SOLUTION: What is the probability of Dead Man's Hand. (Black Aces over Black Eights, the fifth card was unknown)

Question 116346: What is the probability of Dead Man's Hand. (Black Aces over Black Eights, the fifth card was unknown)
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Google found this:
Ask the Wizard! No. 147 (Nov. 22, 2005) - 11:22amI understand you have already answered the probability of getting the "dead man's hand", a two pair of aces and eights, is 0.0609% on April 3, 2005, ...

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Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
First question: How many different 5 card hands are possible from a standard deck of 52 playing cards? Ans: (52 ways to select the first card X 51 ways to select the second card, times 50, times 49, times 48) divided by the number of ways the five cards could be arranged because the order the cards are dealt doesn't matter, which is 5 X 4 X 3 X 2 X 1 or 5! . (See the formula for Combinations - I'm going to use the notation for n things chosen k at a time.

, so for 52 choose 5, we get:

which gives us the total number of 5 card hands.

So in our experiment, we will call a trial a success if we get both black aces and both black 8s plus any other card as the 5th card. If we were only dealing 4 card hands with success being getting both black aces and both black 8s, then there would only be one combination that would be considered a success, so we only have to deal with the number of ways the 5th card could be selected, and that is simply the entire deck of 52 minus the 4 cards represented by the black aces and black 8s. Therefore, out of the total hands, 48 would be considered successes, and the probability would then be

So, let's invert and multiply, then do some cancelling before we multiply the factors to get a single fraction

or roughly 0.00185%

Super-Double-Plus Extra Credit Question 1:
What is the probability of filling the full house on a one-card draw given that you already hold the Aces and Eights? (Hint: You know the values of 5 cards, so 47 remain that you don't know about.)

Question 2:
Your opponent to your right took three cards on the draw. When he passed his discards to the dealer, he accidentally flipped them face up. One of these face up cards was the 8 of hearts. Now what is the probability for filling your full house on a one-card draw?

Hope this helps, and I sure hope to see answers for the extra credit problems if you respond.

John
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