Puzzle. Alice and Bob roll an unbiased 6-sided die until two consecutive rolls are the same. They add up the scores from all of the rolls. If the total is even, Alice wins; if it is odd, Bob wins. Who is more likely to win?

Interested to know if there’s an “aha” solution to this ðŸ™‚ A variant, where they keep going until two consecutive rolls are *different* (i.e., until two different results have been observed) seems easier.

We can track the game by recording the last outcome and the sum. Both of them are either even or odd so we have four possibilities:

EE (the last outcome is even and the sum is even)
OO (the last outcome is odd and the sum is odd)
OE (the last outcome is odd and the sum is even)
EO (the last outcome is even and the sum is odd)

From EE and OO the game can be terminated in E (end of game with even total) while from OE and EO is O (end of game with odd total) that can be reached.
The first roll of the die places us either in EE or in OO (the sum equals the last outcome): it’s apparent that E is favoured.

The probability that the sum of the 2 dices in one roll is even is 1/2. The probability that the sums of the 2 dices in two rolls is even is 1/2 again because the parity changed (with probability 1/2) when the sum of the second roll was odd and remained unchanged when it was even.
If A and B throw the same number the sum would be even which would keep the parity. So it seems they have fair chances but the first
roll is something else. If they roll the same number in the first roll A would win because the sum before would be the even 0.
So A’s chances to win are: P(2 equal numbers in the first roll) + 1/2 * P(no equal numbers in the first roll) = 1/6 + (1/2) * (5/6) = 7/12

Sorry my answer is wrong. If the game hasn’t ended after the first roll the probability that the sum is odd is now 18/30 = 3/5. This gives B an advantage (advantage = higher winning chances when a roll has equal numbers) for the second roll. If the numbers are equal he has won with probability 3/5. Since the probability of a parity change after a roll with 2 different numbers is larger than parity preservation the advantage flips between A and B. It is easy to see that A’s initial advantage is too huge to be leveled during this process, so Alice stillhas the bigger chances to win but I would like to have a value for A’s total winning probability.

After reading the responses above I think I am probably missing something.

How I see it: we know that the last two rolls are the same, so the sum of the last two rolls is even. So the last two rolls don’t affect the parity of the total sum. Then, for any sequence of rolls the probability of sum being even or off is 1/2. The only tricky case not covered by the logic just described is when the first two rolls are the same, so only 2 rolls are made in total. In this case the sum is even. Thus, the chance of rolling even sum is higher due to this corner case.

The problem was created by my colleague Kent Boklan. However, I came up with the following aha solution:

Imagine that Alice and Bob play for a while. Alice can win after two rolls (if they are equal) but Bob can’t. Bob will complain that he doesn’t like it when the game ends early and Alice wins. Alice then makes the following generous offer: “I won’t claim a win if the first two rolls happen to be equal. When that happens, we’ll keep going until two later consecutive rolls match”. Alice and Bob are now playing a slightly different game which is certainly better for Bob than the old one.

Their friend Eve who is watching points out that since nobody will use the first dice roll to call an end to the game, there’s no need to know what it is until the game ends. After the first dice is rolled, Eve covers it with a cloth and reveals it after the game has ended and the other rolls are summed. When Eve uncovers the first dice and adds its score to whatever other total there is, it’s equally likely that it makes the overall total even or odd. Therefore in the new game Alice and Bob have equal chances to win. The old game is better for Alice and she is more likely to win.

## KK:

Alice

19 May 2023, 4:44 pm## James:

Interested to know if there’s an “aha” solution to this ðŸ™‚ A variant, where they keep going until two consecutive rolls are *different* (i.e., until two different results have been observed) seems easier.

21 May 2023, 7:12 pm## Guido:

We can track the game by recording the last outcome and the sum. Both of them are either even or odd so we have four possibilities:

EE (the last outcome is even and the sum is even)

OO (the last outcome is odd and the sum is odd)

OE (the last outcome is odd and the sum is even)

EO (the last outcome is even and the sum is odd)

From EE and OO the game can be terminated in E (end of game with even total) while from OE and EO is O (end of game with odd total) that can be reached.

The first roll of the die places us either in EE or in OO (the sum equals the last outcome): it’s apparent that E is favoured.

PS: the odds for E are 47:35.

26 May 2023, 5:47 am## Gennardo:

The probability that the sum of the 2 dices in one roll is even is 1/2. The probability that the sums of the 2 dices in two rolls is even is 1/2 again because the parity changed (with probability 1/2) when the sum of the second roll was odd and remained unchanged when it was even.

25 June 2023, 2:39 pmIf A and B throw the same number the sum would be even which would keep the parity. So it seems they have fair chances but the first

roll is something else. If they roll the same number in the first roll A would win because the sum before would be the even 0.

So A’s chances to win are: P(2 equal numbers in the first roll) + 1/2 * P(no equal numbers in the first roll) = 1/6 + (1/2) * (5/6) = 7/12

## Gennardo:

Sorry my answer is wrong. If the game hasn’t ended after the first roll the probability that the sum is odd is now 18/30 = 3/5. This gives B an advantage (advantage = higher winning chances when a roll has equal numbers) for the second roll. If the numbers are equal he has won with probability 3/5. Since the probability of a parity change after a roll with 2 different numbers is larger than parity preservation the advantage flips between A and B. It is easy to see that A’s initial advantage is too huge to be leveled during this process, so Alice stillhas the bigger chances to win but I would like to have a value for A’s total winning probability.

25 June 2023, 3:17 pm## Roman:

After reading the responses above I think I am probably missing something.

How I see it: we know that the last two rolls are the same, so the sum of the last two rolls is even. So the last two rolls don’t affect the parity of the total sum. Then, for any sequence of rolls the probability of sum being even or off is 1/2. The only tricky case not covered by the logic just described is when the first two rolls are the same, so only 2 rolls are made in total. In this case the sum is even. Thus, the chance of rolling even sum is higher due to this corner case.

10 July 2023, 11:15 am## Alex Ryba:

The problem was created by my colleague Kent Boklan. However, I came up with the following aha solution:

Imagine that Alice and Bob play for a while. Alice can win after two rolls (if they are equal) but Bob can’t. Bob will complain that he doesn’t like it when the game ends early and Alice wins. Alice then makes the following generous offer: “I won’t claim a win if the first two rolls happen to be equal. When that happens, we’ll keep going until two later consecutive rolls match”. Alice and Bob are now playing a slightly different game which is certainly better for Bob than the old one.

Their friend Eve who is watching points out that since nobody will use the first dice roll to call an end to the game, there’s no need to know what it is until the game ends. After the first dice is rolled, Eve covers it with a cloth and reveals it after the game has ended and the other rolls are summed. When Eve uncovers the first dice and adds its score to whatever other total there is, it’s equally likely that it makes the overall total even or odd. Therefore in the new game Alice and Bob have equal chances to win. The old game is better for Alice and she is more likely to win.

15 July 2023, 9:28 am