Probability Question - Show me the math!

[vassman86]

Okay, simple question, but I want to see the math!


The game is really simple. Using a regular deck of cards, The Dealer offers the following game: "I flip three cards from the deck, and the only way I can win is if any Ace, Jack, or 2 hits. IF they don't hit, then the player wins."

The Dealer flips three cards only. What is the probability of the Dealer winning?

[Ganchrow]

The dealer would need to turn at least 1 of 12 cards in order to win (4 aces + 4 jacks + 4 deuces = 12 cards), leaving 52-12 = 40 "safe" cards (assuming no jokers).

Hence, the probability of a dealer win would be:

= 1 - 40/52 * 39/51 * 38/50 ≈ 55.294%

Or equivalently, using Excel's combinatorial function:

= 1 - COMBIN(40,3) / COMBIN(52,3) ≈ 55.294%

[Justin7]

The dealer would need to turn at least 1 of 12 cards in order to win (4 aces + 4 jacks + 4 deuces = 12 cards), leaving 52-12 = 40 "safe" cards (assuming no jokers).

Hence, the probability of a dealer win would be:

= 1 - 40/52 * 39/51 * 38/50 ≈ 55.294%

Or equivalently, using Excel's combinatorial function:

= 1 - COMBIN(40,3) / COMBIN(52,3) ≈ 55.294%

I never new about the COMBIN function. Is there a "Select" function also, in case I'm too lazy to use multiple Combines?

[Ganchrow]

I never new about the COMBIN function. Is there a "Select" function also, in case I'm too lazy to use multiple Combines?

A "Select" function? I'm unfamiliar with that terminology as it relates to combinatorics.

You'd don't just mean PERMUT(), do you?

Are you referring to some specified function of multiple COMBIN()'s for a first and/or second argument within a given range?

Whatever is it is let me know, I'm sure it could easily be replicated in VBA.

[vassman86]

thanks ganchrow!

[ms61853]

The dealer would need to turn at least 1 of 12 cards in order to win (4 aces + 4 jacks + 4 deuces = 12 cards), leaving 52-12 = 40 "safe" cards (assuming no jokers).

Hence, the probability of a dealer win would be:

= 1 - 40/52 * 39/51 * 38/50 ≈ 55.294%

Nothing I can argue with there, but I am wracking my brain trying to figure why this other approach would be wrong:

The dealer must hit one of the twelve cards, but the other two can be any of the remaining cards in the deck. So the total number of winning hands for the dealer is 12*51*50.

The total number of possible hands is 52*51*50.

12*51*50/52*51*50 = 0.23.

??????

[Ganchrow]

Nothing I can argue with there, but I am wracking my brain trying to figure why this other approach would be wrong:

The dealer must hit one of the twelve cards, but the other two can be any of the remaining cards in the deck. So the total number of winning hands for the dealer is 12*51*50.

The total number of possible hands is 52*51*50.

12*51*50/52*51*50 = 0.23.

The total number of winning hands for the dealer is not 12*51*50.

Number of ways the dealer can get exactly 1 of Ace, Jack or Deuce: combin(3,1)*12*40*39 = 56,160
Number of ways the dealer can get exactly 2 of Ace, Jack or Deuce: combin(3,2)*12*11*40 = 15,840
Number of ways the dealer can get exactly 3 of Ace, Jack or Deuce: combin(3,3)*12*11*10 = 1,320
Number of ways the dealer can get 1 or more of Ace, Jack or Deuce: 56,160 + 15,840 + 1,320 = 73,320

Total number of hand combinations: 52*51*50 = 132,600

Probability(delaer win) = 73,320 / 132,600 ≈ 55.294%

[Ganchrow]

12*51*50

This corresponds to the number of ways the dealer can be dealt a 2, J, or A on the just the 1^st card. The number of ways the dealer can be dealt a 2, J, or A on any card would be:

n(dealt on any card)
= n(dealt on 1^st card) + n(dealt on 2^nd card) + n(dealt on 3^rd card)
- n(dealt on 1^st & 3^rd cards) - n(dealt on 1^st & 2^nd cards) - n(dealt on 2^nd & 3^rd cards)
+ n(dealt on 1^st, 2^nd, & 3^rd cards)
= 12*51*50 + 12*51*50 + 12*51*50 - 12*11*50 - 12*11*50 - 12*11*50 + 12*11*10
= 73,320

which is of course precisely the same figure as that obtained in the above post.