Dice Drama
There are six squares marked 1, 2, 3, 4, 5, 6. You are invited to place your money as you wish on any one square. Three dice are then thrown.
The Rules
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If your number appears on one dice only, you get your money back plus the same amount.
If two dice show your number, you get your money back plus twice the amount you placed on the square.
If your number appears on all three dice, you get your money back plus three times the amount. Of course if the number is not on any of the dice, the operator gets your money.
Now, you might reason: the chance of my number showing on one dice is 1 : 6, but since there are three dice, the chances must be 3 : 6 or 1 :2, therefore the game is a fair one. Of course this is the way the operator of the game wants everyone to reason, because it’s quite fallacious. But is the game favourable to the operator or the player, and, in any case, just how favourable is it ?
The Rules
-------------
If your number appears on one dice only, you get your money back plus the same amount.
If two dice show your number, you get your money back plus twice the amount you placed on the square.
If your number appears on all three dice, you get your money back plus three times the amount. Of course if the number is not on any of the dice, the operator gets your money.
Now, you might reason: the chance of my number showing on one dice is 1 : 6, but since there are three dice, the chances must be 3 : 6 or 1 :2, therefore the game is a fair one. Of course this is the way the operator of the game wants everyone to reason, because it’s quite fallacious. But is the game favourable to the operator or the player, and, in any case, just how favourable is it ?
posted by Rajesh Lal at
1:43 PM
10 Comments:
Ok, so first...
Probability that you will get exactly one die to land on your number:
1/6 * 5/6 * 5/6 * 3 = 75/216
Probability that you will get exactly two dice to land on your number:
1/6 * 1/6 * 5/6 * 3 = 15/216
Probability that you will get exactly three dice to land on your number:
1/6 * 1/6 * 1/6 = 1/216
So, assuming you played 216 games with each outcome happening, making a one dollar bet on each game, you would...
1. gain 75 * 1 = $75
2. gain 15 * 2 = $30
3. gain 1 * 3 = $3
4. lose 125 * 1 = $125
Therefore, in 216 rounds, you would expect to lose $17, and subsequently, you would expect to lose 17/216 dollars each round.
This means that the deviation from a fair game (50%-50%) is .5 - (17/216 * 1/2) which means that the odds are tilted in favor of the operator:
Operator: 53.935%
You: 46.065%
My response did not show up.
this will be short
There are 216 different combinations. Let's say you play 216 times and each combination occurs once.
You paid $216, for a dollar each play.
125 times your number will not show
and you get nothing.
75 times your number shows once and you get $2
15 times your number shows twice and you get $3.
One time your number shows on all three dice, you get $4.
75*2 + 15*3 + 1*4 = $199
Ok, now I see where wolf got 17
216-199 = 17
Cry wolf and Ragknot nailed it ... 'nuff said ....
Confirmed.
EV(expected value)= wager*[-1*(5/6*5/6*5/6) +3*(1/6*1/6*1/6)+ 2*(5/6*1/6*1/6)*3+ 1*(5/6*5/6*1/6)*3]
EV= 1/216*(-125+3+30+75)
EV= -17/216*wager
Cam
It's easier to do it the other way round: the probability of the bank winning is (5/6)³ = 125/216 ≈ 57.87. So the probability of you winning is 1 - (5/6)³ = 91/216 ≈ 42.13%.
Ooooops! I goofed. I should've re-read the question before posting.
kicking the post through.
for 1 die the probability is 1/6
but for 3 dice the probability is 3/18
there for the game is favourable to the operator
It looks like Jude hasn't read the question either!
Strange... 1/6 becomes 3/18 with 3 dices ?
does the mistake come from the fact you added the probabilities, which first is wrong, then :
1/6+1/6+1/6 = (1+1+1)/(6+6+6) which is wrong since 1/6+1/6+1/6= 3(1/6)= 1/2.
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