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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:

enter image description here

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

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    $\begingroup$ Unless I misunderstood the question, the bottom two branches win as well. $\endgroup$ – David Jul 20 '18 at 5:27
  • $\begingroup$ it says you win if your total is 3..doesnt that include only the two I mentioned? $\endgroup$ – David Jul 20 '18 at 5:32
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    $\begingroup$ 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. $\endgroup$ – Hashimoto Jul 20 '18 at 5:35
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    $\begingroup$ The bottom branch should stop imediately as a win. The prob is $\frac 13\frac 12+\frac 13\frac 12+\frac 13=\frac 23$ $\endgroup$ – fleablood Jul 20 '18 at 5:35
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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.

enter image description here

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.

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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 $\frac 16 +\frac 16 +\frac 13=\frac 23$

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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 $1 - 2(1/6) = 2/3$.

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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.

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