Joe dealt 20 cards from a standard 52-card deck, and the number of red cards exceeded the number of black cards by 8. He reshuffled the cards and dealt 30 cards. This time, the number of red cards exceeded the number of black cards by 10. Determine which deal is closer to the 50/50 ratio of red/black expected of fairly dealt hands from a fair deck and why.
r-b = 8
r+b = 20
solve for r and b
find r/(r+b)
r-b = 10
r+b = 30
solve for r and b
find r/(r+b)
To determine which deal is closer to the 50/50 ratio of red and black cards, we need to compare the differences in the number of red and black cards in each deal.
Let's start by finding the difference between the number of red and black cards in the first deal. We know that the number of red cards exceeds the number of black cards by 8.
So, let's set up an equation:
(Number of Red Cards - Number of Black Cards) = 8
Now let's move on to the second deal. We know that the number of red cards exceeds the number of black cards by 10.
Again, we'll set up an equation:
(Number of Red Cards - Number of Black Cards) = 10
To compare the two deals, we need to convert these differences into percentages relative to the total number of cards dealt.
In the first deal, 20 cards were dealt. The total number of cards in a standard deck is 52. So, the percentage difference in the first deal would be:
(8 / 20) * 100 = 40%
In the second deal, 30 cards were dealt from a 52-card deck. The percentage difference in the second deal would be:
(10 / 30) * 100 = 33.33%
Comparing the percentages, we can conclude that the second deal is closer to the expected 50/50 ratio of red and black cards. The second deal had a difference of 33.33% compared to the first deal's 40%. Thus, the second deal is closer to the fair distribution of red and black cards.