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Zophar
Moderator
Joined: 26 Oct 2005
Posts: 3276
Location: East Coast
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 Posted: Mon Nov 26, 2007 3:57 pm Post subject: |
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| Big Slick x13x wrote: |
1 / 17850
52/52 * 3/51 * 2/50 * 1/42(assuming full table)
There is Ten known cards out going into fourth street and you have an equal shot any of the of the remaining 42 cards so you use that number.
I was hoping someone else would post this before me, I gave plenty of time. |
That is what I was thinking in my last post. Because whole cards are known, they have to be removed from the equation. |
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looseandontilt
Pair
Joined: 25 Nov 2007
Posts: 41
Location: Cashiers Window
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| Posted: Mon Nov 26, 2007 4:05 pm Post subject: |
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Its 33%
either bet call or fold
the choice is yours |
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Strasse
Forum Ego
Joined: 16 Nov 2005
Posts: 3571
Location: Austin, TX
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| Posted: Tue Nov 27, 2007 12:41 am Post subject: |
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| Big Slick x13x wrote: |
1 / 17850
52/52 * 3/51 * 2/50 * 1/42(assuming full table)
There is Ten known cards out going into fourth street and you have an equal shot any of the of the remaining 42 cards so you use that number.
I was hoping someone else would post this before me, I gave plenty of time. |
OMG NO. NOOOOOOOOOO. WRONG. NO STOP.
There is a chance that the card you need will end up as someone else's doorcard. You have to take that into consideration when you calculate this. Your equation is the probability that you will get 4 of a kind in 4 cards GIVEN THAT your 4th card isn't dead as someone else's doorcard. If you are talking strictly what if scenarios of getting 4 of a kind in 4 cards, you don't take into account dead cards. Please just understand this, you have no idea how utterly frustrating this is for me. Especially when you have people agreeing with you.  |
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cubbies760
Next Year Is Here
Joined: 19 Oct 2006
Posts: 5903
Location: Suburban Chicago
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| Posted: Tue Nov 27, 2007 12:45 am Post subject: |
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| I suppose to make my original question simple, what are the odds that if I mixed up a deck and spread it all over the floor.....that if I picked a Queen (for example), the next (3) picks would be the remaining Queens. |
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Strasse
Forum Ego
Joined: 16 Nov 2005
Posts: 3571
Location: Austin, TX
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| Posted: Tue Nov 27, 2007 12:57 am Post subject: |
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| cubbies760 wrote: |
| I suppose to make my original question simple, what are the odds that if I mixed up a deck and spread it all over the floor.....that if I picked a Queen (for example), the next (3) picks would be the remaining Queens. |
see my OP |
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Zophar
Moderator
Joined: 26 Oct 2005
Posts: 3276
Location: East Coast
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| Posted: Tue Nov 27, 2007 1:17 am Post subject: |
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It's almost a trick question I think, because it could be answered two ways.
If you are asking what the odds are of being dealt quads in 4 consecutive cards on any random hand, strasse is correct. If it's put as an after the fact question, using a specific hand, I think removing the up cards from the equation is correct. Along the lines of, what are the odds of being dealt XXX, w/no X's up and catching the fourth X. It strictly deals w/the percentage of hands of strasses equation where the up card is known, increasing the odds of it happening.. |
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Strasse
Forum Ego
Joined: 16 Nov 2005
Posts: 3571
Location: Austin, TX
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| Posted: Tue Nov 27, 2007 1:51 am Post subject: |
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| Zophar wrote: |
It's almost a trick question I think, because it could be answered two ways.
If you are asking what the odds are of being dealt quads in 4 consecutive cards on any random hand, strasse is correct. If it's put as an after the fact question, using a specific hand, I think removing the up cards from the equation is correct. Along the lines of, what are the odds of being dealt XXX, w/no X's up and catching the fourth X. It strictly deals w/the percentage of hands of strasses equation where the up card is known, increasing the odds of it happening.. |
Your way of solving it isn't really usable ever though. Because to know that your quad card is live, that means everyone has to have already been dealt 3 cards, in which case you have to already have trips, so the only really usable situation you could get from there is, "what are the odds that I will now get my 4th card?" |
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Big Slick x13x
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Joined: 18 Jun 2006
Posts: 4115
Location: ROK
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| Posted: Tue Nov 27, 2007 3:08 am Post subject: |
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| Strasse wrote: |
| Big Slick x13x wrote: |
1 / 17850
52/52 * 3/51 * 2/50 * 1/42(assuming full table)
There is Ten known cards out going into fourth street and you have an equal shot any of the of the remaining 42 cards so you use that number.
I was hoping someone else would post this before me, I gave plenty of time. |
OMG NO. NOOOOOOOOOO. WRONG. NO STOP.
There is a chance that the card you need will end up as someone else's doorcard. You have to take that into consideration when you calculate this. Your equation is the probability that you will get 4 of a kind in 4 cards GIVEN THAT your 4th card isn't dead as someone else's doorcard. If you are talking strictly what if scenarios of getting 4 of a kind in 4 cards, you don't take into account dead cards. Please just understand this, you have no idea how utterly frustrating this is for me. Especially when you have people agreeing with you. |
Well seriously, I wouldn't consider quads an option. No one will. So why ask for a calculation if you're just going to shoot it down with a hypothetical when it doesn't agree with you? Who the hell would ask what the odds are of hitting quads in 4 cards if one of your cards is visibly dead? No one, that's who. Please some common sense would be super. |
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Zophar
Moderator
Joined: 26 Oct 2005
Posts: 3276
Location: East Coast
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| Posted: Tue Nov 27, 2007 12:57 pm Post subject: |
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| Strasse wrote: |
| Zophar wrote: |
It's almost a trick question I think, because it could be answered two ways.
