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.

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 5*5*5*5/6*6*6*6 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(opp. Winning)=4*4*4*4/6*6*6*6 similarly u can get for other cases

Total probability of opponent winning is 5^46^4 + 4^4/6^4 + 3^46^4 +2^4/6^4 + 1/6^4

So if u subtract this from one u will get ur answer

I hope this helps u

Find:$$P(D_1,D_2,D_3,D_4leq D_5)=sum_{k=1}^6P(D_1,dots,D_4leq D_5mid D_5=k)P(D_5=k)$$where the $D_i$ denote the results of 5 independent die rolls.

Here $P(D_1,dots,D_4leq D_5mid D_5=k)=P(D_1,dots,D_4leq k)$ on base of independence.

The probability of **not** throwing a $6$ with four dice is $left(frac56right)^4=frac{625}{1296}$.

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 $left(frac46right)^4=frac{256}{1296}$.

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

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

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.