIIIIIIIII HEARD THAT!
January 29th, 2018 at 1:06:29 PM
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1275/22100 = 3/52Quote: IbeatyouracesI believe the differences you guys are getting is due to rounding.
Again 22,100 ÷ 1,275 = 17.333333333333333... or 0.0576923077%
and
"the differences you guys are getting is due to rounding" is false.
some are adding dependant events, not following probability rules.
1- [51/52 * 50/51 * 49/50] = 3/52
Wolfram Alpha 2
no rounding.
btw, many can't wrap their head around 22,100 and 1275 (where they come from)
too large numbers and not basic math
some can and are the lucky ones
some can't and are the lucky ones
Sally
I Heart Vi Hart
January 29th, 2018 at 1:08:56 PM
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Quote: mustangsally1275/22100 = 3/52
and
"the differences you guys are getting is due to rounding" is false.
some are adding dependant events, not following probability rules.
1- [51/52 * 50/51 * 49/50] = 3/52
Wolfram Alpha 2
no rounding.
btw, many can't wrap their head around 22,100 and 1275 (where they come from)
too large numbers and not basic math
some can and are the lucky ones
some can't and are the lucky ones
Sally
22,100 = number of possible 3 card flops.
1,275 = number of those flops that have the ace of spades in it.
1,275 ÷ 22,100 = 0.0576923077. Same as 3/52 or 1/17.333333333333...
DUHHIIIIIIIII HEARD THAT!
January 29th, 2018 at 1:28:00 PM
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I don't know why, I really liked this and got a chuckle out of it =).Quote: mustangsally...some can and are the lucky ones
some can't and are the lucky ones
Sally
This would indicate two different approaches converging to the same answer... which would incline me to believe the 3/52 is the correct answer.Quote: Ibeatyouraces22,100 = number of possible 3 card flops.
1,275 = number of those flops that have the ace of spades in it.
1,275 ÷ 22,100 = 0.0576923077. Same as 3/52 or 1/17.333333333333...
Playing it correctly means you've already won.
January 29th, 2018 at 1:35:02 PM
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It's being made more complicated than it is.
DUHHIIIIIIIII HEARD THAT!
January 29th, 2018 at 3:28:33 PM
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one possible way to look at it.Quote: IbeatyouracesIt's being made more complicated than it is.
using combinations as a solution is difficult for most, as most of us do not get it.
one solution is like
roll 2 dice.
What is probability of getting at least one 6?
most will say 1/6 + 1/6 = 2/6
'Yahoo answers' has this many times.
the reason why that is not correct is it overcounts (now we lose almost everyone)
so let's go the other way.
*****
probability of NO 6 on 1 die = 5/6
so 2 dice = 5/6 * 5/6 =25/36
1 - 25/36 = 11/36
close but not = to 1/6 + 1/6 = 6/36 + 6/36 = 12/36
*****
(the 1,1 roll counts both 1s as 2 and not 1)
1,1
1,2
1,3
1,4
1,5
1,6
2,1
3,1
4,1
5,1
6,1
yep, only 11 from 36
as 1,1 is listed only once
Sally
I Heart Vi Hart
