## Poll

1 vote (33.33%) | |||

No votes (0%) | |||

1 vote (33.33%) | |||

1 vote (33.33%) | |||

No votes (0%) |

**3 members have voted**

March 13th, 2020 at 9:24:02 AM
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A fair die is rolled until a 6 appears. My payout in dollars is equal to the number of rolls if no odd numbers were rolled and is zero otherwise. What is the conditional expected value of my payout given that no odd numbers were rolled until the appearance of a six ? You are told that I have played the game and have rolled a 6 without rolling odd numbers. What is the conditional probability that I got a payout of $1 in this situation?

Last edited by: Klopp on Mar 13, 2020

March 13th, 2020 at 9:35:13 AM
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My gut instinct is that this sounds like a TERRIBLE game, but I don't know what the math would be so I could be wrong...haha

March 13th, 2020 at 10:18:00 AM
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Is this a fair game? Not even close - the fair value is 37.5 cents.

The problem is not just that any odd number ends the game with a loss, but if you roll a 6 in either of your first two rolls, you lose money as well.

You get $1 if the first roll is 6; this has probability 1/6

You get $2 if the first roll is 2 or 4, and the second is 6; this has probability 1/3 x 1/6

You get $3 if the first two rolls are 2 or 4, and the second is 6; this has probability (1/3)

and so on

The expected return is 1 x 1/6 + 2 x 1/3 x 1/6 + 3 x (1/3)

= 1/6 x (1 + 2 x 1/3 + 3 x (1/3)

= 1/6 x (1 + 1/3 + (1/3)

= 1/6 x (1 / (1 - 1/3))

= 1/6 x 9/4 = 3/8

If you rolled a 6 without any odd numbers, the only way to win $1 is if your first roll was a 6. Just in case this is a homework problem, I'll leave the rest to you; just figure out the probability of winning $2, $3, $4, and so on. (Hint: what numbers do you need to roll to win $3?)

The problem is not just that any odd number ends the game with a loss, but if you roll a 6 in either of your first two rolls, you lose money as well.

You get $1 if the first roll is 6; this has probability 1/6

You get $2 if the first roll is 2 or 4, and the second is 6; this has probability 1/3 x 1/6

You get $3 if the first two rolls are 2 or 4, and the second is 6; this has probability (1/3)

^{2}x 1/6and so on

The expected return is 1 x 1/6 + 2 x 1/3 x 1/6 + 3 x (1/3)

^{2}x 1/6 + 4 x (1/3)^{3}x 1/6 + ...= 1/6 x (1 + 2 x 1/3 + 3 x (1/3)

^{2}+ ...)= 1/6 x (1 + 1/3 + (1/3)

^{2}+ (1/3)^{3}+ ...)^{2}= 1/6 x (1 / (1 - 1/3))

^{2}= 1/6 x 9/4 = 3/8

If you rolled a 6 without any odd numbers, the only way to win $1 is if your first roll was a 6. Just in case this is a homework problem, I'll leave the rest to you; just figure out the probability of winning $2, $3, $4, and so on. (Hint: what numbers do you need to roll to win $3?)

March 13th, 2020 at 10:59:10 AM
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I don't agree about your answer for the expected payout. Also, the probability I asked about is a CONDITIONAL probability: the probability that the first roll was a six given that no odd numbers were rolled until the appearance of a six.

March 13th, 2020 at 11:55:07 AM
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Quote:KloppI don't agree about your answer for the expected payout. Also, the probability I asked about is a CONDITIONAL probability: the probability that the first roll was a six given that no odd numbers were rolled until the appearance of a six.

What do you get for the expected payout, and how did you get that value?

And I understand that the probability was conditional; it's (the probability of winning $1) divided by (the probability of winning $1 + the probability of winning $2 + the probability of winning $3 + ...)

March 13th, 2020 at 12:03:33 PM
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Here is my interpretation of what the player gets for his $3 to play:

The player rolls a die until any odd number or a six is rolled. If that game ending roll is a 1, 3, or 5, then the players nothing. If that game ending roll is a 6, then the player will win $1 for every roll of the die (including the last roll of a 6).

If this is true, I get an expected win of 37.5 cents. There is a 75% chance of winning nothing. If the player does win, his average number of rolls is 1.5. Thus, the expected win is 0.25 * 1.5 = $0.375 .

In Klopp's last post he mentioned a conditional probability, which seems to indicate there is a misunderstanding here.

The player rolls a die until any odd number or a six is rolled. If that game ending roll is a 1, 3, or 5, then the players nothing. If that game ending roll is a 6, then the player will win $1 for every roll of the die (including the last roll of a 6).

If this is true, I get an expected win of 37.5 cents. There is a 75% chance of winning nothing. If the player does win, his average number of rolls is 1.5. Thus, the expected win is 0.25 * 1.5 = $0.375 .

In Klopp's last post he mentioned a conditional probability, which seems to indicate there is a misunderstanding here.

It's not whether you win or lose; it's whether or not you had a good bet.

March 13th, 2020 at 12:30:47 PM
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There is only one situation where you get $1:

Roll a 6

Probability = 16.67%

Otherwise, your payout is not $1.

Unless the question is seeking a payout of $4 ($4-$3 ante), then there are more situations. One being:

Roll a 2 + 2 + 2 + 6

So, .33 * .33 * .33 * .17

Probability = 0.62%

Roll a 6

Probability = 16.67%

Otherwise, your payout is not $1.

Unless the question is seeking a payout of $4 ($4-$3 ante), then there are more situations. One being:

Roll a 2 + 2 + 2 + 6

So, .33 * .33 * .33 * .17

Probability = 0.62%