Canasta odds question -- Its a tough one

This is a tough one.

In Canasta, there are 108 Cards. Two decks of 52 cards plus 4 jokers. Two players are dealt 15 cards to start the hand and then one card is dealt from the remaining deck pile to start the discard pile. So that is a total of 31 cards dealt right away. Players are then dealt two cards at a time from the deck pile and must discard one into the discard pile before their opponent can draw his two cards. In computer Canasta, and all of the in real life Canasta games I have played the rules state that the only cards you cannot discard are the Red 3's. You must keep those.

So, here is my question. What are the odds of being dealt all of the Red 3's in a game of Canasta, assuming that all of the cards are dealt?

Well, here's an innocent stab by a non-mathematician who hasn't played canasta in more than half a century...

If all the cards are dealt, and you get half of them, then it seems you have a 50% chance of getting any particular card. The chances of getting all four red 3s should be something like (0.5)^4, or 1 in 16.

O.K., math fans, show me where I erred.

Quote: Doc

Well, here's an innocent stab by a non-mathematician who hasn't played canasta in more than half a century...

If all the cards are dealt, and you get half of them, then it seems you have a 50% chance of getting any particular card. The chances of getting all four red 3s should be something like (0.5)^4, or 1 in 16.

O.K., math fans, show me where I erred.

I agree with Doc. But it sure does feel like my opponents get those 3s more often than that...

Quote: magestyayla

So, here is my question. What are the odds of being dealt all of the Red 3's in a game of Canasta, assuming that all of the cards are dealt?

Allow me to attempt to rephrase the question, hopefully correctly.

Q: Two 54-card decks (including two jokers) are shuffled together. A player is given half of them. What is the probability that player got all four red threes?

A: There are 4 red aces and 104 other cards. There is just one way to get all four red threes. There are combin(104,50)= 1.46691 × 10^28 ways the player could get 48 of the other 100 cards. The total number of combinations is combin(108,54)= 2.48578 × 10^30.
combin(104,50)/combin(108,54) = 0.059012.

If you don't dealing with such large numbers, here is an alternative solution. Number the four red threes 1 to 4. The probability the first red three is in the player's stack is 54/108, since he has half the cards. Now remove it. The probability the player has the second red three is 53/107, because the player has 51 cards left, but red three #2 cold be anywhere in the 103 remaining cards. Likewise, the probability the player has the third red three is 52/106, and the fourth red three is 51/105. (54/108) × (53/107) × (52/106) × (51/105) = 0.059012.

Quote: Wizard

Allow me to attempt to rephrase the question, hopefully correctly.

Q: Two 52-card decks are shuffled together. A player is given half of them. What is the probability that player got all four red threes?

A: There are 4 red aces and 100 other cards. There is just one way to get all four red threes. There are combin(100,48)= 9.32066 × 10^28 ways the player could get 48 of the other 100 cards. The total number of combinations is combin(104,52)= 1.58307 × 10^30.
combin(100,48)/combin(104,52) = 0.058877.

If you don't dealing with such large numbers, here is an alternative solution. Number the four red threes 1 to 4. The probability the first red three is in the player's stack is 52/104, since he has half the cards. Now remove it. The probability the player has the second red three is 51/103, because the player has 51 cards left, but red three #2 cold be anywhere in the 103 remaining cards. Likewise, the probability the player has the third red three is 50/102, and the fourth red three is 49/101. (52/104) × (51/103) × (50/102)
× (49/101) = 0.058877.

There are also 4 jokers in play. Does that make it COMBIN(104, 52)/COMBIN(108, 56)=.068536?

Revised---

Well, we now have three suggested solutions, each of which has its advantages.

(1) The Wizard's solution to a modified/reworded problem has the advantage that it appears to be correct for that modified problem.

(2) rdw4potus's suggested solution (for the original problem with the jokers) has the advantage that it appears (to me) to be correct for the original problem.

(3) My solution for the original problem is evidently quite incorrect, but it has the advantage of being a hell of a lot easier to calculate. Why do people put so much emphasis on correctness in the answers to math questions anyway? :-)

I don't understand what's wrong with the Doc's solution. I don't know the rules of canasta, maybe, that's the reason, but in th esimple formulation, given by Wizard, you get two pile of cards, and four threes. Each three ends up in one of the piles with 50% probability, so the probability of getting all four must be 1/16th, or 0.0625.

Another oddity that makes me think that Wizard's solution is incorrect is examining his formula - combin(N-n,N/2-n)/combin(N,N/2) - for the case when n=1. Suppose, one of the 104 cards is marked with a marker. What is the probability of the player getting that card? Usine Wizard's formula combin(107,55)/combin(108,56) = 0.51852. What is so special about this player that makes it more likely for hime to get the card than it is for the other player?