If you are asking what the odds are of being dealt quads in 4 consecutive cards on any random hand, strasse is correct. If it's put as an after the fact question, using a specific hand, I think removing the up cards from the equation is correct. Along the lines of, what are the odds of being dealt XXX, w/no X's up and catching the fourth X. It strictly deals w/the percentage of hands of strasses equation where the up card is known, increasing the odds of it happening.. |
Your way of solving it isn't really usable ever though. Because to know that your quad card is live, that means everyone has to have already been dealt 3 cards, in which case you have to already have trips, so the only really usable situation you could get from there is, "what are the odds that I will now get my 4th card?" |
That's just another context for the question to be asked. A series of questions could be asked along those same lines. After two are dealt, you could ask what are the odds of the next two coming X and X? The odds increase because you already have part of the equation answered.
I think it is usable because it it can easily formed as a question pertaining to a specific scenario. The conditions are already in place and a specific factor is given. You stopped the scenario on 3rd street and gave 4th street an independent action. My concept was the possibility on 4th street of the previous occurence happening, while factoring in variable of 3rd street being now being known.
I want to reiterate that I 100% agree with you on the equation of how to determine the odds of 4 straight being dealt in any random situation. I just feel that when a previously unkown X factor is now known, a slightly different equation can be used.
What it come down to is your equation is the most commonly applicable to the basic question. If you want to agree to disagree beyond that point, that's fine. |
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Strasse
Forum Ego
Joined: 16 Nov 2005
Posts: 3571
Location: Austin, TX
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| Posted: Tue Nov 27, 2007 6:03 pm Post subject: |
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| Big Slick x13x wrote: |
| Strasse wrote: |
| Big Slick x13x wrote: |
1 / 17850
52/52 * 3/51 * 2/50 * 1/42(assuming full table)
There is Ten known cards out going into fourth street and you have an equal shot any of the of the remaining 42 cards so you use that number.
I was hoping someone else would post this before me, I gave plenty of time. |
OMG NO. NOOOOOOOOOO. WRONG. NO STOP.
There is a chance that the card you need will end up as someone else's doorcard. You have to take that into consideration when you calculate this. Your equation is the probability that you will get 4 of a kind in 4 cards GIVEN THAT your 4th card isn't dead as someone else's doorcard. If you are talking strictly what if scenarios of getting 4 of a kind in 4 cards, you don't take into account dead cards. Please just understand this, you have no idea how utterly frustrating this is for me. Especially when you have people agreeing with you. |
Well seriously, I wouldn't consider quads an option. No one will. So why ask for a calculation if you're just going to shoot it down with a hypothetical when it doesn't agree with you? Who the hell would ask what the odds are of hitting quads in 4 cards if one of your cards is visibly dead? No one, that's who. Please some common sense would be super. |
Comprehension would be even better. First of all, I didn't ask for a calculation, cubbies did, I gave the right answer, and then you come in later and answer a different question that was never asked. Obviously nobody would ask what the odds are of hitting 4 of a kind when one of your cards is dead, but when you ask what the odds are of hitting 4 of a kind, having one of your cards dead is an option. On the same note, obviously nobody would ask what the odds are of hitting 4 of a kind in 4 cards is when your first two cards are different, but that is an option, and you have to factor for it, otherwise you would simply be asking what the odds are of getting the 4th card after you already have the 3. |
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HuJwang
Forum Blight
Joined: 20 Aug 2006
Posts: 5337
Location: Halifax, NS
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| Posted: Tue Nov 27, 2007 6:28 pm Post subject: |
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| Strasse's calculation is correct. |
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cubbies760
Next Year Is Here
Joined: 19 Oct 2006
Posts: 5903
Location: Suburban Chicago
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| Posted: Tue Nov 27, 2007 9:08 pm Post subject: |
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Thanks for everyone's effort in regards to my OP. I didn't mean to cause any arguments or frustration. I don't play much Stud, and I was shocked when I saw quads and thought how unlikely it must be to get quads on 4 cards.
I suppose I didn't necessarily need to know the exact odds of it happening, I was mainly interested in a ballpark idea, which all of the calculations gave me. I understand the differences in how the calculations vary depending on how one would interpret what I was asking. |
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Big Slick x13x
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Joined: 18 Jun 2006
Posts: 4115
Location: ROK
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| Posted: Mon Dec 10, 2007 7:28 pm Post subject: |
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| Ok, I'll finally repent for what I did. I did what I did with a sense of humor involved. No one got the movie reference I guess, which is kind of sad. The stuff that was meant to rile up Strasse did exactly that. I only know of one other forumer that realized I wasn't 100% serious in any of my posts. He was kind enough to play along though. So now I'll publicly apologize for turning this thread the wrong direction. |
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Nobody4727
Four of a Kind
Joined: 30 Nov 2007
Posts: 284
Location: God's practical joke
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| Posted: Tue Dec 11, 2007 2:06 am Post subject: |
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The odds of me getting quads on fourth street: 1 in 1,000,000.
The odds of my opponent doing it: 2 to 1. |
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JazzOne
Message Board Junkie
Joined: 17 Apr 2006
Posts: 1376
Location: Texas
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| Posted: Tue Dec 11, 2007 6:12 pm Post subject: |
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| Big Slick x13x wrote: |
| Who the hell would ask what the odds are of hitting quads in 4 cards if one of your cards is visibly dead? No one, that's who. |
Hahahaha, this is hillarious. Actually, however, the odds of hitting quads in four cards are 100%. After all, who the hell would ask about the odds unless he'd already been dealt quads? |
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