wu :: forums « wu :: forums - Card Game Odds Calculation (no poker) » Welcome, Guest. Please Login or Register. Sep 7th, 2024, 10:39am RIDDLES SITE WRITE MATH! Home Help Search Members Login Register
 wu :: forums    riddles    cs (Moderators: Grimbal, towr, Eigenray, Icarus, william wu, ThudnBlunder, SMQ)    Card Game Odds Calculation (no poker) « Previous topic | Next topic »
 Pages: 1 Reply Notify of replies Send Topic Print
 Author Topic: Card Game Odds Calculation (no poker)  (Read 1230 times)
icon
Newbie

Posts: 28
 Card Game Odds Calculation (no poker)   « on: Aug 26th, 2019, 4:33am » Quote Modify

This may not be the right forum, but would most likely need a program to calculate so I thought I'd try it.

This is somewhat different than texam holdem calculator, where best hands are known from start and once 5 cards shown on river, it's pretty simple to calculate odds in your head in seconds...

I play a Russian game called preferans, which consistents of 32 cards in a deck (7 through ace) where each player (in 3 player situation only) is dealt 10 cards and 2 put into the pot (no one can see the cards until highest bet takes them and only then they can see them). You start by placing bets on how many tricks you will take (max 10). Whoever bets highest, will take the pot and will be allowed to discard 2 cards of their choice.

This is the situation:

Player A has:

clubs: Ace, King Jack, 9 and 8
spdes: Nothing
Hearts: Ace
Diamonds: Ace, King, Queen, 7

The second betting part comes when player who won pre-pot will adjust their bet (can only go higher or stay same, to keep it simple) and other 2 or 3 players will accept his bet, or bet against him not taking those tricks.

Player A bets 9 tricks with trump being clubs. Being a friendly game, and since there is a high chance he will take all 10 tricks, player A says (based on his experience only) that there is approximately 70% chance that he will take 9 tricks, 29.99+% chance he will take 10 tricks and approximately 1 in 100,000 chance he will only take 8 tricks.

Here are the possibilities:

1. Player A takes all 10 tricks if either of the players do not hold all 3 remaining clubs
2. Player A takes only 8 tricks if either of the players has both 3 clubs and 4 diamonds.
3. Player A makes 9 tricks if one side has 3 clubs or 4 diamonds

Questions are:

1. How to calculate the odds with so many unknows?
2. Was 70% and 29.99+% even close to mathematical odds?
3. Was 100,000 of player taking 8 tricks even close to mathematical odds?

TIA and hopefully you guys will find interesting
 « Last Edit: Aug 26th, 2019, 4:34am by icon » IP Logged
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Card Game Odds Calculation (no poker)   « Reply #1 on: Aug 26th, 2019, 9:52am » Quote Modify

So, just to get it straight, there are three clubs distributed between the two other players and the pot (together 22 cards). If there's any clubs in the pot player A wins 10 tricks, if both players have at least one club, player A also wins 10 tricks.
A quick program gives 84.4% chance of winning 10 tricks

There's a 19/22 * 18/21 probability that neither cards in the pot are clubs.
Given the clubs aren't in the pot, there is a 10/20*9/19*8/18 probability that the 3 clubs end up with player B (any 3 of his/her 10 cards out of the 20 under consideration)
The probability for player C is the same by symmetry

So the probability A doesn't get 10 tricks is (19/22 * 18/21) * 2 * (10/20 * 9/19 * 8/18) ~= 15.58%, which means the chance he has 10 is 84.32% which is close to the experimental value above. (But I'll admit my first attempt was waaaaay off.)
 « Last Edit: Aug 26th, 2019, 10:10am by towr » IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
icon
Newbie

Posts: 28
 Re: Card Game Odds Calculation (no poker)   « Reply #2 on: Aug 26th, 2019, 10:17am » Quote Modify

on Aug 26th, 2019, 9:52am, towr wrote:
 So, just to get it straight, there are three clubs distributed between the two other players and the pot (together 22 cards). If there's any clubs in the pot player A wins 10 tricks, if both players have at least one club, player A also wins 10 tricks. A quick program gives 84.4% chance of winning 10 tricks   There's a 19/22 * 18/21 probability that neither cards in the pot are clubs. Given the clubs aren't in the pot, there is a 10/20*9/19*8/18 probability that the 3 clubs end up with player B (any 3 of his/her 10 cards out of the 20 under consideration) The probability for player C is the same by symmetry   So the probability A doesn't get 10 tricks is (19/22 * 18/21) * 2 * (10/20 * 9/19 * 8/18) ~= 15.58%, which means the chance he has 10 is 84.32% which is close to the experimental value above. (But I'll admit my first attempt was waaaaay off.)

