FMM14: Difference between revisions
From charlesreid1
No edit summary |
|||
| Line 1: | Line 1: | ||
{{FMM | |||
|title=Which Color Cab | |||
|problem= | |||
Suppose you have a standard deck of 52 cards. You shuffle the pack thoroughly, then turn up cards, one at a time, from the top of the deck. On average, how many cards do you turn over before you see an ace? | |||
NOTE: You have to turn over the ace to see it's an ace, so include the ace itself in the count of cards. | |||
HINT 1: It may be easier to think of "average number of cards" as "expected number of cards". | |||
HINT 2: It may be easier to think of the problem in terms of a deck of N cards and M marked cards. | |||
|solution= | |||
The 4 (M) aces will divide the pack into 5 (M+1) segments, ranging in size from 0 to 48 (N-M). (To clarify: if two aces are side by side, the segment between them is of size zero. If the first card is an ace, the first segment is of size zero. Etc.) | |||
If we can assume that discrete events, like continuous events, are symmetric, then the expected size of the 5 segments is (N-M)/(M+1) = 48/5 = 9.6, meaning we expect the segment of cards in front of the first ace to be of size 9.6. If we include turning over the ace, we expect the number of cards we turn over to be 10.6, which we can interpret to mean "10 cards 40% of the time and 11 cards 60% of the time". If we round to the nearest whole number, we should expect to turn over 11 cards before the first ace. | |||
}} | |||
Revision as of 22:30, 22 November 2019
Friday Morning Math Problem
Which Color Cab
Suppose you have a standard deck of 52 cards. You shuffle the pack thoroughly, then turn up cards, one at a time, from the top of the deck. On average, how many cards do you turn over before you see an ace?
NOTE: You have to turn over the ace to see it's an ace, so include the ace itself in the count of cards.
HINT 1: It may be easier to think of "average number of cards" as "expected number of cards".
HINT 2: It may be easier to think of the problem in terms of a deck of N cards and M marked cards.
| Solution |
|---|
|
The 4 (M) aces will divide the pack into 5 (M+1) segments, ranging in size from 0 to 48 (N-M). (To clarify: if two aces are side by side, the segment between them is of size zero. If the first card is an ace, the first segment is of size zero. Etc.)
If we can assume that discrete events, like continuous events, are symmetric, then the expected size of the 5 segments is (N-M)/(M+1) = 48/5 = 9.6, meaning we expect the segment of cards in front of the first ace to be of size 9.6. If we include turning over the ace, we expect the number of cards we turn over to be 10.6, which we can interpret to mean "10 cards 40% of the time and 11 cards 60% of the time". If we round to the nearest whole number, we should expect to turn over 11 cards before the first ace.
|
Flags
 |
Friday Morning Math Problems weekly math problems
Spider Socks and Shoes FMM1 Sums of Powers of 2 FMM2 Fifty Coins FMM3 The Zeta Monogram FMM4 The Cthulhus Monogram FMM4B Multiplication Logic FMM5 The Termite and the Cube FMM6 Sharing Dump Trucks FMM7 The Flippant Juror FMM8 Bus Routes FMM9 A Robust Bus System FMM9B Square-Free Sequence FMM10 Inferring Rule from Sequence FMM11 Checkerboard Color Schemes FMM12 One-Handed Chords FMM13 First Ace FMM14 Which Color Cab FMM15 Petersburg Paradox Revisited FMM16 A Binomial Challenge FMM17 A Radical Sum FMM18 Memorable Phone Numbers FMM19 Arrange in Order FMM20 A Pair of Dice Games: FMM21
Category:Puzzles · Category:Math Flags · Template:FMMFlag · e |