What are the odds ?

[Pawprint]

Horrible exit hand with 20 mins to go in tonight's GPS Day 1a.

3 way allin.  AA v JJ v AJ on a AJ6 flop.

I'm trying to get my head around the odds of top set vs middle set vs top two pair.

Anyone care to have a go at the numbers ?

[The Camel]

Horrible exit hand with 20 mins to go in tonight's GPS Day 1a.

3 way allin.  AA v JJ v AJ on a AJ6 flop.

I'm trying to get my head around the odds of top set vs middle set vs top two pair.

Anyone care to have a go at the numbers ?

Remote!

Never actually seen this one.

Only seen three sets on the flop once.

[doubleup]

In the old days the AJ wuld've mucked and the JJ  tank-called

[Ironside]

Ok just guessing you didnt have the aa then would of loved to see the aa cracked to a back door flush after that flop

[Oxford_HRV]

this hurts my brain too much, i'm guessing at 60,000-1

[Acidmouse]

Horrible exit hand with 20 mins to go in tonight's GPS Day 1a.

3 way allin.  AA v JJ v AJ on a AJ6 flop.

I'm trying to get my head around the odds of top set vs middle set vs top two pair.

Anyone care to have a go at the numbers ?

This is the EXACT hand that happened to me the night before a sky poker tour event in Leeds, my table had all sky players there.

I actually had the AJ and with that flop won, runner runner flush I had to be called back to tell me I had won the hand lol.

[Pawprint]

Right then,

The odds of being dealt a pocket pair is 17:1

I can't quite decide what the calculation should be for the odds of two players being dealt a pocket pair, but have found a calculation for a 6 handed table (this was 7 handed), so am going to use that.

0 players dealt a pair is simply:
(16/17)^6 = 0.6951 for 6-handed
and
(16/17)^9 = 0.5795 for 9-handed

1 player dealt a pair is:
6*(1/17)*(16/17)^5 = 0.2606 for 6-handed
and
9*(1/17)*(16/17)^8 = 0.3296 for 9-handed

So 2 or more people being dealt a pair will roughly be:
1 - 0.6951 - 0.2606 = 0.0443 for 6-handed (1 in every 22.6 deals)
and
1 - 0.5795 - 0.3296 = 0.0909 for 9-handed (1 in every 11.0 deals)



So, the odds of two players getting dealt a pocket pair is 22:1

The odds of a third player then being dealt one each of those pocket pair cards is 564:1 (2/48 x 2/47)

And finally, the odds of the remaining case pocket pair cards being flopped is 15,180:1 (91,080 combinations of 46 cards being flopped, with 6 combinations of the two case cards being flopped)

Which would make the total odds 22 x 564 x 15,180 : 1 or 188,353,440 : 1


Seems a big number.  Have I missed something obvious ?  (Should I be ignoring the first odds, as the probabilty of the second and third part are already based on the first bit having happened ?)

[TightEnd]

My head hurts Leigh.

[redsimon]

Fold Jacks pre. Saves a lot of hassle

[Cf]

I usually enjoy calculating odds but I think in this case I'd just go with

Remote!

[MintTrav]

Should the 2/48 be 4/48, Leigh?

[theprawnidentity]

As I have a habbit of trying to make maths wayyyyy too simple (and therefore wrong), can we calculate it by saying:

odds of flopping set (-1 out) * odds of flopping set (-1 out) * odds of flopping quads (2 precise cards)

0.06*0.06*0.00102 = 0.000003672

OR

1 in 272,331.2

1 in xxx million sounds right for this exact case of set over set over two pair.

[Tal]

Should the 2/48 be 4/48, Leigh?

We are saying that we know the exact location in the pack of two aces and two jacks and that our AJ man cannot get any of those four cards. What are the chances that he is dealt one of the remaining aces and one of the remaining jacks?

The odds of him being dealt either one of the remaining aces or one of the remaining jacks in the first card are 4/48.

The odds of him being dealt either one of the remaining aces or one of the remaining jacks in the second card, assuming an ace or a jack has gone (to us or someone else) would be 3/47.

So, the idea of both cards being either an ace or a jack is 4/48 x 3/47

However, that isn't what we want to know.

We want the first card to be an ace and the second to be a jack OR
The first to be a jack and the second to be an ace.

So, that's one of the two aces (2/48) and one of the two jacks (2/47)
OR
One of the two jacks (2/48) and one of the two aces (2/47)

So it's 2(2/48 x 2/47) = 1/282

That's for this one AJ bit.

[MintTrav]

Should the 2/48 be 4/48, Leigh?

We are saying that we know the exact location in the pack of two aces and two jacks and that our AJ man cannot get any of those four cards. What are the chances that he is dealt one of the remaining aces and one of the remaining jacks?

The odds of him being dealt either one of the remaining aces or one of the remaining jacks in the first card are 4/48.

The odds of him being dealt either one of the remaining aces or one of the remaining jacks in the second card, assuming an ace or a jack has gone (to us or someone else) would be 3/47.

So, the idea of both cards being either an ace or a jack is 4/48 x 3/47

However, that isn't what we want to know.

We want the first card to be an ace and the second to be a jack OR
The first to be a jack and the second to be an ace.

So, that's one of the two aces (2/48) and one of the two jacks (2/47)
OR
One of the two jacks (2/48) and one of the two aces (2/47)

So it's 2(2/48 x 2/47) = 1/282

That's for this one AJ bit.

Yes. Is that different to what I said?

[titaniumbean]

maths mugs itt


fiddy fiddy brah it happens or it dont!