Hi :).

I need to clarify one thing. There are def no clubs in the pot. I was merely giving the breif dynamics of the game there. The hand I listed is post pot betting. I believe you covered it in tje 2nd part of the reply though.

If I understood your reply correctly, then my 70% approximation based on experience was too low and the mathematical odds are ~15.6%

How would one go to answer request 3 though, since it involves another 4 cards that HAVE to be with the same player that has 3 clubs

Thank you
 IP Logged
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Card Game Odds Calculation (no poker)   « Reply #3 on: Aug 26th, 2019, 10:19am » Quote Modify

Hmm, I must be misunderstanding something, because for the 9 trick probability I get ~80%, which is a lot more than the 15.6% that's left over after the 10-trick probability.
Then again, the cases seem to overlap, cause B or C could have 4 diamonds while at the same time neither B or C has 3 clubs, so A would have 9 and 10 tricks simultaneously.

What is a "trick" anyway?
 IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Card Game Odds Calculation (no poker)   « Reply #4 on: Aug 26th, 2019, 10:22am » Quote Modify

on Aug 26th, 2019, 10:17am, icon wrote:
 I need to clarify one thing. There are def no clubs in the pot. I was merely giving the breif dynamics of the game there. The hand I listed is post pot betting. I believe you covered it in tje 2nd part of the reply though.
No, my calculations (and simulation) are definitely incorrect then.
What are the two cards A discarded then? They have to be taken into account to do the calculations, considering he knows what they are.
 IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
icon
Newbie

Posts: 28
 Re: Card Game Odds Calculation (no poker)   « Reply #5 on: Aug 26th, 2019, 9:04pm » Quote Modify

on Aug 26th, 2019, 10:19am, towr wrote:
 Hmm, I must be misunderstanding something, because for the 9 trick probability I get ~80%, which is a lot more than the 15.6% that's left over after the 10-trick probability. Then again, the cases seem to overlap, cause B or C could have 4 diamonds while at the same time neither B or C has 3 clubs, so A would have 9 and 10 tricks simultaneously.   What is a "trick" anyway?

trick is what  players refer as a round of playing where 1 party takes with high card of the same suit or a trumo card. so, in this situation, because each person has 10 cards, and they bet how many trick they will get, with max being 10.

yes, there is some overlap, for example: the only way player A gets 8 if either player b or c have both 3 clubs and 4 diamonds in same hand. if they are split in any way, then player A gets 9 or 10, depending on how they are split (for example, if player a and b each has any of the clubs or diamonds, then player A gets 10 regardless of what cards they have). If player b has 3 clubs and player c has 4 diamonds, then player 9 will take 9. If b or c have 2 or 1 of the clubs and 4 diamonds, player A takes 9 as well.

The underlined principle of forced move comes into player. For example: If I lead with diamond and player B has 3 clubs but 0 diamonds, KNOWING that player c has 4 diamonds and can take that trick, player B cannot play anything but clubs (being trump).

I can go into more detail about actual game. It's sorta based off bridge, just plauyed by individuals not pairs.

on Aug 26th, 2019, 10:22am, towr wrote:
 No, my calculations (and simulation) are definitely incorrect then. What are the two cards A discarded then? They have to be taken into account to do the calculations, considering he knows what they are.

I think I may have complicated that part without realizing. I mentioned the discard cards just to show the dynamics of the game, and not make it a part of the problem. For our calculations. lets assume that the 2 discarded cards are not clubs or diamonds.
 IP Logged
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Card Game Odds Calculation (no poker)   « Reply #6 on: Aug 27th, 2019, 10:43am » Quote Modify

So basically it comes down to how the 3 clubs and 4 diamonds are distributed among the two players (ten cards each).

