magestyayla
magestyayla
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August 12th, 2010 at 6:55:36 AM permalink
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?
Doc
Doc
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August 12th, 2010 at 7:16:02 AM permalink
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.
rdw4potus
rdw4potus
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August 12th, 2010 at 7:48:31 AM permalink
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...
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
Wizard
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Wizard
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August 12th, 2010 at 8:05:27 AM permalink
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.
It's not whether you win or lose; it's whether or not you had a good bet.
rdw4potus
rdw4potus
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August 12th, 2010 at 8:59:12 AM permalink
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?
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
Doc
Doc
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August 12th, 2010 at 9:00:56 AM permalink
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? :-)
weaselman
weaselman
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August 12th, 2010 at 11:00:06 AM permalink
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?
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konceptum
konceptum
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August 12th, 2010 at 11:09:43 AM permalink
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.
Doc
Doc
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August 12th, 2010 at 11:38:20 AM permalink
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.
rdw4potus
rdw4potus
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August 12th, 2010 at 11:58:27 AM permalink
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.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett

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