UNBELIEVABLE hand, What are the odds?

[KyleH186]

We had a 8 person sit n go tonight, and when it was down to four-handed, we had this ridiculous hand. First to act raised to triple the bb, and the button was a medium stack, and pushed all in. The small blind called, and the blind folded. First to act called. Lots of money in the pot.

Flop was Js 3h 8s...the small blind checked, first to act then pushed all in, and small blind called.

One flipped over AsQs, and the other flipped over AhQd. The button, who is all in, flipped over AcQc, and the big blind, who had folded, flipped AdQh. Screaming ensued.

The turn:
9s. River was a blank.

How probable (or improbable) is it that each of the four players was dealt AQ, two of them were suited, and one of them flushed? Does anyone know the exact odds?

By the way, we are positive the hand was not set up, and this did actually happen.

Thanks

[gaming_mouse]

[ QUOTE ]

How probable (or improbable) is it that each of the four players was dealt AQ, two of them were suited, and one of them flushed? Does anyone know the exact odds?


[/ QUOTE ]

Since you'd be equally surprised if all 4 were suited, we'll change the question to "at least 2 of them were suited".

There are (4 choose 2) + 1 = 7 ways to deal out 4 AQ hands in which at least 2 are suited.

We first count the number of ways that these 4 hands can be dealt to 4 particular players (eg, the players in seats 1-4).

For each of the 7 possible ways to deal out AQ, there are 8 cards left to be dealt to the remaining 4 players. There are

(44 choose 8) = 177232627

ways to select those 8 cards, and for each set of 8 cards selected there are

7!! = 7*5*3*1 = 105

ways to break them into 4 hands of 2 cards each. Thus there are

177232627*105 = 18 609 425 835

possible unordered deals FOR EACH of the 7 AQ unordered deals.

The chance, then, of having the 4 AQ (with at least 2 suited) dealt to the first 4 players is:

7*18 609 425 835/( (52 choose 16)*15!!)

And now, to get the chance that the 4 AQ are dealt to ANY four players, we simply multiply the above by (8 choose 4):

(8 choose 4)*7*18609425835/( (52 choose 16)*15!!) = 4.3408652 * 10^-7, or 1 in 2 303 688.21

Now we have to calculate the chance that one of them flushes. There are two cases: when we have exactly 2 suited cards (6 of 7 times) and when everyone is suited (1 of 7 times).

When we have 2 suited cards, there are:

2*((11 choose 3)*(39 choose 2) + (11 choose 4)*(39 choose 1) + (11 choose 5))

boards that give the suited players a flush, and

2*((11 choose 4)*(39 choose 1) + (11 choose 5))

additional boards that give the unsuited players a flush, for the total chance of:

((2*((11 choose 3)*(39 choose 2) + (11 choose 4)*(39 choose 1) + (11 choose 5))) + (2*((11 choose 4)*(39 choose 1) + (11 choose 5))))/(50 choose 5) = 0.140581283

or about 14% that anyone flushes. When we have 4 suited cards, the chance of a flush is:

(4*((11 choose 3)*(39 choose 2) + (11 choose 4)*(39 choose 1) + (11 choose 5)))/(50 choose 5) = 0.255993128

Thus the overall chance of a flush is:

0.255993128*(1/7) + 0.140581283*(6/7) = 0.157068689

almost 16%.

The final answer to the original question is therefore:

4.3408652 * 10^-7 * 0.157068689, or 1 in 14 666 756.5

Interpreting the results

We need to note, however, that you would have been equally surprised had the same thing happened with AK or AJ. So we should really divide the above answer by 3, bring us down to 1 in 4 888 918.83

It gets more tricky when you try to factor in all the other possible events that would have shocked you just as much: for example, AA, KK, QQ, and JJ all being dealt. 2 AA hands and 2 KK hands. 2 KK hands and 2 QQ hands. The list goes on and on, and each additional shocking event further discounts the probablilty we should really assign to the event you witnessed.

This is a general problem with trying to attach a probability to a rare event that you witness without stipulating beforehand the event you are waiting to see.

[KyleH186]

Thank you so much

[gaming_mouse]

[ QUOTE ]
Thank you so much

[/ QUOTE ]

You're welcome. I found a couple simple errors, though, in the flush calcs. Here are corrections:

((2*((11 choose 3)*(33 choose 2) + (11 choose 4)*(33 choose 1) + (11 choose 5))) + (2*((11 choose 4)*(33 choose 1) + (11 choose 5))))/(44 choose 5) = 0.202252654

(4*((11 choose 3)*(33 choose 2) + (11 choose 4)*(33 choose 1) + (11 choose 5)))/(44 choose 5) = 0.362693461

Changing the chance of a flush to:

(1/7)*0.362693461 + (6/7)*0.202252654 = 0.225172769

Changing the final answer to:

4.3408652 * 10^-7 * 0.225172769, or 1 in 10 230 758.5

The discussion on what the result means still applies.

gm

[Paul2432]

[ QUOTE ]
[ QUOTE ]

How probable (or improbable) is it that each of the four players was dealt AQ, two of them were suited, and one of them flushed? Does anyone know the exact odds?


[/ QUOTE ]

Since you'd be equally surprised if all 4 were suited, we'll change the question to "at least 2 of them were suited".

There are (4 choose 2) + 1 = 7 ways to deal out 4 AQ hands in which at least 2 are suited.

We first count the number of ways that these 4 hands can be dealt to 4 particular players (eg, the players in seats 1-4).

For each of the 7 possible ways to deal out AQ, there are 8 cards left to be dealt to the remaining 4 players.

[/ QUOTE ]

Nice analysis. However, you overlooked the following from the OP:

[ QUOTE ]
We had a 8 person sit n go tonight, and when it was down to four-handed

[/ QUOTE ]

If we modify your calculation to consider just a 4-player table then,

[ QUOTE ]
(8 choose 4)*7*18609425835/( (52 choose 16)*15!!) = 4.3408652 * 10^-7, or 1 in 2 303 688.21

[/ QUOTE ]

becomes,

7 / ( (52 choose 8)*7!!) ~= 1 / 11 billion

Using your numbers in the other post below to factor in the flush, this deal should occur once every 50 billion hands or so.

Paul

[Paul2432]

[ QUOTE ]
How probable (or improbable) is it that each of the four players was dealt AQ, two of them were suited, and one of them flushed? Does anyone know the exact odds?

[/ QUOTE ]

See my other post (and GMs) for the math, but the technically correct answer to this question is 100%.

That is, the odds that an event that has already happened, has actually happened is 100% (assuming you are not a troll or something).

Paul

[KyleH186]

I guess you can say that we saw something that doesn't happen too often then, can't you.

[gaming_mouse]

[ QUOTE ]

Nice analysis. However, you overlooked the following from the OP:

[ QUOTE ]
We had a 8 person sit n go tonight, and when it was down to four-handed

[/ QUOTE ]

[/ QUOTE ]

Wow. I did. Yes, given that there were only 4 people left, this is nearly impossible.

I'd be very suspicious that someone stacked that deck.