Suits but no values - (a card game, not a law firm description)

[ctswitzer]

A deck of 25 cards consists of 5 suits, 5 cards of each suit, no other values. Cards are face down and randomized, and revealed one at a time. The only other rule is that the game ends when at least one card of each suit is revealed (meaning, at most 21 cards will be revealed).

I'm trying to figure out the probability that the game ends in 5 turns, 6, 7, 8, ... 21, but my knowledge of factorials, which I suspect could help here, is limited (B+ in college calculus some years ago).

I know 5: Any card can be drawn first, then the probabilities are 20/24 * 15/23 * 10/22 * 5/21 = 15,000/255,024 = .0588 = 5.88%. So there is a 5.88% chance the game ends when the 5th card is revealed. Easy enough.

By following a similar process I believe there is an 11.75% of the game ending in 6 rounds, but I'm not sure that answer is correct and I'm looking for a formula that might simplify things.

Your thoughts welcome.

 

[I like Serena]

A deck of 25 cards consists of 5 suits, 5 cards of each suit, no other values. Cards are face down and randomized, and revealed one at a time. The only other rule is that the game ends when at least one card of each suit is revealed (meaning, at most 21 cards will be revealed).

I'm trying to figure out the probability that the game ends in 5 turns, 6, 7, 8, ... 21, but my knowledge of factorials, which I suspect could help here, is limited (B+ in college calculus some years ago).

I know 5: Any card can be drawn first, then the probabilities are 20/24 * 15/23 * 10/22 * 5/21 = 15,000/255,024 = .0588 = 5.88%. So there is a 5.88% chance the game ends when the 5th card is revealed. Easy enough.

By following a similar process I believe there is an 11.75% of the game ending in 6 rounds, but I'm not sure that answer is correct and I'm looking for a formula that might simplify things.

Your thoughts welcome.

Welcome to MHB, ctswitzer! :)

In any game the total number of possible outcomes is:
$$25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20$$

Suppose the game ends in 6 rounds.
Then the first way to make that happen is to draw 1-1-2-3-4-5.
More generally, suppose only that the first 2 cards are equal, then there are:
$$25 \cdot 4 \cdot 20 \cdot 15 \cdot 10 \cdot 5$$
favorable outcomes.
Since the 2 equal cards can be in $\binom 6 2$ positions, the probability is:
$$P(\text{game ends in 6}) = \frac{\text{#favorable outcomes}}{\text{#total outcomes}}
=\binom 6 2 \cdot \frac{25 \cdot 4 \cdot 20 \cdot 15 \cdot 10 \cdot 5}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20} \approx 17.6\%$$

 

[ctswitzer]

Excellent. Thanks. Let me see if I understand. Say the game ends in 7 rounds. The possible outcomes are:
$$25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19$$
One way to make this happen is to draw 1-1-1-2-3-4-5, which gets us:
$$25 \cdot 4 \cdot 3 \cdot 20 \cdot 15 \cdot 10 \cdot 5$$
favorable outcomes.

The three equal cards can be in $\binom 7 3$ positions. So the probability is:
$$P(\text{game ends in 7, with 3-of-a-kind}) = \frac{\text{#favorable outcomes}}{\text{#total outcomes}}
=\binom 7 3 \cdot \frac{25 \cdot 4 \cdot 3 \cdot 20 \cdot 15 \cdot 10 \cdot 5}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19} \approx 6.5\%$$
But, the game could also end with there being two pairs rather than a three-of-a-kind. Let's say 1-1-2-2-3-4-5, which gets us:
$$25 \cdot 4 \cdot 20 \cdot 4 \cdot 15 \cdot 10 \cdot 5$$
favorable outcomes. Now help me out here; does this mean that 4 cards can be in $\binom 7 4$ positions? If so, the probability is:
$$P(\text{game ends in 7, with 2 pairs}) = \frac{\text{#favorable outcomes}}{\text{#total outcomes}}
=\binom 7 4 \cdot \frac{25 \cdot 4 \cdot 20 \cdot 4 \cdot 15 \cdot 10 \cdot 5}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19} \approx 8.67\%$$

Then I add 8.67 to 6.5, and there is a 15.17% chance that the game ends in 7 draws.

