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Probability for Powerball and PokerPowerPoint Presentation

Probability for Powerball and Poker

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Probability for Powerball and Poker

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Probability for Powerball and Poker

Extra Chapter 4 stuff

by D.R.S., University of Cordele

- A classic type of problem
- You have various subgroups.
- When you pick 6, what is the probability that you get 2 of this group and 4 of that group?
- Jellybeans: 30 red, 30 yellow, 40 other
- Choose 6. Find P(2 red and 4 yellow)

- Analysis – you must THINK! – “This is a Fundamental Counting Principle situation…
- One event is drawing 2 red out of 30
- The other is drawing 4 yellow out of 40
- FUNDAMENTAL COUNTING PRINCIPLE says to multiply how many ways for each of them.
- Each of these events is modeled by a COMBINATION, because the order doesn’t matter.

- So how do you write it in Combination language?

- Jellybeans: 30 red, 30 yellow, 40 other
- Choose 6. Find P(2 red and 4 yellow)
- Always go back to
- Numerator:
- Denominator:

- Draw 5 cards, what is the probability of exactly 0 aces?
- We can do this with our earlier techniques:
- P(first card not at ace) = ____ / 52, times …
- P(second card not an ace) = ____ / 51, times …
- P(third card not an ace) = ____ / 50, times …
- P(fourth card not an ace) = ____ / 49, times …
- P(fifth card not an ace) = ____ / 48

- P(0 aces out of 5 cards drawn)
- A more sophisticated view
- 5 non-aces out of 52 cards
- How many non-aces are there?

- Numerator: ways to get 5 non-aces:
- Denominator: total 5-card hands:
- P(0 aces) =

- P(exactly 1 ace out of 5 cards drawn)
- Our earlier techniques could do P(≥1 ace)
- But P(=1 ace) would be harder or impossible
- Counting techniques makes it easier
- Choose 1 ace out of 4 aces
- Choose 4 other cards out of 48 non-aces

- P(1 ace) =

- Similiarly for 2 aces, 3 aces, 4 aces:
- P(2 aces) =
- P(3 aces) =
- P(4 aces) =
- Check: P(0) + P(1) + P(2) + P(3) + P(4) must total to exactly 1.000000000000000000. Why?

- Three of a kind
- Choose 1 out of 13 ranks
- Choose 3 out of 4 suits

- One pair
- Choose 1 out of the remaining 12 ranks
- Choose 2 out of the 4 suits

- P(full house) =

- A Flush: five cards all of the same suit
- Choose 1 out of the 4 suits
- Take 5 out of the 13 ranks

- P(flush) =

- Choose 5 out of the 59 white numbers.
- Choose 1 out of the 35 red powerball numbers.
- The Fundamental Counting Principle: Multiply the number of outcomes of the sub-events.
- There are therefore possible ways to play the ticket, not counting the extra PowerPlay “multiplier” option.

- Repeating: possible ways to play the ticket, not counting the extra PowerPlay “multiplier” option.
- This is the number of outcomes in the sample space.
- Therefore this is the denominator in each of our powerball probability calculations.

- You choose 5 out of the 59 white numbers
- All 5 match the 5 winners

- You choose 1 out of the 35 red numbers
- And it matches the winner

- Numerator is
- Denominator as before, (59 C 5)(35 C 1).
- Compare this result to the odds printed on the ticket.

- You choose 5 out of the 59 white numbers
- All 5 match the 5 winners

- You choose 1 out of the 35 red numbers
- And it is one out of the 34 that don’t match the winner

- Numerator is
- Notice we still have 5 out of 5 on the white numbers
- But the Powerball choice is 1 out of 34 losers

- Reconcile this result with the printed odds.

- You choose 5 out of the 59 white numbers
- 5 winners but you picked got 4 of them
- 54 losers and you picked one of those

- You choose 1 out of the 35 red numbers
- And it matches the winner

- Numerator is
- For the $100 via 4 white only with no red match, just change the to a

- Two ways to win $7
- 3 white matches, 2 losers; red is no match
- Another way: 2 white matches, 3 losers, and the red powerball matches

- Match absolutely nothing at all
- 5 out of the 54 losing white numbers
- 1 out of the 34 losing red powerball numbers

- Or match 1 white number only
- 1 out of the 5 winning white numbers
- 4 out of the 54 losing white numbers
- 1 out of the 34 losing red powerball numbers