Bezique

During lockdown My wife and I have been playing this card game.
Background
Bezique is a game for 2 players with 2 packs of cards with the 2's and 6's removed(i.e 64 cards)
Each player is dealt 8 cards and the object is to create melds(e.g 4 aces, K and Q of the same suit etc) by drawing cards from the deck until the deck is exhausted.A new card is drawn by each player after playing a card to a trick(e.g I may play the 8 of spades and my opponent the 10.These cards then take no further part in the game.
The most valuable "meld" is Q of spades *2 + J of diamonds *2
QuestionWhat is the probability of making double bezique?(2*Q spades + 2* J diamonds)?
My logic I do not hold any of the required cards originally
My opponent has 8 cards so the remaining deck is 56 cards.
There are 2*Qs and 2*Jd so my chances of drawing a desired card is 4/56.After each has a turn, there are 54 cards remaining so assuming I draw a desired card and my opponent does not ,my chances of drawing another desired card are 3/54 and then 2/52 and then 1/50.
However,I cannot know what card my opponent has drawn or if she has one or more of the desired cards in the original deal so the probability can never be 100% accurate.
I therefore conclude that it is not possible to answer my original question as there are too many unknowns.
Do we agree?

@venda said
During lockdown My wife and I have been playing this card game.
Background
Bezique is a game for 2 players with 2 packs of cards with the 2's and 6's removed(i.e 64 cards)
Each player is dealt 8 cards and the object is to create melds(e.g 4 aces, K and Q of the same suit etc) by drawing cards from the deck until the deck is exhausted.A new card is drawn by each player ...[text shortened]... that it is not possible to answer my original question as there are too many unknowns.
Do we agree?

So after a bit of more thinking I come to the conclusion that we have here a special problem of the urn problem. Since this is normally formulated using coloured balls our problem is that we have four red balls in a urn containing 64 balls.

The number of permutation is 64! for the pack, if we convert to balls it is 64!/(60!*4!) or (61*62*63*64)/(4*3*2).

Now we have the job to find those cases of the 635376 cases in which the red balls are all on the odd (or even) place.

If we assume a random drawing oder depending on who got the last stich we are really lost as far as I can see

@ponderable said
So after a bit of more thinking I come to the conclusion that we have here a special problem of the urn problem. Since this is normally formulated using coloured balls our problem is that we have four red balls in a urn containing 64 balls.

The number of permutation is 64! for the pack, if we convert to balls it is 64!/(60!*4!) or (61*62*63*64)/(4*3*2).

Now we have t ...[text shortened]... ume a random drawing oder depending on who got the last stich we are really lost as far as I can see

@venda said
Perhaps it's a bit simpler.
At any one time ,I hold 8 cards.
Of those 8 cards the combinations are 2/8 for the Qs leaving 2/6 for the Jd
There are 64 cards in the sample size.
If I enter combin (8,2)*combin(6,2)*combin(64,4)into an xlspreadsheet I get a result starting with 2 billion.
That can't be right can it?

@ponderable said
Since there are only 635376 possible permutations for the 4 b-cards and 60 non-b-cards the probability to get all four b-cards must be considerably higher, since several of those permutations will yield all four b-cards in one hand.

@venda said
During lockdown My wife and I have been playing this card game.
Background
Bezique is a game for 2 players with 2 packs of cards with the 2's and 6's removed(i.e 64 cards)
Each player is dealt 8 cards and the object is to create melds(e.g 4 aces, K and Q of the same suit etc) by drawing cards from the deck until the deck is exhausted.A new card is drawn by each player ...[text shortened]... that it is not possible to answer my original question as there are too many unknowns.
Do we agree?

Another thing that adds significant complexity is the probabilities are going to be updated after each discard from your opponent ( some of the remaining possible permutations for you become impossible after you learn what your opponent discards at each round )

I don't know if I'm going to be able to go much further, but to give you some sense of the probability ( a flag in the ground ) lets examine the probability of drawing atleast 2 Queens, and 2 Jacks on a single draw of 8 cards from the reduced decks

52 cards per deck
2 decks = 104 Cards

Take out 6's and 2's ( one of each of 4 suits per deck ) = lose 16 cards.

88 cards remain.

You draw 8 cards at random

We want to look for all the ways this can be done such that you atleas draw 2 Queens and 2 Jacks ( 4 cards )

Firstly figure out how many ways we can select 4 cards out of 8 card placeholders.

8 slots choose 4

C( 8,4) = 8!/( 4! * ( 8-4)! ) = 70

Now that we've chosen our 4 slots we need the number of way of placing 2 Queens in the 4 slots ( by symmetry the Jacks are automatically placed by placing the Queens or visa versa )

C( 4,2) = 6

That is 6 different ways of placing 2 Queens for each of the "70" 4 slot combinations.

There is 8 Queens and 8 Jacks in 2 decks.

For each of the 70*6 distinct ways to place 2 Queens and 2 Jacks out of 8 cards. there are 8 Choices to place the first Queen, 7 the next, 8 the first Jack and 7 the next.

Now, we have placed 4 cards ( 2 Queens 2 Jacks ), and have to place 4 more cards. So for each of the distinct 70*6*8^2*7^2 ways of having the queens and jacks. We have 84 cards to chose from to fill the first remaining spot, 83 the next ... down to 81 cards to fill the last spot.

So we get:
70*6*8^2*7^2*84*83*82*81 = a god awfully huge number of ways of doing this.

ALL possible ways of drawing 8 cards from 88 cards:

88*87*86*85*84*83*82*81 = another god awfully 'huger' number!

Thus; the probability of getting your win on the first draw of 8 cards is ( notice the cancellation ) :

70*6*8^2*7^2*84*83*82*81 / ( 88*87*86*85*84*83*82*81 ) = ( 70*6*8^2*7^2 )/( 88*87*86*85 ) ≈ 23/1000

So it should absolutely be much much more likely than that if you are then sequentially drawing 36 more cards after the fact.

@joe-shmo said
I don't know if I'm going to be able to go much further, but to give you some sense of the probability ( a flag in the ground ) lets examine the probability of drawing atleast 2 Queens, and 2 Jacks on a single draw of 8 cards from the reduced decks

52 cards per deck
2 decks = 104 Cards

Take out 6's and 2's ( one of each of 4 suits per deck ) = lose 16 card ...[text shortened]... e much much more likely than that if you are then sequentially drawing 36 more cards after the fact.

There are 700 icons in a cathedral. each showing 7 saints and 7 halos.However only "x" saints(and no fewer)have halos in each icon(some saints have more than one halo)All the icons are different.What is "x"

@venda said
Sorry Joe.
I meant all cards up to the sixes(i.e 2,3 4 5 and 6 )making 64 cards and the queens are queen of spades only, the jacks being the jack of diamonds only.However the principal is the same and as you say the number will be enormous and the odds will alter each time cards are drawn from the deck.
Thank you both for your thoughts.
See below for the newspaper one-should be a lot easier!!

@venda said
There are 700 icons in a cathedral. each showing 7 saints and 7 halos.However only "x" saints(and no fewer)have halos in each icon(some saints have more than one halo)All the icons are different.What is "x"

@joe-shmo said
This doesn't seem eaiser... 😞

@venda said
There are 700 icons in a cathedral. each showing 7 saints and 7 halos.However only "x" saints(and no fewer)have halos in each icon(some saints have more than one halo)All the icons are different.What is "x"