Blackjack and specific hands

Hello all,

I am working on probability and specific hands in blackjack.

Phil and Jerry are playing blackjack. Phil is the dealer and they follow standard rules. On the first hand, Jerry gets an 18 (Jack and Eight), but Phil wins with 21 (King and Ace).

What are the odds of Jerry and Phil each getting a queen-ten on the next hand?

Ok, I know that the first number in my work should be 8/48 because there are a total of 8 cards satisfying the condition of being a Queen or a Ten and there are 48 cards left because of two hands being dealt. If we assume they alternate cards between the person receiving and dealer, then the next number should be 7/47 because we have removed either a Queen or a Ten and therefore both numerator and denominator must decrease by one.

Here is where I am perplexed, I thought the next two numbers would be 6/46 and 5/45 for the following:

8/48 x 7/47 x 6/46 x 5/45

However, I have been told that the correct numbers should be: 8/48 x 7/47 x 4/46 x 3/45

I am trying to understand the reasoning of this and believe it may have to do with perhaps the order? or removing the possibility that someone receives both Queens or Tens?

Thanks for any help.

The odds are not what you expected because Jerry and Phil each need a Queen and a 10. So, if the first card for each of them was a Queen, then they'd no longer need a Queen or a 10, but instead only a 10 would work. And a similar scenario would apply if the first cards were a 10 and a Queen, a 10 and a 10, etc.

Thanks for your input.

Question:
If the first player is dealt a Queen, then the second player is dealt a 10...

That would mean the first player now has only a 3/46 chance and then the second player, needing a Queen, would have a 3/45 chance to draw a Queen, as there are only 3 left. This seems to go against the answer: 8/48 x 7/47 x 4/46 x 3/45.

I can see that if a Queen were dealt to the first player (8/48), a Queen to the second (7/47), then the third card drawn must be a 10 to satisfy the conditions (4/46) and finally another 10 (3/45). I guess I am missing how it works out when the first two cards drawn (one per player) are not the same.

Thanks

I think it necessary to work through all possibilities to avoid having to account for a queen or ten previously dealt.

Consider dealing four cards, two each to Jerry and Phil, alternately
Our desired result is achieved when cards are dealt as follows
Q, Q, 10, 10 (J = Q, 10 / P = Q, 10) with prob 4/48 x 3/47 x 4/46 x 3/45
Q, 10, 10, Q (J = Q, 10 / P = 10, Q) with prob 4/48 x 4/47 x 3/46 x 3/45
10, Q, Q, 10 (J = 10, Q / P = Q, 10) with prob 4/48 x 4/47 x 3/46 x 3/45
10, 10, Q, Q (J = 10, Q / P = 10, Q) with prob 4/48 x 3/47 x 4/46 x 3/45

Probabiliy of desired hands is the total of these, which are all equal : (4 x 4 x 3 x 3) / (48 x 47 x 46 x 45)

Total probability = (4 x (4 x 4 x 3 x3)) / (48 x 47 x 46 x 45)