Quote: Doc

(3) My solution for the original problem is evidently quite incorrect, but it has the advantage of being a hell of a lot easier to calculate. Why do people put so much emphasis on correctness in the answers to math questions anyway? :-)

Doc, you might find some interest in reading Street-Fighting Mathematics:

"This engaging book is an antidote to the rigor mortis brought on by too much mathematical rigor, teaching us how to guess answers without needing a proof or an exact calculation. "

You can read the Creative Commons version of the book (PDF format) for free, not that I would recommend doing so unless you REALLY like math. But, some of the ideas are interesting.

konceptum: Thanks for the suggestion. I'll check that out.

weaselman: I admit to my error. I took a shortcut that gave a fair approximation. The error can perhaps be seen this way:

Suppose there had been 55 red 3s in the double deck plus jokers. Would my probability of getting all of them be (0.5)^55? Remember now, I'm only going to get 54 of the cards. Obviously the formula doesn't hold in that extreme case. I failed to take into account that my getting one red 3 gets in the way of my getting a 2nd, 3rd, and 4th red 3 in the same deal.

If the question had been: "What is the probability of my getting the 3 of hearts from deck #1 in my share of the cards on four successive deals?", then maybe my method would actually have given the correct answer.

I confess that I take a lot of short cuts in my calculations, particularly when little is at stake. That's why the book that konceptum mentions might be of interest.



Edit:
weaselman, I did not see the calculations in your post at first. You gave a formula:

Quote: weaselman

combin(107,55)/combin(108,56) = 0.51852

I think you added the value of n=1 when you meant to subtract and that it should instead be:

combin(107,53)/combin(108,54) = 0.50000.

Quote: Doc

I confess that I take a lot of short cuts in my calculations, particularly when little is at stake. That's why the book that konceptum mentions might be of interest.

I think that's a good way to go. Not that the exact math isn't interesting, but it's often not worth the effort or possible to produce. Your 6.25% approximation is within .6% of what appears to be the correct answer. That's close enough for most applications, and also a practical method in the absence of a computer, calculator, or other mathematical aid.

Say I was willing to give you 18:1 that you'd get all 4 red 3s. The approximation is good enough to justify the bet. I suppose if I put the payout at 15:1, you'd have some further thinking to do. But even then, if you know which side your error is on, you're still good to go with the approximation.

Quote: rdw4potus

...Say I was willing to give you 18:1 that you'd get all 4 red 3s. The approximation is good enough to justify the bet.

I admit that my approximation would have led me to believe that I had the advantage there. But... Oh, no. Before taking that bet, I would have been all over it with calculator, spreadsheet, and slide rule. Any wager opportunity that looks that good just has to have some trickery hidden in it somewhere.

I forgot about the jokers. My answer has just been revised.

Well, it seems we are all making minor glitches today. When rdw4potus pointed out that the Wizard had forgotten the jokers, he made errors in his suggested revised formula -- there are 50 cards in the hand that are non-3s, not 52. I didn't catch that, but the Wizard did. Now the Wizard has edited his original post and (I think) has the formulas correct, but left some earlier text that doesn't correspond to the revised equations.

If we all go back and edit our earlier posts to make them correct, then the continuity in the thread discussion will fall apart. Ah, the joys of forum life.

I also found the rdw4potus calculation to be in error. But how is it possible to increase your chances of getting all four 3's by adding additional cards to the population? The Wizards first calculation indicated .05877 and the second calculation .059xxx.

I think it has something to do with the first red 3 you get not getting as much in the way of the others coming into your hand as it did before. If each player were getting half of a deck of 1000 cards, the probability of all red 3s in one hand would be even higher. For a large, sub-infinite deck, maybe my original approximation might work pretty well.

I you have a deck of 4 threes, and deal 4 cards, you have a 100% chance of getting all 4. Add one joker to the deck, deal 4 cards, and your chance of getting 4 threes is less than 100%.

Quote: Doc

I think it has something to do with the first red 3 you get not getting as much in the way of the others coming into your hand as it did before. If each player were getting half of a deck of 1000 cards, the probability of all red 3s in one hand would be even higher. For a large, sub-infinite deck, maybe my original approximation might work pretty well.

Nice explanation. This is the reason that playing BJ with an 8 deck shoe is more favorable to the house than a 2 deck shoe, I suppose.