Continuing this method, I found there is about a 5.2% chance the game ends in 8 draws (1.38% chance of 4-of-a-kind, 2.3% chance of "full house," and 1.5% chance of three pairs).

Going one more level, there is about a 1.62% chance of ending in 9 draws (.15% chance of 5-of-a-kind, .4% chance of 4 of one suit and 2 of another, .61% chance of 2 3-of-a-kinds, .35% chance of a 3-of-a-kind with 2 pairs, and .12% chance of 4 pairs).

 

[ctswitzer]

No, something is still wrong, at least with my 8 and 9 because I could never get up to 100%.

 

[I like Serena]

My apologies.
The probability I gave for game-end-in-6 also includes the probability for game-end-in-5.
This is what happens if you draw 2-3-4-5-1-1.
In that case the game will end 1 turn earlier.
To get the proper probability for game-in-6 the chance for game-in-5 has to be subtracted.

Excellent. Thanks. Let me see if I understand. Say the game ends in 7 rounds. The possible outcomes are:
$$25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19$$
One way to make this happen is to draw 1-1-1-2-3-4-5, which gets us:
$$25 \cdot 4 \cdot 3 \cdot 20 \cdot 15 \cdot 10 \cdot 5$$
favorable outcomes.

The three equal cards can be in $\binom 7 3$ positions. So the probability is:
$$P(\text{game ends in 7, with 3-of-a-kind}) = \frac{\text{#favorable outcomes}}{\text{#total outcomes}}
=\binom 7 3 \cdot \frac{25 \cdot 4 \cdot 3 \cdot 20 \cdot 15 \cdot 10 \cdot 5}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19} \approx 6.5\%$$

Yep!
(Let's continue before considering game end in 5 or 6.)

But, the game could also end with there being two pairs rather than a three-of-a-kind. Let's say 1-1-2-2-3-4-5, which gets us:
$$25 \cdot 4 \cdot 20 \cdot 4 \cdot 15 \cdot 10 \cdot 5$$
favorable outcomes. Now help me out here; does this mean that 4 cards can be in $\binom 7 4$ positions? If so, the probability is:
$$P(\text{game ends in 7, with 2 pairs}) = \frac{\text{#favorable outcomes}}{\text{#total outcomes}}
=\binom 7 4 \cdot \frac{25 \cdot 4 \cdot 20 \cdot 4 \cdot 15 \cdot 10 \cdot 5}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19} \approx 8.67\%$$

Then I add 8.67 to 6.5, and there is a 15.17% chance that the game ends in 7 draws.

That would be $\binom 7 4 \cdot \binom 4 2 \cdot \frac 1 {2!}$ or $\binom 7 2 \cdot \binom 5 2 \cdot \frac 1 {2!}$.

First you select 4 from the 7 cards, then you select 2 of those 4 for the second pair.
After that, you have to divide by 2, because the 2 pairs can be interchanged, leading to double counting. We have to divide by 2 to compensate.

Or alternatively, first you select 2 from the 7 cards, then you select another 2 of the remaining 5, and divide by 2 to compensate for double counting.

Btw, after summing the probabilities for 1-1-1-2-3-4-5 and 1-1-2-2-3-4-5, we have to subtract the probability on game-in-5 and game-in-6 to get the proper probability.

Continuing this method, I found there is about a 5.2% chance the game ends in 8 draws (1.38% chance of 4-of-a-kind, 2.3% chance of "full house," and 1.5% chance of three pairs).

Going one more level, there is about a 1.62% chance of ending in 9 draws (.15% chance of 5-of-a-kind, .4% chance of 4 of one suit and 2 of another, .61% chance of 2 3-of-a-kinds, .35% chance of a 3-of-a-kind with 2 pairs, and .12% chance of 4 pairs).