3 consecutive hands: KK,AA,AK

BillFranklin · #1 03-16-2005, 09:42 PM

what are odds of these 3 hands back to back to back: KK,AA,AK (happened to me today.. couldn't believe it)

is it (6/1326)*(6/1326)*(16/1326)=0.247060189
405 to 1 against.
not sure if this correct, the number seems alittle low)

gaming_mouse · #2 03-16-2005, 10:13 PM

You did the calc right, but the arithmetic wrong.

It's 1 in 4,047,697.87

TaintedRogue · #3 03-20-2005, 10:24 AM

Your correct, however, you didn't answer his question....which he articulated poorly. What he was saying, was: I've played a total of X amount of hands on line since I started and was finally dealt AK,AA,KK three hands in a row.
Now, if he had asked: "What are the odds of me logging into my poker site, getting into a game and being dealt AK, AA & KK the first three hands.....well, then you are correct.
You must also take into consideration that he would have been here posting he phenomenon if he had been dealt AA, KK and the AA again. OR, AA,KK, QQ or etc., so you have to take into consideration all those various hands with extremely high odds against catching in 3 consecutive hands.
However, you did do a good job of pacifying the poster.

MickeyHoldem · #4 03-20-2005, 12:12 PM

so lets assume you see 300 hands during an online session, and we don't worry about what order we get these 3 hands in... so the odds become 6/52c2 * 6/52c2 * 16/52c2 * 6 of seeing these 3 hands back to back in any order.

Now we have 298 chances during the session to see this phenom. occur (it's actually a little less than 298, but we'll use 298 and be farily accurate), so we plug this into excel function BINOMDIST(1,6/52c2 * 6/52c2 * 16/52c2 * 6,298,false) = 0.0004415381435

So you should see this happen about once every 2265 sessions.

I'm not 100%(not even 20%) sure how to calculate the exact number of trials for this problem since the 3 hands have different chances of occuring!!... I got 295.876702 but I'm sure that ain't right!! Anyone???

Siegmund · #5 03-20-2005, 07:51 PM

If you want the odds of precisely KK, AA, and AK, in that order, your calculation is set up right.

More likely, though, what you want is something else. Either the odds of "holding only aces and kings for three hands in a row", which is (28/1296)^3, or the odds of "holding a starting hands this powerful or more so three hands in a row." The best starting hands: AA KK QQ JJ AKs TT AQs AKo. You didn't say whether your AK was suited or not. If so, it's again (28/1296)^3; if you just want AKo or better three times in a row, (50/1296)^3.

These are about 1 in 99,000 and 1 in 17,400 respectively. Uncommon - but common enough that many 2+2 regulars will see this several times a year.

This afternoon, for instance, I sat down at a table was dealt pockets pairs on my second, third, fourth, fifth, and sixth hands. (tens twice, nines, sevens, and fives - no monsters.) Not seen it before. But lots of weird things eventually happen.

jpg7n16 · #6 03-20-2005, 08:11 PM

Shouldn't it be set up as

(4/52)*(3/51)*(4/52)*(3/51)*(4/52)*(4/51) =

4/32,381,583

or 1 in 8,095,395.75

?? Shouldn't that be how it is set-up ??

MickeyHoldem · #7 03-20-2005, 08:35 PM

[ QUOTE ]
Shouldn't it be set up as

(4/52)*(3/51)*(4/52)*(3/51)*(4/52)*(4/51) =

[/ QUOTE ]

(4/52)*(3/51)*(4/52)*(3/51)*(8/52)*(4/51)

That AK gets 8 chances for the first card!

jpg7n16 · #8 03-21-2005, 05:45 PM

I didn't think about that [img]/images/graemlins/smile.gif[/img] thanks!

planB · #9 03-27-2005, 10:37 AM

In a recent sng: KK KK KK KJ (4 hands in a row - all 3 pp held up).

popniklas · #10 03-28-2005, 09:47 AM

</font><blockquote><font class="small">Svar till:</font><hr />
In a recent sng: KK KK KK KJ (4 hands in a row - all 3 pp held up).

[/ QUOTE ]

yikes, that's one nice run of cards. i once won a party poker sng with AA, AA, 44 as my three last hands. opponent started whining like crazy, it was pretty funny. [img]/images/graemlins/smile.gif[/img]