Quote: Headlock

I you have a deck of 4 threes, and deal 4 cards, you have a 100% chance of getting all 4. Add one joker to the deck, deal 4 cards, and your chance of getting 4 threes is less than 100%.

I don't think I follow that. If your deck consists entirely of the four red 3s, and you deal out two hands (as we have been discussing throughout this thread), you have zero chance of getting all four, because you only get two of the cards. I really don't mean for that to be a snide comment, but if you could explain your example again, I will try to follow it.

Edit: On re-reading your post, I think you are now considering dealing just one hand, with either the full 4-card deck or with 4 out of a 5-card deck. If I got it right that time, then you are considering a vastly different scenario than dealing an entire deck evenly between two players, which is what we have been discussing.

Quote: Headlock

I you have a deck of 4 threes, and deal 4 cards, you have a 100% chance of getting all 4. Add one joker to the deck, deal 4 cards, and your chance of getting 4 threes is less than 100%.

I think you need to have at least 8 cards before the qualifying condition can be met. There are 2 players, and each is getting 4 cards. If there are 8 cards to be dealt, you need the exactly correct 4 out of 8 to have all the red 3s. If 10 cards are used, you need 4 red 3s and 1 blank. For 12 cards, you need 4 red 3s and 2 blanks. The extra cards make it easier to have the 4 specific cards you're looking for.

Think of your 3 card poker progressive. It's a lot easier to get the AKQ of spades in a 7 card game than in a 3 card game, right?

Quote: Doc

I don't think I follow that. If your deck consists entirely of the four red 3s, and you deal out two hands (as we have been discussing throughout this thread), you have zero chance of getting all four, because you only get two of the cards. I really don't mean for that to be a snide comment, but if you could explain your example again, I will try to follow it.

Edit: On re-reading your post, I think you are now considering dealing just one hand, with either the full 4-card deck or with 4 out of a 5-card deck. If I got it right that time, then you are considering a vastly different scenario than dealing an entire deck evenly between two players, which is what we have been discussing.

You're right, and rdw4potus recognized the same thing. My example is not the same as dealing two hands. I have to think about it some more.....rdw4potus explanation does make sense.

Say I have a deck of 8 cards, 4 threes and 4 fours. I will deal two hands of four cards each. The first card dealt has a 4 out of 8 chance of being a 3. The second, 3 out of 7, the third 2 out of 6, the fourth 1 out of 5. Multiply those results to get .014286.

Now add two fives to the 8 card deck. I will deal two hands of five cards each. The first card dealt has a 4 out of 10 chance of being a 3, the next 3 out of 9, 2 out of 8, 1 out of 7. Multiply those results and you arrive at .004762.

Quote: Headlock

Say I have a deck of 8 cards, 4 threes and 4 fours. I will deal two hands of four cards each. The first card dealt has a 4 out of 8 chance of being a 3. The second, 3 out of 7, the third 2 out of 6, the fourth 1 out of 5. Multiply those results to get .014286.

Now add two fives to the 8 card deck. I will deal two hands of five cards each. The first card dealt has a 4 out of 10 chance of being a 3, the next 3 out of 9, 2 out of 8, 1 out of 7. Multiply those results and you arrive at .004762.

I agree with your numbers for the 8-card deck. (Remember, though, I'm the guy who posted the very first incorrect answer in this thread!)

For the 10-card deck, what it the chance that you will get a three as your fifth card dealt? Haven't you left that out of your calculation? I think you are considering the probability of getting four 3s in your first four cards from a 10-card deck, but that's not required. If you are allowed to receive all five cards, your probability of getting four 3s is greater.

Quote: Doc

I agree with your numbers for the 8-card deck. (Remember, though, I'm the guy who posted the very first incorrect answer in this thread!)

For the 10-card deck, what it the chance that you will get a three as your fifth card dealt? Haven't you left that out of your calculation? I think you are considering the probability of getting four 3s in your first four cards from a 10-card deck, but that's not required. If you are allowed to receive all five cards, your probability of getting four 3s is greater.

I don't know. I wish the Wizard would show us the way!

Quote: Headlock

Say I have a deck of 8 cards, 4 threes and 4 fours. I will deal two hands of four cards each. The first card dealt has a 4 out of 8 chance of being a 3. The second, 3 out of 7, the third 2 out of 6, the fourth 1 out of 5. Multiply those results to get .014286.

I agree.

Quote: Headlock

Now add two fives to the 8 card deck. I will deal two hands of five cards each. The first card dealt has a 4 out of 10 chance of being a 3, the next 3 out of 9, 2 out of 8, 1 out of 7. Multiply those results and you arrive at .004762.

