Exercise 1.16 [Mitzenmacher and Upfal's textbook]

Consider the following game, played with three six-sided dice. If the player ends with all three dice showing the same number, she wins. The player starts by rolling all three dice. After this first roll, the player can select any one, two, or all of the three dice and re-roll them. After this second roll, the player can again select any of the three dice and re-roll them one final time.

Suppose that the player has this strategy: If all three dice match, the player stops and wins. If two dice match, the player re-rolls the die that does not match; and if no dice match, the player re-rolls them all.

Now, this is how we do it. Clearly we have three cases:

Case 1: When all three dices match the same face in first rolling. So, the player stops and wins. The probability that this happens is 1/36.

Case 2: When only two dices match the same face in first rolling. So, we need to do re-roll the one that doesn't match. Thus we have probability 1/6 to win but if the second rolling doesn't match, then we re-roll it again for third and last rolling. Then we have 5/6*1/6 (This is because we wouldn't do the third rolling unless the second rolling did not match the two dices that are matched and probability that the player gets in third rolling the same face is 1/6). So, in total we have $1/6+5/6*1/6 = 11/36$.

Case 3: When all three dices did not match in the first rolling. Then the player will throw all the three dices again. Thus, we have three ways that could happen: either all match up, so we have probability that the player wins in the second rolling: (6/6*5/6*4/6)*1/36, or only two match up and we have probability to win in the second and third rolling: (6/6*5/6*4/6)*5/12*1/6. Or all three are unmatched and probability that to win in this case is: 2(6/6*5/6*4/6)*1/6. So we sum up all these probabilities in this case 3.

Now, we need to sum up all these three cases to get the probabilities that a player wins in this game.

Is that correct?

Thank you!

  • 2
    $\begingroup$ Your case 2 doesn't include the probability that the initial throw has two matching dice. $\endgroup$ – Arthur Aug 24 '17 at 10:30
  • $\begingroup$ @Arthur So, 11/36 should be multiplied by 5/12 (where 5/12 is the probability that case 2 happens). Thank you Arthur! $\endgroup$ – YOUSEFY Aug 24 '17 at 10:34
  • $\begingroup$ $2(6/6\times5/6\times4/6)\times1/6=5/27,$ which is greater than the chance to win given that the player has already rolled three non-matching numbers twice. If you apply the same kind of logic you used on the other cases, I think you can fix this. $\endgroup$ – David K Aug 24 '17 at 13:03


By throwing $3$ dice let $p$ denote the probability that all faces will be the same and let $q$ denote the probability that all faces are distinct.

Then $p=\frac66\frac16\frac16=\frac1{36}$ and $q=\frac66\frac56\frac46=\frac{20}{36}$.

The probability of winning will take the shape:$$p+(1-p-q)\times\cdots+q\times\cdots=\frac1{36}\times\cdots+\frac{15}{36}\times\cdots+\frac{20}{36}\times\cdots$$

Can you fill in $\cdots$ yourself?


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  • $\begingroup$ Thank you drhab, the way that you explain the answer is much better than I used! $\endgroup$ – YOUSEFY Aug 24 '17 at 19:32
  • $\begingroup$ You are very welcome $\endgroup$ – drhab Aug 24 '17 at 19:41


Round1 - All three Match $\frac{1}{36}$ - Win

Round1 - Two match $\frac{15}{36}$ - Round 2 - All three match $\frac{1}{6}$-Win

Round1 - Two match $\frac{15}{36}$ - Round2 - Two Still match$\frac{15}{36}$ - Round 3 - All three match $\frac{1}{6}$ - Win

Round1 - All three don't match $\frac{20}{36}$ - Round2 Roll all three - all three match $\frac{1}{36}$.

Round1 - All three don't match$\frac{20}{36}$ - Round2 Roll all three - Two match$\frac{15}{36}$ - Round3 - Roll one die - All three match$\frac{1}{6}$-Win

Round1 - All three don't match$\frac{20}{36}$ - Round2 Roll all three - All three don't match$\frac{20}{36}$ - Round3 - Roll all three - All three match $\frac{1}{36}$. - Win

Thus the required probability = $\frac{1}{36} + \frac{15}{36}.\left[\frac{1}{6}+\frac{15}{36}.\frac{1}{6}\right] + \frac{20}{36}.\left[\frac{1}{36}+\frac{15}{36}.\frac{1}{6}+\frac{20}{36}.\frac{1}{36}\right] $

I hope I captured everything.

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  • $\begingroup$ But $\frac1{36}+\frac56+\frac23\neq1$. The events $3$ match in round1, $2$ match in round1, no matches in round1 are mutually exclusive and exhaustive. $\endgroup$ – drhab Aug 24 '17 at 13:12
  • $\begingroup$ I was typing it in a hurry. A lot of typos. Got it fixed should be right $\endgroup$ – Satish Ramanathan Aug 24 '17 at 17:13
  • $\begingroup$ Probability on $2$ match is $3\times\frac{5}{36}$. Three possibilities: MMU,MUM,UMM $\endgroup$ – drhab Aug 24 '17 at 17:21
  • $\begingroup$ Being in your 40s and a 10 year old child at home is enough reason why this could happen. Thanks for correcting. But the idea is sound for the OP to analyze such games. Thank you sir. @drhab $\endgroup$ – Satish Ramanathan Aug 25 '17 at 0:22

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