# What are the odds of winning?

I have a game coming up I am putting in downtown Toronto, I need to know what the odds of winning are so I can plan for the prizes. The game is comprised of 20 cards which have 10 categorize. To win you need to match two cards of the same category. The first selection has no chance of winning.

The second selection has a 1 in 19 chance of winning.

What are the odds of each selection winning after that? I need to know the chance to win selecting 3 cards, 4 cards, 5 cards and then 6 cards?

I'm assuming that once you draw even one matching pair, the game is over and you draw no further cards. Then, as you noted, you have a zero probability of winning after one card, and a $\frac{1}{19}$ probability of winning after two cards.

You therefore draw a third card with probability $\frac{18}{19}$, and win on that card with probability $\frac{2}{18}$, so your probability of winning on the third card is $\frac{18}{19} \times \frac{2}{18} = \frac{2}{19}$.

You draw a fourth card with probability $\frac{18}{19} \times \frac{16}{18} = \frac{16}{19}$, and win on that card with probability $\frac{3}{17}$, so your probability of winning on the fourth card is $\frac{16}{19} \times \frac{3}{17} = \frac{48}{323}$.

You draw a fifth card with probability $\frac{18}{19} \times \frac{16}{18} \times \frac{14}{17} = \frac{224}{323}$, and win on that card with probability $\frac{4}{16}$, so your probability of winning on the fifth card is $\frac{224}{323} \times \frac{4}{16} = \frac{56}{323}$.

You can continue along those lines; the computations are tedious but straightforward.

I will give you simulated answers. You can check some of them against your computations. In my simulation, I draw all the allowed cards. There are two cards of each number 1 though 10. There is a 'win' if there are any redundant numbers among those drawn. From simulating a million games, one can expect to get the probability with three place accuracy. Simulated answers for $n = 3, 4, 5$ agree with @BrianTung's combinatorial results.(+1) I also did $n = 6.$

 cards = c(1:10, 1:10);  n = 3
m = 10^6;  win = logical(m);
for (i in 1:m) {
drawn = sample(cards, n)
win[i] = (length(unique(drawn)) < n) }
mean(win) # probability of winning
## 0.158244
3/19
## 0.1578947  # compare; getting 3 place accuracy

# n = 4
mean(win) # probability of winning
## 0.306402
1/19 + 2/19 + 48/323
## 0.3065015

# n = 5
mean(win) # probability of winning
## 0.479232
1/19 + 2/19 + 48/323 + 56/323
## 0.4798762

# n = 6
mean(win) # probability of winning
## 0.653267  # first simulation
## 0.653276  # second simulation

• Looks good. The probability of winning on (as opposed to by) the sixth card is equal to that of the fifth card—$56/323$—which jibes with your simulation results. – Brian Tung Mar 25 '16 at 18:04