# What is the probability of winning at this game?

you throw a fair, 6 sided dice. If the result is 3 or lower, you lose. If not then you can draw the number of cards the dice shows from a standard 52 deck of cards (if you throw a 5 then you draw 5 cards randomly). You win the game if the 4 aces are among the cards you drew.

What are the odds of winning at this game?

My thinking right now would be to proceed like this:

$\left ( \frac{1}{6} \cdot \frac{1}{\binom{52}{4}}\right )+\left ( \frac{1}{6} \cdot \frac{48}{\binom{52}{5}}\right )+\left ( \frac{1}{6} \cdot \frac{48\cdot 47}{\binom{52}{6}}\right )$

how would you do it?

Thank you

• Seems to be OK. – zoli Oct 4 '15 at 22:52

There is a slight error. The final term reads as $\left(\frac{1}{6}\cdot \frac{48\cdot 47}{\binom{52}{6}}\right)$, but where did those numbers actually come from.

$\frac{1}{6}$ because this is the probability that we are in the scenario of drawing six cards.

$\frac{1}{\binom{52}{6}}$ because we are finding probabilities associated with having drawn a hand of six cards

You have however $48\cdot 47$. This number here should instead be $\binom{48}{2}=\frac{48\cdot 47}{2}$. Why? Because we are trying to count the number of six-card hands which contain all four aces, but since we opted to use combinations for the bottom, we are going the route where order doesn't matter.

We are only interested in the hands which contain four aces, but the other two cards could have been any of the remaining $48$ cards in the deck. Picking which two of those cards to be included in our hand, having picked $J\heartsuit 8\clubsuit$ is the same to us as having picked $8\clubsuit J\heartsuit$.

This brings the final probability to:

$$\left(\frac{1}{6}\cdot \frac{1}{\binom{52}{4}}\right) + \left(\frac{1}{6}\cdot \frac{48}{\binom{52}{5}}\right)+\left(\frac{1}{6}\cdot\frac{\binom{48}{2}}{\binom{52}{6}}\right)$$

If curious about the odds as opposed to the probability, $\textbf{odds}_{win}$ is written as $pr(\text{win}):1-pr(\text{win})$

• thank you, that answer my question :) – spexel Oct 4 '15 at 23:15
• if my friend told me that he won at this game, what are the odds that when he won he rolled a 6? my intuition tell me to just put everything on the same denominator like that: $\frac{1}{1624350} \frac{5}{1624350} \frac{15}{1624350}$ so $[\frac{15}{21}]$ is the probability he won rolling a 6 given that we know he already won. am i right? – spexel Oct 4 '15 at 23:25