# Is my reasoning for this probability question correct?

Consider a game of three. You shuffle a deck of three cards: 1,2,3. You draw cards without replacement until total is 3 or more. You win if your total is 3. What is P[W], the probability that you'll win?

I drew a tree diagram. Below is the number of the card along with its probability:

(1/3)(1/2) + (1/3)(1/2) + 1/3 = 2/3. Is this right?

• Unless I misunderstood the question, the bottom two branches win as well. – David Jul 20 '18 at 5:27
• it says you win if your total is 3..doesnt that include only the two I mentioned? – David Jul 20 '18 at 5:32
• It took me some time to realize the David's are distinct people. Seems right to me, although I would do it like this: (Probability of winning) = (# winning scenarios)/(# of total scenarios) = 2/6 = 1/3. -------- Edit: scratch that. @fleablood seems right. – Hashimoto Jul 20 '18 at 5:35
• The bottom branch should stop imediately as a win. The prob is $\frac 13\frac 12+\frac 13\frac 12+\frac 13=\frac 23$ – fleablood Jul 20 '18 at 5:35

Morning after. Had a chance to go to photoshop.

This should be the tree. Originally you overlooked that the last branch ($3$ is the first card) was a win.

Your new image is better but now it has the branch extending so that you are drawing after you've already won. Obviously once you win, you stop playing.

========

Can't draw but:

Branch 1a: draw 1 (prob $\frac 13$), go on; draw 2 (prob $\frac{1}{2}$) WIN

Branch 1b: draw 1 (prob $\frac 13$), go on; draw 3 (prob $\frac{1}{2}$) lose

Branch 2a: draw 2 (prob $\frac 13$), go on; draw 1 (prob $\frac{1}{2}$) WIN

Branch 2b: draw 2 (prob $\frac 13$), go on; draw 3 (prob $\frac{1}{2}$) lose

Branch 3: draw 3 (prob $\frac 13$) WIN

So you win if Branch 1a or 2a or 3. So $\frac 16 +\frac 16 +\frac 13=\frac 23$

Without drawing a tree, you can reason as follows.

Since the game stops when the accumulated value of all cards drawn (which I will call the "score") is at least $3$, the only possible stopping states are when the score equals $3$, $4$, or $5$. It is not possible to score $6$ or higher because the minimum attainable score after drawing $2$ cards is already $3$; thus at most two cards are ever drawn.

Since the only way to score $4$ is $(1,3)$, and the only way to score $5$ is $(2,3)$, and each of these occur with probability $1/6$, being among equally likely elementary outcomes of drawing two cards from a pile of three without replacement, it follows that the desired probability is $1 - 2(1/6) = 2/3$.

Yes, you are correct. Another way of looking at it is the only way you can win is if 3 is either the first card or the third card. The reason is that if 3 is first you win straight away. If it is not then you have either a one or two in which case a three would be a loss so you must draw the other card (two or one) next leaving the three as the third/last card.

Given your shuffle each card has an equal probability of being in each of the three positions so you have a 2/3 chance of winning.