The chance that player B gets C clubs and D diamonds is
F(C,D) = Choose(3, C) * Choose(4, D) * Choose(13, 10-C-D) / Choose(20, 10) (where Choose(N,K) = N!/[K! * (N-K)!] )
(Which automatically fixes how many player C gets, which is 3-C and 4-D)

So then,
the probability of 8 tricks is F(0, 0) + F(3, 4) = 1/323 ~= 0.0031 (0.31%)
the probability of 9 tricks is F(0, 1..4) + F(3, 1..3) + F(1..3, 0) + F(1..2, 4) = 88/323 ~ = 0.272  (27%)
the probability of 10 tricks is F(1, 1..3) + F(2, 1..3) = 234/323 ~= 0.724 (72 %)
( Where F(x, y..z) is shorthand for F(x, y) + ... + F(x, z) )

Player A seems to have misjudged which option is ~ 70% and which ~30%, but the probabilities are pretty close.
 « Last Edit: Aug 27th, 2019, 10:51am by towr » IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
icon
Newbie

Posts: 28
 Re: Card Game Odds Calculation (no poker)   « Reply #7 on: Aug 27th, 2019, 10:50am » Quote Modify

on Aug 27th, 2019, 10:43am, towr wrote:
 So basically it comes down to how the 3 clubs and 4 diamonds are distributed among the two players (ten cards each).   The chance that player B gets C clubs and D diamonds is   F(C,D) = Choose(3, C) * Choose(4, D) * Choose(13, 10-C-D) / Choose(20, 10) (where Choose(N,K) = N!/[K! * (N-K)!] ) (Which automatically fixes how many player C gets, which is 3-C and 4-D)   So then, the probability of 8 tricks is F(0, 0) + F(3, 4) = 1/323 ~= 0.0031 (0.31%) the probability of 9 tricks is F(0, 1..4) + F(1..3, 0) + F(1..2, 4) + F(3, 1..3) = 88/323 ~ = 0.272  (27%) the probability of 10 tricks is F(1, 1..3) + F(2, 1..3) = 234/323 ~= 0.724 (72 %) ( Where F(x, y..z) is shorthand for F(x, y) + ... + F(x, z) )   Player A seems to have misjudged which option is ~ 70% and which ~30%, but the probabilities are pretty close.

Very interesting. So, chances to get 8 tricks are about 1:300.

I was very surprised that chance to 9 is only 30% vs 70% for 10 tricks. The bet 9 is definitely standard, there, where as many would call it not aggressive or weak, as the penalty for missing a bet is very harsh, vs getting 1 extra trick is weak.

Thank you
 IP Logged
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Card Game Odds Calculation (no poker)   « Reply #8 on: Aug 27th, 2019, 11:00am » Quote Modify

If I understand the explanation on wikipedia correctly, then betting 10 tricks, gives you 10 points if you win (vs 8 points for 9 tricks). And if you lose it costs you 10 points.
If we go with approximate 30/70 odds, then betting 10 is worth 0.7*10 - 0.3*10 = 4, whereas 9 is a virtually guaranteed 8 points. So it's definitely the best choice to go for 9 tricks.

NB The chance for 10 tricks would have to be 90% or higher for it to make sense to make that bet.
 « Last Edit: Aug 27th, 2019, 11:08am by towr » IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
icon
Newbie

Posts: 28
 Re: Card Game Odds Calculation (no poker)   « Reply #9 on: Aug 27th, 2019, 11:10am » Quote Modify

on Aug 27th, 2019, 11:00am, towr wrote:
 If I understand the explanation on wikipedia correctly, then betting 10 tricks, gives you 10 points if you win (vs 8 points for 9 tricks). And if you lose it costs you 10 points. If we go with approximate 30/70 odds, then betting 10 is worth 0.7*10 - 0.3*10 = 4, whereas 9 is a virtually guaranteed 8 points. So it's definitely the best choice to go for 9 tricks.   NB The chance for 10 tricks would have to be 90% or higher for it to make sense to make that bet.

yes, you are absolutely right. even at 70%, it does not justify the risk. I wonder at what % would you think it's justified. my guess 85%+, although in this game, such odds are much harder to manupulate.
 « Last Edit: Aug 27th, 2019, 11:25am by icon » IP Logged
 Pages: 1 Reply Notify of replies Send Topic Print

 Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft => cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »