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Three dices are rolled, and I bet $1 on 6,
If there is no 6 among 3 dices, then I lose $1.
If there is three 6 among 3 dices, then my profit is $12 and collect my original $1 back.
If there is two 6 among 3 dices, then my profit is $2 and collect my original $1 back.
If there is one 6 among 3 dices, then my profit is $1 and collect my original $1 back.
I get no idea on how to determine the expected value for this combination, since only one condition can be met.
Does anyone have any suggestions?
Thanks in advance for any suggestions
Eric
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Hi oem7110;
You can dot product the following two vectors:
[125,75,15,1].[-1,1,2,12] = -8
Your expectation is E(for each throw) = -8 / 216 = -1 / 27dollars
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Bobbym:
Thank you very much for suggestions
I get another similar issue with little different conditions:
Three dices are rolled, and I bet $1 on 4,5,6,7,8 numbers with different profits, so total bets is $5.
If sum of the dices is 4 among 3 dices, then my profit is $62 and collect my original $1 back, but I lose $4 on the other numbers.
If sum of the dices is 5 among 3 dices, then my profit is $31 and collect my original $1 back, but I lose $4 on the other numbers.
If sum of the dices is 6 among 3 dices, then my profit is $18 and collect my original $1 back, but I lose $4 on the other numbers.
If sum of the dices is 7 among 3 dices, then my profit is $12 and collect my original $1 back, but I lose $4 on the other numbers.
If sum of the dices is 8 among 3 dices, then my profit is $8 and collect my original $1 back, but I lose $4 on the other numbers.
I get no idea on how to determine the expected value for this combination, since only one condition can be met, and I must lose bets on another 4 numbers.
Do you have any suggestions?
Thanks in advance for any suggestions
Eric
Last edited by oem7110 (2010-12-08 22:03:11)
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Hi oem7110;
That problem is done in the same way as the other one. With only a slight difference.
We use 2 generating functions:
and
They generate the sample space for us.
You have told me what happens on sums of 4 to 8, what happens on the sums 3,9,10, etc.?
Is it a freeroll or do you lose all bets?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Since I bet $5 for each game on 5 different numbers (betting $1 on each number), if the sum of the dices equal to any numbers between 4 and 8, then I will collect my original $1 back on winning number, but will lose $4 on the other number.
If sum of the dices is 3,9,10,11,12,13,14,15,16,17,18 among 3 dices, then I lose $5 totally.
If sum of the dices is 4 among 3 dices, then my profit is $62 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $62 - $4 = $58.
If sum of the dices is 5 among 3 dices, then my profit is $31 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $31 - $4 = $27.
If sum of the dices is 6 among 3 dices, then my profit is $18 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $18 - $4 = $14.
If sum of the dices is 7 among 3 dices, then my profit is $12 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $12 - $4 = $8.
If sum of the dices is 8 among 3 dices, then my profit is $8 and collect my original $1 back, but I lose $4 on the other numbers, so my net profit is $8 - $4 = $4.
Do you have any suggestions?
Thank you very much for any suggestions
Last edited by oem7110 (2010-12-08 22:54:27)
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Please adjust the payoffs you have 58 for all the numbers!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I just finished on adjust the payoffs for all the numbers.
Do you have any suggestions?
Thank you very much for any suggestions
Last edited by oem7110 (2010-12-08 22:56:29)
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Hi;
Without showing the intervening work to compute the 2 vectors, you then take the dot product of them.
So the E(throwing 3 dice) = -125 / 216 dollars
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Oh I see, (Sum of all [Probability x payoffs]) / Total numbers of combination, right?
I try it.
Thank you very much for your suggestions
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Hi;
Probabilities of each * 216 gives the first vector. Turns out this is a useless move. So just probabilites * payoffs and add them up.
Welcome to the forum!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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. . . . . . . . . . . . .
Edit: I corrected my error . . . Thank you for the heads-up, bobbym . . .
Last edited by soroban (2010-12-09 13:08:00)
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Hi soroban;
Do not mention it, you have helped me many times. Glad to work with you!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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