I assume you're asking what is the probability that the player gets all 4 threes in his 5 cards. For that I get 6/combin(10,5) = 0.023809524. Assuming we are solving the same question, I think your error is in assuming the player gets the 4 threes in the first four cards. They can be in any position in the player's 5 cards.

Quote: Headlock

I don't know. I wish the Wizard would show us the way!

He did. For your modified 10-card-deck problem, the solution is:

P=combin(10-4,5-4)/combin(10,5)= 0.0238095 (assuming I didn't make a calculation error)

which is a bit higher than the probability that you already found for the 8-card-deck problem. You added two fives to the deck, and the Wizard corrected his original post by adding four jokers, also increasing the probability of all four red 3s winding up in the same hand.

Edit: Ooops! Once again posted slow & this time it was the Wizard who posted while I typed.

Quote: Wizard

I agree.



I assume you're asking what is the probability that the player gets all 4 threes in his 5 cards. For that I get 6/combin(10,5) = 0.023809524. Assuming we are solving the same question, I think your error is in assuming the player gets the 4 threes in the first four cards. They can be in any position in the player's 5 cards.

You assume correctly. I'm beginning to comprehend, but I don't see how to solve without using the combin function.

Quote: Headlock

You assume correctly. I'm beginning to comprehend, but I don't see how to solve without using the combin function.

The combin function is just a way to write the calculations more compactly. Try it this way for your 10-card deck:

The probability of getting all four 3s in one hand is the ratio of the number of hands that have all the 3s to the total possible number of hands.

As a first step toward the total possible number of hands, note that there are 10 ways to select the first card times 9 ways for the second card, etc. or 10*9*8*7*6=30,320. Unfortunately, this calculation considers the hand 3s3h4s4h4d as a different hand from 3h3s4h4d4s; i.e., the order of selecting the cards isn't really important, even though I calculated it as if it was.

To correct for this, we consider how many different versions of each hand we have counted, or how many ways the five cards could be arranged: 5 cards that could be first in line * 4 cards that could be second, etc. or 5*4*3*2*1=120 ways. Thus, the real total number of hands you could be dealt (not caring which order you got the cards in) is 30320/120=252. (Note that this is combin(10,5)=252.)

As for the total number of hands that contain all four 3s, well four of the cards have to be the 3s and there are six other cards that can fill the remaining spot. Thus, there are only 6 possible hands that contain all four of the 3s. (Note that this is combin(10-4,5-4)=combin(6,1)=6.)

Finally, the desired probability is "hands with all of the 3s" divided by "all possible hands" or 6/252=0.023809524, as reported before.

Now if you expand the problem to consider a double-deck-plus-jokers of 108 cards, as the problem was originally stated, you can solve it this very same way, but it gets more cumbersome. That's why the combin function is used.

As far as an intuitive reason that the probability improves for getting all the 3s in one hand as the number of cards increases (and the size of your hand with half of the cards increases), I go back to my original suggestion: the spaces available in your hand where you might wind up putting more 3s doesn't get as crowded with the first few 3s you get. Or as stated in rdw4potus's poker-oriented suggestion, there is more room to fit AKQ of spades in a seven card hand than in a three card hand.

Does that help? (Remember now, I'm the guy who got it wrong the first time around.)

Quote: Headlock

You assume correctly. I'm beginning to comprehend, but I don't see how to solve without using the combin function.

The combin function is wonderful! If you were in my UNLV class I would have forced it down your throat every single lecture.

However, if you must, here is how you can solve the problem without it.

Probability of getting the four threes in the first four cards dealt (call them cards C1, C2, C3, and C4) =(4/10)*(3/9)*(2/8)*(1/7)*(6/6)= 0.02381
Probability of getting the four threes in C1, C3, C4, C5 =(4/10)*(6/9)*(3/8)*(2/7)*(1/6) = 0.004761905
Probability of getting the four threes in C1, C2, C3, C5 =(4/10)*(3/9)*(2/8)*(6/7)*(1/6) = 0.004761905
Probability of getting the four threes in C1, C2, C4, C5 =(4/10)*(3/9)*(6/8)*(2/7)*(1/6) = 0.004761905
Probability of getting the four threes in C1, C3, C4, C5 =(4/10)*(6/9)*(3/8)*(2/7)*(1/6) = 0.004761905
Probability of getting the four threes in C2, C3, C4, C5 =(6/10)*(4/9)*(3/8)*(2/7)*(1/6) = 0.004761905

Add them all up and you get 5*0.004761905 = 0.02381.

Wizard:

How can you and I always be typing replies to the same thread at the same time?

I get it now! Thanks Doc and Wizard.