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I have a question - If a player rolls $4$ dice, and the maximum result is the highest number he gets (for example he tosses and gets $1$,$2$,$4$,$6$ the maximum result is $6$). His opponent rolls a single die and if the player's result is higher than his opponent's, he wins. What is the chance of the player to to lose?

So, I can't seem to compute this in my mind and can't see which distribution this is since I don't know what are the results on each die when the player rolls them.

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  • $\begingroup$ With four dice, what's the probability that $M$, the maximum roll, is $≤5$? How about $≤4$ and so on? Use that to compute the probability that $M=6$ and so on. $\endgroup$ – lulu May 19 at 12:32
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Let’s think of the chances when the opponent win Let the opponent got a six on his die then the chances for opponent to win is $$\frac{5\times5\times5\times5}{6\times6\times6\times6}$$ as our player can get from $1$ to $5$ on any die and total cases are $6^4$ So if the opponent rolls $4$ $$P(opponent Winning)= \frac{4\times4\times4\times4}{6\times6\times6\times6}$$similarly you can get for other cases Total probability of opponent winning is $$\frac{5^4}{6^4} + \frac{4^4}{6^4} + \frac{3^4}{6^4} +\frac{2^4}{6^4} + \frac{1}{6^4}$$

So if you subtract this from one you will get your answer I hope this helps you.

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    $\begingroup$ Please learn how to use MathJax. $\endgroup$ – Toby Mak May 19 at 13:34
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    $\begingroup$ I am preparing for jee advanced after I complete my exam I start focusing mainly on this and learn new skills $\endgroup$ – Namburu Karthik May 19 at 14:04
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Find:$$P(D_1,D_2,D_3,D_4\leq D_5)=\sum_{k=1}^6P(D_1,\dots,D_4\leq D_5\mid D_5=k)P(D_5=k)$$where the $D_i$ denote the results of 5 independent die rolls.

Here $P(D_1,\dots,D_4\leq D_5\mid D_5=k)=P(D_1,\dots,D_4\leq k)$ on base of independence.

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The probability of not throwing a $6$ with four dice is $\left(\frac56\right)^4=\frac{625}{1296}$.

This, however, includes throws with no $5$ in., etc.., and so doesn't give the probability of $\max=5$ with four dice.

The probability of not throwing a $5,6$ with four dice is $\left(\frac46\right)^4=\frac{256}{1296}$.

So the probability of $\max=5$ is $\frac{625}{1296}-\frac{256}{1296}=\frac{369}{1296}$.

In general $P(\max=k)=P(\lt k+1)-P(\lt k)$.

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  • $\begingroup$ Hi, sorry but I can;t get it why I should calculate the probability max 5? I mean, If the opponent throws and gets 2 and I get (1,1,1,1), I lose. I he throws 3 and I get (1,1,1,2) I lose. Should I compute it somehow like that since I need my probability to lose? $\endgroup$ – JustEquvilant May 19 at 15:57
  • $\begingroup$ You get a distribution of the value of $\max(d_1,d_2,d_3,d_4)$, which is essentially what your question is asking for. $\endgroup$ – JMP May 19 at 16:01
  • $\begingroup$ $\max=5$ here means the maximum is $5$, not $\le 5$. $\endgroup$ – JMP May 19 at 16:03
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If Player 1 (P1) rolling one die rolls a 6, i'm assuming he wins and Player 2 rolling the 4 dice loses even if manages a 6 in his grouping, if so...

If P1 rolls

6 - then P1 wins 100% of time;

5 - P1 wins 48.2% of time (5x5x5x5 / 6x6x6x6 = 625/1296)

4 - P1 wins 9.87% of time (4x4x4x4 / 6x6x6x6 = 128/1296)

3 - P1 wins 6.25% of time (3x3x3x3 / 6x6x6x6 = 81 /1296)

2 - P1 wins 1.2% of time (2x2x2x2 / 6x6x6x6 = 16 /1296)

1 - P1 wins 1 in 1296.

So assigning each of these a 1/6th chance of happening and then adding those products together, P1 has a 1/6 + 8.03% + 1.64% + 1.04% + .2% + .013% chance of winning, which is about a 27.6% chance of winning.

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