FMM14 - Revision history

Admin at 22:34, 22 November 2019

2019-11-22T22:34:39Z

← Older revision		Revision as of 22:34, 22 November 2019
Line 1:		Line 1:
	{{FMM		{{FMM
	\|title=~~Which Color Cab~~		\|title=First Ace
	\|problem=		\|problem=

Admin at 22:34, 22 November 2019

2019-11-22T22:34:26Z

← Older revision		Revision as of 22:34, 22 November 2019
Line 15:		Line 15:
	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.)		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.		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.

			An alternative way to interpret this is, if we are placing ONE bet on the number of cards that will be turned over, we should place our bet on (N+1)/(M+1), rounded to the nearest whole number.

	}}		}}

Admin at 22:30, 22 November 2019

2019-11-22T22:30:05Z

← Older revision		Revision as of 22:30, 22 November 2019
Line 1:		Line 1:
	~~#REDIRECT [[FMM15]]~~		{{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.

			}}

Admin: Admin moved page FMM14 to FMM15

2019-11-22T22:27:29Z

Admin moved page FMM14 to FMM15

New page

#REDIRECT [[FMM15]]