Unusual BJ hand

I had a rather unusual situation come up while playing blackjack last week. In a single deck game all 3 players and the dealer each got a ten pointer and a 6. I know it has to be long odds for every one to not only get a 16, but to also do it with the same composition (T-6). But what exactly are the odds of this happening?

The way to work it out is.
Player 1) 16 tens 4 sixes. 52 cards then 51 cards. Either (a) ten-six or (b) six-ten.
(a) 16/52 * 4/51
(b) 4/52 * 16/51
Thus P(player 1 getting T6) = 2*16/52*4/51.
Player 2) the numbers are 15 tens 3 sixes 50 and 49 cards - same idea.
Players 3 & 4) 14 2 48 47, 13 1 46 45; same idea.
Multiply the four figures together - simples.

Quote: RenoGambler

I had a rather unusual situation come up while playing blackjack last week. In a single deck game all 3 players and the dealer each got a ten pointer and a 6. I know it has to be long odds for every one to not only get a 16, but to also do it with the same composition (T-6). But what exactly are the odds of this happening?

Ask Djatc, he gets the answer right every time.

Quote: AxelWolf

Ask Djatc, he gets the answer right every time.

50/50

Quote: djatc

50/50

How is that possible? I'm assuming that the OP meant a ten and a six (as opposed to a Jack, Queen or King). But even if it were a 10-value card the odds of three players and the dealer all having a two-card 16 (using all four 6s) in a single-deck game would be astronomical.

Quote: Greasyjohn

How is that possible? I'm assuming that the OP meant a ten and a six (as opposed to a Jack, Queen or King). But even if it were a 10-value card the odds of three players and the dealer all having a two-card 16 (using all four 6s) in a single-deck game would be astronomical.

To be clear, it was 10-value cards, not necessarily 10s. There were face cards included as well. Still an odd thing to see at the table.

With only three people? Not all that oDD at all.

Quote: GWAE

With only three people? Not all that oDD at all.

It would be really odd with more than three players and the dealer, because that would mean there's something fishy going on.

If I did the math correctly there's a 0.00088% chance of this happening in a SD game, or roughly 113,000:1.

Quote: RenoGambler

It would be really odd with more than three players and the dealer, because that would mean there's something fishy going on.

If I did the math correctly there's a 0.00088% chance of this happening in a SD game, or roughly 113,000:1.

i get a 1 in 1,808,985.94 chance
4 hands on the deal all have a 10 value card and a 6

here is my math in Excel in Google
this is a hypergeometric distribution
i did not use that function
I could have made a mistake too in me thinking

https://goo.gl/EGlFoa

hand a would require
16Tens Choose1 * 4Sixes Choose1 / 52totalCards Choose 2
hand b would require
15Tens Choose1 * 3Sixes Choose1 / 50totalCards Choose 2

and so on
then the product of the 4
<<<< >>>>
of course just
hand a
getting one of the 10 value cards should be
16Tens Choose1 * 36others Choose1 / 52totalCards Choose 2 = 576 / 1326 =
0.43438914

added to finish thought:
so 3 other hands matching hand a
would be about 1 chance in 200,998.44
a 10 value card with same same other card

Sally

Quote: mustangsally

i get a 1 in 1,808,985.94 chance
4 hands on the deal all have a 10 value card and a 6

Thanks, Sally! I knew it had to be very improbable. I just wasn't sure how improbable.

I get the same values as mustangsally using integers for a probability of 16/28943775