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The Theoretical Likelihood of an All Red or All Black Deal
Seven cards and the top deck card are dealt face up at the beginning of a standard game of Solitaire. The likelihood that all of these eight cards are all one color is calculable.
The problem simplifies to "What are the odds of dealing 8 all red or 8 all black cards from the top of a stack of shuffled cards"
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The first card that is turned up is either red or black in all cases. The probability of this is 100%.
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The second card is dealt from a deck containing 51 cards, 25 of which will be the same color as the first card that was turned up, and 26 will be the other color. So the odds of dealing another card of the same color as the first is 25 out of 51.
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The third card will then be dealt from a deck of 50 cards, 24 of the same color as the first 2 and 26 of the other color, so the odds of this happening are 24 out of 50.
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And so on for the remaining 5 cards.
The cumulative odds for this happening are therefore:
1 x 25/51 x 24/50 x 23/49 x 22/48 x 21/47 x 20/46 x 19/45
which equals
0.004152 or 1 in 240.8 deals
Using the Mersenne-Twister random number algorithm option and starting with the default seed sequence (so easy to reproduce if anyone is interested) I found the following:
| Number of Deals | Occurrences of All Red or All Black | |
| 1,000 | 4 | 1 in 250 |
| 10,000 | 34 | 1 in 259 |
| 100,000 | 384 | 1 in 260 |
| 1,000,000 | 4193 | 1 in 238 |
| 10,000,000 | 41609 | 1 in 240.3 |
| 100,000,000 | 415141 | 1 in 240.9 |
From these numbers I conclude that ultimately with enough trials, and despite being slow to converge, the tool will approach the theoretical answer to a degree of accuracy that indicates to me that all is well with the program.