"What are the odds?" Of Rolling 7 Doubles in a Row (Dice)

In backgammon, each game begins with each player throwing a single die to determine who goes first. In the event that each player's roll is the same, "then both players roll again until they roll different numbers". The following question was posed to me: what are the odds that each player rolls the same number... 7 times in a row?

There are 6 potential numbers on each die, so over two dice there are 6 * 6 = 36 possible combinations. Additionally, there are 6 possible pairs: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So 6 / 36 = 1 / 6 ~= 0.167% chance of tying on one roll. But when this is replicated 7 times in a row: (1 / 6) ^ 7 = 0.00036%, or 1 in 279,936.

N	Chances	1 in ...
1	16.67%	6
2	2.78%	36
3	0.46%	216
4	0.08%	1,296
5	0.01%	7,776
6	0.002%	46,656
7	0.00036%	279,936
8	0.0001%	1,679,616
9	0.00001%	10,077,696
10	0.000002%	60,466,176

That's far less likely than having a doctor on your flight or getting the same taxi driver twice in a large city, but way more likely than two teams sharing identical box scores on the same day or three people in the same family all being born at the exact same time (on different days).

What if you play backgammon a lot though? How long would you have to play in order to expect to see a streak of 7 dice ties in a row?

The math is explained on this nice "Probability of Runs" calculator, but you would need to play 232,845 games of backgammon until there would be a better than 50/50 chance of seeing a run of ties like this. The average number of games needed until you would see this happen is even larger: 335,922 games of backgammon.