probability of four 2's being dealt in a full deck...please help... - Page 2

gaming_mouse · #11 02-27-2005, 06:33 PM

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Ok, glad we sorted that out. I actually thought that he was asking what the probability of a particular person getting all 4 of them was. Confusion like that seems to happen often in this forum.

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This isn't the greatest example of it, but I find that often people don't really know what their question is. They'll see AA versus KK, and wonder, What are the odds? But they don't really know if they're wondering what the chance of that happening on the next hand is, or what the chance of that happening _given_ that they were dealt AA or KK is, etc. It's almost like running into someone you know at a train station in a foreign country and asking, What are the chances?

gaming_mouse · #12 02-27-2005, 06:35 PM

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Also, note that the OP asked for the answer without using the "choose" function.

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Yeah, didn't read that far.

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GM, you should think about doing an FAQ for the probability forum that explains such things since your are usually the one that ends up answering all of the questions anyway.

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Yes, but what I should REALLY start doing is stop answering so many of these questions and sticking to hand analysis posts, which is a much more profitable activity for me [img]/images/graemlins/smile.gif[/img]

slickpoppa · #13 02-27-2005, 06:37 PM

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Ok, glad we sorted that out. I actually thought that he was asking what the probability of a particular person getting all 4 of them was. Confusion like that seems to happen often in this forum.

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This isn't the greatest example of it, but I find that often people don't really know what their question is. They'll see AA versus KK, and wonder, What are the odds? But they don't really know if they're wondering what the chance of that happening on the next hand is, or what the chance of that happening _given_ that they were dealt AA or KK is, etc. It's almost like running into someone you know at a train station in a foreign country and asking, What are the chances?

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Yeah, like I said in my previous post, you might want to think about including stuff like that in a sticky for this forum. When most people ask questions, they just say "AA v. KK, what are the odds?" without specifying whether they mean "If I have KK, what is the probability of someone else having AA" or "What is the probability of one person having KK and another person having AA."

ojsdaman · #14 02-27-2005, 06:50 PM

thankyou all for your help...i appreciate it...but one thing..i was wondering what the chance of being dealt all 4 twos when each person gets 13 cards...does anyone know the chance that you get 3 of the 4 twos

slickpoppa · #15 02-27-2005, 06:58 PM

The probability of a particular person being dealt 3 twos is:

[(4/52)*(3/51)*(2/50)*(49/49)*(48/48)*(47/47)*(46/46)*(45/45)*(44/44)*(43/43)*(42/42)*(41/41)*(40/40)] * (13!/(10!3!)) = .0518

ojsdaman · #16 02-27-2005, 07:05 PM

slick- i am sorry but it has been so long since i was in school that i forgot what the ! means for the equation...it might be imaginary but i am not sure if i am remembering correctly..if it is imaginary then how do you do the equation without the !

ojsdaman · #17 02-27-2005, 07:08 PM

slick- also one more question...in your explanation of why you use 4! and 9! you phrased it as the ways each can be arranged...for example the 2's can be arranged 4 ways...is that because there are 4 2's and the 9! is because you have 9 other cards in your hand?

slickpoppa · #18 02-27-2005, 07:14 PM

! is the symbol for factorial. 9! = 9*8*7*6*5*4*3*2*1. In general, X! = X*(X-1)*(X-2)*......*1.

Since you don't already know that, it would probably be very diffucult for me to explain other probability concepts to you over this forum. If you are really interested in learning how to calculate things on your own, you should probably buy a book or find a good website that explains everything from scratch.

BruceZ · #19 02-27-2005, 09:35 PM

Here is a simple way to do this problem without combinatorics or factorials:

The probability of any particular card in your hand being the 2 [img]/images/graemlins/club.gif[/img] is 1/52, and since you have 13 cards, the probability that the 2 [img]/images/graemlins/club.gif[/img] is in your hand is 13/52.

Now given that the 2 [img]/images/graemlins/club.gif[/img] is in your hand, each of the other cards in your hand has a probability of 1/51 of being the 2 [img]/images/graemlins/diamond.gif[/img], and since there are 12 other cards in your hand, the probability that the 2 [img]/images/graemlins/diamond.gif[/img] is in your hand is 12/51.

Now given that the 2 [img]/images/graemlins/club.gif[/img] and the 2 [img]/images/graemlins/diamond.gif[/img] are in your hand, each of the other cards in your hand has a probability of 1/50 of being the 2 [img]/images/graemlins/heart.gif[/img], and since there are 11 other cards in your hand, the probability that the 2 [img]/images/graemlins/heart.gif[/img] is in your hand is 11/50.

Now given that the 2 [img]/images/graemlins/club.gif[/img], 2 [img]/images/graemlins/diamond.gif[/img] and the 2 [img]/images/graemlins/heart.gif[/img] are in your hand, each of the other cards in your hand has a probability of 1/49 of being the 2 [img]/images/graemlins/spade.gif[/img], and since there are 10 other cards in your hand, the probability that the 2 [img]/images/graemlins/spade.gif[/img] is in your hand is 10/49.

So all together, the probability that all four 2s are in your hand is the product of the above 4 probabilities, or 13/52 * 12/51 * 11/50 * 10/49 = 0.26% or 1 in 378.6. Since this is the probability that one particular hand has all four 2s, the probability that any of the 4 hands has all four 2s is 4 times this or 1.05% = 1 in 94.7.

I'm working on setting the above to music. [img]/images/graemlins/grin.gif[/img]

BruceZ · #20 02-27-2005, 09:56 PM

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I can figure out the odds that the first 4 cards are 2's

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That probability would be 4/52 * 3/51 * 2/50 * 1/49.

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...but i dont know where to go from there to take in account that you have 13 chances to get the 4 2's.

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You actually have C(13,4) chances to get the four 2s, so you would just multiply by C(13,4). That's exactly what C(13,4) means - it is the number of ways to pick the 4 cards out of 13 which will be the 2s. It is equal to 13*12*11*10/(4*3*2*1). So multiplying this by the probability that the first 4 cards are 2s found above, we get 13*12*11*10/(52*51*50*49) = 0.26% or 1 in 378.6. This is the probabilty that you have all four 4s, so multiply this probability by 4 to get the probability that any of the four players has all four 2s. This would be 1.05% or 1 in 94.7.

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Also in a side question. If you get 3 of the 2's, what is the probability that the person on your right gets the other 2.

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1/3, since each of the other 3 players has an equal probability of getting the remaining 2.

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If you are explaining how to come to this please explain without the "choose" button on the calc.

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See my other post in this thread for how to do it without using choose.

If you understand combinations, the simplest way to do this is to just compute C(48,9)/C(52,13) = 0.26%. The total number of hands is C(52,13), and C(48,9) is the number hands with all four 2s, since we choose the remaining 9 cards from the 48 non-2s.