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gamboolman
11-27-2002, 08:30 PM
Get Q4o in BB, then get Q4o in SB, then get Q4o on button.
They were not the same suits, but 3 in a row seems like it would be pretty high odds??
How do I calculate this?
Thanks,

Bozeman
11-27-2002, 09:42 PM
There are 12 ways to get Q4o (4 suits for the queen, 3 suits for the 4 since if it matches the Q you have Q4s).

There are 1326 possible holdem hands (51*52/2), so the chance that you will get dealt Q4o is 12/1326=0.9%.

The chance that you will be dealt it twice after you got dealt it once, is 12/1326*12/1326=1/12210, so this should happen occasionally to anyone who plays a fair amount of hold'em.

(The reason it is not proper to cube this probability is that you didn't know that this was an unusual situation until you got the hand the second time, it could have been any unsuited, unpaired hand, like 72o. Unless Q4 has some a priori significance to you.)

Craig

11-27-2002, 10:01 PM
First calculate how many possible 2 cards can be dealt, which is 52*51 = 2652. You then divided by 2! (2*1) to get rid of the times where the same 2 cards come, but just in different order (AsKs is the same as KsAs) 2652/2 = 1326. Now figure out how many different possibility are there for Q4o? There are 12:

Qs 4c
Qs 4d
Qs 4h
Qc 4s
Qc 4d
Qc 4h
Qd 4c
Qd 4s
Qd 4h
Qh 4s
Qh 4c
Qh 4d

So the chance of getting Q4o for 1 hand is 12/1326. Now you simply multiply that 3 times to get the chance of getting dealt Q4o for 3 hands in a row:

12/1326 * 12/1326 * 12/1326 =

1728 / 2331473976 = 0.000000741

0.0000741%

Or about 1 out of 1,349,233 (2331473976 / 1728)

11-27-2002, 10:07 PM
I heard this arugement before and always thought you had to do all 3, not just the 2 after the first 1 comes. He ask for the NEXT 3 hands to be Q4o which should make it cubed, no? Where am I going wrong?

Dynasty
11-27-2002, 10:10 PM
Here's another one for you. What are the odds that I get J8o in three consecutive hands on the same day you make this post? It happened today.

Bozeman
11-27-2002, 10:49 PM
Look at it this way: he had to have some hand in the BB (p=1). This one happened to be Q4o, but if he had 62o, and then got it two more times, he'd be posting about that.

I guess a slightly more accurate calculation would be P(suited)*P(same hand two more times)+P(unsuited)*P(same hand two more times)+P(pair)*P(same hand two more times)=4/17*4/1326*4/1326+12/17*12/1326*12/1326 +1/17*6/1326*6/1326=2e-6+5.8e-5+1e-6=6.1e-5 or about 1 in 16000. Approximately the same as just the unsuited probability.

Craig

Ray Zee
11-28-2002, 08:37 AM
to ruin it for everyone. three in a row of something happens regularly in life. we only notice sometimes after twice, but mostly after the third time. so the prob. is one of it happening. its just a matter of where it happens.its always most memorable with aces in poker. and there we start from twice as we get aces once or twice a session. so the chances of getting aces twice in a row is the same as getting them the next hand as we only look for the event after it happens first. can everyone figure out the prob. of getting aces on the next hand. i hope so.

gamboolman
11-28-2002, 02:41 PM
nm

Lin Sherman
11-29-2002, 03:39 AM
If the probability of getting a SPECIFIC HAND once is 1/x, then the probability of getting THAT hand three times in a row is 1/(x^3). However, the probability of getting ANY hand once is 1, so the odds of getting ANY hand three times in a row is 1 x 1/(x^2).

Lin

11-29-2002, 04:59 PM
Len wrote:
If the probability of getting a SPECIFIC HAND once is 1/x, then the probability of getting THAT hand three times in a row is 1/(x^3). However, the probability of getting ANY hand once is 1, so the odds of getting ANY hand three times in a row is 1 x 1/(x^2).

Correct, but the poster wasn't asking for ANY hand he was asking for Q4o, which would be 1 / (12/1326) ^ 3. Why did Bozeman say that you don't cube it because it has no priori significance? This all sounds like a word game to me. If some ask "what's the chance of getting 3 Q4o in the next 3 hands?" It is 1 / (12 / 1326) ^ 3, not 1 * (12 / 1326) ^ 2, correct? The first card is not a certainty. If I'm wrong and you have to square it, not cube, can you point me to a math website that explains this so I can get it right?

Thanks

11-29-2002, 05:04 PM
not 1 * (12 / 1326) ^ 2, correct?

not 1 * 1 / (12/1326) ^ 2

MSchmahl
12-01-2002, 11:22 PM
The "correct" answer is as much psychological as mathematical. This type of question is inspired by some unusual occurrence, and only afterwards does someone say "Wow, what were the chances of that?"

Well to answer the question correctly from a mathematical standpoint, "that" must be specified precisely.

It is correct to answer the question "What are the chances that my next three hands will be Q4o?" with (12/1326)^3, or 0.00000074 (the odds being 1,350,000-to-1 against). But perhaps a better question would be "What are the chances of being dealt the same hand in three particular deals?" (definining "same hand" as same two ranks, suitedness mattering, but the actual suits themselves not mattering: approximately 0.000061; 16,350-to-1 against); or "What are the chances of being dealt Q4o the next two times after it was dealt once?" ((12/1326)^2=0.000082; 12,209-to-1 against).

Perhaps an even better question would be "What are the chances of being dealt three of the same hand in a row sometime during a 4-hour session of hold 'em?" Assuming 150 hands in 4 hours, the answer would be close to 1 in 1000, which is not really all that remarkable. It should happen to you on average once every 4000 hours of play.